Thermal Engineering Pune University MCQs

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Steam Generators – Classification of Boilers”.

1. Which of the following is NOT a fire tube boiler?

a) Cochran Boiler

b) Lancashire Boiler

c) Locomotive Boiler

d) Babcock and Wilcox Boiler

Answer: d

Explanation: In case of fire tube boilers, hot gases flow through tubes and water surrounds the tubes and in case of water tube boilers, water flows through the tubes and hot gases surround the tubes. Cochran, Lancashire, Locomotive are all fire tube boilers while Babcock and Wilcox is a water tube boiler.

2. Which of the following is NOT a valid classification of boilers?

a) Forced circulation and natural circulation

b) High pressure and low pressure

c) Stationary and Portable

d) Single fired and Double fired

Answer: d

Explanation: Boilers are classified into internally fired and externally fired on the basis of firing. There is no such classification as single fired and double fired.

3. Babcock and Wilcox boiler is an internally fired boiler.

a) True

b) False

Answer: b

Explanation: In case of internally fired boiler furnace is located inside the boiler, while in externally fired boilers furnace is outside the boiler. Babcock and Wilcox boiler has its furnace outside, hence it’s a externally fired boiler.

4. Which of the following statement is NOT true about fire tube boilers?

a) Hot gases are inside the tubes and water surrounds them

b) For a given power it occupies more floor area

c) Operating pressure can be as high as 100 bars

d) Not suitable for large power plants

Answer: c

Explanation: The operating pressure for fire tube boilers is limited to 16 bars. For a given power it occupies more floor area, hence it is not suitable for large power plants.

5. The classification of boilers into horizontal, vertical and inclined is done on the basis of their _____

a) pressure

b) method of firing

c) tubes

d) axis

Answer: d

Explanation: Boilers are classified as high and low pressure boilers on the basis of pressure, internally fired and externally fired on the basis of method of firing, water tube and fire tube on the basis of relative position of hot gases and water and as horizontal, vertical and inclined on the basis of their axis.

6. Which of the following is a low pressure boiler?

a) Babcock and Wilcox boiler

b) Benson boiler

c) Lancashire

d) Lamont Boiler

Answer: c

Explanation: Boilers with production of steam below 80 bar of pressure are call low pressure boilers. Lancashire boiler is a low pressure boiler while the rest of given options are high pressure boilers.

7. Which of the following statement is FALSE about boilers?

a) Portable boilers are also called as mobile boilers

b) Lamont boiler is a forced circulation type boiler

c) Cochran boiler is a high pressure boiler

d) Horizontal boilers occupy more space

Answer: c

Explanation: Cochran boiler produces steam at a pressure greater than 80 bar, hence it’s a high pressure boiler. A pump is used to circulate water in Lamont boiler, therefore it is classified as forced circulation boiler. Portable boilers are also called as mobile boilers; they include locomotive type boilers. Horizontal boilers occupy more space but can be easily repaired and inspected.

8. Which of the following statement is TRUE about fire tube and water tube boilers?

a) Fire tube boilers have high risk of bursting than water tube boilers

b) Water tube boilers have high risk of bursting than fire tube boilers

c) Bursting of boilers isn’t possible

d) It depends upon the quality of water fed to the boiler

Answer: b

Explanation: Operating pressure of water tube boilers can be as high as 100 bar whereas for fire tube boilers the operating pressure is limited to 16 bar. Therefore, water tube boilers have high risk of bursting.

9. Which of the following is NOT a forced circulation boiler?

a) Velox boiler

b) Lancashire boiler

c) Lamont boiler

d) Benson boiler

Answer: b

Explanation: Circulation of water is done by use of pumps in case of forced circulation boilers. In case of Lancashire boiler, the circulation of water takes place due to natural convection current, therefore it a natural circulation boiler and not a forced circulation one.

10. Which of the following is NOT true about Babcock and Wilcox boiler?

a) Water tube boiler

b) Externally fired

c) High pressure boiler

d) Single tube

Answer: d

Explanation: Babcock and Wilcox is an externally fired, high pressure and water tube boiler. It is also a multi-tube boiler since it has multiple tubes for the flow of water.

11. Which of the following is a single tube boiler?

a) Cornish boiler

b) Lancashire boiler

c) Benson boiler

d) Cochran boiler

Answer: a

Explanation: In case of single tube boilers, there is only one tube through which hot gases or water passes. Cochran boiler is an example of single tube boiler. In multi-tube boiler there are multiple tubes for passing of water . Lancashire boiler, Benson boiler and Cornish boiler are all examples of multi-tube boilers.

12. In natural circulation boilers, water circulation takes place due to natural convection current produced by the application of heat.

a) True

b) False

Answer: a

Explanation: The given statement is the definition of natural circulation boilers. Lancashire and Babcock and Wilcox boiler are examples of natural circulation boilers.

13. Stationary boilers are used for _____

a) locomotive applications

b) temporary applications

c) power plant steam generation

d) marine application

Answer: c

Explanation: Stationary boilers are used for power plant steam generation. Mobile or portable boilers are used for marine and locomotive applications. For temporary use like in small coal field pits involves use of portable boilers.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Steam Generators – Boilers Selection and Terminology”.

1. Which of the following factors is NOT considered while selecting a boiler?

a) Available floor area

b) Number of tubes in the boiler

c) Available fuel and water

d) The portable load factor

Answer: b

Explanation: While selecting a boiler its working pressure and quality of steam produced, steam generation rate, floor area available, accessibility for repairing parts, erection facilities, portable load factor, fuel and water available, operating and maintenance cost are the factors that are considered.

2. What is the usual geometry of a boiler shell?

a) Torus

b) Cuboidal

c) Cubical

d) Cylindrical

Answer: d

Explanation: Boiler shell consists of one or more steel plates bent into cylindrical form that are riveted or welded together. Cylindrical geometry distributes stresses equally and has more strength and is easy to manufacture.

3. The function of “Setting”  is to confine heat to the boiler and form a passage for gases.

a) True

b) False

Answer: a

Explanation: It is the primary function of “Setting” to confine heat to the boiler and form passage for gases. It is made of brickwork and also provides support to some type of boilers like in Lancashire boiler.

4. _____ is the platform in the furnace upon which fuel is burnt.

a) Shell

b) Grate

c) Setting

d) Heat exchanger

Answer: b

Explanation: Grate is made of cast iron and it is the platform on which the fuel is burnt. The bars in the grate are arranged so that air may reach the fuel for proper combustion. The area on which the fire rests in a wood or coal fired boiler is called the grate surface.

5. What is the chamber formed by the space above the grate and below boiler shell called, where combustion takes place?

a) Grate surface

b) Mounting

c) Furnace

d) Setting

Answer: c

Explanation: Furnace is where the combustion takes place and hot gases are released. Grate surface is the area over the grate available for burning fuel. Mountings like fusible plug, pressure gauges are attached for safe operation. Function of setting is to confine heat to the boiler shell.

6. In boiler terminology, the volume of the shell occupied by water is termed as _____

a) water space

b) steam space

c) wet volume

d) wet space

Answer: a

Explanation: According to boiler terminology the volume of shell occupied by water is called water space. The rest of the volume of the shell is taken up by the steam and is called as steam space.

7. The items that are added to a boiler for safe operation are called _____

a) safety components

b) setting

c) accessories

d) mountings

Answer: d

Explanation: Items such as stop valve, pressure gauges, safety valves, Water level gauges etc. are called mountings. A boiler cannot safely work without mountings. Items that are added to increase the efficiency of the boiler are called as accessories.

8. The items that are added to increase the efficiency of boiler are called _____

a) mountings

b) accessories

c) setting

d) boiler essentials

Answer: b

Explanation: Items such as superheaters, economisers, feed pumps etc. are called accessories and they increase the efficiency of the boiler. Mountings are attached so that the boiler could be operated safely.

9. According to boiler terminology, Formation of steam bubbles on the surface of water is called _____

a) foaming

b) scale

c) lagging

d) bubbling

Answer: a

Explanation: Foaming is the formation of steam bubbles on the surface of water. It happens due to high surface tension of water. Scale is the deposit of medium to extreme hardness occurring on water heating surfaces. Lagging are blocks of asbestos or magnesia insulation wrapped on the outside of a boiler shell.

10. Which of the following according to boiler terminology, is the term used for the deposition of medium to extreme hardness occurring on water heating surfaces because of undesirable condition in the boiler water?

a) Refractory

b) Foaming

c) Scale

d) Lagging

Answer: c

Explanation: The correct answer is scale. Refractory is a heat insulation material. Lagging are blocks of asbestos or magnesia insulation wrapped on the outside of a boiler shell. Foaming is the formation of steam bubbles on the surface of water.

11. According to boiler terminology, the removal of the mud and other impurities of water from the lowest part of the boiler is termed as _____

a) lagging

b) blowing off

c) waste removal

d) foaming

Answer: b

Explanation: The correct answer is blowing off. The removal of mud by blowing off is accomplished by use of a blow off cock or valve. Foaming is the formation of steam bubbles on the surface of water. Lagging are blocks of asbestos or magnesia insulation wrapped on the outside of a boiler shell.

12. What is lagging?

a) The formation of steam bubbles on the surface of water

b) Blocks of asbestos or magnesia insulation wrapped on the outside of a boiler shell

c) The deposit of medium to extreme hardness occurring on water heating surfaces

d) Heat insulation material

Answer: b

Explanation: Lagging are blocks of asbestos or magnesia insulation wrapped on the outside of a boiler shell or steam piping. Refractory is a heat insulation material. Scale is the deposition of medium to extreme hardness occurring on water heating surfaces. Foaming is the formation of steam bubbles on the surface of water.

13. Which of the following is a refractory material?

a) Fire brick

b) Plastic

c) Wood

d) Paper

Answer: a

Explanation: Refractory materials are heat insulation materials and they are supposed to withstand high temperatures. Refractory materials such as fire brick and plastic fire clay are used to make lining combustion chambers in boilers.

14. Furnace is also called fire-box.

a) True

b) False

Answer: a

Explanation: Furnace is chamber where combustion of fuel takes place. It is also called as fire-box. Hot gases originate from furnace. It is the chamber formed by the space above the grate and below the boiler shell.

15. Which of the following is NOT a feature of a good boiler?

a) Light in weight

b) Steam production rate should be as per requirement

c) Capable of quick starting

d) Water and gas circuits should be such as to allow maximum fluid velocity

Answer: d

Explanation: The water and gas circuits should be such as to allow minimum fluid velocity to minimize friction losses. Essentials of a good boiler includes – reliability, less space requirement, light in weight, accessibility to various parts, simple installation, steam production rate as per requirement, capable of quick starting.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Steam Generators – Fire Tube Boilers – 1”.

1. In Simple Vertical boiler, the ash that falls off from the grate falls in the _____

a) sink

b) ash pit

c) ash disposer

d) down-header

Answer: b

Explanation: The correct answer is ash pit. Grate is a mesh on which fuel is burnt, and after combustion the ashes fall down in the ash pit. While cleaning, this ash is removed from the ash pit.

2. What is the highest steam production rate that can be achieved by Simple Vertical boiler?

a) 1000 kg/hr

b) 2500 kg/hr

c) 3000 kg/hr

d) 5000 kg/hr

Answer: b

Explanation: Steam production rate in case of Simple Vertical boiler does not exceed 2500 kg/hr. The pressure of the steam produces is also limited to 7.5 to 10 bar.

3. In a Simple Vertical boiler, two cross tubes are provided with the fire box. Which of the following statement about the orientation of these tubes is correct?

a) Both of them are vertical

b) Both of them are horizontal

c) Both of them are inclined

d) One is horizontal and the other one is vertical

Answer: c

Explanation: Cross tubes are fitted inclined. The presence of cross tubes increases the heating surface. The inclined orientation of the cross tubes ensures effective circulation of the water. For cleaning purpose, hand holes are provided at the end of each of these tubes.

4. What is the purpose of fire door in boilers?

a) It is opened to increase air supply for combustion

b) It is opened to put out the fire, in case of emergency

c) It allows the hot gases to pass safely through chimney

d) It is used to feed fuel for combustion

Answer: d

Explanation: Fire door is provided to feed fuel to the fire. It is provided near the grate. Fire doors must have high resistance to heat.

5. Cochran boiler is a multi-tube boiler.

a) True

b) False

Answer: a

Explanation: Cochran boiler is a multi-tube boiler. It is vertical and has a number of fire tubes arranged horizontally. The height of the boiler is 5.79 m and working pressure is around 6.5 bar usually.

6. What is the shell diameter of a Cochran boiler?

a) 2.00 m

b) 1.50 m

c) 1.00 m

d) 2.75 m

Answer: d

Explanation: The shell diameter of the Cochran boiler is 2.75 m. Its height is 5.79 m. It has a maximum working pressure of 15 bar but is usually operated at around a pressure of 6.5 bar.

7. What is the height of a Cochran boiler?

a) 5.79 m

b) 4.10 m

c) 4.26 m

d) 3.50 m

Answer: a

Explanation: The height if Cochran boiler is 5.79 m. It has a maximum working pressure of 15 bar but is usually operated at around a pressure of 6.5 bar. Steam capacity is around 3500 kg/hr with maximum steam capacity of 4000 kg/hr.

8. What is the maximum working pressure of Cochran boiler?

a) 8 bar

b) 10 bar

c) 15 bar

d) 20 bar

Answer: c

Explanation: The maximum working pressure of Cochran boiler is 15 bar. It has a maximum steam capacity of 4000 kg/hr. It is usually operated around 6.5 bar. It has an efficiency of 70 to 75 % which depends on the fuel used.

9. What is the area of the heating surface of a Cochran boiler?

a) 100 m 2

b) 120 m 2

c) 150 m 2

d) 200 m 2

Answer: b

Explanation: The heating surface of a Cochran boiler has an area of 120 m 2 . Cochran boiler is a vertical multi-tube boiler. It is a fire tube boiler with working pressure around 6.5 bar and Steam capacity of 3500 kg/hr.

10. What is range of efficiency of a Cochran boiler?

a) 50 – 55%

b) 60 – 65%

c) 70 – 75%

d) 80 – 85%

Answer: c

Explanation: The efficiency of a Cochran boiler varies in the range of 70 to 75%, it depends on the fuel used. The steam capacity of a Cochran boiler is 3500 kg/hr and working pressure around 6.5 bar. The maximum steam capacity and working pressure are 4000 kg/hr and 15 bar respectively.

11. Which of the following statements is correct about the smoke box in boilers?

a) It collects smoke and is emptied later after being done operating the boiler

b) It regulates hot gases from tubes to chimney

c) It treats the smoke before it is released to the atmosphere

d) It is used to store ashes

Answer: b

Explanation: The hot gases produced by the combustion of fuel in the furnace pass through the tubes  or go about the tubes containing water  and then they are released to atmosphere through smoke box and chimney.

12. Which of the following boiler mounting is NOT present in a Cochran boiler?

a) Stem stop valve

b) Blow off cock

c) Safety valve

d) Fusible plug

Answer: d

Explanation: Cochran boiler has water level gauge, safety valve, steam stop valve, blow off cock, man hole and a pressure gauge as its mountings. Mountings are provided for the safe operation of the boiler. Cochran boiler doesn’t have a fusible plug.

13. How many flue-tubes does a Cornish boiler has?

a) One

b) Two

c) Three

d) Four

Answer: a

Explanation: Cornish boiler has only one flue tube, it is a single tube boiler. It has a shell of diameter 1.25 to 1.75 m and length ranging from 4 m to 7 m. The steam capacity of a Cornish boiler is 6500 kg/hr.

14. What is the steam capacity of a Cornish boiler?

a) 5500 kg/hr

b) 6500 kg/hr

c) 7500 kg/hr

d) 8500 kg/hr

Answer: b

Explanation: The correct answer is 6500 kg/hr. The pressure of the steam produces is around 10.5 bar. Cornish boiler is a single tube boiler, it has only one tube through which the flue gases pass.

15. Lancashire boiler is easy to operate and has less maintenance and operating costs.

a) True

b) False

Answer: a

Explanation: Lancashire boiler has a simple design, easy to operate and reliable. The maintenance cost and operating cost are also less as compared to other boilers. It is generally used in textile industries and sugar mills.

This set of Thermal Engineering Interview Questions and Answers focuses on “Steam Generators – Fire Tube Boilers – 2”.

1. Which of the following is the correct range of the diameter of shell for a Lancashire boiler?

a) 1 to 2 m

b) 2 to 3 m

c) 3 to 4 m

d) 4 to 5 m

Answer: b

Explanation: The diameter of shell for a Lancashire boiler is 2 to 3 m. The length of shell lies in the range of 7 to 9 m. The shell of Lancashire boiler is placed horizontally over a brickwork. The shell is created using a number of rings of cylindrical form.

2. Which of the following statements is TRUE about Lancashire boiler?

a) The length of its shell is about 10 m

b) It has a steam capacity of 12000 kg/hr

c) It is a water tube boiler

d) Its efficiency is about 50 to 70%

Answer: d

Explanation: The efficiency of Lancashire boiler lies in the range of 50 to 70%. The length of the boiler shell is 7 to 9 m and its diameter ranges from 2 to 3 m. Lancashire boiler has a steam capacity of 9000 kg/hr. It is a fire tube boiler.

3. What is the maximum working pressure of a Lancashire boiler?

a) 10 bar

b) 12 bar

c) 14 bar

d) 16 bar

Answer: d

Explanation: The maximum working pressure of a Lancashire boiler is 16 bar. Its steam capacity is 9000 kg/hr and has an efficiency of 50 to 70%. Lancashire boiler is easy to operate and has low maintenance cost.

4. In Lancashire boiler, an arrangement is made so that burning coal and ashes do not enter the furnace tubes. What is that arrangement called?

a) Low brickwork fire bridge

b) Ash-filter

c) Filtering mesh

d) Furnace separator

Answer: a

Explanation: Furnace is placed at the front end of the furnace tubes. The burning coal and ashes can get in the furnace tubes, hence a low brickwork fire bridge is placed at the back of the gate avert the entry of ashes and burning coal.

5. How many fire tubes  does a Lancashire boiler has?

a) One

b) Two

c) Three

d) Four

Answer: b

Explanation: Lancashire boiler has two tubes also called as furnace tubes inside the cylindrical shell. These tubes are constructed with several rings of cylindrical form. They are covered all around with water and carry hot gases coming from the furnace.

6. What is the function of dampers in Lancashire boiler?

a) They control the flow of hot gases to the chimney

b) They regulate the fire in the furnace

c) They control the flow water to the boiler

d) They control the steam going out of the boiler

Answer: a

Explanation: In case of Lancashire boilers, the flow of hot gases is controlled by dampers. The flow of water to the boiler is controlled by feed check valve. Steam stop valve controls the flow of steam leaving the boiler.

7. The efficiency of Locomotive boiler is around _____

a) 30%

b) 50%

c) 70%

d) 90%

Answer: c

Explanation: The efficiency of Locomotive boiler is 70%. The presence of superheater in the Locomotive boiler increases the efficiency. The placement of superheater tubes is done in the fire tubes with larger diameter.

8. What is the steam capacity of a Locomotive boiler?

a) 3000 kg/hr

b) 5000 kg/hr

c) 7000 kg/hr

d) 9000 kg/hr

Answer: d

Explanation: The steam capacity of a Locomotive boiler is 9000 kg/hr. It has an efficiency of 70%. The coal burning rate is 1600 kg/hr. It is a type of fire-tube boiler.

9. What is the working pressure of a Locomotive boiler?

a) 12 bar

b) 14 bar

c) 16 bar

d) 18 bar

Answer: b

Explanation: The working pressure of a locomotive boiler is 14 bar. Locomotive boilers are generally used in locomotives but may be used as stationary boilers also. Locomotive boilers are small in size but they have high steam generation capacity.

10. In locomotive boiler, why natural draught cannot be obtained?

a) The chimney height required is too low

b) The boiler keeps moving, cannot maintain a constant pressure difference

c) The chimney height required is too high

d) The diameter of chimney required is impractical

Answer: c

Explanation: Natural draught in locomotive boilers cannot be maintained because the chimney height required is too high and it is difficult to make provision of such a high chimney since the boiler has to run on rails.

11. Which of the following statements is FALSE about a Locomotive boiler?

a) It has high steam capacity

b) It is portable

c) Scale formation and corrosion could take place

d) Water spaces can be cleaned easily

Answer: d

Explanation: There are some water spaces that cannot be cleaned properly. A Locomotive boiler has high steam capacity, portable, compact and has low installation cost. It is not bestowed with high overloads due to overheating damage.

12. In case of Locomotive boilers, what purpose the dampers serve?

a) They control the flow of air to grate

b) They control the flow of hot gases

c) They control the flow of steam leaving the boiler

d) They reduce the intensity of fire, if needed

Answer: a

Explanation: In case of Locomotive boilers, dampers are used to control air flow to the grate. Hot gases produced in the fire-box pass though the tubes and reach the smoke box where they are released to the atmosphere through the chimney. In Lancashire boiler, dampers are used to control the flow of hot gases.

13. In Locomotive boilers, the superheater tubes are placed inside the fire tubes.

a) True

b) False

Answer: a

Explanation: Presence of superheater increases the efficiency of the Locomotive boiler. The superheater tubes are placed inside fire tubes having larger diameter. The efficiency of a Locomotive boiler is 70%.

14. Which of the following statements is FALSE about steam dome in Locomotive boilers?

a) It is located at the upper part of the barrel

b) It has a regulator to regulate the steam flow

c) It is the chamber to collect the steam produced

d) From steam dome, the steam flows into the chamber through a steam pipe

Answer: c

Explanation: Steam dome is located at the upper portion of the barrel. It has a dome shape, hence its name. From steam dome the steam is passes through a steam pipe and it ultimately reaches the chamber. There is a regulator to regulate the steam flow from the steam dome.

15. When the locomotive is at rest, there is no exhaust steam available. So, fresh steam from the boiler is used as exhaust steam.

a) True

b) False

Answer: a

Explanation: Fresh steam is used as exhaust steam when the locomotive is at rest. The locomotive boiler has a high steam capacity. Natural draught cannot be obtained in these boilers because the chimney height required is too high and the boiler has to be kept mobile.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Steam Generators – Water Tube Boilers”.

1. Water tube boilers are further classified into horizontal straight tube and bent tube boilers.

a) True

b) False

Answer: a

Explanation: Water tube boilers are further classified into horizontal straight tube  and bent tube boilers . Horizontal straight tube boilers are further classified into longitudinal and cross drum boilers.

2. Which of the following is a horizontal straight tube boiler?

a) Lancashire boiler

b) Babcock and Wilcox boiler

c) Stirling boiler

d) Locomotive boiler

Answer: b

Explanation: Lancashire and locomotive boilers are fire tube boilers. Babcock and Wilcox and Stirling boilers are water tube boilers out of which the former one is a horizontal straight tube boiler and the latter one is a bent tube boiler.

3. What is the correct range of diameter of the drum for a Babcock and Wilcox boiler?

a) 0.61 – 1.02 m

b) 1.22 – 1.83 m

c) 2.87 – 3.52 m

d) 4.89 – 5.66 m

Answer: b

Explanation: Babcock and Wilcox boiler is a water tube  boiler. The diameter and length of the boiler drum are in the range of 1.22 to 1.83 m and 6.096 to 9.144 m respectively.

4. For a Babcock and Wilcox boiler, what is the correct range of size of the water tubes?

a) 1.22 to 1.44 cm

b) 4.98 to 6.02 cm

c) 7.62 to 10.16 cm

d) 12.25 to 15.54 cm

Answer: c

Explanation: The correct answer is 7.62 to 10.16 cm. Babcock and Wilcox boiler has multiple water tubes as it is a multi-tube boiler. These tube join the uptake header with the down take header. These water tubes are made of mild steel.

5. What is the correct range of size of superheater tubes in Babcock and Wilcox boiler?

a) 1.20 to 2.10 cm

b) 2.12 to 2.84 cm

c) 3.84 to 5.71 cm

d) 5.84 to 6.25 cm

Answer: c

Explanation: The correct answer is 3.84 to 5.71 cm. The range of size of water tubes is 7.62 to 10.16 cm. The presence of superheater increases the efficiency of the boiler. Babcock and Wilcox boiler has an efficiency of 60 to 80%.

6. What is maximum working pressure of Babcock and Wilcox boiler?

a) 40 bar

b) 50 bar

c) 60 bar

d) 80 bar

Answer: a

Explanation: The maximum working pressure of a Babcock and Wilcox boiler is 40 bar. Babcock and Wilcox is used when pressure above 10 bar and steam capacity greater than 7000 kg/hr is required. It is a water-tube boiler.

7. What is the maximum steam capacity of a Babcock and Wilcox boiler?

a) 20000 kg/hr

b) 30000 kg/hr

c) 40000 kg/hr

d) 50000 kg/hr

Answer: c

Explanation: The maximum steam capacity of a Babcock and Wilcox boiler is 40000 kg/hr. This boiler is used when a steam capacity greater than 7000 kg/hr is required. The maximum working pressure of the boiler is 40 bar.

8. In Babcock and Wilcox boiler, just before entering the superheater the steam enters _____

a) uptake header

b) down take header

c) main stop valve

d) antipriming pipe

Answer: d

Explanation: The steam is collected in the drum. The steam first enters the antipriming pipe and then it enters the superheater. The steam then reaches the main stop valve. The water from the drum flows to down take header and then to uptake header through the inclined pipes.

9. Babcock and Wilcox boiler is a forced circulation boiler.

a) True

b) False

Answer: b

Explanation: Babcock and Wilcox boiler does not use a pump for circulation of water. The water flows though the down take header, inclined pipes and the uptake header by means of natural convection current. Hence Babcock and Wilcox is not a forced circulation boiler but a natural circulation boiler.

10. Which of the following statements is FALSE for a cross drum Babcock and Wilcox boiler?

a) The alignment of boiler drum is different in cross drum and longitudinal drum boilers

b) Any number of connecting tubes can be used

c) Steam of pressure 100 bar could be generated

d) Steam capacity is 45000 kg/hr

Answer: d

Explanation: Steam having pressure 100 bar and steam capacity of 27000 kg/hr could be achieved with the help of cross drums. Cross drum and longitudinal drum are placed differently with respect to the axis of the water tubes. In case of cross drum boiler, there is no limitation to the number of connecting tubes.

11. Stirling boiler is _____

a) a fire tube boiler

b) a bent-tube  boiler

c) a horizontal straight tube  boiler

d) a high pressure boiler

Answer: b

Explanation: Stirling boiler is a water tube boiler that comes under the category of bent tube boilers. The drums and headers in the boilers are connected by bent tubes. It has a steam capacity of 50000 kg/hr.

12. What purpose do baffle plates serve in a Babcock and Wilcox boiler?

a) They direct the flow of water

b) They direct the flow of steam

c) They direct the flow of hot gases

d) They direct the flow of air to combustion chamber

Answer: c

Explanation: The baffle plates are provided to direct the flow of hot gases around tubes containing water. The water absorbs the heat from the hot gases and the temperature of hot gases is dropped. The hot gases are ultimately directed to the chimney through which they escape to atmosphere.

This set of Thermal Engineering online quiz focuses on “Steam Generators – High Pressure Boilers”.

1. Shell boilers are preferred for high pressure and high output.

a) True

b) False

Answer: b

Explanation: Water tube boilers are preferred for high pressure and high output while shell boilers are preferred for low pressure and low output. Water tube boilers can be erected easily and its parts are easily transportable.

2. Which of the following statements is FALSE about high pressure boilers?

a) Modern high pressure boilers used for power generation have steam capacities even greater than 650 tonnes/hr

b) Water is circulated through a pump in most of the modern high pressure boilers

c) They are mostly water tube boilers

d) One continuous tube is used for water circulation

Answer: d

Explanation: The pressure drop that takes place due to friction in a single large tube is greater than multiple small tubes. Hence, multiple tubes of small diameter are preferred over one continuous large tube. The Steam capacities for modern high pressure boilers vary from 30 to 650 tonnes/hr and even more with a pressure up to 160 bars.

3. Which of the following is NOT a unique feature of high pressure boiler?

a) Water circulation method

b) Tubing type

c) Firing method

d) Improved method of heating

Answer: c

Explanation: Unique features of high pressure boilers are – method of water circulation, type of tubing, improved method of heating. Pumps are used in most of the high pressure boilers for water circulation, i.e. forced circulation method. Multiple tubes instead of one continuous tube is used since the friction loss in the former one is much smaller than in the latter.

4. Which of the following heating methods is NOT used in high pressure boilers?

a) Evaporation of water above critical pressure of steam

b) Decreasing gas velocity 

c) Increasing water velocity inside tube 

d) Mixing of water with superheated steam

Answer: b

Explanation: The overall heat transfer coefficient is increased when the flow velocity of water inside tube is increased and the gas velocity is increased above sonic velocity. Heat is saved by evirating water above critical pressure of the steam. Higher heat transfer coefficient is obtained by heating water by mixing it with superheated steam.

5. Which of the following is FALSE about high pressure boilers?

a) They are compact

b) Heat of combustion utilized more effectively

c) Heat is saved by evaporating water below critical pressure

d) Pumps are used to circulate water

Answer: c

Explanation: High pressure boilers are compact; hence less floor space is required. Since tubes of small diameter are used heat of combustion is used more effectively. Some amount of heat is saved by evaporation of water above critical pressure. High pressure boilers are forced circulation boilers i.e. pumps are used to circulate water.

6. In LaMont boilers, the circulation of water is maintained by a centrifugal pump which is driven by a steam turbine in which steam is fed from the boiler itself.

a) True

b) False

Answer: a

Explanation: LaMont boiler is a forced circulation and high pressure boiler. It uses a pump which driven by a steam turbine which uses the steam from the boiler. An electrically driven pump is also installed for emergency.

7. The main drawback of LaMont boilers is _____

a) limited capacity of evaporating drum

b) bubble formation on the inner surface of heating tubes

c) hindrance in the flow of hot gases

d) radiant superheater decreases the efficiency

Answer: b

Explanation: Formation of bubbles on the inner side of heating tube takes place. These bubbles reduce the heat flow due to its high thermal resistance. This reduces the steam generation. This is drawback is rectified in Benson boiler.

8. Which of the following statements is FALSE about Benson boiler?

a) Bubble formation in the heating tubes is prevented by slow heating

b) It is a forced circulation boiler

c) Oil is used as fuel

d) Feed water is compressed to 235 bar

Answer: a

Explanation: Benson boiler is a forced circulation boiler. To avoid bubble formation in the heating tubes the water is compressed to 235 bar. 235 bar is the critical pressure for water. At this temperature water and steam have same density thereby eliminating the chance of bubble formation.

9. What is the maximum working pressure obtained by a commercial Benson boiler?

a) 300 atm

b) 400 atm

c) 500 atm

d) 600 atm

Answer: c

Explanation: 500 atm is the maximum working pressure obtained from a commercial Benson boiler. Benson boiler having steam generation of tonnes/h are also used. It can also generate steam having a temperature of 650°C.

10. Which of the following statements about Benson boiler is correct?

a) It requires large surface area

b) It is prone to explosions more than other boilers

c) It takes a lot of time to start

d) It is lighter than other boilers

Answer: d

Explanation: Benson boiler can be erected on comparatively small floor area. Since small diameter tubes are used explosion hazards are not severe. It takes lesser time to start. It also has a very small storage capacity which makes it even more safe. Since there are no drums in this boiler, it has 20% less weight than other boilers.

11. Which of the following statements about super-critical is TRUE?

a) It only needs preheater and superheater

b) It only needs economizer and preheater

c) It only needs economizer and superheater

d) It only needs preheater

Answer: a

Explanation: A super-critical boiler does not require an economizer. It only requires a preheater and a superheater. Super critical boilers have large heat transfer rate and have higher thermal efficiency.

12. Which of the following statements about Velox boiler is FALSE?

a) It starts quickly

b) It is compact

c) Combustion rates are low

d) Pressurized air is used

Answer: c

Explanation: High combustion rates are possible in Velox boiler. It is compact. Problem of draught is solved by using pressurized air, low excess air is required. In addition to easy operation is also starts quickly.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Boiler Mountings – 1”.

1. Which of the following statements regarding boiler mountings is TRUE?

a) They ensure safe operation of the boiler

b) They are installed to increase efficiency of the boiler

c) Economizer, superheater and feed pumps are boiler mountings

d) Water level indicator is not a boiler mounting

Answer: a

Explanation: Boiler mountings are installed for safety of a boiler. Economizer, superheater and feed pumps are not boiler mountings, they are boiler accessories. Examples of boiler mountings are water level indicator, steam stop valve, man hole, pressure gauge, safety valve etc.

2. Water level indicator _____

a) indicates the amount of steam in the boiler

b) indicates the amount of water in the boiler

c) indicates the amount of water converted to steam

d) indicates the amount of steam left in the boiler

Answer: b

Explanation: Water level indicator constantly indicates the level of water inside the boiler. The water level in a boiler must not fall below a certain safe-level. If the level of water inside the boiler happens to fall below safe-level, crack and leak can occur.

3. What is the purpose of guard glass in a glass-tube water level indicator?

a) It protects the attendant, if water starts leaking

b) It protects the water gauge, if steam pressure increases the design pressure

c) It protects the attendant, if boiler cracks

d) It protects the attendant, if the gauge glass breaks

Answer: d

Explanation: The guard glass is tough. It doesn’t give shards on breaking. Guard glass covers the gauge glass. If due to sudden rush of water and steam the gauge glass breaks, the guard glass protects the attendant from getting hurt.

4. What is the purpose of pressure gauge?

a) It measures the pressure exerted on the boiler shell from outside

b) It measure the current atmospheric pressure

c) It measures pressure exerted inside the boiler shell

d) It measures hydrostatic pressure of water at the bottom of the boiler shell

Answer: c

Explanation: Pressure gauge measures the pressure exerted inside the boiler shell. Normally, it is manufactured to indicate pressure upto double the maximum working pressure. It measures gauge pressure i.e. pressure above atmospheric pressure.

5. Which of the following statements about compound pressure gauge is TRUE?

a) It measure the hydrostatic pressure exerted by water at the bottom surface

b) It can measure pressures above and below atmospheric pressure on the same dial

c) It is installed at the bottom of the boiler shell

d) It measures absolute pressure

Answer: b

Explanation: A compound pressure gauge is designed such that it can measure pressures above and below atmospheric pressure. It is a pressure gauge, hence mounted on the front top of the shell.

6. The syphon tube in the Bourdon pressure gauge  is filled with _____

a) mercury

b) turpentine oil

c) mineral water

d) condensed water

Answer: d

Explanation: Syphon tube is an integral part of the Bourdon pressure gauge. The syphon tube is completely filled with condensed water. Steam pressurizes the water in the syphon from one side and the water is forced into the elliptical tube.

7. Which of the following statements about Bourdon pressure gauge is FALSE?

a) Double-tube Bourdon gauge is more rigid than the single tube

b) Double-tube Bourdon gauge is more suitable for portable boilers

c) It measures the gauge pressure

d) The movement of the free end of the elliptical tube is proportional to the internal pressure of the boiler

Answer: d

Explanation: The movement of the free end of the elliptical tube is proportional to the difference between internal and external pressure in the tube. Since the external pressure is atmospheric the free end measures gauge pressure.

8. The function of a safety valve is _____

a) to release the excess steam when the steam pressure exceeds the rated pressure

b) to extinguish fire when the temperature reaches impermissible value

c) to release the excess water

d) it cuts off the water supply when water level reaches its maximum limit

Answer: a

Explanation: A safety valve releases excess steam if the steam pressure exceeds the rated pressure. The safety valve automatically opens when the pressure exceeds the limit and closes automatically when the pressure reduces to acceptable value.

9. Which of the following is NOT a type of safety valve?

a) Dead weight safety valve

b) Lever safety valve

c) Spring loaded safety valve

d) High water and low steam safety valve

Answer: d

Explanation: The types of safety valve are – high steam and low water safety valve, dead weight safety valve, spring loaded safety valve and lever safety valve. The purpose of safety valve is to release excess steam when the steam pressure exceeds the design pressure.

10. Which of the following statements about safety valves is FALSE?

a) It is usually mounted on the top of the shell

b) It releases the steam if the steam pressure exceeds the rated pressure

c) Dead weight safety valve is a type of safety valve

d) An attendant is required to operate the valve

Answer: d

Explanation: The safety valve automatically opens once the steam pressure exceeds the rated pressure and it automatically closes when the pressure drops down to permissible value. Safety valves are usually mounted on top of the shell.

11. Which of the following statements about dead weight safety valve is FALSE?

a) It is unsuitable for Locomotive boilers

b) It is suitable for low pressure and low capacity boilers

c) To lift the valve the inside pressure must overcome the valve weight and the dead weights

d) It is suitable for high pressure boilers

Answer: d

Explanation: Large amount of weights will be required to balance high pressure; therefore, it is not suitable for high pressure boilers. It is used in low pressure and low capacity boilers. It cannot bear vibrations and movements, hence not suitable for locomotive boilers.

12. Spring loaded safety valve is suitable for locomotive and marine engines.

a) True

b) False

Answer: a

Explanation: Spring loaded safety valve unlike dead weight safety valve and lever safety valve is unaffected by vibrations and movements in vertical directions. It is therefore suitable for locomotive and marine engines.

13. Which of the following statements about high steam and low water safety valve is FALSE?

a) It can be used on portable boilers

b) Steam escapes out when water level falls below a certain level

c) Steam is released when its pressure exceeds a certain pressure

d) It is generally used in Lancashire and Cornish boiler

Answer: a

Explanation: High steam and low water safety valve is not suitable for locomotive and marine engines. It is designed such that when the steam pressure exceeds a certain pressure limit or when the water level falls below a certain limit the valve opens and the steam is released to atmosphere.

14. In water level indicator, the glass tube that indicated the water level is called _____

a) measurement tube

b) guard glass tube

c) gauge glass tube

d) level tube

Answer: c

Explanation: In a water level Indicator the glass tube that indicates the water level is called gauge glass tube. It is very hard. It produces shards when it breaks which could hurt the attendant. Therefore, the gauge glass is completely covered by guard glass which does not produce shards after breaking.

15. A Steam separator is a boiler mounting.

a) True

b) False

Answer: b

Explanation: A steam separator comes under boiler accessories and not under boiler mountings. Boiler mountings include – safety valves, water level indicators, pressure gauges, steam stop valve, feed check valve etc.

This set of Thermal Engineering Questions and Answers for Freshers focuses on “Boiler Mountings – 2”.

1. What is the function of fusible plug?

a) Protects the boiler from overheating

b) Prevents mixing of water and steam

c) Regulated the flow of steam outside boiler

d) It is opened to increase the flow of air for combustion

Answer: a

Explanation: Overheating damage to the boiler is prevented by fusible plug. Overheating occurs when water level is low. It Is fitted above the fire box. When water level falls below the level of fusible plug, it is exposed to steam and it melts. Steam and water rush to the fire box and extinguish the fire which prevents overheating of boiler.

2. Which of the following statements about fusible plug is FALSE?

a) It extinguishes fire when water level is low

b) It protects the boiler from overheating

c) It should have a very high melting point

d) It needs to be renewed after a period of time

Answer: c

Explanation: The melting point of fusible plug should be less than the temperature of the steam getting generated, since the fusible plug is supposed to melt when it comes in contact with the steam. Fusible plugs should usually be replaced after a period of about 2 years.

3. In fusible plug, the fusible metal is protected from fire by flange on the hollow gun metal plug.

a) True

b) False

Answer: a

Explanation: The fusible metal is to be melted only by the steam and not by the hot gases released from the combustion in the fire-box. The fusible metal is therefore protected from fire by flange on the hollow gun metal plug.

4. Which of the following statements about blow-off cock is FALSE?

a) It is used to discharge the mud that has settled at the bottom

b) It is used to empty the boiler for cleaning

c) Scale and sediments are also removed when the blow-off cock is opened

d) It is used to fill the boiler with water

Answer: d

Explanation: Blow-off cock is used to remove or blow out mud, sediments and scales that had settled down at the bottom of the boiler. Also it is used to empty the boiler when it is required to be cleaned. Feed check valve is used to fill water inside the boiler.

5. In a fusible plug, the substance that melts when it comes in contact with the steam is called _____

a) Melt

b) Meting wax

c) Fusible metal

d) Fusible wax

Answer: c

Explanation: Fusible metal melts when it comes in contact with the steam. Normally fusible plug is completely inside the water. When the water level falls down and the fusible plug is exposed to steam, the fusible metal melts and the fusible plug opens. The steam and water being under pressure rush through the opening and extinguish the fire.

6. What is the function of a feed check valve?

a) It regulates the flow of hot gases from the fire-box

b) It controls the flow of air to the combustion chamber

c) It controls the supply of steam outside the boiler

d) It controls the supply of water to boiler

Answer: d

Explanation: The function of feed check valve is to control the supply of water to the boiler. It also prevents the back flow of water when the pump is switched-off. It is installed slightly below the normal water level in the water space.

7. The part of the feed check valve which moves up and down automatically under the pressure of water on its gun metal seat is called _____

a) Check valve

b) Fusible plug

c) Flange

d) Safety valve

Answer: a

Explanation: Check valve moves up and down automatically under the pressure of water on its gun metal seat. It is designed such that it only allows water flow only in direction. This prevents the back-flow of water from the boiler to the pipe.

8. Which of the following statements about feed check valve is FALSE?

a) It controls the flow of water

b) It only allows flow of water in one direction

c) It is fitted on top of the boiler shell

d) It consists of check valve which automatically moves up and down under pressure of water

Answer: c

Explanation: The feed check valve is mounted in the water space slightly below normal level of water. It controls the flow of water in the boiler and also prevents the back-flow of water from the boiler back to the pipe when the pump is switched off.

9. Which of the following is NOT a boiler mounting?

a) Steam stop valve

b) Man hole

c) Economizer

d) Pressure gauge

Answer: c

Explanation: Steam stop valve, Man hole and pressure gauge are boiler mountings. Economizer is a boiler accessory. Boiler accessories increase the efficiency of the boiler while boiler mountings are attached for safe operation of the boiler.

10. Which of the following statement about stop valve  is FALSE?

a) It regulates the flow of steam

b) Fitted at top of the boiler

c) It connects boiler with the steam pipe

d) The smaller sized valve is called junction valve

Answer: d

Explanation: The smaller diameter valve that connects boiler and steam pipe is called stop valve and the larger one is called junction valve. It is mounted on top of the boiler shell. It is used to regulate the flow of steam outside the boiler.

11. Where is feed check valve mounted?

a) On top of the boiler shell

b) At the bottom of the boiler shell

c) Above the water level to be maintained

d) Slightly below the water level to be maintained

Answer: d

Explanation: The feed check valve is fitted slightly below the normal water level. The valve is used to supply water to the boiler. Being below normal water level there is a chance of back-flow of water to the pipe. The feed check valve is designed such that it prevents water from flowing back.

12. Feed check valve is used to empty the boiler when it is required to be cleaned.

a) True

b) False

Answer: b

Explanation: A feed check valve is used to supply water to the boiler. The boiler is emptied using the blow-off cock. Boilers are emptied for cleaning, inspecting and repairing. Bow-off cock is also used to remove the mud, scales and sediments that have settled down at the bottom of the boiler.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Boiler Accessories – 1”.

1. Boiler Accessories increase efficiency of the steam boilers.

a) True

b) False

Answer: a

Explanation: Boiler accessories are auxiliary plants that help in proper operation of the boiler and also increase its efficiency. Feed pumps, injector, superheater, air preheater, economizer, superheater are some commonly used boiler accessories.

2. Which of the following is NOT a boiler accessory?

a) Feed pump

b) Steam trap

c) Steam separator

d) Mud hole 

Answer: d

Explanation: The correct answer is mud hole . Feed pump, steam trap, steam separator are all boiler accessories whereas a mud hole is a boiler mounting. Boiler mounting increases the efficiency of the boiler.

3. Which of the following statements about feed pump is TRUE?

a) It is used to deliver feed water to the boiler

b) It is used to take out steam from the boiler

c) It is a boiler mounting

d) It delivers air to the combustion chamber

Answer: a

Explanation: Feed pump is used to deliver feed water to the boiler. Reciprocating pump and rotary pump are the commonly used feed pumps. Reciprocating pump is classified into single acting and double acting pump.

4. Which of the following statements about reciprocating feed pumps is FALSE?

a) Water is discharged in alternate strokes in single acting pump

b) Water is discharged in each stroke in double acting pump

c) They are supplied steam from the boiler itself

d) In double acting pump water us displaced by only one side of the piston

Answer: d

Explanation: In a double acting reciprocating pump, water is displaced by both side of the piston. This is the reason why water is discharged in each stroke. In case of single acting pump water is displaced be only one side of the piston and hence water is discharged in alternate strokes.

5. The rotating element of the rotary feed pump is called _____

a) impeller

b) propeller

c) armature

d) propulsor

Answer: a

Explanation: Impeller is the correct answer. Rotary pumps are usually run by an electric motor or small steam turbine. It makes use of the centrifugal force of the rotating impeller for pumping the water.

6. Which of the following is TRUE about an injector?

a) It feeds water into the boiler

b) Kinetic Energy of steam is utilized to increase velocity of feed water

c) It is used where there isn’t enough space to install feed pump

d) It is ideal for high capacity high pressure boilers

Answer: d

Explanation: Injector is used to feed water into a boiler. It is used in place of feed pump when there isn’t enough space to install a feed pump. It is usually used in vertical and locomotive boilers. It is not at all used in case of high capacity and high pressure boilers.

7. Pumping efficiency of an injector is high.

a) True

b) False

Answer: b

Explanation: The pumping efficiency of an injector is low. But its thermal efficiency is high. It is compact and has low initial cost. It cannot be used to force very hot water.

8. What is the function of economizer?

a) It is used to heat the feed water

b) It is used to improve the quality of steam

c) It is used to heat air being supplied for combustion

d) It separates suspended water particles from steam

Answer: a

Explanation: Economizer makes use of waste heat from the hot gases to heat the feed water before suppling it to the boiler. There are two types of economizers – independent type and integral type. Independent type economizer is installed in the pathway of the hot gases. Integral type economizer is situated in the boiler setting itself.

9. Which of the following statements about vertical tube economiser 

 is FALSE?

a) It is used with boilers operating below 25 bars

b) It is an integral type economiser

c) The direction of flow of flue gases is opposite to that of water

d) It consists of a blow-off cock for removal of sediments

Answer: b

Explanation: Green’s economiser is an independent type economiser. It is put into service for boilers of medium pressure range upto about 25 bar. The direction of flow of flue gases around the tubes is opposite to the flow of the water. A blow-off cock is allocated at the end of vertical pipes for removal of sediment deposits.

10. What is the function of an air preheater?

a) It increases the temperature of air before it enters the furnace

b) It increases the temperature of hot gases before they are released into the atmosphere

c) It reduces the temperature of hot gases before they are released into the atmosphere

d) It improves the quality of steam by mixing it with hot air

Answer: a

Explanation: The function of an air preheater is to heat the air that is supplied to the furnace. It helps in the combustion and eases the burning of coal. It consists of plates or tubes with hot gases on one side and air on the other.

This set of Thermal Engineering Interview Questions and Answers for freshers focuses on “Boiler Accessories – 2”.

1. Convective superheater is placed in the furnace and it receives heat from the burning fuel through radiation.

a) True

b) False

Answer: b

Explanation: Convective superheaters receive heat from flue gases. Radiant superheaters are placed in the furnace. The wall tubes of radiant superheater receive heat from the burning fuel through radiant process. Radiant superheaters are usually used where the steam is to be raised to a high superheat temperature.

2. Steam separator reduces the wetness of the steam before it enters the engine or turbine.

a) True

b) False

Answer: a

Explanation: The function of a steam separator is to reduce the wetness of the steam by removing the water particles from the steam before it enters the steam turbine or engine. It is installed on the main pipe delivering the steam as close as possible to the engine.

3. Which of the following is NOT a valid type of steam separator?

a) Baffle  type

b) Tubular type

c) Reverse current type

d) Centrifugal type

Answer: b

Explanation: On the basis of principle of operation steam separators are classified into baffle or impact type, reverse current type and centrifugal type. There is a loss of heat in the steam pipe after the steam leaves the boiler which produces wetness. This wetness could damage the turbine or engine. Steam separator is installed close to the engine to remove the water particles and reduce the wetness of the steam.

4. What is the function of a steam trap?

a) It is used to store stem once the generation is complete

b) It prevents the leakage of steam from the boiler

c) It is used to increase the temperature of the steam

d) It captures the condensed steam from steam pipes and steam separators

Answer: d

Explanation: Steam trap captures the condensed steam from the steam pipes, steam jackets and steam separators. It doesn’t allow steam to escape. Removal of condensed steam is important because wetness could damage the engine or turbine.

5. Which of the following statement about convective superheater is TRUE?

a) It is located in the furnace

b) It absorbs heat from the burning fuel through radiant process

c) It absorbs heat from the flue gases

d) It is used where very high superheat temperature is required

Answer: c

Explanation: Convective superheaters utilize the heat from flue gases. It is radiant superheater that is located in the furnace and it absorbs heat from the burning fuel through radiant process. Radiant superheaters are generally used when a high superheat temperature is required.

6. Which of the following is NOT a valid type of air preheater?

a) Tubular type

b) Plate type

c) Temperature type

d) Storage type

Answer: c

Explanation: There are three types of air preheater – tubular type, storage type and plate type. There is no such category as temperature type. In tubular type air preheaters, the hot gases after leaving the boiler travel through inside of the tubes in a direction opposite to that of air and transfer some part of their heat to the air. In case of plate type air preheaters, the air absorbs the heat from the flue gases rushing through the heater at high velocity on the opposite side of the plate.

7. The degree of preheating does not depend on _____

a) type of fuel

b) initial temperature of air

c) type of fuel burning equipment

d) boiler and furnace ratings

Answer: b

Explanation: The degree of preheating doesn’t depend on initial temperature of air. It depends on type of fuel, type of fuel burning equipment and ratings at which boiler and furnace are operated. An air preheater increases the temperature of the air being supplied to the furnace for better combustion.

8. Which of the following statements about air preheaters is FALSE?

a) It increases the temperature of air being supplied to the furnace

b) Storage type air preheaters cannot be used in larger plants

c) The temperature of the hot gases released through the chimney is kept minimum

d) It facilitates the burning of fuel

Answer: b

Explanation: The temperature of the hot gases released into the atmosphere through the chimney is kept minimum for loss of heat. Storage type air preheaters are widely used in larger plants. An air preheater increases the temperature of the air being supplied to the furnace. It smoothens the burning of fuel and helps in combustion process.

9. What is the function of a superheater?

a) It increases the temperature of the air being supplied to furnace

b) It increases the temperature of steam above saturation temperature

c) It increases the temperature of feed water

d) It increases the temperature of the flue gases leaving the furnace

Answer: b

Explanation: The function of a superheater is to increase the temperature of steam leaving the boiler above saturation temperature. Economiser increases the temperature of the feed water being supplied to the boiler. Air preheater increases the temperature of the air being supplied to the furnace.

10. Which of the following statements about superheaters is TRUE?

a) Condensation losses in the cylinders and steam pipes are reduced

b) Turbine erosion is avoided

c) Increases efficiency of the steam plant

d) Steam consumption of the engine is increased

Answer: d

Explanation: Superheater reduces the steam consumption of the engine or turbine. It eliminates turbine erosion. It also reduces the losses due to condensation in the steam pipes and cylinders. Superheater being a boiler accessory increases the efficiency of the steam plant.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Draught – Definition and Classification”.

1. What is draught?

a) The condition of a dry boiler is called draught

b) It’s a small pressure difference which causes the flow of gases

c) It is a boiler mounting

d) It is a boiler accessory

Answer: b

Explanation: Draught is the small pressure difference which causes the flow of gas to take place. Draught helps to carry air to the fire-box and then take away the gaseous products of combustion out. Sufficient amount of air is necessary for proper combustion.

2. Draught is classified into _____

a) Artificial and natural draught

b) Induced and forced draught

c) Steam jet and mechanical draught

d) Chimney draught and boiler draught

Answer: a

Explanation: Draught is classified into artificial draught and natural draught. Natural draught is also known as chimney draught since it makes use of a chimney of sufficient height to create the required pressure difference.

3. Artificial draught is classified into steam jet and mechanical draught.

a) True

b) False

Answer: a

Explanation: Artificial draught is classified into steam jet and mechanical draught. Steam jet draught is further classified into induced and forced draught. Mechanical draught is classified into induced fan, balanced  and forced fan draught.

4. Natural draught is obtained using _____

a) Economizer

b) Induced fan

c) Chimney

d) Induced and forced fan

Answer: c

Explanation: Chimney is used to obtain natural draught. Chimney creates the necessary pressure difference for draught. Use of chimney also helps in discharging the products of combustion to a height where they won’t be affecting the surroundings.

5. Which of the following is the correct formula for calculating the pressure difference causing the gas flow through the combustion chamber?

a) gH

b) gH

c) gH

d) gH

Answer: a

Explanation: The correct formula for calculating the pressure difference causing the gas flow though the combustion chamber  is ‘gH’.

6. Calculate the draught in mm of water if the chimney height is 30 m, the temperatures of hot gases inside chimney and outside air are 330°C and 30°C respectively. Also, the furnace is supplied with 15 kg of air per kg of coal burnt.

a) 14.67 mm of water

b) 17.64 mm of water

c) 16.22 mm of water

d) 16.19 mm of water

Answer: c

Explanation:

H = 30 m, Tg = 330°C or 603 K, T a = 30°C or 303 K, ma = 15 kg of air/kg of coal

Draught in mm of water, h w = 353*H*\

 

 

 

\} \)

Substituting the respective values

h w = 353*30*\

 

 

 

\} \)

h w = 16.22 mm of water

7. A chimney is 25 m high. The temperature of hot gases inside chimney is 300°C and temperature of air outside chimney is 30°C. If 20 kg of air is supplied to the furnace per kg of coal find the equivalent column of hot gases that would produce the same draught pressure.

a) 20.02 m

b) 22.40 m

c) 24.42 m

d) 22.60 m

Answer: a

Explanation:

H = 25 m, T g = 300°C or 573 K, T a = 30°C or 303 K, m a = 20 kg of air/kg of coal

Height of equivalent column of hot gases , H 1 =H*\

 

\frac{Tg}{Ta}-1\} \)

Substituting the respective values

H 1 =25*\

 

\frac{573}{303}-1\} \)

H 1 = 20.02 m

8. Calculate how much air is used per kg of coal burnt, if the chimney height is 35 m which creates a draught of 20 mm of water. The temperature of hot gases inside the chimney is 330°C and temperature of air outside is 27°C.

a) 29.52 kg of air per kg of coal

b) 21.34 kg of air per kg of coal

c) 22.45 kg of air per kg of coal

d) 19.54 kg of air per kg of coal

Answer: a

Explanation:

H = 35 m, T g = 330°C or 603 K, T a = 27°C or 300 K, h w = 20 mm of water

We know that,

h w = 353*H*\

 

 

 

\}\)

After substituting the respective values

20 = 353*35*\

 

 

\frac{1}{603}\}\)

Solving for m a we get

m a = 29.52 kg of air per kg of coal

9. A 28 m chimney is full of hot gases at temperature 300°C. The air supplied for complete combustion of 1 kg of coal is 18 kg. calculate draught in N/m 2 if the temperature of outside air is 25°C.

a) 140.32 N/m 2

b) 187.54 N/m 2

c) 150.32 N/m 2

d) 146.76 N/m 2

Answer: d

Explanation:

H = 28 m, T g = 300°C or 573 K, T a = 25°C or 298 K, m a = 18 kg of air/kg of coal burnt

draught (in N/m 2 ) = 353*g*H*\

 

 

 

\}\)

substituting the respective values

draught (in N/m 2 ) = 353*9.81*28*\

 

 

 

\}\)

draught = 146.76 N/m 2

10. A chimney 30 m high produces a draught of 16 mm of water. The temperature of hot gases inside the chimney is 300°C. Calculate the temperature of outside air if 20 kg of air is supplied for complete combustion of 1 kg of fuel.

a) 30°C

b) 27°C

c) 25°C

d) 20°C

Answer: b

Explanation:

H = 30 m, h w = 16 mm of water, T g = 300°C or 570 K, m a = 20 kg of air/kg of fuel

we know that,

h w = 353*H*\

 

 

 

\}\)

Substituting the respective values, we get,

16 = 353*30*\

 

 

 

\}\)

Solving for T a we get,

T a = 299.1 ≈ 300 K or 27°C

11. Find the temperature of the hot gases inside the chimney if the draught produced by a 25 m high chimney is 15 mm of water. The temperature of the air outside is 25°C and 18 kg of air is supplied for complete combustion of 1 kg of fuel.

a) 298°C

b) 330°C

c) 456°C

d) 364°C

Answer: d

Explanation:

H = 25, T a = 25°C or 298 K, m a = 18 kg of air/kg of fuel, draught = 15 mm of water

We know that,

h w = 353*H*\

 

 

 

\}\)

Substituting the values

15 = 353*25*\

 

 

 

\}\)

Solving for T g we get,

T g = 637.42 K ≈ 637 K or 364°C

12. Find draught in terms of column of hot gases, if the temperature of hot gases inside the chimney is 350°C. The height of the chimney is 30 m. The temperature of outside air is 25°C. The air supplied for complete combustion of 1 kg of fuel is 20 kg.

a) 29.73 m

b) 32.46 m

c) 45.32 m

d) 30.25 m

Answer: a

Explanation:

H = 30 m, T g = 350°C or 623 K, T a = 25°C or 298 K, m a = 20 kg of air/kg of fuel

We know that, draught in terms of column of hot gases is given by –

H 1 = H*\

 

\frac{Tg}{Ta}-1\} \)

Substituting the values H 1 = 30*\

 

\frac{623}{298}-1\} \)

Therefore, H 1 = 29.73 m

13. A 27 m high chimney produces a draught of 30 m in terms of Column of hot gases. The amount of air supplied for complete combustion of 1 kg of fuel is 20 kg. If the temperature of outside air is 27°C, determine the temperature of the hot gases inside the chimney.

a) 456°C

b) 392°C

c) 423°C

d) 398°C

Answer: b

Explanation:

H = 27 m, H 1 = 30 m , m a = 20 kg of air/kg of fuel, T a = 27°C or 300 K

we know that,

H 1 =H*\

 

\frac{Tg}{Ta}-1\} \)

Substituting the values, we get

30 =27*\(\{(\frac{20}{21}\frac{Tg}{300}-1\} \)

Solving for T g , we get

T g = 665 K or 392°C

14. How much air is used per kg of coal burnt in a boiler having a chimney of height 32 m to create draught of 170 N/m 2 . The temperature of outside air is 25°C and hot gases inside chimney is 330°C.

a) 10.16 kg of air per kg of coal burnt

b) 12.00 kg of air per kg of coal burnt

c) 14.32 kg of air per kg of coal burnt

d) 19.25 kg of air per kg of coal burnt

Answer: a

Explanation:

H = 32 m, draught = 170 N/m 2 , T a = 25°C or 298 K, T g = 330°C or 603 K

The following formula is used to calculate draught in N/m 2 –

draught (in N/m 2 ) = 353*g*H*\

 

 

 

\}\)

Substituting the respective values, we get

170 = 353*9.81*32*\

 

 

 

\}\)

Solving the above equation for ma, we get

m a = 10.16 kg of air per kg of coal burnt

15. Find out the temperature of outside air if a chimney 30 m high creates a draught of 25 m in terms of column if hot gases. The temperature of hot gases is 330°C. 18 kg of air is to be supplied for complete combustion of 1 kg of fuel.

a) 38°C

b) 28°C

c) 25°C

d) 32°C

Answer: a

Explanation:

H = 30 m, H 1 = 25 m of hot gases, T g = 330°C or 603 K, m a = 18 kg of air/kg of fuel

we know that,

H 1 =H*\(\{\frac{ma}{ma +1}\frac{Tg}{Ta}-1\} \)

Substituting the values, we get

25 =30*\(\{\frac{18}{19}\frac{603}{Ta}-1\} \)

Solving for T g , we get

T g = 311.6 K ≈ 311 K or 38°C

Symbols

ρ a – mass density of air

m a – mass of air supplied per kg of fuel

ρ g – mass density of hot gases

h w – draught in terms of column of water

H 1 – draught in terms of column of hot gases

T a – absolute temperature of atmosphere

T g – average absolute temperature of chimney gases

g – gravitational acceleration

H – chimney height

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Natural Draught”.

1. A 25 m high chimney is filled with hot gases of temperature 330°C. The temperature of outside air is 27°C. If 20 kg of air is supplied for complete combustion of 1 kg of fuel, calculate the velocity of hot gases though the chimney.

a) 21.18 m/s

b) 25.65 m/s

c) 18.56 m/s

d) 15.65 m/s

Answer: a

Explanation: H = 25 m, T g = 330°C or 603 K, T a = 27°C or 300 K, m a = 20 kg of air/kg of fuel

we know that,

Draught in terms of column of hot gases, H 1 = H*\

 

\frac{T_g}{T_a} -1\}\)

Substituting the values

H 1 = 25*\

 

\frac{603}{300}-1\}\)

H 1 = 22.86 m

The velocity of hot gases through chimney is given by, V = \(\sqrt{2*g*H_1}\)

V = \(\sqrt{}\)

V = 21.18 m/s.

Symbols:

h – Actual draught in terms of column of hot gases

H 1 – Theoretical draught in terms of column of hot gases

m a – mass of air supplied per kg of fuel

T a – absolute temperature of atmosphere

T g – average absolute temperature of chimney gases

H – chimney height.

2. A chimney has a height 30 m. The ambient temperature and the temperature of hot gases inside the chimney are 27°C and 300°C respectively. Find the velocity of the flue gases passing through the chimney if 50% of the theoretical draught is lost in friction at the grate and passage. It is given that 20 kg of air is required for complete combustion of 1 kg of fuel.

a) 21.35 m/s

b) 20.65 m/s

c) 15.52 m/s

d) 10.32 m/s

Answer: c

Explanation: H = 30 m, T a = 27°C or 300 K, T g = 300°C or 573 K, m a = 20 kg of air/kg of fuel

we know that,

Draught in terms of column of hot gases, H 1 = H*\

 

\frac{T_g}{T_a} -1\}\)

Substituting the values, we get

H 1 = 30*\

 

\frac{573}{300}-1\}\)

H 1 = 24.57 m

Available head, h = 0.5*H1

h = 0.5 * 24.57

h = 12.28 m

Velocity of the flue gases , V = \(\sqrt{2*g*h}\)

V = \(\sqrt{2*9.81*12.28}\)

V = 15.52 m/s.

Symbols:

h – Actual draught in terms of column of hot gases

H 1 – Theoretical draught in terms of column of hot gases

m a – mass of air supplied per kg of fuel

T a – absolute temperature of atmosphere

T g – average absolute temperature of chimney gases

H – chimney height

g – gravitational acceleration

3. The velocity of flue gases though chimney is 23 m/s. The chimney height is 35 m. The temperature of the flue gases is 300°C. The flue gases formed per kg of fuel burnt is 23 kg. Find out the ambient temperature.

a) 30°C

b) 27°C

c) 40°C

d) 37°C

Answer: d

Explanation: V = 23 m/s, H = 35 m, T g = 300°C or 573 K, m a +1 = 23 kg of flue gases/kg of fuel

m a = 22 kg of air/kg of fuel burnt

we know that,

Velocity of flue gases though chimney, V = \(\sqrt{2*g*H_1}\)

Substituting the values

23 = \(\sqrt{2*9.81*H_1}\)

Solving for H 1 , we get

H 1 = 26.96 m

Also,

H 1 = H*\

 

\frac{T_g}{T_a} -1\}\)

Substituting the values

26.96 = 35*\

 

 \frac{573}{T_a}-1\}\)

Solving for T a , we get

T a = 309.6 K ≈ 310 K or 37°C.

4. The velocity and temperature of flue gases passing through a chimney are 20 m/s and 330°C. 18 kg of air is supplied for complete combustion of 1 kg of fuel. The ambient temperature is 27°C. Determine the chimney height 40% of the theoretical draught is lost in friction at the grate and passage.

a) 58.32 m

b) 37.57 m

c) 29.65 m

d) 35.26 m

Answer: b

Explanation: V = 20 m/s, T g = 330°C or 603 K, m a = 18 kg of air/kg of fuel, T a = 27°C or 300 K

we know that,

V = \(\sqrt{2*g*h}\)

Substituting the respective values

20 = \(\sqrt{2*9.81*h}\)

Solving for h, we get

h = 20.38 m

Theoretical draught in terms of column of hot gases, H 1 = \(\frac{h}{60}\)*100

H 1 = \(\frac{20.38}{60}\)*100

H 1 = 33.97 m

Also

H 1 = H*\

 

\frac{T_g}{T_a} -1\}\)

Substituting the respective values

33.97 = H*\

 

\frac{603}{300}-1\}\)

Solving for ‘H’, we get

H = 37.57 m.

5. Which of the following expressions represents the correct ratio of absolute temperature of the chimney gases and absolute temperature of outside air when the discharge through the chimney is maximum.

a) \

 

 

 \

 

 

 \

 

 

 \)

d) \

 

 

 \)

Answer: c

Explanation: The natural draught or the chimney draught is more effective when the maximum weight of hot gases is discharged in a given time.

At this condition the ratio of absolute temperature of hot gases to absolute temperature of outside air bears a specific value, which is

\

 

 

 \)

6. What is the ratio of draught in terms of column of hot gases to the chimney height when the discharge though chimney is maximum?

a) 2.0

b) 1.5

c) 1.0

d) 0.5

Answer: c

Explanation: When the discharge though chimney is maximum, the ratio of draught in terms of column of hot gases to chimney height becomes 1.

That is, H 1 = H

It is also the maximum value of H 1 .

7. The frictional resistance offered to the flow of flue gases by the grate and gas passages result in draught losses.

a) True

b) False

Answer: a

Explanation: Apart from resistance offered by grate and gas flow passage, draught is also lost near the bends in the gas flow circuit. Also presence of friction head in equipments like superheater, economizer etc. causes loss in a draught.

8. The draught produced by the chimney is due to the difference in densities between the column of hot gases inside the chimney and the cold air present outside.

a) True

b) False

Answer: a

Explanation: Draught is classified into natural draught and artificial draught. Natural draught is produced by chimney. It produces draught by virtue of density difference between the hot air column inside the chimney and the cold air outside.

9. A chimney of height 25 m is used to provide natural draught. The ambient temperature is 25°C and the temperature of the hot gases inside the chimney is 352°C. If the discharge though the chimney is maximum determine the mass of air supplied per kg of fuel burnt.

a) 20.55 kg of air/kg of fuel burnt

b) 18.54 kg of air/kg of fuel burnt

c) 15.23 kg of air/kg of fuel burnt

d) 25.21 kg of air/kg of fuel burnt

Answer: a

Explanation: T a = 25°C or 298 K, T g = 352°C or 625 K

It is given that the discharge is maximum, therefore

\

 

 

 \)

Substituting the respective values

\

 

 

 \)

Solving for m a , we get

m a = 20.55 kg of air/kg of fuel burnt.

10. Determine the temperature of hot gases inside the chimney if the temperature of cold air outside is 27°C and 20 kg of air is required for complete combustion of 1 kg of fuel. It is given that the discharge though the chimney is maximum.

a) 335°C

b) 300°C

c) 318°C

d) 357°C

Answer: d

Explanation: T a = 27°C or 300 K, m a = 20 kg of air/kg of fuel

According to the given condition

\

 

 

 \)

Substituting the respective values

\

 

 

 \)

Solving for T g , we get

T g = 630 K or 357°C.

11. A chimney of height 45 m is full of hot gases at temperature 330°C. If the discharge through the chimney is maximum, determine the draught in terms of column of hot gases. The ambient is 25°C.

a) 45 m

b) 35 m

c) 25 m

d) 39 m

Answer: a

Explanation: H = 45 m, T g = 330°C or 603 K, T a = 25°C or 298 K

Since it is given that the discharge though the chimney is maximum.

Draught in terms of column of hot gases, H 1 = H 

H 1 = 45 m.

12. A boiler has a chimney of height 45 m. The ambient temperature is 25°C. The temperature of the hot gases inside chimney is 330°C. Calculate the draught in terms of column of water if it is given that discharge through the chimney is maximum.

a) 40.23 m

b) 35.45 m

c) 26.65 m

d) 24.32 m

Answer: c

Explanation: H = 45 m, T a = 25°C or 298 K, T g = 330°C or 603 K

It is given that discharge though the chimney is maximum, therefore

h w = 176.5 \(\frac{H}{T_a} \)

Substituting the values, we get

h w = 176.5 \(\frac{45}{298} \)

h w = 26.65 m.

Symbols –

h w – draught in terms of column of water

T a – absolute temperature of atmosphere

T g – average absolute temperature of chimney gases

H – chimney height.

13. The temperature of flue gases was observed to be 350°C with a chimney  of 45 m. The same draught was developed using an induced draught fan and the temperature of the flue gases was measured to be 150°C. The mass of air supplied for complete combustion of 1 kg of fuel is 20 kg. The ambient temperature is 30°C. Assuming c p = 1.004 kJ/kg for flue gases calculate the efficiency of chimney.

a) 0.21%

b) 0.56%

c) 1.23%

d) 0.11%

Answer: a

Explanation: H = 45 m, T g1 = 350°C or 623 K, T g2 = 150°C or 423 K, m a = 20 kg of air/kg of fuel

Ta = 30°C or 303 K

Chimney efficiency is given by-

η = \(\frac{H\big\{

 

\frac{T_{g1}}{T_a}-1\big\}*g}{cp

*1000 }*100 \)

Substituting the respective values

η = \(\frac{45\big\{

 

\frac{623}{303}-1\big\}*9.81}{1.004*1000}*100 \)

η = 0.21%.

Symbols –

m a – mass of air supplied per kg of fuel

H 1 – draught in terms of column of hot gases

T a – absolute temperature of atmosphere

T g – average absolute temperature of chimney gases

g – gravitational acceleration

H – chimney height.

14. A chimney is used to develop natural draught of 20 mm of water. The ambient temperature is 30°C. Determine the chimney height if the discharge though the chimney is maximum.

a) 25.36 m

b) 34.33 m

c) 45.32 m

d) 41.32 m

Answer: b

Explanation: h w = 20 m of water, T a = 30°C or 303 K

Since the discharge through the chimney is maximum

h w = 176.5 \(\frac{H}{T_a} \)

Substituting the respective values

20 = \(176.5\frac{H}{303} \)

Solving for H, we get

H = 34.33 m.

15. Determine the ambient temperature if the temperature of flue gases inside the chimney is 363°C and the 20 kg of air is supplied for complete combustion of 1 kg of fuel. The discharge though the chimney is maximum.

a) 27°C

b) 35°C

c) 30°C

d) 25°C

Answer: c

Explanation: T g = 363°C or 636 K, m a = 20 kg of air/kg of fuel

Since the discharge though the chimney is maximum

\

 

 

 \)

Substituting the values

\

 

 

 \)

Solving for T a , we get

T a = 302.86 K ≈ 303 K or 30°C.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Artificial Draught”.

1. Mechanical draught is used in locomotives and small installations.

a) True

b) False

Answer: b

Explanation: Mechanical draught is used in case of central power stations. For locomotives and small installations steam jet draught is used. Steam jet and mechanical draught are types of artificial draught.

2. In case of Induced draught, where is the fan/blower installed?

a) Near the base of chimney

b) At the top of the chimney

c) At the grate

d) At the base of the boiler

Answer: a

Explanation: In induced draught the fan or blower is installed near the base of the chimney. It creates a partial vacuum in the furnace and the combustion products are extracted out of the main flow and they pass through the chimney.

3. Where is fan/blower located in case of forced draught?

a) Near the base of chimney

b) At the top of the chimney

c) At the grate

d) At the base of the boiler

Answer: d

Explanation: The fan/blower is located at or near the base of the boiler in case of forced draught. The furnace is to be sealed properly so that combustion products don’t leak out of the boiler. It creates positive pressure draught.

4. Which of the following statement is TRUE about balanced draught?

a) No fan/blower is installed

b) A fan/blower is installed in the middle of the chimney

c) One fan/blower is installed at the base of the boiler and one near the base of the chimney

d) Two fans/blowers are installed both at the base of the boiler

Answer: c

Explanation: Balanced draught is the combination of forced draught and induced draught. In forced draught the fan/blower is located near or at the base of the chimney. In induced draught the fan/blower is located near the base of the boiler.

5. Which of the following statements is FALSE regarding artificial draught?

a) There is no need of water cooled bearing in forced draught fan

b) Fan size and power required by a forced draught fan are 1/5 to ½ of that required for an induced draught fan

c) Tendency for air leak into the boiler furnace is less in Induced draught fan as compared to forced draught fan

d) In forced draught the fan/blower is installed near the base of the boiler

Answer: c

Explanation: Tendency for air leak into the boiler furnace is less in forced draught than in induced draught. Forced draught fan doesn’t need water cooled bearings. Since forced draught fan handles cold air, its size and power requirement is 1/5 to ½ of that required by an induced draught fan.

6. Calculate power of motor required to drive a forced draught fan which maintains a draught of 45 mm of water the temperatures of flue gases leaving the boiler and air inside the boiler house are 300°C and 25°C respectively. It is given that the air supplied for complete combustion of 1 kg of fuel is 20 kg. 1800 kg of coal is burnt per hour. The efficiency of the fan is 80%. (Take Volume of air at N.T.P as 0.7734 m 3 )

a) 9.56 kW

b) 4.66 kW

c) 5.65 kW

d) 6.32 kW

Answer: b

Explanation:

h = 45 mm of water, T g = 300°C or 573 K, T a = 25°C or 298 K, m a = 20 kg of air/kg of fuel

M = 1800 kg coal/hour, V 0 = 0.7734 m 3 , η = 0.8

Power required to drive a forced draught fan is given by

P = 0.998*10 -8 \

 

 \)

Substituting the values,

P = 0.998*10 -8 \

 

 \)

P = 4.66 kW.

7. A motor rated 5 kW runs a forced draught fan. It maintains a constant draught of 50 mm of water. Mass of air supplied per kg of fuel is 18 kg. Determine the efficiency of the fan is the ambient temperature is 27°C and 1500 kg of coal is consumed per hour respectively. (Take V 0 = 0.7734 m 3 )

a) 63%

b) 78%

c) 84%

d) 59%

Answer: a

Explanation:

P = 5kW, h = 50 mm of water, m a = 18 kg air/kg fuel, T a = 27°C or 300 K, M = 1500 kg coal/hr

V 0 = 0.7734 m 3

We know that,

P = 0.998*10 -8 \

 

 \)

Substituting the values,

5 = 0.998*10 -8 \

 

 \)

Solving for η, we get

η = 0.63 or 63%.

8. A motor which runs an induced draught fan, maintains a constant draught of 55 mm of water. The efficiency of the fan is 91%. The ambient temperature is 30°C and the temperature of the flue gases leaving the boiler is 330°C. The coal consumption rate is 1850 kg/hr. The volume of air at N.T.P. should be taken as 0.7734 m 3 . Find the power of the motor if 20 kg of air is required for complete combustion of 1 kg of fuel.

a) 4.65 kW

b) 7.69 kW

c) 12.65 kW

d) 10.41 kW

Answer: d

Explanation:

h = 55 mm of water, η = 0.91, T a = 30°C or 303 K, T g = 330°C or 603 K, M = 1850 kg/hr

V 0 = 0.7734, m a = 20 kg air/kg fuel

Power of an induced draught fan is given be –

P = 0.998*10 -8 \

 

 \)

Substituting the respective values

P = 0.998*10 -8 \

 

\)

P = 10.41 kW.

9. Determine the ratio of power required to run a forced draught fan to power required to run an induced draught fan, considering same efficiency and same draught. The ambient temperature and temperature of flue gases leaving the boiler are 28°C and 300°C respectively. The coal consumption rate and mass of air supplied for complete combustion of fuel are also same.

a) 0.52

b) 0.23

c) 0.65

d) 0.78

Answer: a

Explanation:

T a = 28°C or 301 K, T g = 300°C or 573 K

According to the given conditions,

\(\frac{Power \, required \, by \, a \, Forced \, draught \, fan}{Power \, required by \, an \, induced \, draught \, fan} = \frac{T_a}{T_g} \)

Therefore,

\(\frac{Power \, required \, by \, a \, Forced \, draught \, fan}{Power \, required by \, an \, induced \, draught \, fan} = \frac{301}{573} \)

\(\frac{Power \, required \, by \, a \, Forced \, draught \, fan}{Power \, required by \, an \, induced \, draught \, fan}\) = 0.52

10. The temperature of the flue gases leaving the boiler is 330°C and the ambient temperature is 27°C. Find the power required by an induced draught fan if the power required by a forced draught fan under the same circumstances  is 3.6 kW.

a) 7.24 kW

b) 8.54 kW

c) 5.78 kW

d) 4.27 kW

Answer: a

Explanation:

Let power required by an Induced draught fan be Pi and Forced draught fan be Pf.

T a = 27°C or 300 K, T g = 330°C or 603K, Pf = 3.6 kW

According to the given conditions,

\(\frac{Pf}{Pi}=\frac{T_a}{T_g} \)

or

Pi = \(\frac{T_g}{T_a}*Pf \)

Substituting the values

Pi = \(\frac{603}{300}*3.6 \)

Pi = 7.24 kW.

11. Which of the following is FALSE about steam jet draught?

a) Low grade fuel can also be used

b) Requires low maintenance

c) Steam jet draught is a type of artificial draught

d) In case of forced steam jet draught, the jet is directed into smoke box

Answer: d

Explanation: Steam jet draught and mechanical draught are both types of artificial draught. Steam jet draught requires low maintenance and it is simple and economical. Various types of low grade fuel can be used with this system. According the location of steam jet, it is classified into Forced and induced stem jet draught. If the jet is before the grate it is forced steam jet draught and if the steam jet is directed into the smoke box it is induced steam jet draught.

12. A motor of power 8.7 kW is used to run an induced draught fan which maintains a draught of 50 mm of water. 19 kg of air is supplied for complete combustion of 1 kg of fuel. The temperature of hot gases 325°C. The efficiency of the fan is 85%. Determine the coal consumption rate. (Take V 0 = 0.7734 m 3 )

a) 1523 kg/hr

b) 1465 kg/hr

c) 1686 kg/hr

d) 1785 kg/hr

Answer: c

Explanation:

P = 8.7 kW, h = 50 mm of water, m a = 19 kg of air/kg of fuel, T g = 325°C or 598 K, V 0 = 0.7734 m 3

η = 0.85

We know that,

P = 0.998*10 -8 \

 

 \)

Substituting the values

8.7 = 0.998*10 -8 \

 

\)

Solving for “M”, we get

M = 1686.47 kg/hr ≈ 1686 kg/hr.

13. A motor of power 4 kW is used to drive a forced draught fan. The draught obtained is 45 mm of water. The temperature of the flue gases leaving the boiler is 300°C. The mass of air supplied per kg of fuel is 15 kg. The volume of air at N.T.P. is 0.7734 m 3 . The coal consumption rate is 1960 kg/hr. Calculate the ambient temperature if the efficiency of the fan is 78%.

a) 25°C

b) 32°C

c) 27°C

d) 30°C

Answer: b

Explanation:

P = 4kW, h = 45 mm of water, T g = 300°C or 573 K, m a = 15 kg of air/kg of fuel, V 0 = 0.7734 m 3

M = 1960 kg/hr, η = 0.78, T a = ?

For forced draught fan,

P = 0.998*10 -8 \

 

 \)

Substituting the respective values

4 = 0.998*10 -8 \

 

\)

Solving for T a , we get

T a = 305.53 K ≈ 305 K or 32°C

14. Calculate the draught in mm of water using the following data.

Power of the motor = 5 kW

Efficiency of the motor = 75%

Volume of the air at N.T.P. = 0.7734 m 3

Mass of air supplied per kg of fuel = 18 kg

Coal consumption rate = 1800 kg/hr

Ambient temperature = 25°C

Temperature of the gases leaving the boiler = 330°C

The motor drives a forced draught fan.

a) 50.31 mm of water

b) 45.32 mm of water

c) 41.32 mm of water

d) 17.65 mm of water

Answer: a

Explanation:

P = 5kW, V 0 = 0.7734 m 3 , η = 0.75, m a = 18 kg of air/ kg of fuel, M = 1800 kg/hr

T a = 25°C or 298 K, Tg = 330°C or 603 K

For a forced draught fan

P = 0.998*10 -8 \

 

 \)

Substituting the values

5 = 0.998*10 -8 \

 

 \)

Solving for “h”, we get

h = 50.31 mm of water.

15. Assuming same efficiency for the same draught and neglecting leakage, power required by induced draught fan is greater than power required by forced draught fan.

a) True

b) False

Answer: a

Explanation:

According to the given condition

\(\frac{Power \, required \, by \, an \, induced \, draught \, fan}{Power \, required \, by \, a \, Forced \, draught \, fan} = \frac{T_g}{T_a} \)

Since T g > T a

Power requirement of an Induced draught fan > Power required by forced draught fan

This set of Basic Thermal Engineering Questions and Answers focuses on “Evaporative Capacity and Equivalent Evaporation”.

1. Which of the following CANNOT be the unit of ‘evaporative capacity’?

a) kg of steam/h

b) kg of steam/h/m 2 of heating surface

c) kg of steam/kg of air supplied

d) kg of steam/kg of fuel fired

Answer: c

Explanation: Evaporative capacity is not expressed as kg of steam per kg of air supplied. It is expressed usually in kg of steam per hour. It is also expressed in kg of steam per hour per square meter of heating surface and kg of steam per kg of fuel fired.

2. Which of the following is the correct formula for factor of evaporation?

a) f = \

 

 f = \

 

 f = \

 

 f = \(\frac{h_{f1}-h}{1257} \)

Answer: a

Explanation: Factor of evaporation is the ratio of heat that is received by 1 kg of water under working conditions to that received by 1 kg of water evaporated from and at 100°C.

Factor of evaporation is calculated by the following formula –

f = \(\frac{h-h_{f1}}{2257} \)

3. Which of the correct formula for calculating equivalent evaporation?

a) m e =\

 

 m e =\

 

 m e =\

 

 m e =\(\frac{

}{2257} \)

Answer: a

Explanation: Amount of water evaporated from water at 100°C to dry and saturated steam at 100°C is called equivalent evaporation.

It is calculated by the following formula

m e =\(\frac{m_a 

}{2257} \)

4. 64 kg of steam is produced at 14 bar pressure having a dryness fraction of 0.82. The feed water temperature in the boiler is 39°C. Determine equivalent evaporation if mass of coal consumed is 8 kg.

a) 5.05 kg of steam/kg of fuel

b) 7.05 kg of steam/kg of fuel

c) 8.05 kg of steam/kg of fuel

d) 10.25 kg of steam/kg of fuel

Answer: c

Explanation: Given,

m s = 64 kg, P = 14 bar, T = 39°C, x = 0.82, mf = 8 kg

We know that, m a = \(\frac{m_s}{m_f} \)

m a = \(\frac{64}{8} \)

m a = 8 kg steam/kg of coal

From steam tables at 14 bar

h f = 830.1 kJ/kg and h fg = 1957.7 kJ/kg

h = h f + x(h fg ) = 830.1 + 0.82 = 2435.414 kJ/kg

h f1 = 4.18 = 4.18 = 163.02 kJ/kg

Equivalent evaporation, m e =\(\frac{m_a

}{2257} \)

m e = \(\frac{8}{2257} \)

m e = 8.05 kg of steam/kg of fuel.

5. What is the unit of equivalent evaporation?

a) kg of steam/kg of fuel fired

b) kg of steam/h/m 2 of heating surface

c) kg of steam/h

d) kg of steam/kg of air supplied

Answer: a

Explanation: The unit of equivalent evaporation is kg of steam per kg of fuel fired. It represents the quantity of steam that can be generated per kg of fuel fired at 100°C and 1 atm pressure. Equivalent evaporation is an important factor while determining the performance of the boiler.

6. A boiler generates 30 kg of steam at 11.5 bar in 1 hour with the consumption of 3 kg of coal. Feed water temperature is 40°C. Calculate equivalent evaporation if the steam is dry and saturated.

a) 10.78 kg of steam/kg of fuel

b) 11.58 kg of steam/kg of fuel

c) 6.32 kg of steam/kg of fuel

d) 5.65 kg of steam/kg of fuel

Answer: b

Explanation: Given,

m s = 30 kg, P = 11.5 bar, m f = 3 kg, T = 40°C

From steam tables, at 11.5 bar

h g = 2781.3 kJ/kg

h = h g = 2781.3 KJ/kg ……………………………………………………………. 

h f1 = 4.18 = 4.18 = 167.2 kJ/kg

We know that, m a = \(\frac{ms}{mf} \)

m a = \(\frac{30}{3} \)

m a = 10 kg steam/kg of coal

Equivalent evaporation, m e =\(\frac{m_a 

}{2257} \)

m e = \(\frac{10}{2257} \)

m e = 11.58 kg of steam/kg of fuel.

7. A boiler produces 12 kg of steam per kg of coal burnt at a pressure of 11 bar. The steam generated is superheated to a temperature of 260°C. Determine the equivalent evaporation if the mean feed water temperature is 30°C.Take specific heat of superheated steam as 2.33 kJ/kg.

a) 10 kg of steam/kg of fuel

b) 12 kg of steam/kg of fuel

c) 15 kg of steam/kg of fuel

d) 17 kg of steam/kg of fuel

Answer: c

Explanation: Given,

m a = 12 kg of steam/kg of fuel, P = 11 bar, T sup = 260°C, T = 30°C, c p = 2.33 kJ/kg-K

From steam tables at 11 bar

T s = 184.1°C

h f = 781.1 kJ/kg and h fg = 1998.5 kJ/kg

h = h f + h fg + c p (T sup -T s ) = 781.1 + 1998.5 + 2.33 = 2956.45 kJ/kg

h f1 = 4.18 = 4.18 = 125.4 kJ/kg

Equivalent evaporation, m e =\(\frac{m_a 

}{2257} \)

m e = \(\frac{12}{2257} \)

m e = 15.05 kg of steam/kg of fuel

m e ≈ 15 kg of steam/kg of fuel.

8. What is the latent heat of vaporization of water at 100°C and 1 atm pressure?

a) 2546 kJ/kg

b) 1254 kJ/kg

c) 1564 kJ/kg

d) 2257 kJ/kg

Answer: d

Explanation: The latent heat of vaporization is the amount of heat energy required to convert a unit mass of a liquid  into vapor without change in its temperature. The latent heat of vaporization of water is 2257 kJ/kg at 100°C and 1 atm pressure.

9. A boiler generates 600 kg of steam, consuming 40 kg of coal. Determine its evaporation capacity in kg of steam per hour if coal is fed to the furnace at the rate of 4 kg per hour.

a) 60 kg of steam/hour

b) 40 kg of steam/hour

c) 50 kg of steam/hour

d) 70 kg of steam/hour

Answer: a

Explanation: Given,

Mass of steam, m s = 600 kg

Mass of coal, m f = 40 kg

Coal feeding rate, R = 4 kg/hr

Evaporation capacity =(m s /m f ) * R

=  * 4

m e = 60 kg of steam/hour

10. A boiler produces dry and saturated steam at 12 bar pressure. Calculate the factor of evaporation if the feed water temperature is 45°C.

a) 0.95

b) 1.15

c) 1.10

d) 0.85

Answer: b

Explanation: Given,

P = 12 bar, T = 45°C

From steam tables, at 12 bar pressure

h f = 798.4 kJ/kg, h fg = 1984.3 kJ/kg

h = h f + h fg = 798.4 + 1984.3 = 2782.7 kJ/kg

h f1 = 4.18 = 4.18 = 188.1 kJ/kg

Factor of evaporation, f=\(\frac{h-h_{f1}}{2257} \)

f=\(\frac{2782.7-188.1}{2257} \)

f=1.15.

11. A boiler generates steam at 12.5 bar pressure. Temperature of the feed water is 35°C. Determine the factor of evaporation if the dryness fraction of the steam is 0.85.

a) 1.32

b) 0.85

c) 1.56

d) 1.04

Answer: d

Explanation: Given,

P = 12.5 bar, T = 35°C, x = 0.85

From steam tables, at 12.5 bar pressure

h f = 806.7 kJ/kg, h fg = 1977.4 kJ/kg

h = h f + x(h fg ) = 806.7 +0.85 = 2487.49 kJ/kg

h f1 = 4.18 = 4.18 = 146.3 kJ/kg

Factor of evaporation, f=\(\frac{h-h_{f1}}{2257} \)

f=\(\frac{2487.49-146.3}{2257} \)

f=1.04.

12. Steam generated by a boiler is superheated to a temperature of 230°C. The feed water temperature is 30°C. Calculate the factor of evaporation if the pressure of the steam generated is 14 bar. Specific heat of superheated steam is 2.34 kJ/kg-K

a) 1.22

b) 1.03

c) 1.52

d) 1.65

Answer: a

Explanation: Given,

P = 14 bar, T = 30°C, T sup = 230°C, c p = 2.34 kJ/kg-K

From steam tables, at 14 bar pressure

h f = 830.1 kJ/kg, h fg = 1957.7 kJ/kg, T sat = °C

h = h f + h fg + c p (T sup – T sat ) = 830.1 + 1957.7 + 2.34 = 2869.7 kJ/kg

h f1 = 4.18 = 4.18 = 125.4 kJ/kg

Factor of evaporation, f=\(\frac{h-h_{f1}}{2257} \)

f=\(\frac{2869.7-125.4}{2257} \)

f=1.22.

13. A boiler generates dry and saturated steam at 10 bar pressure. The evaporative capacity of the boiler is 10 kg of steam per kg of fuel fired. The equivalent evaporation is 11.6 kg of steam per kg of fuel fired. Determine the temperature of feed water.

a) 30°C

b) 26°C

c) 38°C

d) 40°C

Answer: c

Explanation: Given,

P = 10 bar, m a = 10 kg of steam/kg of fuel, m e = 11.6 kg of steam/kg of fuel

From steam tables, At 10 bar

h g = 2776.2 kJ/kg

h = h g = 2776.2 kJ/kg 

We know that,

m e = \(\frac{ma}{2257} \)

Substituting the values, we get

11.6 = \(\frac{10}{2257} \)

Solving for hf1, we get

h f1 = 158.08 kJ/kg

but, h f1 = 4.18

158.08 = 4.18

T = 37.82°C ≈ 38°C

14. The equivalent evaporation of a boiler is 11 kg of steam per kg of fuel fired. The evaporation capacity of the same boiler is 10 kg of steam per kg of fuel fired. Determine enthalpy of 1 kg of steam if the feed water temperature is 40°C.

a) 2750 kJ/kg

b) 2650 kJ/kg

c) 2250 kJ/kg

d) 2850 kJ/kg

Answer: b

Explanation: m e = 11 kg of steam/kg of fuel, m a = 10 kg of steam/kg of fuel, T = 40°C

h f1 = 4.18 = 4.18 = 167.2 kJ/kg

We know that,

m e = \(\frac{ma}{2257} \)

Substituting the values, we get

11 = \(\frac{10}{2257} \)

Solving for h, we get

h = 2649.9 kJ/kg ≈ 2650 kJ/kg.

15. Determine the temperature of feed water of a boiler having factor of evaporation equal to 1.14. The enthalpy of 1 kg of steam is 2700 kJ/kg.

a) 30°C

b) 40°C

c) 35°C

d) 45°C

Answer: a

Explanation: f = 1.14, h = 2700 kJ/kg

We know that,

f=\(\frac{h-h_{f1}}{2257} \)

Substituting the values, we get

1.14=\(\frac{2700-h_{f1}}{2257} \)

Solving for h f1 , we get

h f1 = 127.02 kJ/kg

But, h f1 = 4.18

T = 127.02/4.18

T = 30.39°C ≈ 30°C.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Steam Generators Performance – Boiler Efficiency”.

1. Which of the following is the correct formula for calculating boiler efficiency?

a) η = \

 

 η = \

 

 η = \

 

 η = \(\frac{m_a 

}{2257} \)

Answer: b

Explanation: The correct formula for calculating boiler efficiency is η = \(\frac{m_a 

}{C} \). The expression \(\frac{m_a 

}{2257} \) is used to calculate the equivalent evaporation of a boiler. Considering Superheater, boiler and economizer as a single unit, the calculated efficiency is called overall efficiency.

2. Which of the following is a fixed factor on which boiler efficiency depends?

a) Actual firing rate

b) Humidity of combustion air

c) Condition of heat absorbing surfaces

d) Properties of the fuel burnt

Answer: d

Explanation: The efficiency of a boiler depends on two types of factors – fixed factors and variable factors. Boiler design, Heat recovery equipment, built in loses, rated rate of firing and properties and characteristics of the fuel burnt are all fixed factors. Actual firing rate, humidity of combustion air, condition of heat absorbing surfaces are variable factors.

3. The following listed are some factors on which boiler efficiency depends. Which of the following is a variable factor?

a) Excess air fluctuations

b) Rated rate of firing

c) Boiler design

d) Heat recovery equipment

Answer: a

Explanation: Variable factors include actual firing rate, fuel condition as it is fired, Condition of heat absorbing surfaces, excess air fluctuations, change in draught due to atmospheric conditions and humidity and temperature of combustion air. Rated rate of firing, boiler design and heat recovery equipment are all fixed factors.

4. Boiler efficiency is the ratio of heat actually absorbed by water during generation of steam to the heat supplied in by the fuel in the same time period.

a) True

b) False

Answer: a

Explanation: The given statement is the correct definition of boiler efficiency.

Mathematically,

boiler efficiency = \(\frac{m_a 

}{C} \)

If economizer and superheater are also considered along with boiler, then the efficiency calculated is called overall efficiency.

5. Which of the following is not a heat recovery equipment?

a) Economizer

b) Air preheater

c) Feed water heater

d) Steam separator

Answer: d

Explanation: Steam separator is not a heat recovery equipment. Economizer, superheater, air preheater and feed water heater are all heat recovery equipments. Steam separator is a boiler accessory. Economizer, superheater, air preheater and feed water heater are also categorized under boiler accessories.

6. Following observations were made after the trial of a boiler for 24 hours.

Mass of coal burnt = 1300 kg

Mass of steam generated = 14000 kg

Mean effective pressure of steam = 10 bar

Temperature of the feed water = 40°C

Calorific value of coal = 30000 kJ/kg

Calculate the boiler efficiency if the dryness fraction of the steam generated is 0.95.

a) 90%

b) 80%

c) 70%

d) 60%

Answer: a

Explanation: Given, m f = 1300 kg, m s = 14000 kg, P = 10 bar, T = 40°C, C = 30000 kJ/kg

m a = \(\frac{m_s}{m_f} \)

m a = \(\frac{14000}{1300} \)

m a = 10.77 kg of steam/kg of fuel

At 10 bar, from steam tables

h f = 762.68 kJ/kg, h fg = 2014.44 kJ/kg

x = 0.95

h = h f + x(h fg ) = 762.68 + 0.95 = 2676.4 kJ/kg

h f1 = 4.18 = 4.18 = 167.2 kJ/kg

Boiler efficiency, η=\(\frac{m_a 

}{C}=\frac{10.77}{30000}\) = 0.9 or 90%.

7. A boiler produces 4000 kg of dry and saturated steam in 6 hours. In the same time the mass of coal consumed is 300 kg of coal. The mean pressure of steam generated is 11 bar. The temperature of the feed water is 35°C. Determine the boiler efficiency if the calorific value of the fuel is 30000 kJ/kg.

a) 85.64%

b) 75.65%

c) 95.65%

d) 87.78%

Answer: d

Explanation: Given, m s = 4000 kg, m f = 300 kg, P = 11 bar, T = 35°C, C = 30000 kJ/kg

At 11 bar, from steam tables

h f = 781.1 kJ/kg, h fg = 1998.5 kJ/kg

h = h f + h fg =781.1 + 1998.5 = 2779.6 kJ/kg

h f1 = 4.18  = 4.18 = 146.3 kJ/kg

Boiler efficiency, η=\(\frac{m_s (h-h_{f1}}{mf*C} = \frac{3000}{300*30000}\) = 0.8778 or 87.78%.

8. A boiler generates superheated steam having temperature 270°C and pressure 13 bar. The specific heat of the superheated steam 2.1 kJ/kg-K. The temperature of the feed water is 32°C. The evaporative capacity of the boiler is 8 kg of steam per kg of fuel. The calorific value of the coal is 26000 kJ/kg.

a) 86.65%

b) 70.32%

c) 65.87%

d) 90.80%

Answer: a

Explanation: Given, T sup = 270°C, P = 13 bar, T = 32°C, c p = 2.1 kJ/kg-K, m a = 8 kg of steam/kg of fuel, C = 26000 kJ/kg

At 13 bar, from steam tables

h f = 814.7 kJ/kg, h fg = 1970.7 kJ/kg, T s = 191.6°C

h = h f + h fg + c p (T sup – T s ) = 814.7 + 1970.7 + 2.1 

h = 2950.04

h f1 = 4.18 = 4.18 = 133.76 kJ/kg

Boiler efficiency, η = \(\frac{m_a 

}{C} = \frac{8}{26000}\) = 0.8665 or 86.65%.

9. A boiler produces steam at the rate of 1500 kg per hour at 12.5 bar. The water fed to the economizer is at a temperature of 35°C and it is raised to a temperature of 110°C before feeding it to the boiler. Coal is fired at a rate of 200 kg per hour. The dryness fraction of the steam generated is 0.87. If 15% of the coal remains unburnt and the calorific value of the coal being fired is 36200 kJ/kg, determine the overall efficiency.

a) 58%

b) 65%

c) 78%

d) 82%

Answer: a

Explanation: Given, m s = 1500 kg/h, P = 12.5 bar, T 1 = 35°C, T 2 = 110°C, m f = 200 kg/h, x = 0.87, C = 36200 kJ/kg

Mass of coal actually burnt per hour = 200 \(\frac{85}{100}\) = 170 kg/h

ma = \(\frac{m_s}{m_f} = \frac{1500}{170}\) = 8.82 kg of steam/kg of fuel

At 12.5 bar, from steam tables

h f = 806.7 kJ/kg, h fg = 1977.4 kJ/kg

h = h f + x(h fg ) = 806.7 + 0.87 = 2527.04 kJ/kg

h f1 = 4.18 = 4.18 = 146.3 kJ/kg

Overall efficiency = \(\)\frac{m_a (h-h_{f1}}{C} = \frac{8.2}{36200} = 0.58 or 58%.

10. A boiler having evaporative capacity 8 kg of steam per kg of fuel produces dry and saturated steam at 13 bar pressure. The feed water temperature is 35°C. Determine the calorific value of coal if the efficiency of the boiler is 70%.

a) 35000 kJ/kg

b) 29560 kJ/kg

c) 31000 kJ/kg

d) 30160 kJ/kg

Answer: d

Explanation: Given, m a = 8 kg of steam/kg of fuel, P = 13 bar, T = 35°C, η = 0.7

At 13 bar, from the steam tables

h g = 2785.4 kJ/kg

h = h g = 2785.4 kJ/kg

h f1 = 4.18 = 4.18 = 146.3 kJ/kg

we know that,

η = \(\frac{m_a 

}{C} \)

Substituting the values, we get

0.7 = \(\frac{8}{C} \)

Solving for ‘C’, we get

C = 30161.14 kJ/kg ≈ 30160 kJ/kg.

11. A boiler is fired with fuel having calorific value 30000 kJ/kg. It produces steam at 5 bar. The dryness fraction of the steam generated 0.85. The efficiency of the boiler is 77% and the evaporative e capacity of the boiler is 10 kg of steam per kg of fuel. Determine the temperature of feed water.

a) 35°C

b) 29°C

c) 25°C

d) 37°C

Answer: b

Explanation: Given, C = 30000 kJ/kg, P = 5 bar, x = 0.85, η = 0.77, m a = 10 kg of steam/kg of fuel

At 5 bar, from steam tables

hf = 640.1 kJ/kg, h fg = 2170.4 kJ/kg

h = h f + x(h fg ) = 640.1 + 0.85 = 2431.39 kJ/kg

we know that,

η = \(\frac{m_a 

}{C} \)

Substituting the values, we get

0.77 = \(\frac{10

}{30000} \)

Solving for h f1 , we get

h f1 = 121.39 kJ/kg

But, h f1 = 4.18

121.39 = 4.18

Therefore, T = 29°C.

12. Determine the evaporative capacity of a boiler in kg of steam per kg of fuel fired if the feed water temperature is 40°C, the efficiency of the boiler is 95%, calorific value of the coal fried is 33000 kJ/kg and the steam generated is dry and saturated at 12 bar pressure.

a) 10 kg of steam/kg of fuel

b) 12 kg of steam/kg of fuel

c) 15 kg of steam/kg of fuel

d) 15 kg of steam/kg of fuel

Answer: c

Explanation: Given, T = 40°C, η = 0.95, C = 33000 kJ/kg, P = 12 bar

At 12 bar, from steam tables

h g = 2782.7 kJ/kg

h = h g = 2782.7 kJ/kg

h f1 = 4.18 = 4.18 = 167.2 kJ/kg

we know that,

η = \(\frac{m_a 

}{C} \)

Substituting the values, we get

0.95 = \(\frac{m_a }{33000} \)

Solving for m a , we get

m a = 11.98 kg of steam/kg of fuel ≈ 12 kg of steam/kg of fuel.

13. Determine the enthalpy of 1 kg of steam if the boiler efficiency is 79%, the evaporative capacity of the boiler is 9 kg of steam per kg of fuel fired, the feed water temperature is 40°C and the calorific value of coal is 30000 kJ/kg.

a) 2500 kJ/kg

b) 3000 kJ/kg

c) 2800 kJ/kg

d) 2200 kJ/kg

Answer: c

Explanation: Given, η = 0.79, m a = 9 kg of steam/ kg of fuel, T = 40°C, C = 30000 kJ/kg

h f1 = 4.18 = 4.18 = 167.2 kJ/kg

We know that

η = \(\frac{m_a

}{C} \)

Substituting the values, we get

0.79 = \(\frac{9}{30000} \)

Solving for ‘h’, we get

h = 2800 kJ/kg.

14. Boiler efficiency doesn’t depend on _____

a) calorific value of fuel fired

b) specific heat of steam generated

c) boiler design

d) operation time

Answer: d

Explanation: Boiler efficiency does not depend on operation time. It depends on boiler design, heat recovery equipments, calorific value of fuel fired, specific heat of steam generated, fuel condition, firing rate etc.

15. In a boiler, some part of heat is lost to flue gases.

a) True

b) False

Answer: a

Explanation: The flue gases generated as product of combustion contain dry products of combustion. They also contain steam generated by the combustion of hydrogen in fuel. This heat loss can be reduced by passing the flue gases through the air preheater and economizer.

This set of Thermal Engineering Questions and Answers for Experienced people focuses on “Steam Flow Through Nozzles”.

1. A steam nozzle is a passage of varying cross section through which the kinetic energy of steam is converted into heat energy.

a) True

b) False

Answer: b

Explanation: Steam nozzle is a passage of varying cross section, through which the heat energy of the steam is converted into kinetic energy. The velocity of steam in increased its pressure is decreased. It is usually used to produce high steam jets to drive the steam turbines.

2. Which of the following statements about steam nozzles is FALSE?

a) It converts the heat energy of the steam into kinetic energy

b) It has a varying cross section

c) The smallest section is called throat

d) The pressure at the outlet is more than at the inlet

Answer: d

Explanation: Pressure at the inlet of the steam nozzle is greater than at the outlet. Steam nozzles increase the velocity of the steam passing through by converting the heat energy of the steam into kinetic energy. It has a varying cross section and the smallest section is called throat.

3. The smallest section of a steam nozzle is called _____

a) maw

b) neck

c) throat

d) muzzle

Answer: c

Explanation: Steam nozzle has varying cross section; the smallest section is called throat. Based upon the variation of cross-section nozzles are categorized into diverging nozzles, converging nozzles and converging-diverging nozzles.

4. Which type of nozzle does the following picture represent?

a) Convergent

b) Divergent

c) Convergent-Divergent

d) Subsonic

Answer: b

Explanation: This picture represents a convergent-divergent nozzle. The smallest section of the nozzle is called throat. A convergent-divergent nozzle converges to throat and diverges afterwards. A convergent-divergent nozzle is able to impart higher velocity to steam than a convergent nozzle.

5. The steam flow though nozzle is considered to be _____

a) adiabatic

b) isobaric

c) isothermal

d) isochoric

Answer: a

Explanation: During the expansion no heat is added not rejected from the system hence, the steam flow though through nozzle is considered to be adiabatic. Steam while passing through nozzle loses its pressure and temperature, also the nozzle is of varying cross section, therefore, the flow cannot be isobaric, isothermal or isochoric.

6. Steam at a temperature 200°C and pressure 4 bar enters a steam nozzles and is discharged at 1 bar pressure. Determine the exit velocity of steam if the inlet velocity is negligible.

a) 500 m/s

b) 720 m/s

c) 580 m/s

d) 750 m/s

Answer: b

Explanation: Given, P 1 = 4 bar, T 1 = 200°C, P 2 = 1 bar

From steam tables, At 4 bar and 200°C

h 1 = 2860.5 kJ/kg, s 1 = 7.171 kJ/kg/K

From steam tables, At 1 bar

h f = 417.5 kJ/kg, h fg = 2257.9 kJ/kg

s f = 1.3027 kJ/kg-K, s fg = 6.0657 kJ/kg-K

Since, s 1 = s 2

s 1 = s f +x(s fg )

7.171 = 1.3027 + x

x = 0.967

therefore, h 2 = h f + x(h fg )

h 2 = 417.5 + 0.967

h 2 = 2601.92 kJ/kg

Exit velocity, C 2 = \(\sqrt{2*1000

}\)

C 2 = \(\sqrt{2*1000}\)

C 2 = 719.14 m/s ≈ 720 m/s.

7. Steam entering a nozzle at 15 bar and 350°C is discharged at 1 bar. If the inlet velocity of the steam is 50 m/s, determine the exit velocity.

a) 785 m/s

b) 365 m/s

c) 565 m/s

d) 475 m/s

Answer: c

Explanation: Given, P 1 = 15 bar, T 1 = 350°C, P 2 = 1 bar, C 1 = 50m/s

Using Mollier chart,

h 1 = 3150 kJ/kg and h 2 = 2992 kJ/kg

Exit velocity, C 2 = \(\sqrt{2*1000

+C_1^2}\)

C 2 = \(\sqrt{2*1000+50^2}\)

C 2 = 564.36 m/s ≈ 565 m/s.

8. Dry and saturated steam enters a nozzle at 11 bar and leaves at 2 bar. What minimum exit area should the nozzle have if the steam flow of 2 kg/s is to be maintained?

a) 0.00365 m 2

b) 0.02156 m 2

c) 0.00312 m 2

d) 0.00228 m 2

Answer: d

Explanation: Given, P 1 = 11 bar, P 2 = 2 bar, m = 2 kg/s

Using Mollier chart,

h 1 = 2780 kJ/kg, h 2 = 2480 kJ/kg, v g2 = 0.885 m 3

We know that,

C 2 = \(\sqrt{2*1000

}\)

C 2 = \(\sqrt{2*1000}\)

C 2 = 774.6 m/s

Exit area, A 2 = \(\frac{m*v_{g2}}{C_2} = \frac{2*0.885}{774.6}\) = 0.00228 m 2

9. Final velocity of steam passing thorough a nozzle is found out to be 586 m/s. The pressure of steam at inlet is 15 bar and is dry and saturated. The exit pressure is 2 bar. Determine the dryness fraction of the steam when it leaves the nozzle.

a) 0.79

b) 0.96

c) 0.88

d) 0.81

Answer: b

Explanation: Given, C 2 = 586 m/s, P 1 = 15 bar, P 2 = 2 bar

From Steam tables,

At 15 bar:

h 1 = h g = 2789.9 kJ/kg

At 2 bar:

h f2 = 504.7 kJ/kg, h fg2 = 2201.6 kJ/kg

We know that,

C 2 = 44.72\(\sqrt{

} \)

Substituting the values

586 = 44.72\(\sqrt{

} \)

Solving for ‘h 2 ’, we get

h 2 = 2618.19 kJ/kg

we know that

h 2 = h f2 + x(h fg2 )

2618.19 = 504.7 + x 2 

Solving for x 2 , we get

x 2 = 0.96.

10. Dry saturated steam enters a steam nozzle at 12 bar and 200°C and leaves at 2 bar. If the exit area is 0.00197 m 2 , determine the mass flow rate of steam.

a) 1 kg/s

b) 2 kg/s

c) 3 kg/s

d) 4 kg/s

Answer: b

Explanation: Given, P 1 = 12 bar, T 1 = 200°C, P 2 = 2 bar, A 2 = 0.00197 m 2

From steam tables, at 12 bar and 200°C

h 1 = 2814.4 kJ/kg, s 1 = 6.587 kJ/kg-K

From steam tables, at 2 bar

h f2 = 504.7 kJ/kg, h fg2 = 2201.6 kJ/kg

s f2 = 1.530 kJ/kg-K, s fg2 = 5.597 kJ/kg-K

v g2 = 0.8854 m 3 /kg

We know that

s 1 = s 2

s 1 = s f2 + x 2 (s fg2 )

6.587 = 1.530 + x2

x 2 = 0.90

h 2 = h f2 + x 2 (h fg2 )

= 504.7 + 0.9

= 2486.14 kJ/kg

We know that,

C 2 = 44.72\(\sqrt{

}\)

Substituting the values

C 2 = 44.72\(\sqrt{}\)

C 2 = 810.23 m/s

Mass flow rate, m = \(\frac{A_2*C_2}{xv_g}\)

m = \(\frac{0.00197*810.23}{0.9*0.8854}\)

m = 2 kg/s.

11. Which of the following equations represent the correct critical pressure ratio?

a) \

 

 

^\frac{n}{n-1} \)

b) \

 

 

^\frac{n}{n-1} \)

c) \

 

 

^\frac{n}{n-1} \)

d) \

 

 

^\frac{n}{n+1} \)

Answer: c

Explanation: Critical pressure ratio represents the ratio of throat pressure to inlet pressure for maximum discharge through nozzle. The correct equation for critical pressure ratio is-

\

 

 

 

^\frac{n}{n-1} \)

12. Calculate the throat pressure if the dry and saturated steam enters a nozzle at 15 bar. Consider the discharge through the nozzle maximum.

a) 8.66 bar

b) 7.58 bar

c) 5.65 bar

d) 2.36 bar

Answer: a

Explanation: P 1 = 15 bar, x 2 = 0.95

n = 1.135 

According to the given condition,

\

 

 

^\frac{n}{n-1} \)

\

 

 

^\frac{1.135}{1.135-1} \)

Therefore, P 2 = 8.66 bar.

13. Steam enters a nozzle at 14 bar and 200°C. Determine the discharge if the discharge through the nozzle is maximum. Take the area of throat as 0.002 m 2 .

a) 3.95 kg/s

b) 4.18 kg/s

c) 2.36 kg/s

d) 1.65 kg/s

Answer: b

Explanation: Given, P 1 = 14 bar, T 1 = 200°C, x 2 = 0.85, A 2 = 0.002 m 2

n = 1.3 

From steam tables, at 14 bar and 200°C

v g1 = 0.1429 m 3 /kg

According to the given condition

m max = A\(\sqrt{n

 

 

^\frac{n+1}{n-1}} \)

m max = 0.002\(\sqrt{1.3{14*10^5}{0.1429})

 

^\frac{1.3+1}{1.3-1}} \)

m max = 4.18 kg/s.

14. The maximum mass flow through a steam nozzle is independent of _____

a) Initial pressure

b) Initial density

c) Final pressure

d) Throat area

Answer: c

Explanation: The equation below represents the maximum mass flow through a nozzle.

m max = A\

 

 

 

, quality and throat area. It is independent of final pressure and quality  of steam.

15. Determine the velocity of steam at throat if the discharge through the nozzle is maximum. Steam enters the nozzle in dry and saturated state and at 11 bar.

a) 455.45 m/s

b) 625.32 m/s

c) 545.87 m/s

d) 325.65 m/s

Answer: a

Explanation: Given, P 1 = 11 bar

n = 1.135 

From steam tables, at 11 bar

v g1 = 0.17739 m 3 /kg

According to the given conditions

C 2 = \(\sqrt{2

 

}\)

C 2 = \(\sqrt{2

 

}\)

C 2 = 455.45 m/s.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Nozzle Efficiency”.

1. The final velocity obtained after passing the steam through a nozzle is less than the calculated one. Which of the following is NOT a valid reason for the same?

a) Friction between steam and nozzle surface

b) Steam not being superheated

c) Shock loses

d) Internal friction of steam

Answer: b

Explanation: The final velocity of the steam after passing it through a nozzle is reduced due to the internal friction of the steam, friction between steam and nozzle surface. Shock loses are also responsible for reduction of final velocity. The quality of steam being passed has no effect on the enthalpy drop.

2. Which of the following is NOT an effect of frictional loses in a convergent-divergent nozzle?

a) Enthalpy drop is increased

b) The expansion is not isentropic

c) The final dryness fraction of the steam is increased

d) The specific volume of steam is increased

Answer: a

Explanation: The enthalpy drop is reduced due to the frictional loses in a convergent-divergent nozzle. The expansion is not isentropic. The dryness fraction of the steam is increased i.e. quality of the steam is improved because the steam absorbs heat generated by friction. As the steam becomes more dry, the specific volume of the steam is increased.

3. Presence of friction in a convergent-divergent nozzle, decreases the final velocity of the steam and increases the dryness fraction of the steam.

a) True

b) False

Answer: a

Explanation: Final velocity of the steam is reduced due to shock loses and presence of friction in the nozzle. Friction between the nozzle surface and steam and the internal friction of the steam generates heat. This heat is absorbed by the steam and the dryness fraction of the steam is increased.

4. Determine the velocity coefficient if the efficiency of a convergent-divergent steam nozzle is 92%. The inlet velocity of steam is negligible.

a) 0.959

b) 0.120

c) 0.465

d) 0.542

Answer: a

Explanation: Given, η = 0.92

When inlet velocity is negligible, the velocity coefficient is the square root of the nozzle efficiency.

Velocity coefficient = \(\sqrt{η} = \sqrt{0.92}\) = 0.959.

5. Dry and saturated steam at 12 bar expands through a convergent-divergent nozzle having 90% efficiency to 2 bar. Determine the actual exit velocity of the steam. Neglect the initial velocity of steam.

a) 451.20 m/s

b) 754.62 m/s

c) 650.32 m/s

d) 856.96 m/s

Answer: b

Explanation: P 1 = 12 bar, η = 0.90, P 2 = 2 bar

At 12 bar, from steam tables

h 1 = h g1 = 2782.7 kJ/kg, s 1 = s g1 = 6.519 kJ/kg-K

At 2 bar, from steam tables

h f2 = 504.7 kJ/kg, h fg2 = 2201.6 kJ/kg

s f2 = 1.530 kJ/kg-K, s fg2 = 5.597 kJ/kg-K

s 1 = s 2

s 1 = s f2 + x 2 (s fg2 )

6.519 = 1.530 + x 2 

x 2 = 0.891

h 2 = h f2 + x 2 (h fg2 )

= 504.7 +0.891

= 2466.32 kJ/kg

C 2 = 44.72\(\sqrt{η

}\)

C 2 = 44.72\(\sqrt{0.90}\)

C 2 = 754.62 m/s.

6. Initial velocity of the steam entering a convergent-divergent steam nozzle is 60 m/s. Steam expands from 15 bar and 250°C to 1 bar. Determine the exit velocity of steam if nozzle has an efficiency of 95%.

a) 973.08 m/s

b) 654.32 m/s

c) 785.45 m/s

d) 900.75 m/s

Answer: a

Explanation: P 1 = 15 bar, T 1 = 250°C, η = 0.95, P 2 = 1 bar, C 1 = 60 m/s

At 15 bar and 200°C, from steam tables

h 1 = 2923.5 kJ/kg, s 1 = 6.710 kJ/kg-K

At 1 bar, from steam tables

h f2 = 417.5 kJ/kg, h fg2 = 2257.9 kJ/kg

s f2 = 1.303 kJ/kg-K, s fg2 = 6.057 kJ/kg-K

s 1 = s 2

s 1 = s f2 + x 2 (s fg2 )

6.70 = 1.303 + x 2 

x 2 = 0.89

h 2 = h f2 + x 2 (h fg2 )

= 417.5 + 0.89

= 2427.03 kJ/kg

C 2 = \(\sqrt{2*1000*η

+C_1^2} \)

C 2 = \(\sqrt{2*1000*0.95+60^2} \)

C 2 = 973.08 m/s.

7. Dry and saturated steam enters a convergent-divergent steam nozzle at 11 bar and leaves at 3 bar. Determine the nozzle efficiency if the actual enthalpy drop is 200 kJ/kg.

a) 78.56%

b) 96.36%

c) 90.35%

d) 85.27%

Answer: d

Explanation: P 1 = 11 bar, P 2 = 3 bar

Actual enthalpy drop = 200 kJ/kg

At 11 bar, from steam tables

h 1 = h g = 2779.7 kJ/kg

s 1 = s g = 6.550 kJ/kg-K

At 3 bar, from steam tables

h f2 = 561.5 kJ/kg, h fg2 = 2163.2 kJ/kg

s f2 = 1.672 kJ/kg-K, s fg2 = 5.319 kJ/kg-K

s 1 = s 2

s 1 = s f2 + x 2 (s fg2 )

6.550 = 1.672 + x 2 

x 2 = 0.917

h 2 = h f2 + x 2 (h fg2 )

= 561.5 + 0.917

= 2545.15 kJ/kg

Isentropic enthalpy drop = h 1 – h 2 = 2779.7 – 2545.15 = 234.55 kJ/kg

Nozzle efficiency, η = \(\frac{Actual \, enthalpy \, drop}{Isentropic \, enthalpy \, drop} = \frac{200}{234.55} \)= 0.8527 or 85.27%.

8. Dry and saturated steam at 10 bar enters a convergent-divergent nozzle and leaves at 1 bar. If 8% of heat drop is lost in friction find the percentage decrease in final velocity. Neglect initial velocity.

a) 5.65%

b) 3.87%

c) 2.14%

d) 4.08%

Answer: d

Explanation: P 1 = 10 bar, P 2 = 1 bar, η = 1.0 – 0.08 = 0.92

At 10 bar, from steam tables

h 1 = h g = 2776.2 kJ/kg

s 1 = 6.583 kJ/kg-K

At 1 bar, from steam tables

h f2 = 417.5 kJ/kg, h fg2 = 2257.9 kJ/kg

s f2 = 1.303 kJ/kg-K, s fg2 = 6.057 kJ/kg-K

s 1 = s 2

s 1 = s f2 + x 2 (s fg2 )

6.583 = 1.303 + x 2 

x 2 = 0.872

h 2 = h f2 + x 2 (h fg2 )

h 2 = 417.5 + 0.872

h 2 = 2386.39 kJ/kg

Isentropic Enthalpy drop, h d = h 1 – h 2 = 2776.2 – 2386.39 = 389.81 kJ/kg

C 2 = 44.72\(\sqrt{h_d}\) = 44.72\(\sqrt{389.81}\) = 882.93 m/s

C 2 ’ = 44.72\(\sqrt{η

}\) = 44.72\(\sqrt{0.92}\) = 846.88 m/s

% decrease in exit velocity = \(\frac{C_2-C_2′}{C_2}\) *100 = \(\frac{882.93- 846.88}{882.93}*100\) = 4.08%.

9. Steam at 10 bar superheated to 200°C enters a convergent-divergent steam nozzles and leaves at 2 bar. Determine the dryness fraction of the steam discharged if the nozzle efficiency is 93%.

a) 0.931

b) 0.865

c) 0.758

d) 0.963

Answer: a

Explanation: P 1 = 10 bar, T 1 = 200°C, P 2 = 2 bar, η = 0.93

At 10 bar and 200°C, from steam tables

h 1 = 2826.8 kJ/kg

s 1 = 6.692 kJ/kg-K

At 2 bar, from steam tables

h f2 = 504.7 kJ/kg, h fg2 = 2201.6 kJ/kg

s f2 = 1.530 kJ/kg-K, s fg2 = 5.597 kJ/kg-K

s 1 = s 2

s 1 = s f2 + x 2 (s fg2 )

6.692 = 1.530 + x 2 

x 2 = 0.922

h 2 = h f2 + x 2 (h fg2 )

h 2 = 504.7 + 0.922

h 2 = 2534.58 kJ/kg

We know that,

η=\(\frac{h_1-h_2′}{h_1-h_2}\)

Substituting the values

0.93=\(\frac{2826.8-h_2”}{2826.8-2534.58}\)

h 2 ’ = 2555.04 kJ/kg

But, h 2 ’ = h f2 + x 2 (h fg2 )

2555.04 = 504.7 + x 2 ’ 

x 2 ’ = 0.931.

10. Find the mass flow rate if dry and saturated steam at 12 bar pressure is discharged at 2 bar by a convergent-divergent nozzle. The nozzle efficiency and exit area of the nozzle are 95% and 0.00256 m 2 .

a) 1.5 kg/s

b) 2.5 kg/s

c) 3.0 kg/s

d) 4.0 kg/s

Answer: b

Explanation: P 1 = 12 bar, P 2 = 2 bar, η = 0.95, A 2 = 0.00256 m 2

At 12 bar, from steam tables

h 1 = h g1 = 2782.7 kJ/kg, s 1 = s g1 = 6.519 kJ/kg-K

At 2 bar, from steam tables

h f2 = 504.7 kJ/kg, h fg2 = 2201.6 kJ/kg

s f2 = 1.530 kJ/kg-K, s fg2 = 5.597 kJ/kg-K

v g2 = 0.88540 m 3 /kg

s 1 = s 2

s 1 = s f2 + x 2 (s fg2 )

6.519 = 1.530 + x 2 

x 2 = 0.891

h 2 = h f2 + x 2 (h fg2 )

= 504.7 +0.891

= 2466.32 kJ/kg

C 2 = 44.72\(\sqrt{η

}\)

C 2 = 44.72\(\sqrt{0.95}\)

C 2 = 775.30 m/s

We know that,

η=\(\frac{h_1-h_2′}{h_1-h_2}\)

Substituting the values

0.95=\(\frac{2782.7-h_2”}{2782.7-2466.32} \)

h 2 ’ = 2482.14 kJ/kg

But, h 2 ’ = h f2 + x 2 (h fg2 )

2482.14 = 504.7 + x 2 ’ 

x 2 ’ = 0.898

Mass flow rate, m = \(\frac{A_2*C_2}{x_2’*v_2g} = \frac{0.00256*775.30}{0.898*0.88540}\) = 2.5 kg/s.

11. Nozzle efficiency is the ratio of isentropic enthalpy drop to actual enthalpy drop, between the same pressures.

a) True

b) False

Answer: b

Explanation: Nozzle efficiency is defined as the ratio of actual enthalpy drop to isentropic enthalpy drop between the same pressures.

Mathematically,

η=\(\frac{Actual \, Enthalpy \, drop}{Isentropic \, Enthapy \, drop} = \frac{h_d’}{h_d}\)

12. A convergent-divergent steam nozzle is supplied with steam at 15 bar. It discharges the steam at 1 bar. Determine the exit velocity if the velocity coefficient of the given nozzle is 0.95.

a) 785.42 m/s

b) 654.32 m/s

c) 908.90 m/s

d) 854.65 m/s

Answer: c

Explanation: P 1 = 15 bar, P 2 = 1 bar, Velocity coefficient = 0.95

At 15 bar, from steam tables

h 1 = 2789.9 kJ/kg, s 1 = 6.441 kJ/kg-K

At 1 bar, from steam tables

h f2 = 417.5 kJ/kg, h fg2 = 2257.9 kJ/kg

s f2 = 1.303 kJ/kg-K, s fg2 = 6.057 kJ/kg-K

s 1 = s 2

s 1 = s f2 + x 2 (s fg2 )

6.441= 1.303 + x 2 

x 2 = 0.848

h 2 = h f2 + x 2 (h fg2 )

= 417.5 + 0.848

= 2332.2 kJ/kg

C 2 = 44.72\(\sqrt{

}\) = 44.72\(\sqrt{}\) = 956.74 m/s

The actual exit velocity, C 2 ’ = Velocity coefficient * C 2

C 2 ’ = 0.95 * 956.74

C 2 ’ = 908.90 m/s.

13. Steam at 14 bar and 250°C enters a convergent-divergent steam nozzle and is discharged at 2 bar. Determine the nozzle efficiency if the velocity coefficient is 0.95. Neglect the initial velocity of steam.

a) 95.65%

b) 90.25%

c) 99.65%

d) 85.32%

Answer: b

Explanation: Velocity coefficient is the square root of nozzle efficiency, if the inlet velocity is negligible.

Using this relation,

Nozzle efficiency =  2

= 0.95 2

= 0.9025 or 90.25%.

14. Choose the most appropriate statement regarding velocity coefficient.

a) It can be zero

b) It can be greater than one

c) It should strictly lie between zero and one

d) It is square of nozzle efficiency

Answer: c

Explanation: Velocity coefficient is the ratio of actual exit velocity to exit velocity considering the flow isentropic.

Mathematically,

Velocity Coeffiecint=\(\frac{Actual \, Exit \, Velocity}{Exit \,Velocity} = \frac{C_2′}{C_2} \)

Since the actual exit velocity can never be equal to the exit velocity considering flow to be isentropic, velocity coefficient lies strictly between zero and one.

15. Dry and saturated steam at 11 bar pressure enters a convergent-divergent nozzle and is discharged at 2 bar. Determine the actual enthalpy drop if he efficiency of the nozzle is 96%.

a) 254.56 kJ/kg

b) 321.32 kJ/kg

c) 288.14 kJ/kg

d) 245.63 kJ/kg

Answer: c

Explanation: Given, P 1 = 11 bar, P 2 = 2 bar, η = 0.96

At 11 bar, from steam tables

h 1 = h g = kJ/kg

s 1 = s g = 6.550 kJ/kg-K

At 2 bar, from steam tables

h f2 = 504.7 kJ/kg, h fg2 = 2201.6 kJ/kg

s f2 = 1.530 kJ/kg-K, s fg2 = 5.597 kJ/kg-K

s 1 = s 2

s 1 = s f2 + x 2 (s fg2 )

6.550 = 1.530 + x 2 

x 2 = 0.897

h 2 = h f2 + x 2 (h fg2 )

h 2 = 504.7 + 0.897

h 2 = kJ/kg

Enthalpy drop, h d = h 1 – h 2 = 2779.7 – 2479.54 = 300.16 kJ/kg

Actual enthalpy drop, h d ’ = ηhd = 0.96 * 300.15 = 288.14 kJ/kg.

This set of Advanced Thermal Engineering Questions and Answers focuses on “Steam Nozzles – Super Saturated Flow and Wilson’s Line”.

1. What is Wilson line?

a) It is an isothermal line, at which the condensation completes

b) Saturation line of water is also called Wilson line

c) It represents the limiting condition of undercooling at which the condensation begins

d) It represents an Isobaric line, at which the condensation commences

Answer: c

Explanation: At Wilson line the normal thermal equilibrium conditions are restored. It represents the limiting condition of undercooling at which the condensation begins. Wilson line is important while dealing with supersaturated expansion.

2. Considering metastable expansion to be in effect, the flow of wet steam though a convergent-divergent nozzle shall result in a discharge slightly less than the calculated one.

a) True

b) False

Answer: b

Explanation: The flow of wet steam through a convergent-divergent nozzle results in a discharge slightly greater than the calculated one. Converging part is small and steam velocity being high, eludes the possibility of normal condensation. This rapid expansion produces supersaturated state. Steam is undercooled to a temperature less than the saturation temperature corresponding to its pressure. This results in increase in density and hence increases in discharge.

3. The sudden expansion of dry saturated steam in the absence of dust, doesn’t lead to condensation until its density is about 8 times that of the saturated vapor of the same pressure.

a) True

b) False

Answer: a

Explanation: Prof Wilson proved this fact though series of experiments. This delayed condensation reduces the enthalpy drop and also the final condition of the steam is improved. This type of expansion in steam nozzles is also called metastable expansion.

4. Calculate the enthalpy drop steam enters a convergent-divergent nozzle at 11 bar and 250°C and leaves at 1 bar. The expansion is to be considered metastable.

a) 427031.34 kJ/kg

b) 55648.32 kJ/kg

c) 323654.25 kJ/kg

d) 158469.21 kJ/kg

Answer: a

Explanation: Given, P 1 = 11 bar, T 1 = 250°C, P 2 = 1 bar

At 11 bar and 250°C, from steam tables

v g1 = 0.2108 m 3 /kg

Since the steam entering is in superheated state,

n = 1.3

Enthalpy drop, h d = \(\frac{n}{n-1}\)(P 1 )(v g1 ){1-\

 

^\frac{n-1}{n}\)}

h d = \(\frac{1.3}{1.3-1}\)(11*10 5 )\

 

^\frac{1.3-1}{1.3}\big\} \)

h d = 427031.34 kJ/kg.

5. Which of the following the correct formula for calculating the exit velocity of steam through nozzle considering the expansion to be metastable.

a) C 2 = \

 

 

 

 C 2 = \

 

 

 

 C 2 = \

 

 

 

 C 2 = \(\sqrt{2*\frac{n}{n-1}

{1-

 

^\frac{n-1}{n}}}

\)

Answer: c

Explanation: When the expansion of steam is metastable the relation Pv n = C is used. Some enthalpy drop is lost and hence we use the following formula for calculating the exit velocity of the steam.

C 2 = \(\sqrt{2*\frac{n}{n-1}

{1-

 

^\frac{n-1}{n}}}

\)

6. Which of the following is the correct formula for calculating the supercooled temperature of steam, when steam metastable expansion of steam takes place inside a steam nozzle. (T 2 – Supercooled temperature, T 1 – Inlet temperature of steam)

a) \

 

 

^\frac{n}{n-1} \)

b) \

 

 

^\frac{n+1}{n} \)

c) \

 

 

^\frac{n}{n-1} \)

d) \

 

 

^\frac{n}{n-1} \)

Answer: d

Explanation: The correct formula for calculating the supercooled temperature is as follow –

\

 

 

^\frac{n}{n-1} \)

T 2 is the supercooled temperature.

7. Steam at 12 bar and 300°C enters a convergent-divergent steam nozzle and leaves at 3 bar. If the expansion of the steam is metastable, calculate the supercooled temperature.

a) 165°C

b) 143°C

c) 154°C

d) 120°C

Answer: b

Explanation: Given, P 1 = 12 bar, T 1 = 300°C or 573 K, P 2 = 3 bar

We know that,

\

 

 

^\frac{n}{n-1} \)

Substituting the values, we get

\

 

 

^\frac{1.3-1}{1.3} \)

Therefore, T 2 = 416.19 K or 143.19°C ≈ 143°C.

8. Steam at 20 bar and 300°C enters a convergent-divergent steam nozzle and is discharged at 3 bar. Considering the effects of supersaturation calculate the degree of undercooling.

a) 36.65°C

b) 45.54°C

c) 23.32°C

d) 62.87°C

Answer: a

Explanation: Given, P 1 = 20 bar, T 1 = 300°C or 573K, P 2 = 3 bar

At 3 bar, from steam tables

T sat = 133.5°C

n=1.3

Supercooled temperature, T 2 = T 1 \

 

^\frac{n}{n-1}\) = 573\

 

^\frac{1.3-1}{1.3}\) = 369.85 K or 96.85°C

Degree of undercooling = T sat – T 2 = 133.5 – 96.85 = 36.65°C.

9. Calculate the degree of supersaturation if steam at 10 bar 250°C is discharged at 2 bar by a convergent-divergent nozzle. Consider metastable steam expansion and negligible inlet velocity.

a) 5.65

b) 3.11

c) 2.15

d) 2.36

Answer: b

Explanation: Given, P 1 = 10 bar, T 1 = 250°C or 523 K, P 2 = 2 bar

Index of expansion, n = 1.3

Supercooled temperature, T 2 = T 1 \

 

^\frac{n}{n-1}\) = 523\

 

^\frac{1.3-1}{1.3}\) = 360.74 K or 87.74 °C

Saturation pressure corresponding to 87.74°C, P sat = 0.64309 bar 

Degree of supersaturation = \(\frac{P_2}{P_{sat}} = \frac{2}{0.64309}\) = 3.11.

10. Steam at 15 bar pressure and 250°C enters a convergent-divergent steam nozzle and leaves at 4 bar. The mass flow rate is 5 kg/s. Calculate the exit area of the nozzle if the expansion is metastable.

a) 0.00542187 m 2

b) 0.0029147 m 2

c) 0.036521 m 2

d) 0.003265 m 2

Answer: b

Explanation: Given, P 1 = 15 bar, T 1 = 250°C or 523 K, P 2 = 4 bar, m = 5 kg/s

At 15 bar and 250°C, from steam tables

v g1 = 0.1520 m 3 /kg

Exit velocity, C 2 = \(\sqrt{2*\frac{n-1}{n+1}

{1-

 

^\frac{n+1}{n}}} \)

C 2 = \(\sqrt{2*\frac{1.3}{1.3-1}

\big\{1-

 

^\frac{1.3-1}{1.3}\big\}}\)

C 2 = 720.74 m/s

We know that,

v g2 = \

 

^\frac{1}{n}\)*v g1 = \

 

\frac{1}{1.3}\)*0.1520 = 0.42015 m 3 /kg

Mass flow rate, m = \(\frac{A_2*C_2}{v_{g2}} \)

5 = \(\frac{A2*720.74}{0.42015} \)

A 2 = 0.0029147 m 2

11. What is the value of index of expansion for supersaturated steam. (Pv n =C, n is the index of expansion)

a) 1.300

b) 1.013

c) 1.135

d) 1.000

Answer: a

Explanation: In case of supersaturated steam the index of expansion is assumed to be same as that for superheated steam i.e. 1.3.

Therefore, supersaturated steam follows the following law –

Pv 1.3 = Constant.

12. Which of the following statement about metastable expansion of steam through a steam nozzle is FALSE?

a) It increases the discharge through the nozzle

b) Steam is undercooled to a temperature less than that corresponding to its pressure

c) Density of steam is increased

d) Exit velocity of the steam is increased

Answer: d

Explanation: Metastable expansion of steam produces a supersaturated state where steam is undercooled to a temperature less than that corresponding to its pressure. It has been observed after a series of experiments that discharge through the nozzle is increased as the density is increased. The exit velocity is reduced due to reduction in enthalpy drop.

13. The point on the h-s diagram at which the condensation beings and normal conditions of thermal equilibrium are restored is said to be on _____

a) wilson line

b) saturation vapor curve

c) constant dryness fraction line

d) constant pressure line

Answer: a

Explanation: Wilson line is the locus of points at which the limit of undercooling is reached and the condensation commences. At this state the normal conditions of thermal equilibrium are also restored. Wilson line is not conventionally shown in a Mollier chart.

14. Which of the following is NOT an effect of supersaturation in steam flowing through a convergent-divergent nozzle?

a) Increase in specific volume of steam

b) Increase in the entropy of steam

c) Improvement in dryness fraction

d) Increase in the enthalpy drop

Answer: d

Explanation: Enthalpy drop is reduced due to supersaturation in nozzles. Besides specific volume, entropy of the steam also increases due to supersaturation. The quality i.e. dryness fraction is also improved.

15. Determine the exit pressure of steam if the steam enters a nozzle at 11 bar and 300°C. At the outlet it is observed that the specific volume of the steam is 0.86804 m 3 /kg. The expansion of steam is metastable.

a) 4 bar

b) 3 bar

c) 2 bar

d) 1 bar

Answer: c

Explanation: Given, P 1 = 11 bar, T 1 = 300°C, v g2 = 0.86804 m 3 /kg

At 11 bar and 300°C, from steam tables

v g1 = 0.2339 m 3 /kg

we know that,

P 1 v g1 n = P 2 v g2 n

Substituting the values

11 1.3 = P 2  1.3

P 2 = 2 bar.

This set of Thermal Engineering Interview Questions and Answers for Experienced people focuses on “Relationship Between Area, Velocity and Pressure in Nozzle Flow”.

1. Which of the following expressions correctly represents the relationship between cross-sectional area of a nozzle, fluid velocity and specific volume?

a) \

 

 

 

 \

 

 

 

 \

 

 

 

 \(\frac{dA}{A}+\frac{dv}{v}=\frac{dC}{C} \)

Answer: a

Explanation: The correct relationship between cross-sectional area, fluid velocity and specific volume is represented by the following equation –

\

 

 

 

 in any two parameters in the equation above predicts the nature of change in the third parameter.

2. Which of the following expressions is correct? 

a) \

 

 

 

 

 \

 

 

 

 

 \

 

 

 

 

 \(\frac{1}{γ}\frac{dA}{A}=\frac{dP}{P}\big\{\frac{1-M^2}{M^2} \big\} \)

Answer: c

Explanation: The correct expression relating the pressure, cross-sectional area of nozzle/diffuser and Mach number is –

\(\frac{dA}{A}=\frac{1}{γ} \frac{dP}{P}\big\{\frac{1-M^2}{M^2} \big\} \)

or

\(\frac{dA}{A}= \frac{1}{γ} \frac{dP}{P}{\frac{Cs^2}{C^2 -1}} \)

where, C s – sonic velocity

C – fluid velocity

3. What is Mach number?

a) It is the ratio of sonic velocity of a fluid at N.T.P. to the local sonic velocity of the same fluid

b) It is the ratio of fluid velocity to sonic velocity of the same fluid at N.T.P.

c) It is the ratio of local sonic velocity to fluid velocity

d) It is the ratio of fluid velocity to local sonic velocity

Answer: d

Explanation: Mach number is the ratio of fluid velocity to the local sonic velocity. It is the ratio of the same quantity and hence is dimensionless. The fluid velocity is called subsonic is Mach number is less than one and supersonic if the Mach number is greater than one.

4. In case of accelerated flow, when the pressure decreases along the flow direction and Mach number is less than one, it corresponds to _____

a) Convergent part of a nozzle

b) Divergent part of a nozzle

c) Throat of a nozzle

d) Convergent part of a diffuser

Answer: a

Explanation: The flow is accelerated, hence it’s a nozzle. Since \(\frac{dP}{P}\) is negative and Mach number is less than one, for the following equation to hold –

\(\frac{dA}{A} = \frac{1}{γ} \frac{dP}{P} \big\{\frac{1-M^2}{M^2} \big\} \)

L.H.S should also be negative. This implies that \(\frac{dA}{A}\) should be negative, which corresponds to convergent part of the nozzle.

5. Which of the following statements regarding the Mach number is TRUE, when the fluid reaches the throat of a nozzle?

a) It becomes unity

b) It is less than one

c) It is greater than one

d) Mach number is not defined at throat of a nozzle

Answer: a

Explanation: At the throat of the nozzle, there is no change in cross-sectional are of the nozzle i.e. \(\frac{dA}{A}\)=0. Therefore, from the following equation –

\(\frac{dA}{A}=\frac{1}{γ} \frac{dP}{P} \big\{\frac{1-M^2}{M^2} \big\} \)

It is evident that the R.H.S. should also be zero. This implies that the Mach number must be one. The fluid velocity attains the value of local sonic velocity at the throat.

6. Which of the following conditions corresponds to divergent part of a nozzle?

a) M < 1 and \

 

 M < 1 and \

 

 M > 1 and \

 

 M > 1 and \(\frac{dP}{P}\) > 0

Answer: c

Explanation: Consider the following equation –

\(\frac{dA}{A}=\frac{1}{γ}\frac{dP}{P} \big\{\frac{1-M^2}{M^2} \big\} \)

For a nozzle \(\frac{dP}{P}\)<0, and for divergent part, \(\frac{dA}{A}\)>0. For the above equation to hold M > 1.

Therefore, M > 1 and \(\frac{dP}{P}\) < 0 is the correct answer.

7. A decelerated flow, having fluid velocity greater than the local sonic velocity corresponds to _____

a) Convergent part of a nozzle

b) Divergent part of a nozzle

c) Convergent part of a diffuser

d) Divergent part of a diffuser

Answer: c

Explanation: For a diffuser, \(\frac{dP}{P}\)>0. It is given that fluid velocity is greater than the local sonic velocity i.e. M > 1. Therefore, according to the following equation –

\(\frac{dA}{A}=\frac{1}{γ}\frac{dP}{P} \big\{\frac{1-M^2}{M^2} \big\} \)

8. Which of the following conditions correspond to divergent type diffuser?

a) M < 1 and \

 

 M < 1 and \

 

 M < 1 and \

 

 M > 1 and \(\frac{dA}{A}\)<0

Answer: a

Explanation: Diffuser promotes decelerated flow, \frac{dP}{P}>0. Consider the following equation –

\(\frac{dA}{A}=\frac{1}{γ}\frac{dP}{P} \big\{\frac{1-M^2}{M^2} \big\} \)

For the divergent part, \frac{dA}{A}>0. For the above equation to hold under the listed conditions, the Mach number must be less than one. Hence, M < 1 and \(\frac{dA}{A}\)>0 is the correct answer.

9. The purpose of a steam injector is to force water into the boiler under pressure.

a) True

b) False

Answer: a

Explanation: A steam injector is used to force water into the boiler under pressure. It employees the principle of steam nozzles. It makes use of the kinetic energy of a steam jet for increasing the pressure and velocity of a corresponding amount of water.

10. Air at 18 bar and 100°C enters a convergent nozzle. Assume the flow to be isentropic and calculate the sonic velocity. Take adiabatic index equal to 1.4.

a) 353.40 m/s

b) 321.56 m/s

c) 360.87 m/s

d) 400.32 m/s

Answer: a

Explanation: P = 18 bar, T = 100°C or 373 K, γ = 1.4

Critical pressure, P* = P\

 

^\frac{γ}{γ-1}\) = 18\

 

^\frac{1.4}{1.4-1}\) = 9.51 bar

T 1 = T\

 

^\frac{γ-1}{γ}\) = \

 

^\frac{1.4-1}{1.4}\) = 310.83 K

Sonic velocity, C s = \(\sqrt{γR

} = \sqrt{1.4*287*310.83}\) = 353.40 m/s.

11. Air enters a frictionless adiabatic horizontal nozzle at 12 bar and 167°C with inlet velocity 50 m/s and leaves at 3 bar. Take adiabatic index equal to 1.4 and c p = 1.005 kJ/kg-K.

a) 654.78 m/s

b) 321.75 m/s

c) 552.45 m/s

d) 456.87 m/s

Answer: c

Explanation: P 1 = 13 bar, T 1 = 167°C or 440 K, C 1 = 50 m/s, P 2 = 3 bar, c p = 1.005 kJ/kg-K, γ = 1.4

We know that, T 2 = T 1 \

 

^\frac{γ-1}{γ}\) = 440 \

 

^\frac{1.4-1}{1.4}\) = 289.40 K

Applying the energy equation at inlet and outlet of the nozzle, we get

m[h 1 +\(\frac{c_1^2}{2}\)+Z 1 *g]+Q=m[h 2 +\(\frac{c_2^2}{2}\)+Z 2 *g]+W

Q = 0, W = 0, Z 1 = Z 2

h 1 +\(\frac{c_1^2}{2}\)=h 2 +\(\frac{c_2^2}{2}\)

c 2 2 =2(h 1 -h 2 )+c 1 2

C 2 = \(\sqrt{2

+c_1^2} \)

C 2 = \(\sqrt{2c_p

+c_1^2} \)

C 2 = \(\sqrt{2*1.005*10^3 +50^2} \)

C 2 = 552.45 m/s.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Classification of Steam Turbines”.

1. Steam turbine produces useful work in the form of rotation of turbine shaft, by extracting thermal energy from pressurized steam.

a) True

b) False

Answer: a

Explanation: Steam turbines are one of the most important prime movers for power generation. In a steam turbine the potential energy of the steam is transformed into kinetic energy and it is then it is transformed into the mechanical energy of rotation of the turbine shaft.

2. According to the number of pressure stages, steam turbines are classified into _____

a) single cylinder and multi-cylinder

b) single stage and multi-stage

c) mono stage and multi-stage

d) axial and radial

Answer: b

Explanation: On the basis of number of stages, steam turbines are classified into single stage and multi stage turbines. Single stage turbines have only one pressure stage but can have multiple velocity stages. They are used for small power capacities like driving centrifugal compressors or blowers etc. Multistage turbine is a combination of impulse and reaction turbines and they are made for a wide range of power capacities.

3. According to the direction of steam flow, steam turbines are classified into _____

a) axial and radial

b) uniaxial and multi-axial

c) upstream and downstream

d) forward and backward

Answer: a

Explanation: According to the direction of steam flow, turbines are classified into axial and radial turbines. In case of axial turbine, the direction of steam flow is parallel to the axis of the turbine. In radial turbines steam flow direction is perpendicular to the axis of turbine.

4. Turbines with separate rotor shafts for each cylinder placed parallel to each other are known as _____

a) multi-rotor turbines

b) multiaxial turbines

c) single-cylinder turbines

d) multi-utility turbine

Answer: b

Explanation: Multiaxial turbines are multi-cylinder turbines with separate rotor shafts for each cylinder placed parallel to each other. Multi-cylinder turbines in which the rotors are mounted on one shaft and coupled to a single generator are known as single-shaft turbines.

5. On the basis of method of governing, steam turbines are classified into turbines with _____

a) diffuser governing and nozzle governing

b) throttle governing and nozzle governing

c) impulse governing and reaction governing

d) throttle governing and diffuser governing

Answer: b

Explanation: In case of turbines with throttle governing, fresh steam enters through one or more simultaneously operated throttle valves. in turbines with nozzle governing, fresh steam enters through two or more consecutively opening regulators.

6. Which of the following is NOT a valid type of steam turbines, based on heat drop process?

a) Condensing turbines with generator

b) Back pressure turbines

c) Topping turbines

d) Supercritical turbines

Answer: d

Explanation: According to heat drop process, steam turbines are classified into mixed pressure turbines, low pressure turbines, back pressure turbines, topping turbines, back pressure turbines with steam extraction from intermediate stages at specific pressure, condensing turbines with generator, condensing turbines with one or two intermediate stage extractions at specific pressures.

7. Which of the following statement regarding condensing turbines with generators is TRUE?

a) Exhaust steam is supplied to intermediate stages

b) Steam is not extracted from intermediate stages

c) The latent heat of exhaust steam during condensation is utilized in feed water heating

d) Steam at a pressure less than atmospheric is directed to a condenser

Answer: d

Explanation: In condensing turbines with generators, steam at a pressure less than atmospheric pressure is directed to a condenser. Also steam is extracted from intermediate stages for feed water heating. The number of extractions lying from 2-3 is as high as 8-9. In these turbines the latent heat of exhaust steam during condensation is lost completely.

8. Which of the following statements regarding topping turbines is TRUE?

a) Exhaust steam is utilized in low and medium condensing turbines

b) Always operate at low initial conditions 

c) It has highest efficiency among all the turbines

d) Exhaust steam is lost completely

Answer: a

Explanation: Topping turbines are similar to backpressure turbines, the difference being – in case of topping turbines, the exhaust steam is utilized in low and medium condensing turbines. Usually topping turbines are operated at high initial conditions i.e. high steam pressure and temperature.

9. Which of the following turbines have inlet pressure of steam greater than 225 ata?

a) Medium pressure turbines

b) High pressure turbines

c) Low pressure turbines

d) Supercritical turbines

Answer: d

Explanation: Supercritical turbines handle steam having pressure greater than 225 ata. Low pressure use steam at a pressure between 1.2 to 2 ata, medium pressure turbines upto 40 ata and high pressure turbines above 40 ata.

10. Stationary turbines with variable speed cannot be used to drive _____

a) turbo-blowers

b) air-circulators

c) pumps

d) ships

Answer: d

Explanation: Stationary turbines could be of constant speed or variable speed. The variable speed stationary turbines are used to drive turbo-blowers, air-circulators, pumps etc. A non-stationary turbine with variable speed is required to drive locomotives like ship and railways.

11. Which of the following statements about steam turbines is FALSE?

a) Thermal efficiency of steam turbine is higher than that of steam engine

b) Internal lubrication is not required

c) Flywheel installation is not necessary

d) It has a non-uniform rate of power generation

Answer: d

Explanation: The rate power generation is uniform and hence flywheel installation is not necessary. Since there are no rubbing parts, internal lubrication is not required. The thermal efficiency of steam engine is less than that of steam turbine.

12. In Simple impulse turbine the pressure of the steam falls gradually.

a) True

b) False

Answer: b

Explanation: In case of simple impulse turbine the expansion of steam takes place in the nozzle, the pressure is not affected as steam moves over the blades. The gradual fall in pressure is observed in case of reaction turbines.

13. Which of the following statements about reaction turbines is TRUE?

a) Steam pressure drops suddenly

b) The complete expansion of the steam takes place inside nozzle

c) Steam pressure is not altered as the steam moves over the blades the turbine

d) Steam pressure gradually drops as the steam moves over the blades of the turbine

Answer: d

Explanation: In reaction turbines, continuous expansion of the steam takes place as it passes over the blades of the turbine and therefore there is a gradual fall in the pressure of the steam during the expansion. Steam pressure is not altered when the steam expansion takes place in a simple impulse turbine.

This set of Thermal Engineering Quiz focuses on “Steam Turbines – Methods of Reducing Rotor Speed”.

1. How many set of nozzles are there in a simple impulse turbine?

a) one

b) two

c) three

d) four

Answer: a

Explanation: There is only one set of nozzles in a simple impulse turbine. The complete expansion of steam takes place at once. The pressure of the steam doesn’t alter when the steam moves over the blades of the turbine.

2. In steam turbine terminology, the loss of energy due to higher exit velocity of steam is called _____

a) velocity loss

b) energy loss

c) kinetic loss

d) carry over loss

Answer: d

Explanation: The loss of energy due to higher exit velocity of steam in steam turbines is called carry over loss or leaving loss. This type of energy loss is observed in simple impulse turbines. The expansion takes place completely in one set of nozzles and there is only one set of blades, therefore, the exit velocity of the steam is high.

3. Which of the following statements about De laval turbine is FALSE?

a) It has only one set of nozzles

b) Impulse blades are attached to the rotors

c) It has multiple rings of moving blades

d) Exit velocity of steam is very high

Answer: c

Explanation: De laval turbine is a simple impulse turbine. It has only one set of turbines and one set of blades, impulse type attached to the rotor. The exit velocity of steam is high, about 3.3 per cent of the nozzle outlet velocity.

4. The wheel or rotor speed in case of De laval turbine is low as only one ring of moving blade absorbs the kinetic energy of the steam.

a) True

b) False

Answer: b

Explanation: Since the expansion of steam takes place completely at once in De laval turbine, and only one ring of blades is present, the wheel or rotor speed is too high. It varies from 25000 to 30000 r.p.m. Some energy is lost as the exit velocity of steam is high.

5. Which of the following statements about multi-stage reaction turbine is FALSE?

a) Pressure drop takes place gradually over the fixed blades

b) Fixed blades act as nozzles

c) Moving blades absorb kinetic energy of the steam

d) Pressure is not altered while the steam is moving over the moving blades

Answer: d

Explanation: In case of a multi-stage reaction turbine, gradual pressure drop takes place in both moving and fixed blades. Fixed blades acting as nozzles increase the velocity of the steam. The moving blades absorb the kinetic energy of the steam.

6. Which of the following statements is FALSE about velocity compounding?

a) Steam is expanded through only one set of nozzles

b) Low cost and high efficiency

c) The kinetic energy of the steam is absorbed in stages

d) The fixed blades re-direct the direction of velocity

Answer: b

Explanation: Velocity compounding is done to reduce the rotor speed. The kinetic energy of the steam is absorbed in stages. Complete expansion of the steam takes place at once in the nozzles. The initial cost is low due to lesser number of stages, and its efficiency is also low.

7. Which of the following statements is FALSE about pressure compounding?

a) Steam pressure is reduced when steam moves over the moving blade

b) Multiple set of nozzles is used

c) Rotor speed is not very high

d) Steam expansion takes place in steps

Answer: a

Explanation: Pressure compounding is one of the methods to reduce rotor speed. In this method steam is expanded in steps. After every set of nozzles there is a set of moving blades to absorb the kinetic energy gained by the steam after expansion. Steam while moving through moving blades does not experience change in pressure.

8. Which of the following turbine uses pressure compounding method for reduction of rotor speed?

a) Curtis turbine

b) Rateau and Zoelly turbine

c) Curtis and Moore turbine

d) De laval turbine

Answer: b

Explanation: Pressure compounding is used in Rateau and Zoelly turbine for reduction of rotor speed. In this method the steam is expanded in steps and after each expansion the kinetic energy of the steam is absorbed by a set of moving blades.

9. Which of the following turbines do not resort to any method of rotor speed?

a) De laval turbine

b) Curtis turbine

c) Rateau and Zoelly turbine

d) Curtis and Moore turbine

Answer: a

Explanation: De laval turbine is a simple impulse turbine. It has only one set of nozzles and only one set of moving blades. The complete expansion of steam takes place in the single set of nozzles and the steam pressure remains unaltered while the steam passes through the set of moving blades.

10. Pressure velocity compounding is used in _____

a) Rateau and Zoelly turbine

b) Curtis and Moore turbine

c) De laval turbine

d) Curtis turbine

Answer: b

Explanation: Curtis and Moore turbine employs pressure velocity compounding. The pressure drop is divided into stages and the velocity is also compounded in these stages. A set of nozzles is fixed at the start of each stage.

11. Identify the incorrect statement regarding impulse turbine.

a) Blade manufacturing of reaction turbine is difficult than that of impulse turbine

b) Impulse turbine makes use of profile type blades whereas reaction turbine uses aerofoil type

c) Pressure drop occurs in only nozzles in impulse turbines but also occurs in moving blades reaction turbine

d) Impulse turbine can generate more power than impulse turbine

Answer: d

Explanation: Not much power can be generated with impulse turbine, but with reaction turbine power generation is better. Impulse turbine has profile type blades and reaction turbines have aerofoil type. Aerofoil blade manufacturing is difficult than profile type blade.

12. Which of the following turbines has the least efficiency?

a) Curtis and Moore turbine

b) Rateau and Zoelly turbine

c) De laval turbine

d) Curtis turbine

Answer: c

Explanation: De laval turbine is a single stage turbine. Curtis turbine, Rateau and Zoelly turbine and Curtis and Moore turbine are all multi-stage turbines. De laval turbine has the maximum carry over loss, therefore, minimum efficiency.

13. In a condensing steam turbine the pressure of the steam leaving the turbine should be equal to _____

a) inlet pressure

b) condenser pressure

c) atmospheric pressure

d) boiler pressure

Answer: b

Explanation: In a turbine the pressure of the steam during expansion is reduced to condenser pressure. And if the turbine is no-condensing type then the pressure is reduced to atmospheric. The inlet pressure of the steam is the equal to the boiler pressure.

14. Identify the incorrect statement about blades.

a) Fixed profile type blade redirects the steam

b) Steam pressure is dropped in moving as well as fixed aerofoil type blades

c) Fixed aerofoil type blades act as nozzles

d) Steam pressure us dropped in moving profile type blades

Answer: d

Explanation: Steam pressure is not affected when the steam is passed through the profile type blades irrespective of them being fixed or moving. In case of aerofoil type blades, steam is pressure is dropped in moving as well as fixed blades. Fixed aerofoil blades act as nozzles.

15. In pressure velocity compounding, the total drop in steam pressure is divided into stages and velocity obtained in each stage is also compounded.

a) True

b) False

Answer: a

Explanation: Nozzles are fixed at the start of each stage and a set of fixed blades and moving blades is installed in each stage for velocity compounding. The pressure remains unaltered during each stage. The method of pressure velocity compounding is used in Curtis and Moore turbine.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Impulse Turbines – 1”.

1. Which of the following statements regarding impulse turbine is correct?

a) Only velocity of flow performs work on the blade

b) Only velocity of Whirl performs work on the blade

c) Axial component of velocity is called velocity of whirl

d) Relative velocity of steam to moving blade at the inlet is always less than the relative velocity at outlet

Answer: b

Explanation: The steam is impinged on the blades at the nozzle angle. Only the tangential i.e. velocity of whirl performs work on the blades. The axial component i.e. velocity of flow causes the steam to flow through the turbine and has no contribution in work. The relative velocity of stem at exit is always less than  the relative velocity at the entrance.

2. Which of the following expressions represent the correct formula for calculating the blade efficiency?

a) \

 

 \

 

 \

 

 \(\frac{2

}{C_1}\)

Answer: c

Explanation: Blade efficiency or diagram efficiency is the ratio of work done on the blade to energy supplied to the blade.

Mathematically,

η blade = \(\frac{ṁ

*C_bl}{\frac{ṁC_2^2}{2}} = \frac{2

*C_bl)}{C_1^2}\)

Symbols –

C bl – Linear velocity of moving blade

C 1 – Absolute velocity of steam entering the moving blade

C w1 – Velocity of whirl at the entrance of the moving blade

C 2 – Absolute velocity of steam leaving the moving blade

C r2 –Relative velocity of steam to moving blade at exit

C w2 – Velocity of whirl at the exit of the moving blade

ṁ – Rate of flow of steam.

3. The ratio of relative velocity of steam to the moving blade at exit to the relative velocity of steam to the moving at entrance is called _____

a) reduction co-efficient

b) blade velocity co-efficient

c) friction loss

d) factor of reduction

Answer: b

Explanation: Blade velocity co-efficient  is the ratio of relative velocity of steam to the moving blade at the exit to relative velocity of moving blade at the entrance. As the steam passes over the blades of the turbine the velocity is reduced due to the friction present on the surface of the blades.

4. Blade speed ratio is the _____

a) ratio of blade speed to steam speed at entrance of the moving blade

b) ratio of steam speed at entrance of the moving blade to blade speed

c) ratio of blade speed to steam speed at exit of the moving blade

d) ratio of steam speed at exit of the moving blade to blade speed

Answer: a

Explanation: Blade speed ratio  is the ratio of blade speed to the steam speed at the entrance of the moving blade.

Mathematically,

ρ = \(\frac{C_{bl}}{C_1} \)

For maximum efficiency blade speed ratio should be half of the cosine of the nozzle angle.

5. What is the maximum possible value of blade efficiency?

a) sin⁡α 2

b) cos⁡α 2

c) sin⁡β 2

d) cos⁡β 2

Answer: b

Explanation: The maximum value of blade efficiency is obtained when the blade speed ratio  is half the cosine of the nozzle angle. The maximum possible value of blade efficiency is cos⁡α 2 considering the blades to be symmetrical and no friction in fluid passage.

6. What is the optimum blade speed ratio for maximum blade efficiency in a multi-stage turbine? 

a) \

 

 \

 

 \

 

 \(\frac{2*cos β}{n} \)

Answer: a

Explanation: The optimum blade speed ratio  for maximum blade efficiency is given by –

\(\frac{cos⁡α}{2*n} \)

Also, in this case the ratio of total work to the work done in the last row is 2 n . The maximum efficiency is cos⁡α 2 .

7. Which of the following statements regarding velocity compounded impulse turbine is FASLE?

a) Maximum stage efficiency decreases with the with the increase in number of moving blades’ rows

b) It has low efficiency and high steam consumption

c) Optimum value of blade speed ratio for maximum efficiency increases with the with the increase in number of moving blades’ rows

d) It requires comparatively small number of stages

Answer: c

Explanation: As the number of moving blades’ rows increases the optimum value of blade speed ratio for maximum efficiency decreases. A velocity compounded impulse turbine requires comparatively less number of stages, due to relatively large heat drop. Usually, more than two rows are not preferred.

8. Steam issues from the nozzle with a velocity of 1000 m/s in a De Laval turbine. The nozzle angle is 20°, mean blade velocity is 350 m/s. The inlet and outlet angles of the blades are equal. The mass of the steam flowing through the turbine per second is 0.3 kg. Calculate the tangential force on the blades. Take blade velocity coefficient as 0.8.

a) 451 N

b) 256 N

c) 318 N

d) 196 N

Answer: c

Explanation: C bl = 350 m/s, C 1 = 1000 m/s, α = 20°, K = 0.8, ṁ = 0.3 kg/sec, θ = Ф

Steps of Construction –

Select a suitable scale and draw a line LM to represent C bl .

At point M make angle of 20°  and cut length MS to represent the velocity 1000 m/s. Produce L to meet the perpendicular drawn from S at P. Thus inlet triangle is completed.

By measurement: θ = 30.4°, C r1 = 681.7 m/s

θ = Ф = 30.4°

C r2 = KC r1 = 0.8 = 545.36 m/s

At point L, make an angle of 30.4° Ф and cut the length LN to represent C r2 . Join MN. Produce M to meet the perpendicular drawn from N at Q. Thus outlet triangle is completed.

By measurement: C w1 = 939.69 m/s and C w2 = 120.38 m/s

Tangential force on the blades = ṁ (C w1 + C w2 ) = 0.3 = 318.021 N ≈ 318 N

9. Velocity of the steam leaving the nozzle is 1200 m/s. The nozzle angle is 20° and the mean blade velocity is 500 m/s. The inlet and outlet angles of blades are equal. The mass of the steam flowing through the turbine per second is 0.4 kg. Considering the blades to be frictionless determine the Power Developed.

a) 251.30 kW

b) 150.21 kW

c) 123.65 kW

d) 321.56 kW

Answer: a

Explanation: C bl = 500 m/s, C 1 = 1200 m/s, α = 20°, ṁ = 0.4 kg/sec, θ = Ф

Steps of Construction –

Select a suitable scale and draw a line LM to represent C bl .

At point M make angle of 20°  and cut length MS to represent the velocity 1200 m/s. Produce L to meet the perpendicular drawn from S at P. Thus inlet triangle is completed.

By measurement: θ = 33°, C r1 = 749.91 m/s

θ = Ф = 33°

C r2 = C r1 = 749.91 m/s

At point L, make an angle of 33° Ф and cut the length LN to represent C r2 . Join MN. Produce M to meet the perpendicular drawn from N at Q. Thus outlet triangle is completed.

By measurement: C w1 = 1127.6 m/s and C w2 = 128.9 m/s

Power Developed = ṁ (C w1 + C w2 ) C bl = 0.4500 = 251300 W or 251.3 kW

10. Steam is issued from the nozzle at a velocity of 1000 m/s. The mean blade velocity is 450 m/s and the nozzle angle is 25°. The inlet and outlet angles of blades are equal. The mass of the steam flowing through the turbine per hour is 1000 kg. Calculate the blade efficiency if the blade velocity coefficient is 0.85.

a) 88%

b) 95%

c) 50%

d) 76%

Answer: d

Explanation: C bl = 450 m/s, C 1 = 1000 m/s, α = 25°, K = 0.85, θ = Ф, ṁ = 1000 kg/hr or 0.278 kg/sec

Steps of Construction –

Select a suitable scale and draw a line LM to represent C bl .

At point M make angle of 25°  and cut length MS to represent the velocity 1000 m/s. Produce L to meet the perpendicular drawn from S at P. Thus inlet triangle is completed.

By measurement: θ = 43°, C r1 =621.95 m/s

θ = Ф = 43°

C r2 = KC r1 = 0.85 = 528.66 m/s

At point L, make an angle of 43° Ф and cut the length LN to represent C r2 . Join MN. Produce M to meet the perpendicular drawn from N at Q. Thus outlet triangle is completed.

By measurement: C w1 = 906.31 m/s and C w2 = 66.36 m/s 

Blade efficiency = \(\frac{2

C_{bl}{C_1^2} = \frac{2450}{1000^2}\) = 0.76 or 76%.

11. If the blade and nozzle efficiencies are 75% and 85% respectively, determine the stage efficiency.

a) 75.65%

b) 45.32%

c) 63.75%

d) 55.32%

Answer: c

Explanation: η blade = 0.75, η nozzle = 0.85

η stage = η blade *η nozzle = 0.75*0.85= 0.6375 or 63.75%.

12. If for a certain simple impulse turbine, the blade inlet and outlet angles are 35° and 30° respectively and the relative velocity of steam to moving blade at entrance and exit are 350 m/s and 320 m/s respectively, determine the axial thrust on the bearings. The turbine has a steam flow rate of 20 kg/s.

a) 850 N

b) 815 N

c) 750 N

d) 456 N

Answer: b

Explanation: C r1 = 350 m/s, θ = 35, C r2 = 320 m/s, Ф = 30°, ṁ = 20 kg/s

C f1 = c r1 *sin⁡θ = 350*sin⁡35 = 200.75 m/s

C f2 = c r2 *sin⁡Ф = 320*sin⁡30 = 160 m/s

Axial Thrust = ṁ (C f1 – C f2 ) = 20  = 815 N.

13. A simple impulse turbine has a mean blade speed of 250 m/s. The nozzle angle is 20° and the steam velocity from the nozzle is 600 m/s. The turbine uses 3000 kg/h of steam. The blade outlet angle is 25°. Determine the amount of energy converted to heat by blade friction if the blade friction factor is 0.86.

a) 65.32 kJ

b) 30.50 kJ

c) 45.32 kJ

d) 78.32 kJ

Answer: b

Explanation: C bl = 250 m/s, α = 20°, C 1 = 600 m/s, ṁ=3000 kg/h or 0.833 kg/s, K = 0.86, Ф = 25°

Steps of Construction –

Select a suitable scale and draw a line LM corresponding to 250 m/s (C bl ).

At M, at an angle of 20° draw a line MS having a length of corresponding to 600 m/s (C 1 ). Join L and S. The inlet triangle is completed.

By measurement –

C r1 = 374.96 m/s

C r2 = K(C r2 ) = 0.86 = 322.46 m/s

To complete the outlet triangle, draw a line of length corresponding to 322.46 m/s (C r2 ) i.e. LN at an angle of 25° at L. Join MN.

Amount of energy converted to heat by blade friction = ṁ (C r1 2 -C r2 2 ) = 0.833 (374.96 2 -322.46 2 ) = 30499.92 J or 30.5 kJ.

14. Steam is issued from the nozzles at 600 m/s in a simple impulse turbine. The nozzle angle is 25° and the mean blade velocity is 350 m/s. The blade velocity coefficient is 0.88. Determine the blade speed ratio.

a) 0.325

b) 0.583

c) 0.654

d) 0.785

Answer: b

Explanation: C 1 = 600 m/s, α = 20°, C bl = 350 m/s, K = 0.88

Blade speed ratio  = \(\frac{C_{bl}}{C_1}\) = 350/600 = 0.583.

15. Nozzle angle of a simple impulse turbine is 25°. Determine the blade speed ratio if the turbine works at maximum efficiency. The Turbine blades are to be assumed frictionless.

a) 0.4532

b) 0.2136

c) 0.5465

d) 0.6987

Answer: a

Explanation: α = 25°

Since the turbine efficiency is maximum,

Blade speed ratio  = \(\frac{cos⁡α}{2} = \frac{cos⁡25}{2}\) = 0.4532.

This set of Thermal Engineering MCQs focuses on “Impulse Turbines – 2”.

1. In a simple impulse turbine, it is observed that 22.4 kJ energy is lost to blade friction. The uses steam at a rate of 1 kg/s. If the velocity of steam relative to the moving blade at an entrance is 300 m/s, determine the blade velocity coefficient.

a) 0.325

b) 0.456

c) 0.867

d) 0.951

Answer: c

Explanation: Given, H = 22.4 kJ or 22400 J, C r1 = 300 m/s, ṁ = 1 kg/s

We know that,

Energy converted to friction = ṁ (C r1 2 – C r2 2 )

22400 = 1(300 2 – C r2 2 )

C r2 = 260 m/s

Blade velocity coefficient  = \(\frac{C_{r2}}{C_{r1}} = \frac{260}{300}\) = 0.86.

2. The maximum possible efficiency of a De-laval turbine is 88.3%. Determine the nozzle angle.

a) 15°

b) 20°

c) 25°

d) 30°

Answer: b

Explanation: η max = 0.883

We know that,

η max = cos ⁡ α 2

0.883 = cos ⁡ α 2

α = cos -1 ⁡\(\sqrt{0.883}\)

α = 20°.

3. In a De laval turbine, steam leaves nozzle with a velocity of 1300 m/s. The nozzle angle is 25° and the mean blade speed is 500 m/s. The inlet and outlet blade angles are equal. Find out work done per kg of steam if the blades of the turbine are frictionless.

a) 678.205 kJ

b) 798.562 kJ

c) 365.458 kJ

d) 456.321 kJ

Answer: a

Explanation: C 1 = 1300 m/s, α = 25°, C bl = 500 m/s, θ = Ф

Steps for Construction –

Select a suitable scale and draw line LM corresponding to 500 m/s (C bl ).

At M, cut a length at S corresponding to 1300 m/s (C 1 ) at an angle of 25° . Join LS.

Produce L to meet perpendicular from S at P. The inlet triangle is complete.

By measurement –

C r1 = 872.81 m/s, θ = 39°

Since the blades of the turbine are considered to be frictionless

C r2 =C r1 = 872.81 m/s

Also, Ф = θ = 39°

At L, cut a length at N corresponding to 872.81 m/s (C r2 ) at an angle of 39°. Join MN.

Produce M to meet perpendicular from N at Q. The outlet triangle is complete.

By measurement –

C w1 = 1178.20 m/s and C w2 = 178.21 m/s

Work done per kg of steam = (C w1 + C w2 )*C bl = *500 = 678205 J or 678.205 kJ.

4. In a De-Laval turbine, steam with absolute velocity of 950 m/s is supplied through a nozzle. The mean blade velocity is 300 m/s. If the turbine is working at its maximum possible efficiency, determine the work done per kg of steam.

a) 180 kJ

b) 200 kJ

c) 1705 kJ

d) 1805 kJ

Answer: a

Explanation: C 1 = 950 m/s, C bl = 300 m/s

We know that, when turbine functions at maximum efficiency-

Work done per kg of steam = 2*C bl 2 = 2*300 2 = 180000 J or 180 kJ.

5. In a single row impulse turbine, the absolute velocity of steam leaving the turbine is 900 m/s. The nozzle angel is 20°. The mean blade speed is 300 m/s. The inlet blade angle is equal to the outlet blade angle. Determine the angle made by the discharging steam with the tangent of the wheel at the exit of moving blade. Neglect friction on turbine blades.

a) 27°

b) 36°

c) 51°

d) 90°

Answer: c

Explanation: C 1 = 900 m/s, α = 20°, C bl = 300, θ = Ф, K = 1

Steps for construction –

Select a suitable scale and draw line LM corresponding to 300 m/s (C bl ).

At M, at an angle of 20° cut a length corresponding to 900 m/s at S. Join LS.

Produce L to meet perpendicular from S at P. The inlet triangle is complete.

By measurement –

C r1 = 626.55 m/s, θ = 29°

Since the blades of the turbine are considered to be frictionless

C r2 = C r1 = 626.55 m/s

Also, Ф = θ = 29°

At L, cut a length at N corresponding to 626.55 m/s (C r2 ) at an angle of 29°. Join MN.

Produce M to meet perpendicular from N at Q. The outlet triangle is complete.

By measurement, β = 51°.

6. In a single row impulse turbine, steam leaves the nozzle at 600 m/s. The nozzle angle and the mean blade speed is 25° and 200 m/s. The blade velocity coefficient is 0.8. Determine the blade outlet angle if the steam leaves the turbine blades axially.

a) 54°

b) 60°

c) 66°

d) 74°

Answer: a

Explanation: C 1 = 600 m/s, α = 25°, C bl = 200 m/s, K = 0.8, β = 90°

Steps for construction –

Select a suitable scale and draw line LM corresponding to 200 m/s (C bl ).

At M, at an angle of 25° cut a length corresponding to 600 m/s at S. Join LS.

Produce L to meet perpendicular from S at P. The inlet triangle is complete.

By measurement –

C r1 = 427.18 m/s, θ = 36°

C r2 = KC r1 = 0.8 = 341.74 m/s

Through M, draw a ray perpendicular to LM.

Join L to a point N on ray through M such that the length LN should correspond to 341.74 m/s.

By measurement, Ф = 54°.

7. In a simple impulse turbine, steam leaves nozzle at 1200 m/s. The mean blade speed and nozzle angle is 300 m/s and 20° respectively. The inlet and outlet blade angles are equal. The blades of the turbine are frictionless. Determine the stage efficiency if the nozzle efficiency is 85%.

a) 58.65%

b) 62.32%

c) 75.45%

d) 84.37%

Answer: a

Explanation: C 1 = 1200 m/s, C bl = 300 m/s, α = 20°, θ = Ф, K=1, η nozzle = 0.85

Steps for Construction –

Select a suitable scale and draw line segment LM corresponding to C bl .

At M, at an angle of 20° , cut a length at S corresponding to C 1 . Join LS.

Produce L to meet perpendicular from S at P. The inlet triangle is complete.

By measurement –

θ = 26°, C r1 = 923.81 m/s

Since the turbine blades are frictionless,

C r2 = C r1 = 923.81 m/s

Also, Ф = θ = 26°

At L, at an angle of 26° Ф cut a length at N corresponding to C r2 . Join MN.

Produce M to meet perpendicular from N at Q. The outlet triangle is complete.

By measurement –

C w1 = 1127.63 m/s and C w2 = 527.61 m/s

Blade efficiency, η blade = \(\frac{2*C_{bl}

}{C_1^2} = \frac{2*300*}{1200^2}\) = 0.69

Stage efficiency, η stage = η blade * η nozzle = 0.69*0.85 = 0.5865 or 58.65%.

8. A simple impulse turbine has one ring of moving blades running at 400 m/s. The steam velocity leaving the nozzle is 1300 m/s. If the nozzle angle is 25°, determine the blade inlet angle.

a) 15°

b) 20°

c) 25°

d) 35°

Answer: d

Explanation: C bl = 400 m/s, C 1 = 1300 m/s, α = 25°

Steps for Construction-

Select a suitable scale and draw line segment LM corresponding to C bl .

At M, at an angle of 25° , cut a length at S corresponding to C 1 . Join LS.

Produce L to meet perpendicular from S at P. The inlet triangle is complete.

By measurement –

θ = 35°.

9. In a single stage impulse turbine nozzle angle is 20° and the mean blade speed is 300 m/s. Steam leaves the nozzle at a velocity of 700 m/s. Determine the exit velocity of the steam if the turbine blades are equiangular and frictionless.

a) 154.65 m/s

b) 195.65 m/s

c) 246.30 m/s

d) 321.32 m/s

Answer: c

Explanation: C bl = 300 m/s, C 1 = 700 m/s, α = 20°, θ = Ф, K=1

Steps for Construction –

Select a suitable scale and draw line segment LM corresponding to C bl .

At M, at an angle of 20° , cut a length at S corresponding to C 1 . Join LS.

Produce L to meet perpendicular from S at P. The inlet triangle is complete.

By measurement –

θ = 34°, C r1 = 430.50 m/s

Since the turbine blades are frictionless,

C r2 = C r1 = 430.50 m/s

Also, Ф = θ = 34°

At L, at an angle of 34° Ф cut a length at N corresponding to C r2 . Join MN.

Produce M to meet perpendicular from N at Q. The outlet triangle is complete.

By measurement –

C 2 = 246.30 m/s.

10. In a single stage impulse turbine, the exit velocity of the steam is 90 m/s at an angle of 45° from the tangential direction. The mean blade speed is 200 m/s. The blades are equiangular and velocity coefficient of the blades is 0.8. Determine the velocity of steam leaving the nozzles.

a) 472.32 m/s

b) 535.50 m/s

c) 624.84 m/s

d) 784.36 m/s

Answer: b

Explanation: C 2 = 90 m/s, β = 45°, C bl = 200 m/s, θ = Ф, K = 0.8

Steps for Construction-

Select a suitable scale and draw line segment LM corresponding to C bl .

At M, at an angle of 45°  cut a length at N corresponding to C 2 . Join LN.

Produce M to meet perpendicular from N at Q. The outlet triangle is complete.

By measurement –

Ф = 14°, C r2 = 271.22 m/s

C r2 =\

 

 

Ф cut a length at S corresponding to 339.02 m/s. Join MS.

Produce L to meet perpendicular from S at P. The inlet triangle is complete.

By measurement,

C 1 = 535.50 m/s.

11. Which of the following statement regarding work done in impulse turbines is TRUE, if the blade speed ratio is equal to zero?

a) Maximum possible work done takes place

b) Work done becomes zero

c) Nothing can be stated about work done

d) work done per kg of steam is equal to

Answer: b

Explanation: When the blade speed ratio is zero, the work done becomes zero since the distance travelled by the blade is zero. In this case the torque on the blade is maximum in spite of work done being zero.

12. Which of the following statement about efficiency of impulse turbines is TRUE, when the blade speed ratio is made equal to half of cosine of the nozzle angle?

a) Efficiency is minimum and is equal to \

 

 Efficiency is maximum and is equal to \

 

 Efficiency is minimum and is equal to cos⁡ α 2

d) Efficiency is maximum and is equal to cos⁡ α 2

Answer: d

Explanation: When the blade speed ratio i.e. \(\frac{C_{bl}}{C_1}\) is equal to \(\frac{cos⁡ α}{2}\), efficiency attains its maximum value i.e. cos⁡ α 2 . This value of efficiency corresponds to a turbine having equiangular and frictionless blades. Blade speed ratio is denoted by the Greek letter ‘ρ’.

13. Which of the following statement about the work done per kg of steam in impulse turbines is TRUE, when the blade speed ratio is equals \

 

a) Work done per kg of steam is maximum and is equal to 2C 1 2

b) Work done per kg of steam is minimum and is equal to 2C 1 2

c) Work done per kg of steam is maximum and is equal to 2C bl 2

d) Work done per kg of steam is minimum and is equal to 2C bl 2

Answer: c

Explanation: The maximum value of work done per kg of steam is attained i.e. 2C bl 2 , when the blade speed ratio, is equal to \

 

 works at highest possible efficiency.

14. Work done per kg of steam in an impulse turbine when the blade speed ratio  is equal to one is _____

a) 0

b) C bl 2

c) 2C bl 2

d) 2C bl (C f1 +C f2 )

Answer: a

Explanation: When blade speed ratio , \(\frac{C_{bl}}{C_1}\) is equal to one the work done per kg of steam is zero even though the distance travelled by the blade is maximum. Also the work done per kg of steam is zero also when \(\frac{C_{bl}}{C_1}\) is zero.

15. In an impulse turbine, the axial force on the blades is zero if _____

a) the whirl components of the inlet and outlet absolute velocities are equal

b) the flow components of the inlet and outlet absolute velocities are equal

c) the exit velocity of steam is axial

d) the flow and whirl component of inlet velocity are equal

Answer: b

Explanation: The axial force is given by –

F axial = ṁ (C f1 -C f2 )

Where, C f1 and C f2 are the flow components of inlet and outlet absolute velocities.

If, C f1 =C f2

F axial =0.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Reaction Turbines – 1”.

1. The magnitude of velocity of steam relative to moving blade in case of an impulse turbine _____

a) always remains constant as steam glides over the blades

b) depending on friction present on the blades, increases or remains constant as the steam glides over the blades

c) depending on friction present on the blades, decreases or remains constant as the steam glides over the blades

d) depending on friction present on the blades, increases or decreases as the steam glides over the blades

Answer: c

Explanation: In Impulse turbines the magnitude of velocity of steam relative to the moving blade either remains contant or decreases slightly due to presence of friction on the blades. The ratio of final relative velocity of steam to initial is called the blade velocity coefficient .

2. In case of reaction turbines, the magnitude of velocity of steam relative to moving blade increases as the steam progresses.

a) True

b) False

Answer: a

Explanation: As the steam flows through the reaction turbine blades, it is continuously expanded. As a result the magnitude of velocity of steam relative to the moving blades increases as the steam flows over the blade.

3. Which of the following expressions for degree of reaction (R d ) of a reaction turbine stage is correct?

a) R d =\

 

 R d =\

 

 R d =\

 

 R d =\(\frac{Heat \, drop \, in \, the \, stage}{Heat \, drop \, in \, fixed \, blades}\)

Answer: a

Explanation: The degree of reaction (R d ) of a reaction turbine stage is defined as the ratio of heat drop in moving blade to the total heat drop in the stage.

Mathematically,

R d =\(\frac{Δh_m}{Δh_f+Δh_m} \)

where, Δh m = heat drop in moving blades and

Δh f = heat drop in the fixed blades.

4. Which of the following is the correct formula for calculating the degree of reaction of a reaction turbine?

a) R d =\

 

 R d =\

 

 R d =\

 

 R d =\(\frac{2

}{C_{bl}

} \)

Answer: b

Explanation: We know that,

R d =\(\frac{Heat \, drop \, in \, moving \, blades}{Heat \, drop \, in \, the \, stage}\)

Since, Heat drop in a moving blade = \(\frac{C_{r2}^2-C_{r1}^2}{2}\), and

Heat drop in the stage = C bl (C w1 + C W2 )

Therefore, R d =\(\frac{C_{r2}^2-C_{r1}^2}{2C_{bl} 

} \)

5. The degree of reaction of a Parson’s reaction turbine is _____

a) 0%

b) 25%

c) 50%

d) 100%

Answer: c

Explanation: In Parson’s reaction turbine the degree of reaction is 50%. Also in a Parson’s reaction turbine the shape of moving and the fixed blades is same. Due to this the velocity diagram for the blades of this turbine are symmetrical.

6. Which of the following statements does NOT hold true for a Parson’s reaction turbine?

a) α = Ф

b) θ = β

c) C 1 = C r2

d) C r1 = C r2

Answer: d

Explanation: Parson’s reaction turbine has same shape for moving and fixed baldes. Which implies,

θ = β and α = Ф

Also, C 1 = C r2 and C 2 = C r1 , which results in a symmetrical velocity diagram.

7. What should be the ratio of blade speed to absolute velocity of steam leaving the nozzle for a Parson’s reaction turbine to work at maximum efficiency?

a) cos⁡α

b) cos⁡β

c) sin⁡α

d) sin⁡β

Answer: a

Explanation: In a Parson’s reaction turbine, the degree of reaction is 50% and the moving and fixed blades are symmetrical. This turbine works at maximum possible efficiency when the ratio of blade speed to absolute velocity of steam leaving the nozzle is equal to cosine of nozzle angle.

8. The maximum value of efficiency for a Parson’s reaction turbine is _____

a) \

 

 \

 

 \

 

 \(\frac{2^2}{1+^2} \)

Answer: d

Explanation: For a Parson’s reaction turbine, the value of maximum efficiency is \(\frac{2^2}{1+^2} \). This value is achieved when the ratio of blade speed to absolute velocity of steam leaving the nozzle is equal to cosine of nozzle angle.

9. Which of the following velocity diagram ccorresponds to a Parson’s reaction turbine?

a) 

b) 

c) 

d) 

Answer: b

Explanation: Parson’s reaction turbine has symmetrical moving and fixed blades which results in a symmetrical velocity diagram. The following properties are observed in a velocity diagram of a Parson’s reaction turbine.

θ = β and α = Ф,

C 1 = C r2 and C 2 = C r1

Therefore, the correct answer is

10. The axial force on the blades in Parson’s reaction turbine is always zero.

a) True

b) False

Answer: a

Explanation: The velocity diagram of a parson’s reaction turbine is symmetrical, which results in equal flow components of absolute inlet and outlet velocities.

i.e. C f1 = C f2

Therefore, axial force on blades = ṁ (C f1 – C f2 ) = 0.

11. The following data refers to a particular stage of a Parson’s reaction turbine:

Mean blade speed = 90 m/s

Velocity of steam leaving the nozzle = 120 m/s

Nozzle angle = 20°

Mass flow rate of steam = 0.8 kg per second

Calculate the tangential force on the blade.

a) 203.54 N

b) 108.42 N

c) 65.32 N

d) 195.45 N

Answer: b

Explanation: C bl = 90 m/s, α = 20°, C 1 = 120 m/s, ṁ = 0.8 kg/s

Steps of Construction:

Select a suitable scale and draw a line LM to represent C bl .

At point M, at an angle of 20° , cut a length MS to represent the velocity C 1 . Join LS. Produce L to meet the perpendicular drawn from S at P. Thus inlet triangle is completed.

For a Parson’s turbine:

Ф = α = 20°

and C r2 =C 1 = 120 m/s

At L, at an angle of 20° Ф, cut a length LN to represent C r2 . Join MN.

Produce M to meet the perpendicular drawn from N at Q. Thus outlet triangle is completed.

By measurement:

C w1 = 112.76 m/s and C w2 = 22.76 m/s

Tangential force = ṁ(C w1 + C w2 ) = 0.8 = 108.42 N.

12. In a single stage Parson’s reaction turbine, the mean blade velocity is observed to be 60 m/s and the nozzle angle to be 20°. Determine the the entrance angle of moving blade if the absolute velocity of the steam leaving the nozzle is 120 m/s.

a) 38°

b) 56°

c) 23°

d) 28°

Answer: a

Explanation: C bl = 60 m/s, α = 20°, C 1 = 120 m/s

Steps of Construction:

Select a suitable scale and draw a line LM to represent C bl .

At point M, at an angle of 20° , cut a length MS to represent the velocity C 1 . Join LS. Produce L to meet the perpendicular drawn from S at P. Thus inlet triangle is completed.

By measurement:

θ = 38°.

13. In a reaction turbine, the fixed blades and the moving blades are of the same shape but reversed in direction. The mean blade speed is 30 m/s. The absolute velocity of the steam leaving the nozzle is 90 m/s and the absolute velocity of steam leaving the moving blade is 64 m/s. Determine the nozzle angle.

a) 20°

b) 25°

c) 30°

d) 35°

Answer: b

Explanation: Given, C bl = 30 m/s, C 0 = C r1 = 64 m/s, C 1 = 90 m/s

Steps of Construction:

Select a suitable scale and draw a line LM to represent C bl .

At point M, draw an arc of length corresponding to C 1 . At point L, draw an arc of length corresponding to length 64 m/s, cutting the previous arc at S. The inlet triangle is complete.

By measurement:

α = ∠LMS = 25°.

14. In a Parson’s reaction turbine, the outlet angle of blade is 25°. Calculate the work done per second by the blades if the blade mean blade speed is 60 m/s and the absolute velocity of the steam leaving the nozzle is 120 m/s. The mass flow rate of steam is 0.1 kg per second.

a) 756 W

b) 854 W

c) 945.12 W

d) 1065 W

Answer: c

Explanation: Given, α = Ф = 25°, C bl = 60 m/s, C 1 = 120 m/s, ṁ = 0.1 kg/s

Steps of Construction:

Select a suitable scale and draw a line LM to represent C bl .

At point M, at an angle of 25° , cut a length MS to represent the velocity C 1 . Join LS. Produce L to meet the perpendicular drawn from S at P. Thus inlet triangle is completed.

For a Parson’s turbine:

Ф = α = 25°

and C r2 = C1 = 120 m/s

At L, at an angle of 25° Ф, cut a length LN to represent C r2 . Join MN.

Produce M to meet the perpendicular drawn from N at Q. Thus outlet triangle is completed.

By measurement:

C w1 = 108.76 m/s and C w2 = 48.76 m/s

Work done per second = ṁ (C w1 + C w2 ) C bl = 0.1  60 = 945.12 W.

15. In a single stage reaction turbine, the mean blade velocity is observed to be 200 m/s and the nozzle angle is 20°. The absolute value of the steam leaving the nozzle is 350 m/s. The factor with which the relative velocity of steam changes wit respect to the moving is 2.1. If the moving blade inlet and outlet aangles are equal then calculate the degree of reaction of the turbine.

a) 50%

b) 66%

c) 75%

d) 84%

Answer: b

Explanation: Given, C bl = 200 m/s, α = 20°, C 1 = 350 m/s, K = 2.1, θ = Ф

Steps of Construction:

Select a suitable scale and draw a line LM to represent C bl .

At point M, at an angle of 20° , cut a length MS to represent the velocity C 1 . Join LS. Produce L to meet the perpendicular drawn from S at P. Thus inlet triangle is completed.

By measurement:

θ = 43° and C r1 = 175.91 m/s

It Is givne that, Ф = θ = 43° and C r2 = KC r1 = 2.1  = 369.41 m/s

At L, at an angle of 43° Ф, cut a length LN to represent C r2 . Join MN.

Produce M to meet the perpendicular drawn from N at Q. Thus outlet triangle is completed.

By measurement:

C w1 = 328.89 m/s and C w2 = 70.70 m/s

Degree of Reaction, R d =\(\frac{C_{r2}^2-C_{r1}^2}{2C_{bl}

} = \frac{369.41^2-175.91^2}{2*200}\) = 0.66 or 66%.

This set of Thermal Engineering Multiple Choice Questions & Answers focuses on “Reaction Turbines – 2”.

1. Which of the following graphs shows correct variation of efficiency of a Parson’s reaction turbine with respect to blade speed ratio?

a) 

b) 

c) 

d) 

Answer: b

Explanation: The correct variation of efficiency of Parson’s turbine with respect to the blade speed ratio (C bl /C 1 ) is shown by the following graph:

The maximum value of efficiency is reached when the blade speed ratio is equal to cos⁡α.

2. Calculate the mean blade velocity of a Parson’s reaction turbine, if the steam leaves the nozzle with an absolute velocity of 320 m/s. Take the nozzle angle as 20° and the turbine works at maximum efficiency.

a) 156 m/s

b) 189 m/s

c) 256 m/s

d) 301 m/s

Answer: d

Explanation: Given, C 1 = 320 m/s, α = 20°

We know that, when a reaction turbine works at maximum efficiency

C bl /C 1 = cos⁡α

C bl = cos⁡α*C 1 = cos⁡20*320 = 300.7 m/s ≈ 301 m/s.

3. The nozzle angle of a Parson’s reaction turbine is 25°. Determine the maximum possible efficiency for the turbine.

a) 75.62%

b) 84.56%

c) 90.19%

d) 93.21%

Answer: c

Explanation: Given, α = 25°

Maximum possible value of efficiency for a Parson’s reaction turbine

η max = \(\frac{2^2}{1+^2} \)

η max = \(\frac{2^2}{1+^2} \) = 0.9019 or 90.19%.

4. The efficiency of steam turbine is greater than that of gas turbines.

a) True

b) False

Answer: a

Explanation: Steam turbines work on Rankine cycle while the gas turbines on Brayton cycle. Rankine cycle is more efficient and close to Carnot cycle than Brayton cycle working between the same maximum and minimum temperatures.

5. The following data refer to a particular stage of a Parson’s reaction turbine:

Speed of the turbine = 1400 r.p.m.

Mean diameter of the rotor = 1 meter

Isentropic enthalpy drop = 8.530 kJ/Kg

Blade outlet angle = 20°

Speed ratio = 0.8

Determine the stage efficiency.

a) 56.32%

b) 65.75%

c) 78.74%

d) 84.79%

Answer: c

Explanation: Given, N = 1400 r.p.m., D = 1 m, h d = 8.530 kJ/Kg, Ф = 20°, ρ = 0.8

C bl = \(\frac{πDN}{60} = \frac{π*1*1400}{60}\) = 73.30 m/s

we know that,

ρ = \(\frac{C_{bl}}{C_1}\) 0.8 = \(\frac{73.30}{C_1}\) C 1 = 91.63 m/s

For Parson’s reaction turbine,

α = Ф = 20°

Steps of Construction:

Select a suitable scale and draw a line LM to represent C bl .

At point M, at an angle of 20° , cut a length MS to represent the velocity C 1 . Join LS. Produce L to meet the perpendicular drawn from S at P. Thus inlet triangle is completed.

For a Parson’s turbine:

C r2 = C 1 = 91.63 m/s

and Ф = 20°

At L, at an angle of 20° Ф, cut a length LN to represent C r2 . Join MN.

Produce M to meet the perpendicular drawn from N at Q. Thus outlet triangle is completed.

By measurement:

C w1 + C w2 = 98.91 m/s

η stage =\(\)\frac{C_{bl}

}{h_d} = \frac{73.30}{8530} = 0.7874 or 78.74%.

6. In a reaction turbine, the fixed blades and moving blades are of the same shape but reversed in direction. The angle of the discharging tip is 20°. The mean blade velocity is 200 m/s. The axial velocity of flow of steam is half of the mean blade velocity. Determine the inlet angles of blades.

a) 45°

b) 53°

c) 63°

d) 71°

Answer: b

Explanation: Given, C bl = 200 m/s, α = 20°

C f1 = C f2 = 0.5(C bl ) = 0.5 = 100 m/s

Steps of construction:

Select a suitable scale and draw a line LM to represent C bl .

Draw line 1-2 parallel to line LM at a distance corresponding to 100 m/s from line LM.

At M, draw a line at an angle of 20°  to cut the line 1-2 at S. Join LS.

Produce L to meet the perpendicular drawn from S at P. Thus inlet triangle is completed.

By measurement:

θ = 53° Ф.

7. A 50% reaction turbine, having symmetrical velocity triangles, runs at 1500 r.p.m. The exit angles of the blades are 20°. The speed ratio is 0.8 and the diameter of the rotor blades is 1 meter. Determine the power developed if the mass flow rate of steam is 0.785 kg per second.

a) 6.534 kW

b) 7.654 kW

c) 8.425 kW

d) 9.612 kW

Answer: a

Explanation: Given, N = 1500 r.p.m., α = 20°, ρ = 0.8, D = 1 m, ṁ = 0.785 kg/s

C bl = \(\frac{π*D*N}{60} = \frac{π*1*1500}{60}\) = 78.54 m/s

We know that,

ρ = \(\frac{C_{bl}}{C_1} \)

0.8 = \(\frac{78.54}{C_1} \)

C 1 = 78.54/0.8 = 98.18 m/s

Steps of Construction:

Select a suitable scale and draw a line LM to represent C bl .

At point M, at an angle of 20° , cut a length MS to represent the velocity C 1 . Join LS. Produce L to meet the perpendicular drawn from S at P. Thus inlet triangle is completed.

For a Parson’s turbine:

C r2 = C 1 = 98.18 m/s

and Ф = α = 20°

At L, at an angle of 20° Ф, cut a length LN to represent C r2 . Join MN.

Produce M to meet the perpendicular drawn from N at Q. Thus outlet triangle is completed.

By measurement:

C w1 + C w2 = 105.98 m/s

Power Developed = ṁ

*C_bl = 0.785*78.54 = 6534.08 W or 6.534 kW.

8. In a steam turbine the turbine blade’s tip experiences more erosion than any other part.

a) True

b) False

Answer: a

Explanation: In case of high and intermediate pressure turbines, the intermediate pressure stages’ steam is wet, which results in the corrosion and erosion of turbine blades. Also turbine blades are subjected to high centrifugal stresses. The effect of moisture is most eminent when it exceeds 10 per cent. Centrifugal force results in concentration of water particles in the outer annulus and the tip speed is greater than the root speed, hence tip experiences more erosion than any other part.

9. Which of the following is the correct expression to calculate stage efficiency in steam turbines?

a) \

 

 \

 

 \

 

 \(\frac{Net \, work \, done \, on \, shaft \, per \, stage \, per \, kg \, of \, steam}{Total \, adiabatic \, heat \, drop}\)

Answer: b

Explanation: Stage efficiency gives the idea of loses in one stage only. It covers all kind of losses in the stage including losses in nozzles, diaphragms, blades and discs.

Therefore, the correct answer is –

η stage =\(\frac{Net \, work \, done \, on \, shaft \, per \, stage \, per \, kg \, of \, steam}{Adiabatic \, heat \, drop \, per \, stage}\)

10. Which of the following is the correct expression for internal efficiency in steam turbines?

a) \

 

 \

 

 \

 

 \(\frac{Heat \, converted \, into \, useful \, work}{Net \, work \, done \, on \, shaft \, per \, stage \, per \, kg \, of \, steam}\)

Answer: a

Explanation: Internal efficiency similar to stage efficiency, but unlike stage efficiency it covers the whole turbine.

The expression for internal efficiency is –

η internal = \(\frac{Heat \, converted \, into \, useful \, work}{Total \, adiabatic \, heat \, drop}\)

11. Which of the following is the correct formula for overall turbine efficiency in steam turbines?

a) \

 

 \

 

 \

 

 \(\frac{Heat \, converted \, into \, useful \, work}{Adiabatic \, heat \, drop \, per \, stage}\)

Answer: a

Explanation: Overall turbine efficiency covers all kind of internal and external losses such as, leakage, radiation, bearing friction etc.

The expression for overall turbine efficiency is –

η overall = \(\frac{Work \, delivered \, at \, the \, turbine \, cooupling \, in \, heat \, units \, per \, kg \, of vsteam}{Total \, adiabatic \, heat \, drop}\)

12. Which of the following is the correct expression for net efficiency in steam turbines?

a) \

 

 \

 

 \

 

 \(\frac{Thermal \, efficiency \, on \, the \, Carnot \, cycle}{Brake \, thermal \, efficiency}\)

Answer: a

Explanation: Net efficiency is also called as efficiency ratio. It is the ratio of brake thermal efficiency to thermal efficiency on the Rankine cycle.

Therefore,

η net = \(\frac{Brake \, thermal \, efficiency}{Thermal \, efficiency \, on \, the \, Rankine \, cycle}\)

13. Which of the following is the correct definition for ‘Adiabatic Power’?

a) It is the power developed at the rim

b) It is the actual power transmitted by the turbine

c) It is the power based on the total internal steam flow and adiabatic heat drop

d) It is the power developed by the blades

Answer: c

Explanation: Adiabatic power is based on the total internal steam flow and adiabatic heat drop. Adiabatic power is calculated assuming that the expansion of the steam is isentropic or adiabatic.

Adiabatic power = ṁ * .

14. Which of the following is the correct definition of ‘Shaft Power’?

a) It is the power developed by the blades

b) It is the power developed at the rim

c) It is the power based on the total internal steam flow and adiabatic heat drop

d) It is the actual power transmitted by the turbine

Answer: d

Explanation: Shaft power is the actual power transmitted by the turbine. Shaft power takes all possible types of power losses into account including, reheating due to friction, nozzle losses, blade friction, pressure variation through channels etc.

15. Which of the following statements regarding a Reaction turbine is FLASE?

a) The relative velocity of steam with respect to moving blade increases as the steam glides over the turbine blade

b) The degree of reaction for a reaction turbine is defined as the ratio of heat drop in moving blade to heat drop in the stage

c) If the degree of reaction of a reaction turbine is 50% then, the moving blade and the fixed blades have the same shape

d) The angles of receiving tips of fixed and moving blades are always equal

Answer: d

Explanation: In a reaction turbine, the angles of receiving tips of moving and fixed blades are not necessarily equal. They are equal in case of Parson’s reaction turbine, which is a special case of reaction tubines. In Parson’s reaction turbine the shape of moving and fixed blades is same and degree of reaction is 50%.

This set of Thermal Engineering Question Bank focuses on “Reheat Factor and Energy Losses in Steam Turbines”.

1. What is ‘state point’?

a) It is that point on h-s diagram which represents the condition of steam at that instant

b) It is that point on P-v diagram which represents the condition of steam at that instant

c) It is that point on T-s diagram which represents the condition of steam at that instant

d) It is that point on P-s diagram which represents the condition of steam at that instant

Answer: a

Explanation: State point is that point on h-s diagram which represents the condition of steam at that instant. Therefore, to know the initial state point, one must have the knowledge of the initial condition of the steam entering the nozzle.

2. The curve traced by joining all the state points is called _____

a) state point curve

b) state point locus

c) state point diagram

d) state point graph

Answer: b

Explanation: State point on a h-s diagram represents the condition of the steam at that particular instant. The curve traced by joining all the state points is called ‘state point locus’. Using state point locus, the condition of steam at any point from initial to final state can be determined.

3. Which of the following is the correct expression for reheat factor?

a) \

 

 \

 

 \

 

 \(\frac{Isentropic \, enthalpy \, drop}{Adiabatic \, heat \, drop} \)

Answer: a

Explanation: Reheat factor is the ratio of cumulative heat drop to isentropic enthalpy drop.

Therefore, the correct answer is –

Reheat factor = \(\frac{Cumulaive \, heat \, drop}{Isentropic \, enthalpy \, drop} = \frac{h_{cum}}{h_{adi}} \)

4. The value of reheat factor doesn’t depend on _____

a) stage efficiency

b) initial condition of steam

c) final pressure

d) atmospheric temperature

Answer: d

Explanation: The value of reheat factor depends on stage efficiency, final pressure and initial pressure and condition of steam. It doesn’t depend on the atmospheric temperature. Reheat factor is calculated using the expression \(\frac{Cumulaive \, heat \, drop}{Isentropic \, enthalpy \, drop} \).

5. Which of the following statements regarding reheat factor is TRUE?

a) It is always less than 1

b) It is always greater than 1

c) It is always equal to or less than 1

d) It might be less than one

Answer: b

Explanation: Reheat factor is the ratio of cumulative heat drop (h cum ) to isentropic  enthalpy drop (h adi ). Isentropic  enthalpy drop is always less than the cumulative heat drop, hence reheat factor is always greater than 1.

6. The process of draining steam from the turbine, at specific points during its expansion for heating the feed water is called _____

a) governing

b) reheating

c) bleeding

d) recycling

Answer: c

Explanation: Bleeding is the process of draining of steam from the turbine, during its expansion, for heating the feed water supplied to the boiler. The steam temperature drops when it comes in contact with the feed water and thus is condensed. The condensed steam is taken to hot well.

7. Which of the following in an implication of bleeding process?

a) efficiency and power developed both are increased

b) efficiency is decreased and power developed is increased

c) efficiency and power developed both are decreased

d) efficiency is increased and the power developed is decreased

Answer: d

Explanation: As a result of bleeding hotter water is supplied to the boiler due to which the efficiency of the boiler is increased. But small amount of steam is withdrawn from the turbine, which results in reduction of turbine work. Hence, power developed is reduced.

8. Which of the following losses does not come under ‘Internal losses in steam turbines’?

a) Losses due to wetness of steam

b) Losses in regulating valve

c) Leaving velocity losses

d) Mechanical losses

Answer: d

Explanation: Mechanical losses doesn’t come under internal losses; it comes under external losses. Losses due to wetness of steam, losses in regulating valve and exit velocity losses are all internal losses. External losses also include losses due to leakage of steam from the labyrinth gland seals.

9. Which of the following statements is TRUE regarding the losses the steam turbines?

a) Impingement losses are a type of internal losses

b) Losses due to shrouding are a type of internal losses

c) Mechanical losses are external losses

d) Losses in exhaust piping are a type of external losses

Answer: d

Explanation: Losses in exhaust piping are characterized under internal losses. Internal losses also include impingement losses and losses due to shrouding. Mechanical losses and losses due to leakage come under external losses.

10. Losses directly connected with the steam conditions while it is flowing through the turbine are called internal losses.

a) True

b) False

Answer: a

Explanation: Internal losses are directly connected with the steam conditions while it is flowing throughthe turbine. Internal losses include losses in regulating valves, losses in nozzles, losses in moving blades, exit velovity losses etc.

11. Losses which do not influence steam conditions are called external losses.

a) True

b) False

Answer: a

Explanation: External losses are the losses that do not influence the steam conditions. External losses are further classified as mechanical losses and losses due to leakage of steam from the labyrinth gland seals.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Steam Turbine Governing and Control”.

1. What is the objective of steam turbine governing?

a) To keep turbine load constant irrespective of the turbine speed

b) To keep turbine speed constant irrespective of the load

c) To enhance the output of the turbine

d) To reduce the losses in the turbine

Answer: b

Explanation: The objective of steam turbine governing is to keep turbine speed constant irrespective of the load on the turbine shaft. The performance of a steam turbine is affected when the load changes. Turbine governing is important because in practical conditions the load frequently varies.

2. Which of the following is NOT a method of steam turbine governing?

a) By-pass governing

b) Throttle governing

c) Steam governing

d) Nozzle governing

Answer: c

Explanation: Throttle governing, by-pass governing and nozzle governing are all different methods of steam turbine governing while there is no method called steam governing. The mentioned three methods are used to keep the turbine speed constant at varying load condition.

3. Which of the following statements about throttle governing is wrong?

a) It cannot be used in small turbines

b) It has a relatively simpler mechanism

c) Steam is throttled whenever the load falls below the design load

d) Its initial cost is less

Answer: a

Explanation: Throttle governing is used particularly for small turbines. Its cost is less and it has a simple mechanism. In throttle governing the steam is throttled whenever the load falls below the design load to maintain turbine speed constant.

4. In case of throttle governing, the beat valve is operated using oil supply. The oil is supplies using a _____

a) centrifugal pump

b) servomotor

c) reciprocatory pump

d) diaphragm pump

Answer: b

Explanation: The double beat valve in the throttle governing mechanism is operated by supply of oil. This supply of oil is carried out by an oil servomotor. When the steam turbine gains speed this valve is closed and supply to the nozzle is reduced.

5. The oil servomotor in case of throttle governing mechanism is controlled by a _____ governor.

a) centrifugal

b) reciprocatory

c) diaphragm

d) gear

Answer: a

Explanation: In throttle governing mechanism, the oil servomotor is controlled by a centrifugal governor. The servomotor supplies oil for operating the double beat valve. The closing of the valve reduces the supply to the nozzle.

6. The relationship between steam consumption and load in throttle governing is given by _____

a) Wilson’s line

b) Wilkinson’s line

c) Willan’s line

d) Wilber’s line

Answer: c

Explanation: Willan’s line gives the relationship between the steam consumption and load in throttle governing. In throttle governing the steam is throttled whenever the load falls below the design load. It is used mostly in small turbines.

7. Which of the following is the correct equation for Willan’s line?

a) m s 0 =KM+m s

b) m s 0 =M/K+m s

c) m s =KM+m s 0

d) m s =M/K+m s 0

Answer: c

Explanation: Willan’s line gives the relationship between steam consumption and load in case of throttle governing. It’s given by –

m s =KM+ m s 0

where, m s – Steam consumption at any load M 

m s 0 – Steam consumption at no load 

M – Any load 

K – Constant.

8. Carrying out throttle governing at low loads reduces the turbine efficiency.

a) True

b) False

Answer: a

Explanation: The turbine efficiency is significantly reduced when throttle governing is carried out at low loads. To avoid this, nozzle governing is used. Governing by means of nozzle control is more efficient than throttle governing.

9. In nozzle governing, under full load condition what can be said about regulating valves?

a) They are fully open

b) They are fully closed

c) They are partially open

d) Depends on the input

Answer: a

Explanation: In nozzle governing, around 3 to 5 or even more groups of nozzles are employed and the supply of steam to the nozzles is controlled by regulating valves. Under full load condition the regulating valves are fully open, allowing maximum flow through the nozzles.

10. In nozzle governing, when the load on the turbine deviates from the design load, steam supply to a group of nozzles could be varies to restore the original speed.

a) True

b) False

Answer: a

Explanation: The objective of steam turbine governing it to maintain turbine speed constant irrespective of loading. In case of nozzle governing, the steam supply to a group of nozzles is varied when the load deviates from the design value to keep the turbine speed constant.

11. Which of the following statements is FALSE regarding nozzle-governing in steam turbines?

a) It can only be applied to the first stage of the turbine

b) Nozzle governing is more efficient than throttle governing

c) It can also be employed for large turbines having an impulse stage followed by an impulse reaction turbine

d) It is cannot be employed for simple impulse turbine

Answer: d

Explanation: Nozzle governing is well suited for simple impulse turbine. it is more efficient than throttle governing and nozzle control can be applied only on the first stage of the turbine. Also it is used for large turbines having an impulse stage followed by an impulse reaction turbine.

12. Which of the following method of steam turbine governing is ideal for reaction turbines?

a) Throttle governing

b) Nozzle governing

c) By-pass governing

d) Combination of throttle and by-pass governing

Answer: d

Explanation: In case of reaction turbines, pressure drop is required in case of moving blades and therefore, nozzle control governing is not possible. For this reason, combination of throttle and by-pass governing is considered ideal for reaction turbines.

13. Which of the following statements is FALSE about by-pass governing?

a) There is curving observed in Willan’s line

b) When the by-pass valve is opened efficiency is reduced

c) Opening of by-pass valve reduces the work output of lower stages

d) By-pass valve opens after throttle valve has opened to a definite threshold amount

Answer: b

Explanation: The opening of by-pass valve reduces the effieincy of the turbine, but the work output of the lower stages is increased as the supply of steam has increased. By-pass valve opens only after the throttle valve has opened a definite amount.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Classification of Steam Condensers”.

1. Steam condenser converts steam into water.

a) True

b) False

Answer: a

Explanation: Steam condenser is a device which converts steam into water or in which steam is condensed into water. The heat rejected by the steam during condensation is absorbed by cooling water.

2. Steam condensers help in maintaining high back pressure on the exhaust side of the piston of a steam engine.

a) True

b) False

Answer: b

Explanation: Steam condensers maintain very low back pressure at the exhaust side of the piston of a steam engine . This improves the quality of expansion of steam, steam expands to a greater extent and turbine efficiency is increased.

3. Sub-atmospheric pressure is also called _____

a) Gauge pressure

b) Absolute pressure

c) Vacuum pressure

d) High altitude pressure

Answer: c

Explanation: Sub-atmospheric pressure is vacuum pressure. Sub-atmospheric pressure is measured as the pressure depression below atmospheric pressure. Absolute pressure is measure taking perfect vacuum as reference.

4. Which of the following is NOT an element of a steam condensing plant?

a) Supply of cooling water

b) Turbine

c) Condenser

d) Wet air pump

Answer: b

Explanation: A steam condensing plant consists of a condenser, supply of cooling water, hot well, wet air pump and an arrangement for recooling of cooling water . Hot well is where the condensate is discharged.

5. Discharging condensate to hot well improves plant efficiency because _____

a) condensate is not wasted

b) condensate cools down faster in the hot well

c) feed water for boiler is taken from the hot well

d) it increases the pressure in the hot well

Answer: c

Explanation: When the condensate is discharged into the hot well, the temperature of the water inside the hot well increases as it absorbs heat from the hot condensate. Since, the feed water temperature for the boiler is increased, the overall plant efficiency is increased.

6. Steam condensers are classified into _____

a) jet condensers and fast condensers

b) jet condensers and surface condensers

c) high condensers and surface condensers

d) slow condensers and jet condensers

Answer: b

Explanation: Condensers are mainly classified into two categories – Jet condensers and surface condensers. In case of jet condensers, the cooling water comes in direct contact with the exhaust steam while in surface condensers the cooling water and exhaust steam does not come in direct contact.

7. Cooling water is sprayed into the exhaust steam in _____

a) steam condensers

b) surface condensers

c) jet condensers

d) jet condensers and high condensers

Answer: c

Explanation: Usually in jet condensers, the cooling water is sprayed into the exhaust steam. This provokes rapid condensation of the steam. It can be observed that in jet condensers cooling water comes in direct contact with the exhaust steam.

8. Which of the following statements is TRUE about single-pass condenser?

a) It is a type of jet Condenser

b) The flow of water is in one direction only

c) The water flows in one direction through some tubes and returns through the remainder

d) It is not used to condense steam

Answer: b

Explanation: A single-pass condenser is a type of surface condenser. In this water flows in only one direction through the tubes. The cooling water does not come in direct contact with the exhaust steam.

9. Which of the following statements is FLASE about steam condensers?

a) Jet condensers are cheaper than surface condensers

b) Cooling water comes direct contact with the exhaust steam in jet condensers

c) Cooling water does not come in direct contact with exhaust steam in surface condensers

d) Jet condensers are further classified into single-pass and double-pass condensers

Answer: d

Explanation: Single-pass and Doubles-pass condensers belong to the category of surface condensers. In both type of condensers, the steam passes around tubes containing cooling water. The cooling water does not come in direct contact with the steam.

10. Which of the following statements is FALSE about double-pass condensers?

a) The water flows in one direction through some tubes and returns through the remainder

b) Exhaust steam flows through the tubes, while cooling water passes over the outer surface of the tubes

c) It is a type of surface condenser

d) Cooling water doesn’t come in contact with the exhaust steam

Answer: b

Explanation: Double-pass condensers are a type of surface condensers. Cooling water flows through the tubes and exhaust steam passes over the outer surface of the tubes. The water in the tubes flows in one direction through some tubes and returns through the remainder.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Jet Condensers”.

1. Which of the following is NOT a type of jet condenser?

a) Parallel flow type

b) counter flow type

c) Ejector type

d) Central flow type

Answer: d

Explanation: Jet condensers are classified into parallel flow type, counter flow type and ejector flow type condensers. Central flow type condensers are not jet condensers but surface condensers. In jet condensers the exhaust steam comes in direct contact with the cooling water.

2. Parallel flow type condensers are further classified into _____

a) high speed type and low speed type condensers

b) high steam type and low steam type condensers

c) high level type and low level type condensers

d) high capacity and low capacity type condensers

Answer: c

Explanation: Parallel flow type condensers are a type of jet condensers. Further parallel flow condensers are classified into high level type and low level type. In this type of steam condenser, water and exhaust steam both enter the condenser from the top.

3. Which of the following statement regarding parallel flow type jet condensers is FALSE?

a) They are classified into Low level type and high level type

b) Condensate is collected at the bottom

c) Exhaust steam and cooling water enter the condenser from top

d) Exhaust steam and cooling water enter from opposite directions

Answer: d

Explanation: Exhaust steam and cooling water enter from opposite directions in case of counter-flow type steam condensers and not parallel-flow type. In parallel flow type jet condensers, cooling water and exhaust steam enter from the top and the condensate is collected at the bottom.

4. Counter-flow type jet condensers are classified into – Low-level type and High level type jet condensers.

a) True

b) False

Answer: a

Explanation: The given statement is correct. Jet condensers are classified into parallel flow type and counter flow type jet condensers. Both these categories are further sub-divided into low level type and high level type jet condensers.

5. What is the function of baffle plate in parallel flow type low level jet condenser?

a) It keeps cooling water and exhaust steam from mixing

b) It ensures proper mixing of cooling water and exhaust steam

c) It collects condensate over it for extraction

d) It helps condenser wall withstand the inside pressure

Answer: b

Explanation: In low level parallel flow type jet condensers, the steam enters from the top of the condenser. The cooling water is sprayed over the steam. A baffle plate is provided to ensure proper mixing of the cooling water and steam.

6. Where is condensate extraction pump provided in case of low level parallel flow type jet condenser?

a) At the bottom of the condenser

b) At the top of the condenser

c) Near the baffle plate

d) Near the steam entry point

Answer: a

Explanation: In a low level parallel flow type jet condenser, the steam comes in direct contact with the cooling water. The proper mixing is ensured using a baffle plate. The condensate is collected at the bottom of the condenser. The condensate extraction pump is hence located at the bottom of the condenser.

7. Which of the following statements regarding low level counter flow type jet condensers is FALSE?

a) Cooling water is broken into jets by perforated trays

b) Condensate is collected at the bottom

c) Steam is discharged near the bottom and moves upwards

d) The air suction pump is placed near the bottom of the condenser

Answer: d

Explanation: In low level counter flow type jet condensers, the air suction pump is placed at the top of the condenser. Steam is discharged near the bottom and it moves upwards. Cooling water is discharged from the top, it is broken into water jets by perforated trays.

8. Which of the following condensers is also called barometric condenser?

a) Low level parallel flow type jet condenser

b) Low level counter flow type jet condenser

c) High level counter flow type jet condenser

d) High level parallel flow type jet condenser

Answer: c

Explanation: A high level counter flow type jet condenser is also called as barometric condenser. The shell of this condenser is placed at an approximate height of 10.363 m above hot well. The need of an extraction pump is thus obviated.

9. Which of the following condensers does not require a condensate extraction pump?

a) Low level parallel flow type jet condenser

b) Low level counter flow type jet condenser

c) High level counter flow type jet condenser

d) Central flow type surface condenser

Answer: c

Explanation: High level counter flow type jet condensers also called as barometric condensers, they don’t need a condensate extraction pump as their shells are placed at a height of about 10.363 m above hot well.

10. High level counter flow type jet condenser requires an injection pump to pump cooling water inside the condenser shell 

.

a) True

b) False

Answer: a

Explanation: High level counter flow type jet condenser requires high pressure injection of water in the condenser shell. When water under required pressure isn’t available, an injection pump is used to pump the cooling water.

11. Which of the following statements regarding ejector condenser is FALSE?

a) It is a type of jet condenser

b) Exhaust steam passes through a non-return valve

c) Kinetic energy of cold water decreases as it passes through the truncated cones

d) Pressure of cold water drops as the water progresses through the truncated cones

Answer: c

Explanation: In an ejector condenser, cold water passes through a number of truncated cones and as it passes its kinetic energy increases and pressure drops. Due to this pressure drop, exhaust steam is drawn along with the cold water. The exhaust steam passes through a non-return valve to ensure that the steam doesn’t flow back. The exhaust steam and the cold water mix and finally pass through a diverging cone, where a part of its kinetic energy is converted into pressure energy, the mixture is then discharged into hot well.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Surface Condensers”.

1. Which of the following is NOT a type of surface condenser?

a) Regenerative type

b) Evaporative type

c) Inverted-flow type

d) Ejector type

Answer: d

Explanation: Ejector condenser is a type of jet condenser and not a surface condenser. Surface condensers are classified into down-flow type, central-flow type, inverted-flow type, regenerative type and evaporative type.

2. Which of the following condensers is also called cross-surface condenser?

a) Inverted-flow type

b) Evaporative type

c) Down-flow type

d) Central-flow type

Answer: c

Explanation: In down-flow type surface condensers the direction of steam flow is perpendicular to the direction of flow of water, hence they are also called cross-surface condensers. Exhaust stream passes over a series of tubes having cold water flow in it, and thus condenses.

3. Radial flow of steam is observed in _____

a) down-flow type surface condenser

b) central-flow type surface condenser

c) regenerative type surface condenser

d) evaporative Surface condenser

Answer: b

Explanation: In central-flow type surface condenser, the suction pipe of air extraction pump is located at the center of the tubes. This results in radial flow of the steam. In this case the pressure drop of steam is less.

4. Which of the following condensers has air suction located at the top?

a) Down-flow type surface condenser

b) Central-flow type surface condenser

c) Inverted-flow type surface condenser

d) Evaporative condenser

Answer: c

Explanation: Inverted-flow type surface condenser has air suction at the top. The steam enters the condenser shell from the bottom, it rises up and then flows down. The condensate is collected at the bottom.

5. In Inverted-flow type surface condensers, where is the pipe of condensate extraction pump located?

a) At the top of the condenser shell

b) At the bottom of the condenser shell

c) Near the middle portion of the condenser shell

d) Between the middle and the top portion of the condenser shell

Answer: b

Explanation: The steam enters the condenser from the bottom, it rises to the top and then flows down to the bottom getting condensed the whole time. The condensate is collected at the bottom of the condenser shell; hence the pipe of condensate extraction pump is located at the bottom of the condenser shell.

6. In regenerative type surface condensers, the condensate before being used as feed water for the boiler is heated using the entering exhaust steam.

a) True

b) False

Answer: a

Explanation: In case of regenerative type surface condensers, the condensate is passed though the entering exhaust steam so as to heat it before it goes back to the boiler as feed water. A regenerative method of heating of condensate is used in this condenser.

7. _____ works best when the availability of cooling water is limited.

a) Down-flow type surface condenser

b) Regenerative type surface condenser

c) Inverted-flow type surface condenser

d) Evaporative condenser

Answer: d

Explanation: In evaporative condensers, the quantity of cooling water required is less as the circulating water is caused to evaporate under small partial pressure. Hence, it will work best when the availability of cooling water is low or limited.

8. In down-flow type surface condensers, the plate that separates the water box into two sections is called _____

a) baffle plate

b) separating plate

c) partitioning plate

d) divider

Answer: a

Explanation: Baffle plate acts as the partition between the two sections in the water box. Cold water enters into the condenser through one section and leaves through the other. Baffle plate ensures that the inlet water doesn’t come in contact with the water leaving the water box.

9. Which of the following statements is FALSE about down-flow type surface condensers?

a) Cold water flows through the tubes and the exhaust steam flows around the tubes

b) The condenser shell is usually made cylindrical

c) Cold water doesn’t come in direct contact with the exhaust steam

d) It is also called barometric condenser

Answer: d

Explanation: High level counter flow type jet condenser is also known as barometric condenser. Down-flow type surface condensers are also called cross-surface condenser since the flow of exhaust steam is at right angle to the direction of the water flow.

10. Where is the suction pipe of the air suction pump located in central-flow type surface condensers?

a) At the centre of the tubes

b) At the top of condenser shell

c) At the bottom of the condenser shell

d) Near the pipe of condensate extraction pump

Answer: a

Explanation: In central-flow type surface condensers, the suction pipe of the air suction pump is located at the centre of the tubes. Due to this placement of the pipe of air suction pump, the exhaust stema flows radially.

This set of Thermal Engineering Questions and Answers for Entrance exams focuses on “Steam Condensers – Sources and Effects of Air Leakage”.

1. Which of the following statements is not a correct reason for inefficiency in surface condensers?

a) Air leakage

b) High resistance faced by the steam while entering

c) Condensate undercooling

d) Circulating water passing through the condenser almost smoothly

Answer: d

Explanation: Circulating water passes through the condenser with high friction and at a velocity not consistent with high efficiency, this results in increased inefficiency in surface condensers. Since the pressure inside the condenser is less than atmospheric, air leaks into the condenser. This limits the amount of work done by unit mass of steam, thus, results increases inefficiency.

2. Surface condensers require more power for water pumping than jet condensers.

a) True

b) False

Answer: a

Explanation: For water pumping, surface condensers require more power than jet condensers. Also the floor area required by surface condensers is more than that of jet condensers. Air pumping power required by jet condensers is more than that of surface condensers.

3. Which of the following statements is FALSE regarding air leakage in condensers?

a) The sources of air leakage can be found by putting soap water at infiltration prone points

b) The sources of air leakage can be found out by putting pepprament oil on suspected joints

c) The source of air leakage, if large, can be found out by passing candle flame over possible openings

d) Thermometer and vacuum gauge readings become steady when leakage is present in a condenser

Answer: d

Explanation: In case of presence of air leakage, when the plant is running and temperature and pressure conditions reach a steady state, the thermometer and vacuum gauge readings record a fall. To locate the sources of air leakage, soap water, pepprament oil or a candle flame can be used.

4. Which of the following methods is the correct one for using pepprament oil to detect the source of air leakage?

a) Put the pepprament oil on the suspected joint and wait for the bubbles

b) Put the pepprament oil on the suspected joint and wait for it to get sucked

c) Put the pepprament oil on the suspected joint and check for pepprament odour at the discharge of air ejector

d) Put the pepprament oil on the suspected joint, pepprament oil being coloured will change the colour of condensed water

Answer: c

Explanation: Pepprament oil has a peculiar odour, this property is used to identify the source of air leakage in the condensers. It is put on the suspected joint and if the joint leaks, the oil will be sucked in the condenser and pepprament odour will be observed at the discharge of air ejector.

5. Which of the following statement about condensers is FALSE?

a) Jet condensers are more suitable for high capacity plants owing to their high vacuum efficiency

b) Surface condensers require more cooling water than jet condensers

c) As compared to surface condensers, jet condensers require smaller floor space

d) The manufacturing cost of surface condensers is greater than that of jet condensers

Answer: a

Explanation: vacuum efficiency of surface condensers is higher than that of jet condensers, that’s why surface condensers are more suitable for high capacity plants. In comparison with jet condensers, surface condensers require more cooling water, larger floor space.

6. What is the effect of air leakage in condensers on thermal efficiency of the steam power plant?

a) Thermal efficiency gets lowered

b) Thermal efficiency increases

c) Thermal efficiency remains unchanged

d) It increases or decreases depending upon the degree of leak

Answer: a

Explanation: Air leakage reduces thermal efficiency of a condenser. The leaked air increases back-pressure on the prime mover which results in loss of heat drop and a result the thermal efficiency of the steam power plant is lowered.

7. How does air leakage in a condenser affect the requirement of cooling water?

a) Air leakage reduces the amount of cooling water required

b) Air leakage increases the amount of cooling water required

c) Air leakage doesn’t affect the amount of cooling water required

d) Amount of cooling water required increases or decreases depending upon the rate of air leakage

Answer: b

Explanation: The partial pressure of steam is lowered by the air leakage. It reduces the saturation temperature of steam, therefore, its latent heat increases. Hence, increased amount of cooling water is required.

8. How does air leakage in condensers affect the heat transfer rate?

a) Heat transfer is reduced

b) Heat transfer is increased

c) Heat transfer rate is not affected by air leakage

d) Heat transfer rate increases or decreases depending on the rate of air leakage

Answer: a

Explanation: Air being poor conductor of heat reduces the heat transfer rate in the condenser. The heat transfer from the vapour is reduced. To tackle this problem, the surface area of tubes of a surface condenser is increased.

9. What effect does air leakage in a condenser has on corrosion?

a) Air leakage reduces corrosion

b) Air leakage increases corrosion

c) Air leakage does not affect corrosion

d) Corrosive action increases or decreases depending upon the degree of leak

Answer: b

Explanation: Air leakage introduces air in the condenser. The inside of a condenser is already humid due to presence of water. Hence, the presence of air increases the corrosive action. Corrosion on the tube surface reduces the heat transfer rate.

10. Which of the following is not an effect of air leakage in condensers?

a) Reduced thermal efficiency of steam power plant

b) Increased requirement of cooling water

c) Increase in corrosive action in condenser

d) Increased heat transfer in condenser

Answer: d

Explanation: Air leakage in the condenser reduces the thermal efficiency of the steam power plant. It increases the amount of cooling water required and also increases the corrosive action in the condenser. The heat transfer rate in the condenser is reduced.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Condenser and Vacuum Efficiency”.

1. What type of pump can be used for obtaining maximum vacuum in condensers?

a) Air pump

b) Centrifugal pump

c) Steam pump

d) Gear pump

Answer: a

Explanation: Air pump can be used to obtain a desired level of vacuum in the condensers. It extracts air and other non-condensable gases. Air pumps are classified into wet air pumps and dry air pumps. Wet air pumps are used to remove a mixture of condensate and non-condensable gases while dry air pump removes the air only.

2. What instrument is used to measure vacuum in the condensers.

a) Thermometer

b) Vacuum gauge

c) Barometer

d) Pressure transducer

Answer: b

Explanation: Vacuum measurement in condensers is done using vacuum gauge. In case of condensers, vacuum pressure refers to the pressure below atmospheric pressure. For calculation purposes the barometric reading is taken to be 760 mm of Hg.

3. Which of the following is the correct definition of vacuum efficiency?

a) It is the ratio of actual vacuum to atmospheric pressure

b) It is the ratio of atmospheric pressure to actual vacuum

c) It is the ratio of maximum obtainable vacuum to actual vacuum

d) It is the ratio of actual vacuum to maximum obtainable vacuum

Answer: d

Explanation: Vacuum efficiency is defined as the ratio of the actual vacuum to the maximum obtainable vacuum.

Mathematically,

η vacuum = \(\frac{Actual \, vacuum}{Maximum \, obtainable \, vacuum} \)

4. Which of the following expressions is the correct one for calculating the vacuum efficiency of a condenser?

a) \

 

 \

 

 \

 

 \(\frac{Absolute \, pressure \, of \, steam-Barometer \, Pressure}{Actual \, vacuum} \)

Answer: b

Explanation: Vacuum efficiency is the ratio of the actual vacuum to the maximum obtainable vacuum.

Therefore,

η vacuum = \(\frac{Actual \, vacuum}{Maximum \, obtainable \, vacuum} \)

but Maximum obtainable vacuum = Barometer pressure – Absolute pressure of steam

\(\frac{Actual \, vacuum}{Barometer \, Pressure-Absolute \, pressure \, of \, steam} \)

5. Which of the following is the correct definition of condenser efficiency?

a) It is the ratio of the difference between the outlet and the inlet temperatures of cooling water to the difference between the temperature corresponding to the vacuum in the condenser and inlet temperature of cooling water

b) It is the ratio of the difference between the temperature corresponding to the vacuum in the condenser and inlet temperature of cooling water to the difference between the inlet and outlet temperature of cooling water

c) It is the ratio of the difference between the atmospheric pressure and the absolute pressure inside the condenser to the difference between the atmospheric pressure and the absolute partial pressure of the steam

d) It is the ratio of the difference between the actual pressure inside the condenser and the partial pressure of the steam to the atmospheric pressure

Answer: a

Explanation: Condenser efficiency is defined as the ratio of the difference between the outlet and the inlet temperatures of cooling water to the difference between the temperature corresponding to the vacuum in the condenser and inlet temperature of cooling water.

6. Which of the following is the correct expression for calculating condenser efficiency?

a) \

 

 \

 

 \

 

 \(\frac{Temp. \, corresponding \, to \, vacuum \, in \, the \, condenser-Inlet \, temperature \, of \, cooling \, water}{Rise \, in \, temperature \, of \, cooling \, water} \)

Answer: a

Explanation: The correct expression for condenser efficiency is –

η condenser = \(\frac{Rise \, in \, temperature \, of \, cooling \, water}{Temp. \, corresponding \, to \, vacuum \, in \, the \, condenser-Inlet \, temperature \, of \, cooling \, water}\)

It can also be written as

η condenser = \(\frac{Rise \, in \, temperature \, of \, cooling \, water}{Temp. \, corresponding \, to \, the \, absolute \, pressure \, in \, the \, condenser-Inlet \, temperature \, of \, cooling \, water} \)

7. The inlet and outlet temperatures of cooling water to a certain condenser are recorded to be 30°C and 40°C respectively. If the absolute pressure in the condenser is given to be 0.11 bar, determine efficiency.

a) 45.21%

b) 75.65%

c) 95.32%

d) 56.46%

Answer: d

Explanation: Given, t wi = 30°C, t wo = 40°C

From steam tables corresponding to 0.11 bar

t s = 47.71°C

Therefore,

η condenser = \

 

η condenser = \(\frac{t_{wo} – t_{wi}}{t_s-t_{wi}}\) x 100

Substituting the values,

η condenser = \(\frac{40 – 30}{47.71-30}\) x 100 = 56.46%.

8. In a condenser, the temperature of steam entering in the condenser is observed to be 30°C. The condenser vacuum is recorded to be 670 mm of Hg. If the barometer reading is 760 mm of Hg, determine the vacuum efficiency.

a) 92.01%

b) 84.25%

c) 31.65%

d) 67.51%

Answer: a

Explanation: Actual vacuum = 670 mm of Hg

Using steam tables, At 30°C

Partial pressure of steam, P = 0.04242 bar = 0.04242/0.001333 mm of Hg

= 31.82 mm of Hg

Now, Maximum obtainable vacuum = 760 – 31.82 = 728.18 mm of Hg

Vacuum efficiency = η vacuum = \(\frac{Actual \, vacuum}{Maximum \, obtainable \, vacuum} \)

= 670 / 728.18

= 0.9201 or 92.01%.

9. The outlet temperature of cooling water to a condenser is observed to be 42°C. Determine the inlet temperature of cooling water if the absolute pressure inside the condenser is 0.1 bar. Take condenser efficiency as 76%.

a) 12.36°C

b) 29.87°C

c) 36.21°C

d) 40.65°C

Answer:

Explanation: Given, t wo = 42°C, η condenser = 0.76, P = 0.1 bar

Using steam tables, corresponding to 0.1 bar

t s = 45.83°C

We know that,

η condenser = \(\frac{t_{wo} – t_{wi}}{t_s-t_{wi}} \)

Substituting the values,

0.76=\(\frac{42- t_{wi}}{45.83-t_{wi}} \)

Solving the equation for t wi we get,

t wi = 29.87°C.

10. The following data refers to a steam condenser –

Inlet temperature of cooling water = 36°C

Absolute pressure inside the condenser = 0.12 bar

efficiency = 78%

Determine the outlet temperature of cooling water.

a) 42.37°C

b) 39.12°C

c) 46.49°C

d) 35.21°C

Answer: c

Explanation: Given, t wi = 36°C, P = 0.12 bar, η condenser = 0.78

Using steam tables, corresponding to 0.12 bar

t s = 49.45°C

We know that,

η condenser = \(\frac{t_{wo} – t_{wi}}{t_s-t_{wi}} \)

Substituting the values,

0.78 = \(\frac{t_{wo} – 36}{49.45-36} \)

Solving for two we get,

t wo = 46.49°C.

This set of Thermal Engineering Problems focuses on “Steam Condensers – Dalton’s Law of Partial Pressure”.

1. Dalton’s law of partial pressure states that the total pressure exerted by a mixture of gas and vapour is the sum of partial pressure of the gas and partial pressure of the vapour at the common temperature.

a) True

b) False

Answer: a

Explanation: The given statement is the correct one for Dalton’s law of partial pressure.

Therefore, at a particular temperature ‘t’ –

Mathematically,

P = Pa + Ps

Where, P = Total pressure inside the container

Pa = Partial pressure of the gas 

Ps = Saturation pressure of water at temperature t

2. Which of the following expression is the correct one for calculating total mass of mixture present in a container? 

a) m s \

 

 m s \

 

 m s \

 

 m s \(\{1+\frac{v_a}{v_g}\}\)

Answer: b

Explanation: The correct expression for calculating the total mass of mixture  present inside a container is m s \(\{1+\frac{v_g}{v_a}\}\).

where, m s = Mass of vapour inside the container

v g = Specific volume of saturated water vapour

v a = Specific volume of air

v g and v a at a particular temperature ‘t’.

3. Which of the following expression is the correct one for calculating total mass of mixture present in a container? 

a) m a \

 

 m a \

 

 m a \

 

 m a \(\{1-\frac{v_a}{v_g}\}\)

Answer: c

Explanation: The correct expression for calculating the total mass of mixture  present inside a container is m a \(\{1+\frac{v_a}{v_g}\}\).

where, m a = Mass of air inside the container

v g = Specific volume of saturated water vapour

v a = Specific volume of air

v g and v a at a particular temperature ‘t’.

4. Given specific volume of steam as 54.25 m 3 /kg and volume of the container as 190 m 3 , calculate the mass of steam present inside the container.

a) 3.0 kg

b) 3.5 kg

c) 4.0 kg

d) 4.5 kg

Answer: b

Explanation: Given, v = 54.25 m 3 /kg, V = 190 m 3

Mass of steam = \(\frac{V}{v} = \frac{190}{54.25}\) = 3.5 kg.

5. In a conatainer, having a mixture of steam and air, the specific volumes of air and steam are 42 m 3 /kg and 55 m 3 /kg. Calculate the total mass of the mixture present inside the container if the mass of steam inside the container is 4.23 kg.

a) 9.77 kg

b) 10.23 kg

c) 12.85 kg

d) 15.64 kg

Answer: a

Explanation: Given, v a = 42 m 3 /kg, v g = 55 m 3 /kg

We know that,

Total mass of mixtue = m s \(\{1+\frac{v_g}{v_a}\}\) = 4.23\(\{1+\frac{55}{42}\}\) = 9.77 kg.

This set of Thermal Engineering Questions and Answers for Campus interviews focuses on “Steam Condensers – Quantity of Cooling Water Required”.

1. Which of the following is the correct expression for finding out the quantity of cooling water required in jet condensers?

a) m w =\

 

 m w =\

 

 m w =\

 

 m w =\(\frac{m_s\{xh_{fg}+c_{pw}

\}}{c_{pw} 

}\)

Answer: d

Explanation: The correct formula for calculating the quantity of cooling water required in case of jet condensers is –

m w =\(\frac{m_s\{xh_{fg}+c_{pw}

\}}{c_{pw} 

}\)

where,

m w = Mass of cooling water required 

m s = Mass of steam condensed 

t s = Saturation temperature of steam 

t w1 = Temperature of cooling water at inlet 

t w2 = Temperature of cooling water at outlet 

c pw = Specific heat of water at constant pressure

x = Dryness fraction of steam entering the condenser

h fg = Latent heat of steam entering the condenser .

2. Which of the following is the correct expression for finding out the quantity of cooling water required in surface condensers?

a) m w =\

 

 m w =\

 

 m w =\

 

 m w =\(\frac{m_s\{h_{fg}+c_{pw}

\}}{c_{pw}

}\)

Answer: c

Explanation: The correct formula for calculating the quantity of cooling water required in case of surface condensers is –

m w =\(\frac{m_s\{xh_{fg}+c_{pw}

\}}{c_{pw}

}\)

where,

m w = Mass of cooling water required 

m s = Mass of steam condensed 

t s = Saturation temperature of steam 

t c = Temperature of condensate leaving the condenser

t w1 = Temperature of cooling water at inlet 

t w2 = Temperature of cooling water at outlet 

c pw = Specific heat of water at constant pressure

x = Dryness fraction of steam entering the condenser

h fg = Latent heat of steam entering the condenser .

3. Which of the following statement s about the given expression is TRUE?

m w =\

 

 It is used to calculate mass of cooling water required in jet condensers

b) While deriving this expression, it is assumed that all the heat lost by steam is gained by cooling water

c) t s is he saturation temperature of steam corresponding to the condenser vacuum.

d) c pw represents specific heat of water at constant volume

Answer: b

Explanation: Heat lost by the steam = m s {xh fg +c pw (t s -t c )}

Heat gained by water = m w {c pw (t w2 -t w1 )}

Assuming that all the heat lost by steam is gained by cooling water, we get the required expression-

m w =\(\frac{m_s\{xh_{fg}+c_{pw}

\}}{

}\).

4. Determine the amount of cooling water required, if the condenser  deals with 15000 kg of steam per hour. The steam is dry and has a latent heat of 2432 kJ per kg. The temperature of condensate is 20°C is less than the steam temperature. Also, the rise is cooling water temperature is 10°C. Take specific heat of water at constant pressure as 4.186 kJ/kg-°C.

a) 154265 kg/hr

b) 901476 kg/hr

c) 456279 kg/hr

d) 874325 kg/hr

Answer: b

Explanation: m s = 15000 kg/hr, x = 1, h fg = 2432 kJ/kg, t s – t c =20°C, t w1 – t w2 = 10°C, c pw = 4.186 kJ/kg-°C

Quantity of cooling water required –

m w =\(\frac{m_s\{xh_{fg}+c_{pw}

\}}{c_{pw} 

} \)

Substituting the values, we get –

m w =\(\frac{15000\{1+4.186\}}{4.186} \)

m w = 901476.35 kg/hr ≈ 901476 kg/hr.

5. A jet condenser is designed to handle 12000 kg of dry and saturated steam per hour. The steam enters the condenser at 38°C , and the condensate leaves the condenser at 32°C. If the inlet temperature of cooling water is 10°C, find out the mass of cooling water required in kg per hour. Take latent heat of steam as 2300 kJ/kg and specific heat at constant pressure as 4.186 kJ/kg-°C.

a) 302973 kg/hr

b) 246185 kg/hr

c) 876215 kg/hr

d) 741963 kg/hr

Answer: a

Explanation: m s = 12000 kg/hr, x = 1, t s = 38°C, t w2 = 32°C, T w1 = 10°C, h fg = 2300 kJ/kg, c pw = 4.186 kJ/kg-°C

Quantity of cooling water required –

m w =\(\frac{m_s\{xh_{fg}+c_{pw}

\}}{c_{pw}

} \)

Substituting the values, we get –

m w =\(\frac{12000\{1+4.186\}}{4.186} \)

m w = 302973 kg/hr.

6. A surface condenser is supplied with 10000 kg of wet steam per hour. the dryness fraction of the steam is 0.98. The latent heat of steam is 2000 kJ/kg. The quantity of cooling water required by the condenser is 950000 kg per hour. determine the rise in cooling water temperature, if the difference between the temperature of steam entering and condensate leaving the condenser is 10°C. Take c pw = 4.186 kJ/kg-K.

a) 5°C

b) 10°C

c) 15°C

d) 20°C

Answer: a

Explanation: m s = 10000 kg/hr, x = 0.98, h fg = 2000 kJ/kg, m w = 950000 kg/hr, t s – t c = 10°C, c pw = 4.186 kJ/kg-K

For surface condenses, the quantity of cooling water required is given by –

m w =\(\frac{m_s \{xh_{fg}+c_{pw} 

\}}{c_{pw} 

} \)

substituting the respective values,

950000=\(\frac{10000\{0.98+4.186\}}{4.186

} \)

Therefore,

(t w2 – t w1 ) = 5.03°C ≈ 5°C.

7. In a jet condenser, the difference between the temperature of steam  entering the condenser and the condensate leaving the condenser is 4°C. The amount of cooling water required and the mass of steam being supplied to the condenser are 900000 kg/hr and 12000 kg/hr. Take the latent heat of water to be 2300 kJ/kg and the specific heat at constant pressure to be 4.186 kJ/kg-K. If the temperature of cooling water entering the condenser is 26°C, determine the temperature of cooling water leaving the condenser.

a) 35.65°C

b) 33.38°C

c) 45.54°C

d) 28.32°C

Answer: b

Explanation: t s – t w2 = 4°C, x = 1, m w = 900000 kg/hr, m s = 12000 kg/hr, h fg = 2300 kJ/kg, c pw = 4.186 kJ/kg-K

t w1 = 26°C

We know that,

m w =\(\frac{m_s\{xh_{fg}+c_{pw}

\}}{c_{pw}

}\)

Substituting the values, we get-

900000=\(\frac{12000\{1+4.186\}}{4.186

} \)

Therefore,

t w2 = 33.38°C.

8. Steam  having temperature 40°C and dryness fraction 0.96 enters a jet condenser. Temperature of condensate is observed to be 35°C. If the temperature rise in the cooling water is 5°C, determine the ratio of amount of cooling water required to mass of steam condensed. (Take c pw = 4.186 kJ/kg-K)

a) 88

b) 99

c) 111

d) 222

Answer: c

Explanation: h fg = 2400 kJ/kg, t s = 40°C, x = 0.96, t c = t w2 = 35°C, (t w2 – t w1 ) = 5°C, c pw = 4.186 kJ/kg-K

We know that, for jet condensers

Quantity of cooling water required, m w

m w =\(\frac{m_s\{xh_{fg}+c_{pw}

\}}{c_{pw}

}\)

\(\frac{m_w}{m_s}=\frac{xh_{fg}+c_{pw}

\}}{c_{pw}

}\)

Substituting the values,

\(\frac{m_w}{m_s} =\frac{\{0.96+4.186\}}{4.186}\)

\(\frac{m_w}{m_s}\) = 111.08 ≈ 111.

9. Following observations were made after carefully analyzing a jet condenser.

Quantity of cooling water required = 700000 Kg/h

Latent heat = 2500 kJ/Kg

Steam entering the condenser = 15000 kg/h

Temperature of steam = 36°C

Temperature of condensate = 30°C

Determine the rise in temperature of cooling water if the steam entering is dry and saturated.

(c pw = 4.186 kJ/kg-K)

a) 8.93°C

b) 9.93°C

c) 10.93°C

d) 12.93°C

Answer: d

Explanation: m w = 700000 kg/h, h fg = 2500 kJ/kg, m s = 15000 kg/h, t s = 36°C, t w2 = 30°C

We know that,

m w =\(\frac{m_s\{xh_{fg}+c_{pw}

\}}{c_{pw}

}\)

Substituting the values,

700000=\(\frac{15000\{1+4.186\}}{4.186

}\)

Therefore,

(t w2 – t w1 ) = 12.93°C.

10. Dry and saturated steam  at certain temperature is condensed to 32°C in a surface condenser. The ratio amount cooling water required to the amount of steam condensed is 112. Determine the temperature of the steam entering the condenser, the a temperature of cooling water rises by 5°C. (Take c pw = 4.186 kJ/kg-K)

a) 52.55°C

b) 42.55°C

c) 32.55°C

d) 22.55°C

Answer: b

Explanation: x = 1, h fg = 2300 kJ/kg, t c = 32°C, \(\frac{m_w}{m_s}\) = 112, (t w2 – t w1 ) = 5°C

For a surface condenser, we know that –

m w =\(\frac{m_s \{xh_{fg}+c_{pw} 

\}}{c_{pw} 

} \)

\(\frac{m_w}{m_s}=\frac{\{xh_{fg}+c_{pw}

\}}{c_{pw}

} \)

Substituting the values, we get –

112=\(\frac{\{1+4.186

\}}{4.186} \)

t s = 42.55°C.

11. Calculate the minimum amount of cooling water required if 3516240 kJ of heat is to be absorbed per hour and the rise in temperature should not be more than 7°C. Take c pw = 4.186 kJ/kg-K)

a) 110000 kJ/h

b) 120000 kJ/h

c) 130000 kJ/h

d) 140000 kJ/h

Answer: b

Explanation: (t w2 – t w1 ) = 7°C, Heat to be absorbed = 3516240 kJ/h

Amount of heat absorbed by cooling water = m w *c pw (t w2 -t w2 )

Therefore,

m w *c pw (t w2 -t w2 ) = 3516240

Substituting the values,

m w  = 3516240

m w = 120000 kJ/h

12. Calculate the amount of heat absorbed by the cooling water, if it is supplied at the rate of 100000 kg/h and the rise in it s temperature is observed to be 5°C. (Take c pw = 4.186 kJ/kg-K)

a) 2093000 kJ/h

b) 4653247 kJ/h

c) 5478210 kJ/h

d) 5462400 kJ/h

Answer: a

Explanation: (t w2 – t w1 ) = 5°C, m w = 100000 kJ/h

Amount of heat absorbed = m w * c pw * (t w2 – t w1 )

= 100000*4.186*5

= 2093000 kJ/h.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Steam Condensers – Cooling Towers”.

1. What is the function of a cooling tower in a power plant?

a) It cools the hot water coming back from the condenser

b) It cools the hot water being supplied to the condenser

c) It heats the cold water coming back from the condenser

d) It heats the cold water being supplied to the condenser

Answer: a

Explanation: The cooling water that is passed through the condenser, for condensing steam coming out of the turbine, gets heated and in order to cool that water down, cooling towers are used. So that the water can be reused again.

2. Which of the following phenomenon is used to cool water in a cooling tower?

a) Evaporation

b) Radiation

c) Condensation

d) Conduction

Answer: a

Explanation: In a cooling tower, hot water is allowed to fall drop by drop and air is constrained to move in the opposite direction. This results in some part of water getting evaporated. Due to this the rest of the water is cooled.

3. Humidity of air affects the cooling of water in a cooling tower.

a) True

b) False

Answer: a

Explanation: Humidity of air affects the evaporation rate thus affecting the cooling of water. Also, temperature of air, size and height of tower and velocity of air entering the tower affect the cooling of water.

4. The cooling of water is affected by degree of uniformity in descending water.

a) True

b) False

Answer: a

Explanation: Ideally, degree of uniformity of descending water should be high. Also, air should be accessible to all parts of the tower and there should be proper arrangement of plates for efficient cooling of water.

5. Based on the material, with which the towers are made, cooling towers are classified into _____

a) Timber towers, Concrete towers and Alloy duct type

b) Induced and forced draught type

c) Induced and natural draught type

d) Timber towers, Concrete towers and Steel duct type

Answer: d

Explanation: Cooling towers are classified into timber towers, concrete towers and steel duct type, based on the material with which they are made. Timber towers are rarely used. Concrete towers are widely used owing to their huge capacity.

6. Which of the following statements about timber towers is False?

a) Timber towers have longer life than concrete and steel duct type towers

b) Timber towers have high maintenance charges

c) Timber towers have limited cooling capacity

d) Timber towers are rarely used

Answer: a

Explanation: Timber towers on exposure to sun, wind, water etc. rot easily, hence, they have shorter life than concrete and steel duct type towers. They have high maintenance charges and limited cooling capacity. Due to the mentioned reasons, they are rarely used.

7. Which of the following statements about concrete towers is FALSE?

a) Concrete towers have large capacity (of the order 5 x 10 3 m 3 /h)

b) Longer life than timber towers

c) Concrete towers have good stability under air pressure

d) Concrete towers require frequent maintenance

Answer: d

Explanation: Concrete towers have towers are known are known to have large capacity, sometimes of the order 5 x 10 3 m 3 /h. They have a longer life than timber towers and have higher stability under air pressure. They don’t require low maintenance.

8. Which of the following statements about duct type cooling towers is TRUE?

a) They are widely used in modern power plants

b) Their water handling capacity is small

c) On continued exposure to wind, water and sun, they rot

d) They have short life

Answer: b

Explanation: Duct type cooling towers have long life, but small capacity. Owing to the small capacity they are rarely used in the modern power plants. They can operate either on forced, induced or natural draught.

9. Which of the following statements regarding cooling towers is FALSE?

a) Concrete towers have longer life than timber tower

b) Timber towers have high maintenance charges

c) Duct-type cooling towers have small capacity

d) Cooling towers are used to cool down the water being supplied to the boiler of the power plant

Answer: d

Explanation: Cooling towers are used to cool down the water being passed though the condenser. The discharge of cooling towers is collected in the cooling pond and reused as cooling water in the condenser.

10. Which type of cooling tower is shown in the image below?

a) Induced draught cooling tower

b) Forced draught cooling tower

c) natural draught cooling tower

d) Forced or induced depending upon the sense of rotation of fan

Answer: a

Explanation: The given figure represents an induced draught cooling tower. The location of fan in case of induced draught cooling towers is a the top opening of the tower. In case of forced draught, the fan is located at the bottom. Natural draught cooling towers doesn’t have a fan.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Reciprocating Steam Engine – Definition and Classification”.

1. Which of the following statements are correct regarding reciprocating steam engines?

a) It is a form of heat engine

b) It converts heat energy to mechanical work

c) It has high thermal efficiency

d) It is used in locomotives, ships etc

Answer: c

Explanation: Steam engines have low thermal efficiency. They are used particularly where small powers and low speeds are required. They are mainly used in ships and locomotives. A steam engine  converts heat energy into mechanical work due to the action of steam in the piston .

2. According to the class of service, reciprocating steam engines are classified into _____

a) Stationary, Marine, Locomotive and Pumping

b) Stationary and Moving

c) Portable and Compound

d) Manual, Automatic and Centrifugal

Answer: a

Explanation: Based on the class of service, reciprocating steam engines are classified into four categories – stationary, marine, locomotive and pumping . Stationary reciprocating steam engines are used in the applications where the engine is not moved. Locomotive steam engines are used in trains etc., marine engines in marine applications like ships.

3. According to the speed of rotation, reciprocating steam engines are classified into _____

a) Vertical, horizontal and inclined

b) Stationary, locomotive and marine

c) High speed, Medium speed and low speed

d) Slow, fast and very fast

Answer: c

Explanation: Based on the speed of rotation, reciprocating steam engines are classified into three categories – High speed, Medium speed and low speed steam engines. High speed steam engines have speed above 300 r.p.m., medium speed steam engines have speed between 125 r.p.m. and 300 r.p.m. and low speed steam engines have speed below 125 r.p.m.

4. A vertical reciprocating steam engine having its cylinder above the crankshaft is known as _____

a) Horizontal inverted engine

b) Inclined inverted engine

c) Horizontal engine

d) Vertical inverted engine

Answer: d

Explanation: According to the arrangement of cylinder in a reciprocating steam engine, they are classified into vertical, horizontal and inclined steam engines. A common form of vertical steam engine, in which the cylinder is located above the crankshaft is known as vertical inverted steam engine.

5. Based on the arrangement of cylinder, reciprocating steam engines are classified into _____

a) Vertical, inclined and horizontal

b) Straight-up and straight-down

c) Straight and inverted

d) Horizontal, vertical, horizontal inverted and vertical inverted

Answer: a

Explanation: According to the arrangement of cylinder, reciprocating steam engines are classified into vertical, horizontal and inclined. Vertical reciprocating steam engines have their cylinder in vertical positon while, horizontal steam engines have it in horizontal alignment. Inclined steam engine has its cylinder at an angle.

6. Which of the following is NOT a type of reciprocating steam engine ?

a) Simple D valve

b) Corliss valve

c) Piston valve

d) Cylinder valve

Answer: d

Explanation: Reciprocating steam engines, based on their valve design, are classified into – simple D valve, balance plate valve, piston valve, riding cut-off valve, poppet valve and Corliss valve. There is no such type as cylinder valve.

7. According to the action of steam upon the piston, reciprocating steam engines are classified as expansive engines and non-expansive engines.

a) True

b) False

Answer: b

Explanation: Based on the action of steam upon the piston, reciprocating engines are classified into single acting and double reciprocating steam engines. Expansive and non-expansive classification of steam engines is based on the nature of expansion.

8. Steam acts only on one side of the piston in _____

a) Single acting reciprocating pump

b) Double acting reciprocating pump

c) Compound reciprocating steam engine

d) High speed reciprocating steam engine

Answer: a

Explanation: In case of single acting reciprocating steam engine, steam acts on one side of the piston. The other side of the piston is open to atmosphere. But in case of double acting reciprocating steam engine, steam acts on both the sides of the reciprocating steam engine, alternatively.

9. According to the range of expansion of steam, reciprocating steam engine are classified into _____

a) Expansive and non-expansive steam engines

b) Simple steam engine and compound steam engines

c) Condensing and non-condensing steam engines

d) Single acting and double acting steam engines

Answer: b

Explanation: Reciprocating steam engines, based on the expansion of steam, are classified into 2 categories – Simple steam engine and compound steam engine. In case of simple steam engine, the expansion of steam takes place in a single cylinder, while in case of compound steam engine, the expansion of steam takes place in multiple steps, in multiple cylinders.

10. Based on the type of exhaust, reciprocating steam engines are classified into _____

a) Condensing and non-condensing steam engines

b) Single acting and double acting steam engines

c) Simple and compound steam engines

d) Locomotive and marine steam engines

Answer: a

Explanation: Reciprocating steam engines, based on the exhaust type, are classified into condensing and non-condensing steam engines. Condensing steam engines are those in which the absolute exhaust pressure is below atmospheric pressure. While the exhaust pressure of non-condensing steam engines is at or above atmospheric pressure.

11. According to the nature of expansion, reciprocating steam engines are classified into _____

a) Condensing and non-condensing

b) Radial and non-radial

c) Expansive and non-expansive

d) Single acting and double acting

Answer: c

Explanation: According to the nature of expansion, reciprocating steam engines are classified into two categories – Expansive and Non-expansive steam engines. Single acting and double acting are types of reciprocating steam engines, based on the action of steam upon the piston. Classification into condensing and non-condensing is based on the type of exhaust.

12. Expansive reciprocating steam engines are used for all industrial purposes for the power development.

a) True

b) False

Answer: a

Explanation: Expansive type reciprocating steam engines are used for power development . The non-expansive reciprocating steam engines are used only in limited cases. Such as direct acting pumps, rolling mills etc.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Reciprocating Steam Engine – Parts, Working and Terminology – 1”.

1. The engine cylinder of a reciprocating steam engine is made of _____

a) Cast-iron

b) Chromium

c) Aluminum

d) Speed steel

Answer: a

Explanation: In a reciprocating steam engine, the cylinder is made of cast-iron. It is bored out perfectly true. In case of small engines, it is integral with the frame. It is produced by casting.

2. The engine cylinder of a reciprocating steam engine is produced using casting operation. The casting of the cylinder consists of two chamber, namely _____

a) Steam chamber and condensing chamber

b) Cylinder chamber and valve chamber

c) Top chamber and bottom chamber

d) back chamber and front chamber

Answer: b

Explanation: The engine cylinder of a reciprocating steam engine is casted, the casting consists of two chambers, cylinder chamber and valve chamber. Steam ports connect the cylinder ends with the steam chest.

3. Which of the following statement about steam ports  is FALSE?

a) They connect the engine cylinder with the steam chest

b) The steam ports should have large cross-section

c) Steam reaches the engine cylinder after passing through the steam ports

d) The purpose of steam ports is to throttle the steam as it passes through

Answer: d

Explanation: The engine cylinder is connected at its end with the steam chest through steam ports. Steam is supplied to the cylinder through the steam ports. The cross-sectional area of the steam ports should be large, so that it can admit enough steam into the cylinder. If the steam port is not large enough the steam will be throttled or wiredrawn.

4. In an engine cylinder , the end of cylinder through which the piston rod passes is known as _____

a) crank end

b) front end

c) top end

d) cover end

Answer: a

Explanation: Crank end or back end or bottom end, is that end of the engine cylinder through which the piston rod passes. Therefore, crank end is the correct answer. The other end of the cylinder is covered fully by a cover and it is known as cover end or front end or top end.

5. Which of the following statement about stuffing box  is TRUE?

a) Stuffing box is the casing which engine assembly is placed

b) Stuffing box guides the crank shaft

c) Stuffing box provides a steam-tight hole for the piston rod to pass through, to prevent steam leakage

d) Stuffing box serves no purpose during engine operation

Answer: c

Explanation: Piston rod passes through the crank end of the cylinder, at this end the piston rod must pass through a steam-tight hole, to prevent leakage of steam. Stuffing box does this job. The piston passes through the steam-tight hole in stuffing box, ensuring that no steam is leaked.

6. The name of the non-conducting material covering the engine cylinder, of a reciprocating steam engine, to reduce condensation of the steam is called _____

a) graphite

b) ceramide

c) leading

d) lagging

Answer: d

Explanation: As the steam expands, it cools down as tends to condense. To avoid this the steam engine cylinder is steam-jacketed. Covering the cylinder with lagging , the steam condensation is further reduced.

7.Steam is admitted into the engine cylinder from a reservoir of steam called as _____

a) Stuffing box

b) Steam chest

c) Eccentric

d) Cross-head

Answer: b

Explanation: Steam chest acts as a reservoir of steam and allows the steam into engine cylinder during the admission stroke of the piston. It is closed with a cover called steam chest cover. It is cast integral with the cylinder.

8. What is the function of a piston in a reciprocating steam engine?

a) It converts the pressure energy of the steam to its reciprocating movement

b) It converts its reciprocating energy to pressure energy of the steam

c) It acts a reservoir to store steam

d) It reheats the steam before entering into the steam chest

Answer: a

Explanation: The function of the piston in a reciprocating steam engine is to covert the pressure energy to its reciprocating movement. It moves inside the cylinder, end to end. Piston is made of forged steel, cast steel or cast iron.

9. Which of the following statements about the piston rings is FALSE?

a) Piston rings prevent leakage of steam

b) Piston rings reduce the friction between the piston and wall

c) Piston rings are generally made of cast iron

d) Piston rings are fixed to the engine cylinder

Answer: d

Explanation: The function of piston rings is to prevent leakage of steam and to reduce the friction between the piston and the wall. They are usually made of cast iron. Piston rings are fixed to the piston and not the engine cylinder.

10. Stuffing box is made of _____

a) Chromium

b) Aluminum

c) Cast iron

d) Cast steel

Answer: c

Explanation: Stuffing box is made of cast iron. A stuffing box is used to prevent the leakage of fluid where a sliding or rotating piece passes through a vessel containing fluid at a certain pressure. In case of reciprocating steam engine, it used to prevent the leakage of steam.

11. Cross-head of a reciprocating steam engine is usually made of wrought iron, cast steel or forged steel.

a) True

b) False

Answer: a

Explanation: Cross-head connects the outer end of the piston rod to the small end of the connecting rod. It is made of wrought iron, forged steel or cast steel. Cross-heads of small size are usually forged as one with the piston rod.

12. The end of the piston rod is joined to the connecting rod by _____

a) Cross-head

b) D-slide valve

c) Stuffing box

d) Steam chest

Answer: a

Explanation: Cross-head slides inside the guide. It connects the piston rod to the connecting rod. It receives the reciprocating motion from the piston rod. In some engines the guide is cast with the frame of the cylinder.

13. Which of the following statements about cross-head of a reciprocating steam engine is FALSE?

a) It connects the piston rod with the connecting rod

b) Piston rod in some engines is joined to the cross-head by screwing

c) In some engines cotter joint is used to join piston rod with the cross-heads

d) Knuckle-joint is used to join cross-head and piston

Answer: d

Explanation: The piston rod is either screwed to the cross-head or a cotter joint is used to join the two. In some engines the piston rod is forged to the cross-head end. The reciprocating piston rod and the oscillating connecting rod are joined by the cross-head.

14. What is the function of a connecting rod in a reciprocating steam engine?

a) It converts the circular motion of the crankshaft into the reciprocating motion of the piston rod

b) It converts the reciprocating motion of the crankshaft into reciprocating motion of the piston rod

c) It converts the reciprocating motion of the piston rod and cross-head into circular motion of the crankshaft

d) It converts the circular motion of the piston rod and cross-head into circular motion of the crankshaft

Answer: c

Explanation: The connecting rod in a reciprocating steam engine, joined to the piston rod end by cross-head, converts the reciprocating motion of the piston rod and cross-head into circular motion of the crankshaft. Connecting rod performs oscillatory motion.

15. The connecting rod of a reciprocating steam engine is made by casting.

a) True

b) False

Answer: b

Explanation: The connecting rod of a reciprocating steam engine is made of steel forging of rectangular, I, H or circular section. one end of the connecting rod is connected to the crank and the other to the cross-head.

This set of Thermal Engineering Assessment Questions and Answers focuses on “Reciprocating Steam Engine – Parts, Working and Terminology – 2”.

1. In a reciprocating steam engine, the connecting rod is connected to the crank by means of a pin called _____

a) Crank pin

b) Gudgeon pin

c) Wrist pin

d) U-pin

Answer: a

Explanation: The connecting rod is joined to the crank by a crank pin. The end of the connecting rod connected to the crank is also called big end, and the end connected to the cross-head is called small end.

2. One of the ends of the connecting rod is connected to the cross-head by means of a pin known as crank pin.

a) True

b) False

Answer: b

Explanation: The end of the connecting rod connected to the cross-head is called small end. The connection at the small end is done by means of a pin known as gudgeon pin or wrist pin. The connection of the other end to the crank is done by crank pin.

3. Which of the following statements about connecting rod is FALSE?

a) It converts the reciprocating motion of the piston rod into circular motion of the crankshaft

b) The length of the connecting rod is half the length of the piston stroke

c) It is connected to the cross-head by means of a gudgeon pin or a wrist pin

d) It is connected to the crank by means of a crank pin

Answer: b

Explanation: The length of the connecting rod isn’t half of the piston stroke, but varies from about 2 to 6 times the length of the piston stroke. The length of the connecting rod is measured from the center of the cross-head pin to the center of the crank pin.

4. What is the function of a crank shaft?

a) It converts the reciprocating motion of the piston into rotary motion of the crankshaft

b) It converts the rotary motion of the crankshaft into reciprocating motion of the piston

c) It converts the rotary motion of the piton into reciprocating motion of the crankshaft

d) It converts the reciprocating motion of the crank shaft into rotary motion of the piston

Answer: a

Explanation: Crank converts the reciprocating motion of the piston into rotary motion of the crankshaft. Radius of the crankshaft is defined as the distance between the axis of the crankshaft and crank pin.

5. Which of the following statements regarding the radius of a crank  is true?

a) Thrice the radius of the crank is equal to the stroke of the piston

b) One-third the radius of the crank is equal to the stroke of the piston

c) Twice the radius of the crank is equal to the stroke of the piston

d) half the radius of the crank is equal to the stroke of the piston

Answer: c

Explanation: The distance measured between the axis of the crankshaft and the crank pin is called radius of the crank. Twice the radius of the crank is equal to the stroke of the piston. Once the crank starts rotating it can be used to drive a variety of systems like generators, pulleys etc.

6. Which of the following statements about bearings  is FALSE?

a) Split-type bearings are used

b) Bearings need lubrication

c) The portion of the crankshaft in the bearing is called journal

d) Bearings are used to stop the rotation of the crankshaft

Answer: d

Explanation: Bearings are used to support and facilitate the rotation of shaft. Split-type bearings are used, as they are easy to replace. Bearings require lubrication for proper functioning. The bearings may be lubricated by ring oiling, needle lubrication, wick lubrication, drop lubrication or forced lubrication.

7. In a reciprocating steam engine, what is the function of an eccentric?

a) It converts the rotary motion of the D-slide valve into the reciprocating motion of the crankshaft

b) It converts the reciprocating motion of the D-slide valve into the rotary motion of the crankshaft

c) It converts the reciprocating motion of the crankshaft into the rotary motion of the D-slide valve

d) It converts the rotary motion of the crankshaft into the reciprocating motion of the D-slide valve

Answer: d

Explanation: The function of the eccentric, in a reciprocating steam engine, is to convert the rotary motion of the crankshaft into the reciprocating motion of the D-slide valve. The eccentric is fixed on the crankshaft.

8. The cylindrical grooved pulley of the eccentric, which is fixed on the crankshaft is called _____

a) Strap

b) Sheaf

c) Sheave

d) Eccentric pulley

Answer: c

Explanation: The cylindrical grooved pulley of the eccentric is called sheave. It is fixed on the crankshaft by means of a key. It is hollowed at two places to save material and reduce weight.

9. Flywheel is used in a reciprocating steam engine to maintain constant angular motion.

a) True

b) False

Answer: a

Explanation: A flywheel produces avoids the jerks and maintains a steady rotation of the crankshaft. Flywheel stores a part of energy, when the engine develops power, and returns this energy to the crankshaft during idle period of the cycle to maintain constant angular motion.

10. Which of the following component of a reciprocating steam engine keeps the speed of the crankshaft constant over a period, when there is variation of load on the engine?

a) Flywheel

b) Governor

c) Eccentric

d) Piston

Answer: b

Explanation: The correct answer is governor. It increases or decreases the supply of steam in the engine cylinder with the corresponding increase or decrease in load, in order to maintain a constant speed of the crankshaft.

11. Which of the following is the correct formula for calculating swept volume ? 

a) \(\frac{π}{4} \) L 2 D

b) \(\frac{π}{4} \) D 2 L

c) \(\frac{π}{4} \) L 3

d) \(\frac{π}{4} \) D 3

Answer: b

Explanation: Swept volume is defined as the total volume swept by a piston, when it moves from one dead center to another dead center. It is given by the following expression –

Swept Volume = \(\frac{π}{4} \) D 2 L

Where, D – Diameter of the piston

L – Piston Stroke.

12. The steam pressure that acts on the exhaust side of the piston is known as _____

a) Cylinder Pressure

b) Base Pressure

c) Piston Pressure

d) Back Pressure

Answer: d

Explanation: The correct answer is back pressure. It acts on the piston from the exhaust side. The net work done on a piston during a cycle is reduced due to presence of back pressure. Ideally it should be zero.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Reciprocating Steam Engine – Theoretical Indicator Diagram”.

1. An indicator diagram shows _____

a) The variation of temperature of the steam and its pressure during the cycle of operations

b) The variations of temperature and volume of steam during the cycle of operation

c) The variations of steam pressure and its volume in the cylinder during the cycle of operations

d) The variation of temperature of steam and its specific volume during the cycle of operations

Answer: c

Explanation: The plot of the variations of steam pressure and volume in the cylinder during the cycle of operations is called an indicator diagram. Indicator diagrams useful in estimating the performance of a reciprocating steam engine.

2. In a theoretical or hypothetical indicator diagram the expansion is assumed to be _____

a) Parabolic

b) Linear

c) Exponential

d) Hyperbolic

Answer: d

Explanation: Theoretical or hypothetical indicator diagram is based on some assumptions. One of these assumptions is that the expansion process is hyperbolic.

Mathematically, the following relationship holds true between pressure  and volume  –

PV = Constant.

3. In the given indicator diagram, Point 1 is called _____

a) Point of cut-off

b) Point of release

c) Point of flow

d) Point of admission

Answer: a

Explanation: At point 1 steam supply is cutoff and hence is called the point of cut-off. The point 5 is called the point of admission and the point 2 is called the point of release. During the process point 5 to point 1 steam is admitted inside the engine cylinder.

4. Which of the following processes is the expansion process? 

a) Process 5 to 1

b) Process 1 to 2

c) Process 2 to 3

d) Process 3 to 4

Answer: b

Explanation: Process 1 to 2 represents the expansion process. The process is hyperbolic. During the expansion process the steam expands in the engine cylinder to the end of the stroke. Point 1 and 2 are called point of cut-off and point of release respectively.

5. With reference to the given image, steam is admitted in the engine cylinder at _____

a) Point 5

b) Point 1

c) Point 2

d) Point 3

Answer: a

Explanation: At point 5 steam is admitted in the engine cylinder. The steam admitted has same pressure as the boilers. Point 5 is also called as point of admission. The expansion of steam starts at point 1.

This set of Thermal Engineering Question Paper focuses on “Reciprocating Steam Engine – Actual Indicator Diagram and Diagram Factor”.

1. The actual indicator diagram is different from the hypothetical one due to _____

a) timings of the valve action and pressure drop due to friction in the admission port

b) limited amount of steam

c) Excess amount of steam

d) Low quality valves installation

Answer: a

Explanation: Timings of the valve action and the pressure drop due to friction in the admission port reshapes the hypothetical indicator diagram into actual indicator diagram. Unlike hypothetical indicator diagram the actual one doesn’t have sharp corners.

2. The rounding observed in the actual indicator diagram is due to valve action.

a) True

b) False

Answer: a

Explanation: Under practical conditions, the processes cannot be started or terminated instantaneously as a result the actual indicator diagram roundedness. Hence, valves are the reason for this roundedness and the given statement is correct.

3. What is diagram factor?

a) It is the ratio of the area of theoretical indicator diagram and the area of the actual indicator diagram

b) It is the ratio of area of the actual indicator diagram and the area of the theoretical indicator diagram

c) It is the sum of areas of actual and theoretical indicator diagram

d) It is the ratio of the difference between the areas of the theoretical and actual indicator diagram and the theoretical indicator diagram.

Answer: b

Explanation: The diagram factor is defined as the ratio of the area of actual indicator diagram and the area of the theoretical indicator diagram. Mathematically,

Diagram factor = \(\frac{Area \, of \, actual \, indicator \, diagram}{Area \, of \, theoretical \, indicator \, diagram} \)

4. Given that the diagram factor for a particular reciprocating steam engine is 0.70 and the theoretically plotted indicator diagram has an area of 1700 kJ. Determine the area of the actual indicator diagram for the steam engine.

a) 11700 kJ

b) 11800 kJ

c) 11900 kJ

d) 12000 kJ

Answer: c

Explanation: Given, Diagram factor = 0.70, area of the theoretical indicator diagram = 1700 kJ

We knew that,

Diagram factor = \(\frac{Area \, of \, actual \, indicator \, diagram}{Area \, of \, theoretical \, indicator \, diagram} \)

Substituting the values, we get

0.7 = \(\frac{Area \, of \, actual \, indicator \, diagram}{1700} \)

Therefore,

Area of the actual indicator diagram = 11900 kJ.

5. The actual expansion of the steam is not isentropic due to _____

a) leakage of heat through cylinder walls

b) limited amount of steam

c) No heat leakage

d) boiler inefficiency

Answer: a

Explanation: There is an undesirable heat transfer through the cylinder wall during expansion. Due to this the expansion process is not isentropic. The efficiency of expansion process can be increased by limiting the temperature and pressure range through which the steam falls.

6. Calculate the diagram factor if the areas of theoretical and actual and theoretical indicator diagrams are 1400 kJ and 1800 kJ respectively.

a) 0.51

b) 0.52

c) 0.68

d) 0.78

Answer: d

Explanation: Given, area of actual indicator diagram = 1400 kJ, area of theoretical indicator diagram = 1800 kJ

We know that,

Diagram factor = \(\frac{Area \, of \, actual \, indicator \, diagram}{Area \, of \, theoretical \, indicator \, diagram} \)

Substituting the values, we get

Diagram factor = \(\frac{1400}{1800} \)

Therefore,

Diagram factor = 0.78.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Internal Combustion Engines – Heat Engines”.

1. A heat engine is a device which transforms the _________ of a fuel into thermal energy.

a) Electrical energy

b) Chemical energy

c) Mechanical energy

d) Solar Energy

Answer: b

Explanation: Chemical energy is produce in a fuel due to chemical reaction. This chemical energy is converted into thermal energy by means of a heat engine.

2. Which of the following is not an external combustion engine?

a) Steam engine

b) Steam turbine

c) Wankel engine

d) Stirling engine

Answer: c

Explanation: Steam engine, steam turbine, closed cycle steam turbine and Stirling engine are examples of external combustion engine. Wankel engine is an example of internal combustion engine.

3. Which of the following internal combustion engine have reciprocating mechanism?

a) Wankel engine

b) Jet engine

c) Rocket

d) Diesel engine

Answer: d

Explanation: Diesel engine works on the reciprocating mechanism. Wankel engine and jet engine works on rotary mechanism while rocket has no mechanism.

4. The thermal energy transformed by heat engine is used to produce _________

a) Thermal work

b) Electrical work

c) Laser action

d) Mechanical work

Answer: d

Explanation: Mechanical work is produce from thermal energy. Thermal energy is transforms from chemical energy of a fuel by a heat engine.

5. Which of the following engine works on rotary mechanism?

a) Steam engine

b) Steam turbine

c) Stirling engine

d) Diesel engine

Answer: b

Explanation: Steam engine, Stirling engine and diesel engine works on reciprocating mechanism while steam engine works on rotary mechanism.

6. Which of the following is not principal use of Stirling engine?

a) Locomotive

b) Power in space

c) Experimental

d) Vehicles

Answer: a

Explanation: The principal uses of Stirling engine are experimental, power in space and vehicles. Locomotive is principal use of steam engine and diesel engine.

7. What are the principal uses of gas engine?

a) Missiles and aircraft

b) Electrical power and aircraft

c) Industrial and electrical power

d) Marine and locomotive

Answer: c

Explanation: The principal uses of gas engine are industrial and electrical power. Marine, locomotive and electrical power is principal uses of diesel engine. Missiles are principal use of rocket engine.

8. The earliest internal combustion engine is credited to ___________

a) Frenchman Lenior

b) Brayton

c) Atkinson

d) Christian Huygens

Answer: d

Explanation: The earliest internal combustion Huygens gun power engine is credited to Christian Huygens in the year 1680.

9. Theoretically the efficiency of ideal otto cycle is equal to ___________

a) Diesel cycle

b) Brayton cycle

c) Dual cycle

d) Stirling cycle

Answer: b

Explanation: The theoretical efficiency of ideal otto cycle is same as Brayton cycle because brayton cycle is actually improvised otto cycle.

10. Which of the following heat engine part is manufactured by method of forging?

a) Crankcase

b) Bearing

c) Crankshaft

d) Piston rings

Answer: c

Explanation: Crankcase, bearing and piston rings are the parts of heat engine manufactured by the process of casting whereas crankshaft is the manufactured by the process of forging.

11. What material is used to manufacture bearing?

a) Cast iron

b) Aluminum alloy

c) White metal

d) Silicon

Answer: c

Explanation: White metal and leaded bronze materials are used to manufacture bearings in heat engine. Cast iron, aluminum alloy and silicon are used to manufacture cylinder, piston, piston rings and crankcase.

12. Internal combustion engine works based on rotary mechanism is __________

a) Petrol engine

b) Jet engine

c) Diesel engine

d) Gas engine

Answer: b

Explanation: Petrol engine, diesel engine and gas engine are based on reciprocating mechanism but jet engine is based on rotary mechanism.

This set of IC Engines Multiple Choice Questions & Answers  focuses on “Two-Stroke Engine”.

1. In a two stroke engine, the working cycle is completed in two revolutions of the crankshaft.

a) True

b) False

Answer: b

Explanation: In a two stroke engine, the working cycle is completed in one revolution of the crankshaft.

2. A two stroke cycle engine gives _____________ the number of power strokes as compared to the four stroke cycle engine, at the same engine speed.

a) half

b) same

c) double

d) four times

Answer: c

Explanation: None.

3. A two stroke cycle engine occupies larger floor area than a four stroke cycle engine.

a) True

b) False

Answer: b

Explanation: A two stroke cycle engine occupies less floor area than a four stroke cycle engine.

4. A two stroke engine gives _____________ mechanical efficiency than a four stroke cycle engine.

a) higher

b) lower

c) equal

d) none of the mentioned

Answer: a

Explanation: As compared to a four stroke cycle engine, the mechanical efficiency of two stroke engine gives higher.

5. The two stroke cycle engine have lighter flywheel.

a) True

b) False

Answer: a

Explanation: None.

6. Thermal efficiency of a two stroke cycle engine is _____________ a four stroke cycle engine.

a) equal to

b) less than

c) greater than

d) none of the mentioned

Answer: b

Explanation: None.

7. In a petrol engine, the mixture has the lowest pressure at the __________

a) beginning of suction stroke

b) end of suction stroke

c) end of compression stroke

d) none of the mentioned

Answer: b

Explanation: In a petrol engine, only at the end of the suction stroke, the mixture has the lowest pressure.

8. In compression ignition engines, swirl denotes a __________

a) haphazard motion of the gases in the chamber

b) rotary motion of the gases in the chamber

c) radial motion of the gases in the chamber

d) none of the mentioned

Answer: b

Explanation: In compression ignition engines, swirl denotes a rotary motion of the gases in the chamber as swirl is always related to rotary motion.

9. The injector nozzle of a compression ignition engine is required to inject fuel at sufficiently high pressure in order to __________

a) inject fuel in a chamber of high pressure at the end of the compression stroke

b) inject fuel at a high velocity to facilitate atomization

c) ensure that penetration is not high

d) all of the mentioned

Answer: d

Explanation: The injector nozzle of a compression ignition engine is required to inject fuel at a sufficiently high pressure in order to

i) inject fuel in a chamber of high pressure at the end of compression stroke.

ii) inject fuel at a high velocity to facilitate atomization.

iii) ensure that penetration is not high.

10. Which of the following engines will have a heavier flywheel than the remaining ones?

a) 30 kW four stroke petrol engine running at 1500 r.p.m

b) 30 kW two stroke petrol engine running at 1500 r.p.m

c) 30 kW two stroke diesel engine running at 750 r.p.m

d) 30 kW four stroke diesel engine running at 750 r.p.m

Answer: a

Explanation: The flywheel in a four stroke petrol engine is heavier than in a two stroke petrol engine and there is no flywheel in a diesel engine.

This set of IC Engine Questions and Answers for Campus interviews focuses on “Advantages of Four Stroke Engines”.

1. In a four stroke engine, the working cycle is completed in __________

a) one revolution of the crankshaft

b) two revolution of the crankshaft

c) three revolution of the crankshaft

d) four revolution of the crankshaft

Answer: b

Explanation: In a four stroke engine, the working cycle is completed in two revolution of the crankshaft while while in a two stroke engine, the working cycle is completed in one revolution.

2. Theoretically, a four stroke engine should develop ____________ power as that of a two stroke cycle engine.

a) half

b) same

c) double

d) four times

Answer: a

Explanation: None.

3. In a four stroke cycle engine, the sequence of operation is __________

a) suction, compression, expansion and exhaust

b) suction, expansion, compression and exhaust

c) expansion, compression, suction and exhaust

d) compression, expansion, suction and exhaust

Answer: a

Explanation: Four stroke cycle engine works on four strokes, i.e- suction, compression, expansion and exhaust.

4. In a four stroke cycle petrol engine, the pressure inside the engine cylinder during the suction stroke is ____________ the atmospheric pressure.

a) equal to

b) below

c) above

d) none of the mentioned

Answer: b

Explanation: None.

5. In a four stroke cycle petrol engine, the inlet valve __________

a) opens at top dead center and closes at bottom dead center

b) opens at 20º before top dead center and closes at 40º after bottom dead center

c) opens at 20º after top dead center and closes at 20º before bottom dead center

d) may open or close anywhere

Answer: b

Explanation: None.

6. In a four stroke cycle petrol engine, the compression __________

a) starts at 40º after bottom dead center and ends at 30º before top dead center

b) starts at 40º before bottom dead center and ends at 30º after top dead center

c) starts at bottom dead center and ends at top dead center

d) may start and end anywhere

Answer: a

Explanation: None.

7. In a four stroke cycle petrol engine, the charge is ignited at __________

a) 30º before top dead center

b) 30º after top dead center

c) 30º before bottom dead center

d) 30º after bottom dead center

Answer: a

Explanation: None.

8. In a four stroke cycle petrol engine, the expansion __________

a) starts at top dead center and ends at bottom dead center

b) starts at 30º before top dead center and ends at 50º before bottom dead center

c) starts at 30º after top dead center and ends at 50º after bottom dead center

d) may start and end anywhere

Answer: b

Explanation: In a four stroke cycle petrol engine, the expansion starts at 30º before top dead center and ends at 50º before bottom dead center while the compression starts at 40º after bottom dead center and ends at 30º before top dead center.

9. The exhaust valve in a four stroke cycle petrol engine __________

a) opens at 50º before bottom dead center and closes at 15º after top dead center

b) opens at bottom dead center and closes at top dead center

c) opens at 50º after bottom dead center and closes at 15º before top dead center

d) may open or close anywhere

Answer: a

Explanation: The exhaust valve in a four stroke cycle petrol engine opens at 50º before bottom dead center and closes at 15º after top dead center while the inlet valve, opens at 20º before top dead center and closes at 40º after bottom dead center.

10. The inlet valve of a four stroke cycle internal combustion engine remains open for __________

a) 130º

b) 180º

c) 230º

d) 270º

Answer: c

Explanation: None.

11. In a four stroke cycle the minimum temperature inside the engine cylinder occurs at the __________

a) beginning of suction stroke

b) end of suction stroke

c) beginning of exhaust stroke

d) end of exhaust stroke

Answer: a

Explanation: None.

12. In a four stroke cycle petrol engine, the charge is compressed when both the valves  are closed.

a) True

b) False

c) Both the valves are open

d) None of the mentioned

Answer: a

Explanation: None.

13. In a four stroke cycle diesel engine, the inlet valve __________

a) opens at 20º before top dead center and closes at 40º after bottom dead center

b) opens at 20º after top dead center and closes at 20º before bottom dead center

c) opens at top dead center and closes at bottom dead center

d) may open or close anywhere

Answer: a

Explanation: None.

This set of Automobile Engineering Multiple Choice Questions & Answers  focuses on “Ignition System”.

1. Which instrument is used for adjusting the ignition timing?

a) Accurate clock

b) Tachometer

c) Stopwatch

d) Stroboscopic light

Answer: d

Explanation: Stroboscope is used for adjusting the ignition timing. It is the light-emitting device which is connected to the spark plug of the cylinder for setting the ignition timing.

2. Which of the following firing order is commonly used for the four-cylinder vertical engine?

a) 1-2-3-4

b) 3-4-1-2

c) 4-3-2-1

d) 1-3-4-2

Answer: d

Explanation: 1-3-4-2 firing order is commonly used for a four-cylinder engine. For the four-cylinder engine, the firing order is decided such that a power stroke occurs every 180 degrees.

3. What is the dwell time?

a) It is the time for which the contact breaker points remain closed

b) It is the time for which the contact breaker points remain open

c) The time during which inlet and exhaust valves open

d) The time during which inlet and exhaust valves close

Answer: a

Explanation: The dwell time is the time for which the contact breaker points remain closed. Dwell is the time for which the coil turned on.

4. Which of the following is not considered while deciding the optimum firing order of the engine?

a) Engine vibration

b) Engine cooling

c) Development of the backpressure

d) Engine configuration

Answer: d

Explanation: Engine vibration, engine cooling, and development of back pressure are considered while deciding the optimum firing order of the engine. The reduced vibration is required for optimum firing order of the engine.

5. Why is the Dwell meter used?

a) To set the spark plug gap

b) To set the ignition advance

c) To set the contact breaker gap

d) To set ignition timing

Answer: c

Explanation: The dwell meter used for setting the contact breaker gap. It uses the dwell angle as the parameter to set the contact breaker gap. To set the ignition timing stroboscope is used.

6. Which of the following is the disadvantage of the magneto ignition system?

a) Magneto ignition system has a poor quality of spark during starting

b) Magneto ignition system occupies more space

c) Magneto ignition system has more maintenance problems

d) Magneto ignition system is used largely in four wheels

Answer: a

Explanation: During starting, the quality of the spark is poor due to low speed. Magneto ignition system is mainly used in racing cars and two-wheelers. The main advantage is that it does not require a battery.

7. For how many times the contact breaker must make and break the circuit for a four-cylinder engine operating at N rpm?

a) N times

b) 1.5N times

c) N/2 times

d) 2N times

Answer: d

Explanation: The cam is having the same number of lobes as there are engine cylinders. They are used with two contact breakers in parallel.

8. Which of the following is the wrong statement?

a) In the exhaust, retarded timing causes burning of the hydrocarbons

b) The retarded timing improves fuel economy

c) The retarded timing requires the small opening for correct burning of the fuel

d) The exhaust gas temperature becomes higher due to retarded time

Answer: b

Explanation: The retarded timing reduces fuel economy. Due to spark retardation, the fuel burns slowly and after some time than it should be burned.

9. On which condition does the vacuum advance mechanism shift the ignition point?

a) Under part-load operation

b) Under full load operation

c) Under no-load operation

d) Under sudden acceleration

Answer: a

Explanation: The vacuum advance mechanism shifts the ignition point under part-load operation. The adjustment system is designed so that its operation results in the part-load advance curve.

10. The optimum spark timing must be retarded for the engine operating with a rich mixture.

a) True

b) False

Answer: b

Explanation: The optimum spark timing must be advanced for the engine operating with a rich mixture. The larger throttle opening size means better mixing and combustion.

This set of IC Engines Multiple Choice Questions & Answers  focuses on “Requirements Of Ignition System”.

1. The function of the ignition system is to _____________ the flame propagation process.

a) stop

b) initiate

c) balance

d) none of the mentioned

Answer: b

Explanation: None.

2. The total enthalpy required to cause the flame to be self sustaining and promote ignition, is given by the product of the surface area of the spherical flame and the enthalpy per unit area.

a) True

b) False

Answer: a

Explanation: None.

3. A spark can be caused by applying a sufficiently low voltage between two electrodes separated by a gap.

a) True

b) False

Answer: b

Explanation: In fact, a spark can be caused by applying a sufficiently high voltage between two electrodes separated by a gap.

4. An ignition process obeys the law of ______________

a) conservation of mass

b) conservation of energy

c) conservation of momentum

d) none of the mentioned

Answer: b

Explanation: An ignition process obeys the law of conservation of energy as energy is only produced there.

5. The pressure, temperature and density of the mixture between the spark plug electrodes have a considerable influence on the __________ required to produce a spark.

a) voltage

b) current

c) mass

d) none of the mentioned

Answer: a

Explanation: The pressure, temperature and density of the mixture between the spark plug electrodes have a considerable influence on the voltage required to produce a spark and mass and current are dependent on voltage.

6. If the spark energy exceeds 40 mJ and the duration is longer than 0.5 ms, reliable ignition is obtained.

a) True

b) False

Answer: a

Explanation: None.

7. If the resistance of the deposits on the spark plug electrodes is sufficiently high, the loss of electrical energy through these deposits may prevent the spark discharge.

a) True

b) False

Answer: a

Explanation: None.

8. As air is poor conductor of electricity an air gap in an electric circuit acts as a high _____________

a) conductor

b) resistance

c) ignition

d) none of the mentioned

Answer: b

Explanation: As air is poor conductor of electricity, due to this an air gap in an electric circuit acts as a high resistance and conductor and ignition are required before conduction.

9. A spark is produced to ignite a homogeneous air fuel mixture in the combustion chamber of an engine, it is called spark ignition system.

a) True

b) False

Answer: a

Explanation: None.

10. An ignition system should provide a good spark between the _____________ of the plugs at the correct timing.

a) cathodes

b) anodes

c) electrodes

d) none of the mentioned

Answer: c

Explanation: An ignition system should provide a good spark between the electrodes of the plugs at the correct timing as only electrodes are used in ignition system and cathodes and anodes are used in battery.

11. An ignition system should function efficiently over the entire range of engine speed.

a) True

b) False

Answer: a

Explanation: None.

12. An ignition system should be heavy, effective and reliable in service.

a) True

b) False

Answer: b

Explanation: Actually, an ignition system should be light, effective and reliable in service.

13. An ignition system should be compact and easy to maintain.

a) True

b) False

Answer: a

Explanation: None.

14. An ignition system should be cheap and convenient to handle.

a) True

b) False

Answer: a

Explanation: None.

15. The interference from the low voltage source should not affect the functioning of the radio and television receivers inside an automobile.

a) True

b) False

Answer: b

Explanation: In fact, the interference from the high voltage source should not affect the functioning of the radio and television receivers inside an automobile.

This set of Automobile Engineering Multiple Choice Questions & Answers  focuses on “Fuel Injection Systems”.

1. In which of the following fuel injector is used?

a) Compression-ignition engines

b) Spark-ignition engines

c) Steam engines

d) Sterling engines

Answer: a

Explanation: In compression – ignition  engines, only air is compressed in the cylinder and by means of fuel – injection system fuel is injected into the cylinder. In CI engines, compressed air temperature plays a main role in combustion.

2. At which stroke the fuel is injected in the CI engine?

a) Suction stroke

b) Compression stroke

c) Expansion stroke

d) Exhaust stroke

Answer: b

Explanation: Fuel injection in the CI engines start at the end of the compression stroke. Fuel is injected directly into the combustion chamber.

3. Which of the following is the port fuel – injection system?

a) D – MPFI

b) L – MPFI

c) GDI

d) TBI

Answer: b

Explanation: L – MPFI system uses port fuel – injection. In this, the fuel metering is regulated by the speed of the engine and the amount of air that actually enters the engine. D – MPFI system uses manifold injection. GDI system uses direct injection. TBI system uses throttle body injection.

4. What is the range of pressure is achieved in the injection pump?

a) 120 – 200 bar

b) 10 – 20 bar

c) 400 – 500 bar

d) < 10bar

Answer: a

Explanation: The purpose of the fuel injection pump is to increase the pressure of the quantity of fuel in the range of 120 – 200 bars and deliver it at the correct instant to the injector.

5. In pneumatic governor, which of the following control the amount of vacuum applied to the diaphragm?

a) Spring

b) Accelerator pedal

c) Butterfly valve

d) Lever

Answer: a

Explanation: The vacuum applied to the diaphragm is controlled by the accelerator pedal. The diaphragm is in contact with the fuel pump control rack.

6. A six-cylinder, four-stroke diesel engine develops a power of 200 kW at 2000 revolution per minute. The brake specific fuel consumption is 0.3 kg / kW – h. What is the time for injection if the injection takes place over 20° crank angle?

a) 1 millisecond

b) 2 milliseconds

c) 1.5 milliseconds

d) 1.66 milliseconds

Answer: d

Explanation: Power output / cylinder = 200 / 6 = 33.33 kW. Fuel consumption / cylinder = 33.33 * brake specific fuel consumption = 33.33 * 0.3 = 9.99 kg / h. Time for injection =  /  = 1.667 * 10 -3 sec = 1.66 milliseconds.

7. What is the spray cone angle in the Pintle nozzle?

a) 60°

b) 15°

c) 25°

d) 45°

Answer: a

Explanation: Spray cone angle of a Pintle nozzle is 60°. The main advantage of this nozzle is that it avoids weak injection and dribbling. Also, it prevents the carbon deposition on the nozzle hole.

8. Which of the following is not the part of the injector assembly?

a) Needle valve

b) Nozzle

c) Diaphragm

d) Compressor spring

Answer: c

Explanation: The injector assembly consists of a needle valve, nozzle, compressor spring, and an injector body. The diaphragm is a part of the injection pump governor. A nozzle is used to atomize the fuel and break up into small particles.

9. Which of the following is not the function of the fuel injection system?

a) Time the fuel injection

b) Filter the fuel

c) Atomize the fuel to fine particles

d) Control the engine speed

Answer: d

Explanation: The functions of the fuel injection system are to filter the fuel, to time the fuel injection, to break up the fuel into small particles.

10. The cold start injector provides a lean air-fuel ratio.

a) True

b) False

Answer: b

Explanation: The cold start injector gives a rich air-fuel ratio. Its objective is to supply the engine with additional fuel to make the rich mixture in cold temperatures where the air density is increased and additional fuel is required.

This set of Automotive Engine Auxiliary Systems Multiple Choice Questions & Answers  focuses on “Fuel Injection System”.

1. How are the unit injectors operated in a Common rail fuel injection system?

a) Crankshaft

b) Rocker arms

c) Belt

d) Motor

Answer: b

Explanation: The common rail fuel injection system the unit injectors are operated by rocker arms and springs similar to the engine valves. A linkage connects the control racks of all the unit injectors, so that fuel injection in all the cylinders may be equal and simultaneously controlled.

2. Which of the following is used to return the excess fuel in the rail to the fuel tank?

a) Float valve

b) Relief valve

c) Hand pump

d) Governor

Answer: b

Explanation: The excess fuel in the rail is returned to the tank by the means of a relief valve present in the circuit. Then again the fuel is taken from fuel tank by the feed pump and is supplied at low pressure through a filter, to the low-pressure common rail and therefore, to all the unit injectors.

3. Which of the following can be removed by using a hand priming lever without turning the engine on in an Individual pump fuel injection system?

a) Air bled out

b) Temperature

c) Oil

d) Coolant

Answer: a

Explanation: The pump is provided with a hand priming lever so that the diesel oil can be forced into the system and the air bled out without turning the engine. The fuel is passed through a filter and hence to the fuel injection pump.

4. Which among the following is provided in an injection pump to provide automatic speed control?

a) Filter

b) Feed pump

c) Governor

d) Injector

Answer: c

Explanation: The injection pump contains a governor which provides automatic speed control, relative to any set position of the accelerator pedal. Any excess fuel after the lubrication of the injector nozzle is returned to the fuel tank. This system can be seen in the Swaraj Mazda SL engine.

5. Which of the following types of injection is used in the Common rail fuel injection system?

a) Solid injection

b) Air injection

c) Both solid and air injection

d) Oil injection

Answer: a

Explanation: A solid injection system, only the liquid fuel is injected due to which the common rail fuel injection system and individual pump fuel injection system are an example of solid injection.

6. Along with compressed air which of the following is injected inside the cylinder?

a) Antifreeze

b) Coolant

c) Liquid fuel

d) both coolant and liquid fuel

Answer: c

Explanation: In an air injection system liquid fuel is injected along with compressed air. The air injection is less reliable and less efficient. Whereas the coolant is added inside the radiator tank to cool the engine during a long run.

7. What is the pressure range in MPa of the compressed air that has been injected in an air injection?

a) 5

b) 7

c) 6

d) 8

Answer: b

Explanation: The air injection system has a liquid fuel injected with compressed air at 7MPa due to this reason it has been obsolete. This makes the air injection system less reliable, less efficient and requires a compressor all the time.

8. How much percentage of the power output of the engine is used during the injection process in an air injection system?

a) 5

b) 10

c) 15

d) 20

Answer: b

Explanation: The air injection system is less reliable, less efficient and requires an air compressor for supplying 7MPa of compressed air with the fuel which consumes up to 10% of the power output of the engine.

9. Which of the following is the correct flow of fuel in an individual pump fuel injection system?

a) Fuel tank – filter – feed pump – injector

b) Fuel tank – filter – injector- feed pump

c) Fuel tank – feed pump – filter – injector

d) Filter- feed pump – fuel tank – injector

Answer: c

Explanation: The fuel flows from the fuel tank to the feed pump where the pressure is increased and then the fuel is filtered by the filter and then it is sent to the injector. Therefore the correct sequence of the fuel flow is fuel tank – feed pump – filer – injector.

10. What is the main function of a good fuel filter concerning the injector?

a) Faulty spraying

b) Leakage

c) Increased fuel consumption

d) Filter moisture present

Answer: d

Explanation: The abrasive matter would also cause faulty spraying and leakage in the injector thus resulting in increased fuel consumption whereas a good fuel filter should absorb any moisture and dirt present in the fuel.

11. Which of the following is used from the engine to drive a gear-driven injection pump?

a) Crankshaft

b) Camshaft

c) Motor

d) Alternator

Answer: b

Explanation: The injection pump is gear driven from the engine camshaft so that it is driven at half the engine speed. The injection pump consists of a governor which provides automatic speed control, relative to any set position of the accelerator pedal.

12. What is the range of the cranking compression pressure in the range of KP in a diesel engine?

a) 1000

b) 2000

c) 3000

d) 10000

Answer: c

Explanation: The cranking compression pressure on a diesel engine is about 3000KPa. Compression pressure is the pressure that is used to compress the charge inside the cylinder and rise it to the ignition temperature.

13. What is the range in MPa of pressure injection in diesel engines?

a) 100

b) 200

c) 300

d) 400

Answer: b

Explanation: In diesel engines, the injection pressure can reach as High as 200MPa. This high pressure is required to atomize the fuel properly for complete combustion so that the emissions can be controlled to a greater extent.

This set of IC Engines Multiple Choice Questions & Answers  focuses on “Engine Cooling System”.

1. The gasoline engine requires much ____________ air than a diesel engine.

a) less

b) more

c) equal

d) none of the mentioned

Answer: b

Explanation: As a gasoline engine works on a mixture of air and gasoline, therefore the gasoline engine requires much more air than a diesel engine.

2. The turbocharged diesel engine requires ____________ cooling air than naturally aspired diesel engines.

a) less

b) more

c) equal

d) none of the mentioned

Answer: a

Explanation: As the turbocharged engine produces less heat so, the turbocharged diesel engine requires less cooling air than naturally aspired diesel engines.

3. The heat flow to water jackets, on an average, is about ____________ for large engines.

a) 500 kJ/kW-h

b) 4200 kJ/kW-h

c) 5700 kJ/kW-h

d) none of the mentioned

Answer: b

Explanation: The heat flow to water jackets, on an average is about 4200 kJ/kW-h for large engines and it is about 500-5700 kJ/kW-h for small engines.

4. The heat flow to water jackets, on average, is about ____________ for small engines.

a) 500-6200 kJ/kW-h

b) 400-4200 kJ/kW-h

c) 500-5700 kJ/kW-h

d) none of the mentioned

Answer: c

Explanation: The heat flow to water jackets, on average, is about 500-5700 kJ/kW-h for small engines and is about 4200 kJ/kW-h for large engines.

5. What is the outlet cooling water temperature for large engines?

a) about 50ºC

b) 60 to 65ºC

c) 80ºC

d) none of the mentioned

Answer: a

Explanation: The outlet cooling water temperature for large engines is about 50ºC and the outlet cooling water temperature for medium engines is 60 to 65ºC.

6. What is the outlet cooling water temperature for medium engines?

a) about 50ºC

b) 60 to 65ºC

c) 80ºC

d) none of the mentioned

Answer: b

Explanation: The outlet cooling water temperature for medium engines is 60 to 65ºC and the outlet cooling water temperature for small engines is 80ºC.

7. What is the outlet cooling water temperature for small engines?

a) about 50ºC

b) 60 to 65ºC

c) 80ºC

d) none of the mentioned

Answer: c

Explanation: The outlet cooling water temperature for small engines is 80ºC while the outlet cooling water temperature for medium engines is 60 to 65ºC and the outlet cooling water temperature for small engines is 80ºC.

8. In thermo-syphon system, the radiator should be kept well above the engine, to provide a height for natural circulation.

a) True

b) False

Answer: a

Explanation: None.

This set of Automobile Engineering Multiple Choice Questions & Answers  focuses on “Lubrication Systems”.

1. Which type of lubrication system is used in two-stroke engine?

a) Mist lubrication system

b) Wet sump lubrication system

c) Dry sump lubrication system

d) Splash lubrication system

Answer: a

Explanation: The mist lubrication system is used where crankcase lubrication is not suitable. In a two-stroke engine, as the charge is compressed in the crankcase, it is not possible to have the lubricating oil in the sump.

2. In most automobiles, which lubrication system is commonly used?

a) Splash system

b) Pressure system

c) Petrol system

d) Gravity system

Answer: b

Explanation: Pressure type lubrication is used commonly in most automobiles. In this system, oil is drawn in from the sump and forced to all the main bearings of the crankshaft through distributing channels.

3. The pressure inside the lubrication system is controlled by ______

a) Oil pump

b) Oil filter

c) Relief valve

d) Supply voltage

Answer: c

Explanation: The pressure inside the lubrication system is controlled by a relief valve. In the dry-sump lubrication system, if the filter is clogged, the pressure relief valve opens permitting oil to bypass the filter and reaches the supply tank.

4. What is the need for crankcase ventilation?

a) To cool cylinder

b) To cool crankcase

c) To cool piston

d) To remove blow-by

Answer: d

Explanation: The gas inside the cylinder gets past the piston rings and then it enters the crankcase. It contains water and sulphuric acid (H 2 SO 4 ).

5. Why are the detergents used as oil additives?

a) To increase fire point

b) To prevent foaming

c) To prevent sludge formation

d) To reduce viscosity

Answer: c

Explanation: The detergents are added to the oil as additives to increase the detergency. Detergency means property of an oil to clean the engine deposits.

6. Which of the following is the most important property of the lubricant?

a) Density

b) Thermal conductivity

c) Viscosity

d) Melting point

Answer: c

Explanation: Viscosity is the most important property of the lubricant. The viscosity of the oil at a given temperature and pressure must be compatible to ensure the hydrodynamic lubrication.

7. Which of the following viscosity indices shows the larger changes in viscosity with temperature?

a) 50

b) 100

c) 45

d) 10

Answer: d

Explanation: The high viscosity index number indicates relatively a smaller change in viscosity with temperature. A low viscosity number for given oil indicates a relatively large change of viscosity.

8. Which of the following of oils is multi-grade oil?

a) SAE 10W 30

b) SAE 25W

c) SAE 10

d) SAE 20W

Answer: a

Explanation: The SAE designation for multi-grade oils includes two viscosity. The first number indicates the relative flowability and the second number indicates the resistances to film break down.

9. Which of the following oils has the highest viscosity?

a) SAE 20

b) SAE 10

c) SAE 50

d) SAE 45

Answer: c

Explanation: Higher the SAE grade, higher the viscosity. Since SAE 50 has 50 grade hence it as highest viscosity amongst the other oils. SAE 50 is single non -winter-grade oil as there is no ‘W’ in the designation.

10. The vegetable oils are used as lubricants in the automobiles.

a) True

b) False

Answer: b

Explanation: The mineral oils are used as lubricants in the automobiles. The mineral oils are having properties such as to reduce the friction and wear between two metal contact and moving parts, to provide the sealing the action.

This set of IC Engines Multiple Choice Questions & Answers  focuses on “Combustion In S.I. Engine”.

1. Swirl is the rotational flow of charge within the cylinder about the axis.

a) True

b) False

Answer: a

Explanation: None.

2. The _____________ is defined by the parallel portion of the piston and cylinder head which almost touch each other as the piston approaches T.D.C.

a) turbulence

b) swirl

c) quench area

d) none of the mentioned

Answer: c

Explanation: The quench area is defined by the parallel portion of the piston and cylinder head which almost touch each other as the piston approaches T.D.C. while turbulence consists of randomly dispersed vortices of different sizes which become superimposed into the air and petrol mixture flow stream.

3. ____________ consists of randomly dispersed vortices of different sizes which become superimposed into the air and petrol mixture flow stream.

a) Turbulence

b) Swirl

c) Quench area

d) None of the mentioned

Answer: a

Explanation: Only turbulence consists of randomly dispersed vortices of different sizes which become superimposed into the air and petrol mixture flow stream while quench area is defined by the parallel portion of the piston and cylinder head which almost touch each other as the piston approaches T.D.C.

4. The amount of vortex activity,and the disintegration of others, _____________ the turbulent flow with rising engine speed.

a) increases

b) decreases

c) remains same

d) none of the mentioned

Answer: a

Explanation: None.

5. The degree of turbulence increases _____________ with the piston speed.

a) indirectly

b) directly

c) linearly

d) none of the mentioned

Answer: b

Explanation: The degree of turbulence is directly proportional to piston speed, hence it increases directly with the piston speed.

6. Turbulence decreases the heat flow to the cylinder wall and in the limit excessive turbulence may extinguish the flame.

a) True

b) False

Answer: b

Explanation: In fact, turbulence increase the heat flow to the cylinder wall and in the limit excessive turbulence may extinguish the flame.

7. The flame propagation velocities range from _____________

a) 10 to 15 m/s

b) 15 to 70 m/s

c) 20 to 80 m/s

d) 30 to 90 m/s

Answer: a

Explanation: None.

8. When ignition occurs the nucleus of the flame spreads with the whirling or rotating vortices in the form of ragged burning crust from the initial spark plug ignition site.

a) True

b) False

Answer: a

Explanation: None.

9. Swirl ratio is defined as the ratio of air rotational speed to crankshaft rotational speed.

a) True

b) False

Answer: a

Explanation: None.

10. Abnormal combustion knock produced by surface ignition in SI engines is not harmful as normal combustion knock.

a) True

b) False

Answer: b

Explanation: In fact, abnormal combustion knock produced by surface ignition in SI engines is more harmful than normal combustion knock.

This set of IC Engines Multiple Choice Questions & Answers  focuses on “Engine Performance Parameters”.

1. The ratio of indicated thermal efficiency to the corresponding air standard cycle efficiency is called ___________

a) net efficiency

b) efficiency ratio

c) relative efficiency

d) overall efficiency

Answer: c

Explanation: Relative efficiency is defined as the ratio of indicated thermal efficiency to the corresponding air standard cycle efficiency.

2. Compression ratio of I.C. engine is ___________

a) the ratio of volumes of air in cylinder before compression stroke and after compression stroke

b) volume displaced by piston per stroke and clearance volume in cylinder

c) ratio of pressure after compression and before compression

d) swept volume/cylinder volume

Answer: a

Explanation: None.

3. The air standard efficiency of an otto cycle compared to a diesel cycle for the given compression ratio is?

a) same

b) less

c) more

d) more or less depending on power rating

Answer: c

Explanation: None.

4. The calorific value of gaseous fuels is expressed in terms of ___________

a) Kcal

b) Kcal/kg

c) Kcal/m 2

d) Kcal/m 3

Answer: d

Explanation: None.

5. Indicated power of a 4-stroke engine is equal to ___________

a) pLAN

b) 2pLAN

c) pLAN/2

d) 4pLAN

Answer: c

Explanation: Indicated power of a 4-stroke engine is equal to pLAN/2.

where p = mean effective pressure,

L = stroke

A = area of piston and

N = rpm of engine.

6. If the intake air temperature of I.C. engine increases its efficiency will ___________

a) increase

b) decrease

c) remain same

d) none of the mentioned

Answer: b

Explanation: None.

7. All heat engines utilize ___________

a) low heat value of oil

b) high heat value of oil

c) net calorific value of oil

d) calorific value of fuel

Answer: a

Explanation: None.

8. An engine indicator is used to determine the following ___________

a) speed

b) temperature

c) volume of cylinder

d) m.e.p and I.H.P

Answer: d

Explanation: None.

9. Fuel oil consumption guarantees for I.C. engine are usually based on ___________

a) low heat value of oil

b) high heat value of oil

c) net calorific value of oil

d) calorific value of fuel

Answer: b

Explanation: None.

10. The thermal efficiency of a diesel cycle having fixed compression ratio, with increase in cut-off ratio will ___________

a) increase

b) decrease

c) be independent

d) none of the mentioned

Answer: b

Explanation: None.

This set of Engineering Chemistry Multiple Choice Questions & Answers  focuses on “Octane and Cetane Number”.

1. Which of the following compound is considered for calculating the octane number?

a) n-heptane

b) n-hexane

c) iso-octane

d) iso-butane

Answer: c

Explanation: The octane number of a fuel is numerically taken as the percentage of iso-octane in a mixture of iso-octane and n-heptane. The octane number of iso-octane is 100 and of n-heptane is 0.

2. Which of the following compound readily goes under the process of knocking?

a) 2-ethly butane

b) n-heptane

c) benzene

d) toluene

Answer: b

Explanation: n-heptane is a straight chain hydrocarbon which knocks readily and hence its octane number is zero. Presence of straight chain hydrocarbon in gasoline reduces its rate of combustion.

3. Which of the following compound is considered for calculating the cetane number?

a) α-methyl naphthalene

b) n-hexane

c) iso-octane

d) cetane molecule

Answer: d

Explanation: Cetane, C 16 H 34 , is a saturated hydrocarbon that has a very short ignition lag as compared to any commercial diesel fuel. Its cetane number is 100. Cetane number is the percentile of cetane in a mixture of cetane and α-methyl naphthalene.

4. What should be the cetane number of middle speed diesel engine?

a) At least 35

b) At least 25

c) At least 65

d) At least 45

Answer: a

Explanation: The cetane number of high speed diesel engine should be at least 45, that of middle speed diesel engine should be at least 35 and that of low speed diesel engine should be at least 25.

5. The knocking characteristic of gasoline fuel are expressed in terms of cetane number.

a) True

b) False

Answer: b

Explanation: The knocking characteristics of diesel fuel are expressed in terms of cetane number and that of gasoline fuel is expressed in terms of octane number. The molecule with the highest octane number has a least cetane number.

6. Which of the following substance is used to decrease knocking in diesel fuel?

a) Tetra ethyl lead

b) Benzene

c) Sodium hydroxide

d) Acetone peroxide

Answer: d

Explanation: The knocking in diesel fuel can be reduced by adding substances like ethyl nitrite, ethyl nitrate, isoamyl nitrate etc. This process also increases the cetane number of diesel fuel.

7. Which of the following is used to reduce the surface tension in diesel fuel?

a) Chemical additives

b) Inhibitors

c) Poly hydrocarbon

d) Dopes

Answer: a

Explanation: Chemical additives are used to reduce surface tension thus promoting the formation of finer spray. Inhibitors are used to delay or prevent gum formation. Poly hydrocarbons are used to preserve the fluid properties.

8. By which process does the knocking starts in diesel engine?

a) Due to sudden spontaneous combustion of last portion of fuel

b) Due to delay in spontaneous combustion of last portion of fuel

c) Due to the rise in temperature of diesel engine

d) Due to the parts of diesel engine

Answer: b

Explanation: Due to the presence of impurities in diesel, its spontaneous combustion of the last portion of fuel takes some time. Due to this their produces an uneven sound in the engine which is called knocking.

9. Which of the following exhaust gas produces least harmless components after its combustion?

a) Carbon monoxide

b) Ethyne

c) NO x

d) N 2

Answer: d

Explanation: Gases produced by the compression ignition engines are said to be exhaust gases. This engines operate with air-fuel ratios of 14:1 on load and 70:1 on idling.

10. Which of the following has the highest cetane number?

a) n-heptane

b) n-hexane

c) n-pentane

d) n-butane

Answer: a

Explanation: n-heptane has maximum branched atom due to which it has the highest cetane number of all. This fuel is good for diesel engines.

This set of IC Engines Multiple Choice Questions & Answers  focuses on “Dissociation”.

1. Dissociation is the disintegration of burnt gases at __________ temperatures.

a) low

b) high

c) constant

d) none of the mentioned

Answer: b

Explanation: None.

2. In lean mixtures no dissociation takes place at _____________ temperatures.

a) low

b) high

c) constant

d) none of the mentioned

Answer: a

Explanation: None.

3. In rich mixtures dissociation is prevented by the available ____________

a) CO

b) CO 2

c) O 2

d) CO and O 2

Answer: d

Explanation: None.

4. Power output is maximum at stoichiometric ratio where there is ____________

a) dissociation

b) no dissociation

c) rich mixture

d) none of the mentioned

Answer: b

Explanation: None.

5. Maximum dissociation is for ____________

a) lean mixture

b) rich mixture

c) stoichiometric

d) none of the mentioned

Answer: c

Explanation: Maximum dissociation is for stoichiometric mixtures while in lean mixtures, generally no dissociation takes place.

6. With dissociation, peak temperature is obtained ____________

a) at the stoichiometric ratio

b) when the mixture is slightly lean

c) when the mixture is slightly rich

d) none of the mentioned

Answer: c

Explanation: None.

7. With dissociation the exhaust gas pressure ____________

a) decreases

b) increases

c) no effect

d) increases upto certain air fuel ratio and then decreases

Answer: a

Explanation: None.

8. For a given compression ratio, as the mixture is made progressively rich from lean the mean effective pressure ____________

a) increases

b) decreases

c) initially increases then decreases

d) remains more or less same

Answer: c

Explanation: None.

9. When the mixture is lean ____________

a) efficiency is less

b) power output is less

c) maximum temperature and pressure are higher

d) all of the mentioned

Answer: b

Explanation: Power output is less when the mixture is lean.

10. For a compressor process with variable specific heat the peak temperature and pressure are ____________

a) lower

b) higher

c) no effect

d) none of the mentioned

Answer: a

Explanation: None.

This set of IC Engines Multiple Choice Questions & Answers  focuses on “Supercharging”.

1. Supercharging of S.I engines is employed only for aircraft and racing car engines.

a) True

b) False

Answer: a

Explanation: Supercharging requires when speed is taken into consideration so, supercharging of S.I engines is employed only for aircraft and racing car engines.

2. ____________ in supercharging pressure increases the tendency to detonate and pre-ignite.

a) Decrease

b) Increase

c) Unpredictable

d) None of the mentioned

Answer: a

Explanation: Detonation and pre-ignition are increased by an increase in supercharging pressure.

3. The supercharged petrol engines have a lower fuel consumption than naturally aspirated engines.

a) True

b) False

Answer: b

Explanation: Actually, supercharged petrol engines have a greater fuel consumption than naturally aspirated engines.

4. Increased intake pressure and temperature reduces ignition delay and decreases flame speed.

a) True

b) False

Answer: b

Explanation: In fact, increased intake pressure and temperature reduce ignition delay and increases flame speed.

5. The increased flame speeds make the petrol engine more sensitive to fuel-air ratio and engine cannot run on strong mixtures without knock.

a) True

b) False

Answer: b

Explanation: Actually, the increased flame speeds make the petrol engine more sensitive to fuel-air ratio and engine cannot run on weak mixtures without knock.

6. Rich mixtures are used to control detonation, which further increases the specific fuel consumption of the engine.

a) True

b) False

Answer: a

Explanation: None.

7. The ignition timings and thermal load on the engine affect the knock limit of CI engine.

a) True

b) False

Answer: b

Explanation: No, in fact the ignition timings and thermal load on the engine affect the knock limit of SI engine.

8. The ignition must be retarded at high intake pressure and temperature.

a) True

b) False

Answer: a

Explanation: None.

9. The power developed by the turbocharger is not sufficient to drive the compressor, and overcome its mechanical friction.

a) True

b) False

Answer: b

Explanation: In fact, the power developed by the turbocharger is sufficient to drive the compressor, and overcome its mechanical friction.

10. The loss in piston work due to the early opening of the exhaust valve is less than offset by better charging and scavenging of the engine.

a) True

b) False

Answer: b

Explanation: No, in fact, the loss in piston work due to the early opening of the exhaust valve is more than offset by better charging and scavenging of the engine.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Otto Cycle  Constant Volume Cycle”.

1. Which cycle is idealized cycle for the spark ignition internal combustion engines?

a) Otto cycle

b) Diesel cycle

c) Dual cycle

d) Bryton cycle

Answer: a

Explanation: An internal combustion engine that uses spark ignition for combustion. An air standard otto cycle is the idealized cycle for the spark ignition internal combustion engines.

2. What is the efficiency of the otto cycle?

a) 1 – \

 

 1 – r γ-1

c) r γ-1

d) 1 – \(\frac{1}{r^{γ+1}} \)

Answer: a

Explanation: Efficiency of otto cycle = 1- γ-1 .compression ratio, r=(v 1 /v 2 ).

Therefore efficiency of otto cycle = 1 – \(\frac{1}{r^{γ-1}} \).

3. An air standard otto cycle consist of ________

a) One constant pressure process and three adiabatic process

b) One constant pressure process, one constant volume process and two adiabatic process

c) Two constant volume process and two adiabatic process

d) Two constant pressure process and two adiabatic process

Answer: c

Explanation: The sequence of processes in otto cycle is isentropic compression, isochoric heat addition, isentropic expansion and isochoric heat rejection. Therefore an air standard otto cycle consist of two constant volume and two adiabatic process.

4. Otto cycle is air standard cycle of _____________

a) S.I. engine

b) C.I. engine

c) Both SI and CI engine

d) Neither SI nor CI engine

Answer: a

Explanation: An internal combustion engine that uses spark ignition for combustion is called SI engine. An air standard otto cycle is idealized for the spark ignition internal combustion engine.

5. In otto cycle heat rejection occurs at __________

a) Reversible constant volume process

b) Reversible constant pressure process

c) Irreversible constant volume process

d) Irreversible constant pressure process

Answer: a

Explanation: Otto cycle consist of two isochoric process and two adiabatic process. Heat rejection process is isochoric process.

6. Which of the following is not a use of gasoline blend like gasoline mixed with tetraethyl lead in internal combustion engine?

a) Increases the octane rating fuel

b) Allows engine to operate at high compression ratio

c) Avoid auto ignition of fuel

d) Allows engine to operate at low compression ratio

Answer: d

Explanation: Gasoline blend in IC engine increases the octane rating fuel and allows engine to operate at high compression ratio. Gasoline blend is use to avoid auto ignition in internal combustion engine.

7. How the efficiency of SI engine is affected by change in specific heat ratio of working fluid?

a) The efficiency of SI engine increases with increase in specific heat ratio of working fluid

b) The efficiency of SI engine decreases with increase in specific heat ratio of working fluid

c) The efficiency of SI engine increases with decrease in specific heat ratio of working fluid

d) The efficiency of SI engine does not affected by change in specific heat ratio of working fluid

Answer: a

Explanation: Efficiency = 1 – \

 

 

 decreases hence efficiency increases.

8. The compression ratio (r k ) of otto cycle is equal to __________

a) r k = volume of the cylinder at the beginning of the compression/volume of the cylinder at the end of the compression

b) r k = volume of the cylinder at the end of the compression/ volume of the cylinder at the beginning of the compression

c) r k = clearance volume/ volume of the cylinder at the beginning of the compression

d) r k = volume of the cylinder at the end of the compression/clearance volume

Answer: a

Explanation: r k = \(\frac{swept \, volume+clearance \, volume}{clearance \, volume} \)

r k = volume of the cylinder at the beginning of the compression/volume of the cylinder at the end of the compression.

9. What is the value of working fluid in otto cycle?

a) 1.0

b) 1.2

c) 1.4

d) 1.8

Answer: c

Explanation: In air standard otto cycle working fluid is air. The specific heat ratio of air is equal to 1.4.

10. For same compression ratio and heat supplied

a) Otto cycle is more efficient

b) Diesel cycle is more efficient

c) Dual cycle is more efficient

d) Both diesel and otto cycle are equally efficient

Answer: a

Explanation: Efficiency of otto cycle = 1 – \(\frac{1}{r^{γ-1}} \)

Efficiency of diesel cycle = 1 – \(\frac{ρ^γ-1}{r^{γ-1}}\).

11. What is the efficiency of an ideal air standard otto cycle if the compression ratio is 8.5?

a) 67.5%

b) 57.5%

c) 48.5%

d) 47.5%

Answer: b

Explanation: Efficiency of otto cycle = 1 – \(\frac{1}{r^{γ-1}} \)

Efficiency of otto cycle = 1 – \(\frac{1}{8.5^{1.4-1}} \) = 57.5%.

12. What is the efficiency of an ideal air standard otto cycle if the clearance volume is 10% of the swept volume?

a) 39.8%

b) 39.2%

c) 61.7%

d) 61.1%

Answer: c

Explanation: r = \(\frac{swept \, volume+clearance \, volume}{clearance \, volume} \)

r = 1.1/0.1=11

Efficiency of otto cycle = 1 – \(\frac{1}{r^{γ-1}} \)

Efficiency of otto cycle = 1 – \(\frac{1}{11^{1.4-1}} \)=0.6167=61.7%.

13. In an ideal otto cycle works between minimum and maximum temperature of 300K and 1800K. What is the compression ratio?

a) 4

b) 6

c) 8

d) 5

Answer: b

Explanation: Efficiency of otto cycle = 1 – \(\frac{1}{r^{γ-1}} \)

r = T max /T min = 6.

14. What is the efficiency of an ideal air standard otto cycle if the clearance volume is 20% of the swept volume?

a) 52.1%

b) 50.1%

c) 55.1%

d) 51.1%

Answer: d

Explanation: r = \(\frac{swept \, volume+clearance \, volume}{clearance \, volume} \)

r = 1.2/0.20=6

Efficiency of otto cycle = 1 – \(\frac{1}{r^{γ-1}} \)

Efficiency of otto cycle = 1 – \(\frac{1}{6^{1.4-1}} \)=0.511=51.1%.

15. What is the efficiency of an ideal air standard otto cycle if the compression ratio is 7?

a) 51.5%

b) 54%

c) 52.1%

d) 50%

Answer: b

Explanation: Efficiency of otto cycle = 1 – \(\frac{1}{r^{γ-1}} \)

Efficiency of otto cycle = 1 – \(\frac{1}{7^{1.4-1}} \)=54%.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Diesel Cycle  Constant Pressure Cycle”.

1. Which cycle is idealized cycle for the compression ignition engines?

a) Otto cycle

b) Diesel cycle

c) Dual cycle

d) Bryton cycle

Answer: b

Explanation: Compression ignition engine that uses fuel injection to compressed air for combustion. An air standard diesel cycle is the idealized cycle for the compression ignition engines.

2. In diesel cycle heat rejection occurs at ___________

a) Reversible constant volume process

b) Reversible constant pressure process

c) Irreversible constant volume process

d) Irreversible constant pressure process

Answer: a

Explanation: Diesel cycle consist of one isobaric process, one isochoric process and two adiabatic process. Heat rejection process is isochoric process.

3. In diesel cycle heat addition occurs at ___________

a) Reversible constant volume process

b) Reversible constant pressure process

c) Irreversible constant volume process

d) Irreversible constant pressure process

Answer: b

Explanation: Diesel cycle consist of one isobaric process, one isochoric process and two adiabatic process. Heat rejection process is isobaric process.

4. In diesel engine fuel is ignited by ___________

a) Fuel injection

b) Spark

c) Heat resulting from compressing air

d) Heat resulting from compression of air fuel mixture

Answer: a

Explanation: Compression ignition engine that uses fuel injection to compressed air for combustion. An air standard diesel cycle is the idealized cycle for the compression ignition engines.

5. The thermal efficiency of diesel cycle having fixed compression ratio, with increase in cut-off ratio.

a) Increases

b) Decreases

c) Be independent

d) Data insufficient

Answer: b

Explanation: Efficiency of diesel cycle = 1 – \(\frac{ρ^γ-1}{r^{γ-1}} \)

Increase in cut-off ratio will increase in term \(\frac{ρ^γ-1}{r^{γ-1}} \) therefore decrease in efficiency of diesel cycle.

6. The thermal efficiency of diesel cycle having fixed cut-off ratio, with increase in compression ratio.

a) Increases

b) Decreases

c) Be independent

d) Data insufficient

Answer: a

Explanation: Efficiency of diesel cycle = 1 – \(\frac{ρ^γ-1}{r^{γ-1}} \)

Increase in compression ratio will decrease in term \(\frac{ρ^γ-1}{r^{γ-1}} \) therefore increase in efficiency of diesel cycle.

7. An air standard diesel cycle consist of ___________

a) One constant pressure process and three adiabatic process

b) One constant pressure process, one constant volume process and two adiabatic process

c) Two constant volume process and two adiabatic process

d) Two constant pressure process and two adiabatic process

Answer: b

Explanation: The sequence of processes in diesel cycle is isentropic compression, isobaric heat addition, isentropic expansion and isochoric heat rejection. Therefore an air standard diesel cycle consist of one constant volume, one constant pressure and two adiabatic process.

8. The compression ratio (r k ) of diesel cycle is equal to ___________

a) r k = volume of the cylinder at the beginning of the compression/volume of the cylinder at the end of the compression

b) r k = volume of the cylinder at the end of the compression/ volume of the cylinder at the beginning of the compression

c) r k = clearance volume/ volume of the cylinder at the beginning of the compression

d) r k = volume of the cylinder at the end of the compression/clearance volume

Answer: a

Explanation: r k = \(\frac{swept \, volume+clearance \, volume}{clearance \, volume} \)

r k = volume of the cylinder at the beginning of the compression/volume of the cylinder at the end of the compression.

9. For same peak pressure and heat input _________

a) Otto cycle is more efficient

b) Diesel cycle is more efficient

c) Dual cycle is more efficient

d) Both diesel and otto cycle are equally efficient

Answer: b

Explanation: Efficiency of otto cycle = 1 – \(\frac{1}{r^{γ-1}}\)

Efficiency of diesel cycle = 1 – \(\frac{ρ^γ-1}{r^{γ-1}} \).

10. What is the value of working fluid in diesel cycle?

a) 1.0

b) 1.2

c) 1.4

d) 1.8

Answer: c

Explanation: In air standard diesel cycle working fluid is air. The specific heat ratio of air is equal to 1.4.

11. What is the efficiency of diesel cycle if γ and T denotes the specific heat ratio and temperature respectively?

a) 1 – \

 

 1 – \

 

 1 – \

 

 1 – \(\frac{T4-T1}{} \)

Answer: b

Explanation: Q addition = m cp 

Q rejection = m cv 

Efficiency of diesel engine = \(\frac{Q1-Q2}{Q1} \)

Efficiency of diesel engine = 1 – \(\frac{T4-T1}{γ} \)

12. Scavenging in diesel cycle means___________

a) Air used for combustion under pressure

b) Forced air for cooling the cylinder

c) Burnt air containing products of combustion

d) Air used for forcing burnt gases out of engines cylinder during exhaust stroke

Answer: d

Explanation: Scavenging is defined as forcing burnt gases out of engines cylinder during exhaust stroke. Therefore scavenging in diesel cycle means air used for forcing burnt gases out of engines cylinder during exhaust strokes.

13. A diesel engine is usually more efficient than a spark ignition engine because ___________

a) Diesel being a heavier hydrocarbon, release more heat per kg than gasoline

b) The air standard efficiency of diesel engine is higher than the otto cycle, at a fixed compression ratio

c) The compression ratio of diesel engine is higher than petrol engine

d) Self ignition temperature of diesel engine is higher than that of gasoline

Answer: c

Explanation: Diesel engine actually have a higher peak temperature than petrol engines. A diesel engine is usually more efficient than a spark ignition engine because the compression ratio of diesel engine is higher than petrol engine.

14. The combustion in compression ignition engine is __________

a) Homogenous

b) Heterogeneous

c) Laminar

d) Turbulent

Answer: b

Explanation: In compression ignition engine fuel injection system is used. Therefore fuel is injected to compressed air and hence air fuel mixture is heterogeneous.

15. Which of the following is compressed in diesel engine?

a) Air

b) Air and fuel

c) Air and lubricating oil

d) Fuel

Answer: a

Explanation: In compression ignition engine fuel injection system is used. Therefore fuel is injected to compressed air and hence air is only compressed in diesel engine.

This set of Thermal Engineering written test Questions & Answers focuses on “Dual Combustion Cycles”.

1. What is the other name of dual cycle?

a) Mixed cycle

b) Otto cycle

c) Diesel cycle

d) Joule Cycle

Answer: a

Explanation: Dual cycle is combination of otto cycle and diesel cycle. Dual cycle is also known as mixed cycle.

2. Dual cycle is combination of__________

a) Stirling and diesel cycle

b) Otto and diesel cycle

c) Stirling and otto cycle

d) Otto and Brayton cycle

Answer: b

Explanation: Dual cycle is combination of otto cycle and diesel cycle. Dual cycle is also known as mixed cycle.

3. What is the efficiency of dual cycle?

a) 1-\

 

 1-\

 

 

 1-\

 

 

 1-\(\frac{1}{r^{γ-1}}\frac{

}{

+γr_p } \)

Answer: d

Explanation: Efficiency of dual cycle = 1-\(\frac{1}{r^{γ-1}}\frac{

}{

+γr_p } \)

Efficiency of otto cycle = 1 – \(\frac{1}{r^{γ-1}} \)

Efficiency of diesel cycle = 1 – \(\frac{ρ^γ-1}{r^{γ-1}} \)

4. Dual cycle consist of __________

a) Two constant pressure and two adiabatic process

b) Two constant volume and two adiabatic process

c) One constant pressure process, two constant volume process and two adiabatic process

d) One constant pressure process, one constant volume process and two adiabatic process

Answer: c

Explanation: The sequence of processes in dual cycle is isentropic compression, isochoric heat addition, isobaric heat addition, isentropic expansion and isochoric heat rejection. Therefore a dual cycle consist of two constant volume. One constant pressure and two adiabatic process.

5. In a dual cycle heat rejection takes place ______

a) At constant volume process

b) At constant pressure process

c) First at constant volume then at constant pressure process

d) First at constant pressure then at constant volume process

Answer: a

Explanation: The sequence of processes in dual cycle is isentropic compression, isochoric heat addition, isobaric heat addition, isentropic expansion and isochoric heat rejection. Therefore, heat rejection takes place at constant volume process.

6. In a dual cycle heat addition takes place ________

a) At constant volume process

b) At constant pressure process

c) First at constant volume then at constant pressure process

d) First at constant pressure then at constant volume process

Answer: c

Explanation: The sequence of processes in dual cycle is isentropic compression, isochoric heat addition, isobaric heat addition, isentropic expansion and isochoric heat rejection. Therefore, heat addition takes place first at constant volume then at constant pressure process.

7. Most high speed compression ignition operates on ________

a) Diesel cycle

b) Otto cycle

c) Brayton cycle

d) Dual cycle

Answer: d

Explanation: Compression ratio of dual cycle is 12-16, also dual cycle is combination of diesel cycle and otto cycle. Therefore, most high speed compression ignition operates on dual cycle.

8. What is the efficiency of dual cycle if compression ratio and cut-off ratio is equal to 15 and 1.4 respectively?

a) 91%

b) 90.8%

c) 90.6%

d) 90%

Answer: d

Explanation: Efficiency of dual cycle=1-\(\frac{1}{r^{γ-1}}\frac{

}{

+γr_p } \)

Efficiency of dual cycle= 1-\(\frac{1}{15^{1.4-1}}\frac{

}{+1.4×10.71} \)

Efficiency of dual cycle = 90.6%

9. What is the value of working fluid in dual cycle?

a) 1.0

b) 1.2

c) 1.4

d) 1.8

Answer: c

Explanation: In air standard dual cycle working fluid is air. The specific heat ratio of air is equal to 1.4.

10. The compression ratio  of dual cycle is equal to ________

a) r = volume of the cylinder at the beginning of the compression/volume of the cylinder at the end of the compression

b) r = volume of the cylinder at the end of the compression/ volume of the cylinder at the beginning of the compression

c) r = clearance volume/ volume of the cylinder at the beginning of the compression

d) r = volume of the cylinder at the end of the compression/clearance volume

Answer: a

Explanation: compression ratio of dual cycle =\(\frac{swept \, volume+clearance \, volume}{clearance \, volume} \)

r=volume of the cylinder at the beginning of the compression/volume of the cylinder at the end of the compression

11. The cut-off ratio  of dual cycle is equal to ________

a) Volume at which fuel supply cutoff to volume at which fuel supply start

b) Volume at which fuel supply throughout the process to volume at which fuel supply cutoff

c) Volume at which fuel supply cutoff to volume of fuel supply throughout the process

d) Volume at which fuel supply start to volume at which fuel supply cutoff

Answer: a

Explanation: ρ = \(\frac{Volume \, at \, which \, fuel \, supply \, cutoff}{Volume \, at \, which \,fuel \,supply \,start} \)

Cutoff ratio of dual cycle is equal to Volume at which fuel supply cutoff to volume at which fuel supply start.

12. What is the efficiency of dual cycle if compression ratio and cut-off ratio is equal to 13 and 1.4 respectively?

a) 68%

b) 63%

c) 60%

d) 65%

Answer: b

Explanation: Efficiency of dual cycle=1-\(\frac{1}{r^{γ-1}}\frac{

}{

+γr_p } \)

Efficiency of dual cycle= 1-\(\frac{1}{13^{1.4-1}}\frac{

}{+1.4×9.28} \)

Efficiency of dual cycle = 63%

13. What is the relation between compression ratio, cutoff ratio and expansion ratio?

a) r e = ρ×r

b) r e = ρ/r

c) r e = 1/ρr

d) r e = r/ρ

Answer: d

Explanation: r e = \(\frac{V5}{V4} = \frac{V1}{V4} \)

r e = \(\frac{V1}{V3}×\frac{V3}{V4} \)

r e = \(\frac{V1}{V2}×\frac{V3}{V4} \)

r e = r×

r e = r/ρ

14. Determine the value of expansion ratio if compression ratio and cutoff ratio is equal to 15 and 1.4 respectively.

a) 10.5

b) 21

c) 11

d) 10.7

Answer: d

Explanation: r e = r/ρ

r e = 15/1.4

r e = 10.7

15. Determine the value of cutoff ratio if compression ratio and expansion ratio is equal to 15 and 11 respectively.

a) 1.4

b) 1.36

c) 1.5

d) 1.3

Answer: b

Explanation: r e = r/ρ

ρ = 15/11

ρ =1.36

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Brayton Cycle”.

1. Gas turbine works on _____

a) Dual cycle

b) Otto cycle

c) Brayton cycle

d) Diesel cycle

Answer: c

Explanation: Brayton cycle is used in gas power plant. It is used in open system that is gas turbine.

2. The compression ratio  of dual cycle is equal to _________

a) r = volume of the cylinder at the beginning of the compression/volume of the cylinder at the end of the compression

b) r = volume of the cylinder at the end of the compression/ volume of the cylinder at the beginning of the compression

c) r = clearance volume/ volume of the cylinder at the beginning of the compression

d) r = volume of the cylinder at the end of the compression/clearance volume

Answer: a

Explanation: compression ratio of dual cycle =\(\frac{swept \, volume+clearance \, volume}{clearance \, volume} \)

r=volume of the cylinder at the beginning of the compression/volume of the cylinder at the end of the compression

3. For same maximum pressure and heat input, most efficient cycle is _________

a) Dual cycle

b) Rankine cycle

c) Brayton cycle

d) Diesel cycle

Answer: c

Explanation: Efficiency of brayton cycle is directly proportional to pressure ratio. Therefore for maximum pressure and heat input brayton cycle is most efficient.

4. Which of the following methods cannot be used to increase the efficiency of brayton cycle?

a) Decreasing pressure ratio

b) Increasing pressure ratio

c) Regeneration

d) Heat exchanger

Answer: a

Explanation: Heat exchanger and regeneration is used to increase the efficiency of brayton cycle. Also, efficiency of brayton cycle is directly proportional to pressure ratio.

5. In a brayton cycle heat addition takes place __________

a) At constant volume process

b) At constant pressure process

c) First at constant volume then at constant pressure process

d) First at constant pressure then at constant volume process

Answer: b

Explanation: The sequence of processes in brayton cycle is isentropic compression, isobaric heat addition, isentropic expansion and isobaric heat rejection. Therefore heat addition takes place at constant pressure process

6. In a brayton cycle heat rejection takes place ____

a) At constant volume process

b) At constant pressure process

c) First at constant volume then at constant pressure process

d) First at constant pressure then at constant volume process

Answer: b

Explanation: The sequence of processes in brayton cycle is isentropic compression, isobaric heat addition, isentropic expansion and isobaric heat rejection. Therefore heat rejection takes place at constant pressure process.

7. In brayton cycle regenerator is placed between __________

a) Turbine and heat exchanger

b) Turbine and intercooler

c) Compressor and intercooler

d) Compressor and heat exchanger

Answer: d

Explanation: In regenerative brayton cycle a regenerator is placed between compressor and heat exchanger. The exhaust from turbine before entering in heat exchanger and then compressed in compressor.

8. Reheating will _______ the efficiency of brayton cycle.

a) Decreases

b) Increases

c) Zero

d) Remain same

Answer: a

Explanation: Reheating increases the net work output. Reheating necessarily decreases the efficiency.

9. If regeneration is also used along with reheating the efficiency will _____

a) Decreases

b) Increases

c) Zero

d) Remain same

Answer: b

Explanation: Regeneration in the brayton cycle increases efficiency by recovery of waste heat. Hence if regeneration is also used along with reheating the efficiency will increase.

10. What is the effect of regeneration in brayton cycle work output?

a) Work output increases

b) Work output decreases

c) Work output becomes zero

d) Work output doesn’t change

Answer: d

Explanation: Regeneration in the brayton cycle increases efficiency by recovery of waste heat. It doesn’t change the net work output.

11. In how many stages the compression process is divided in intercooling?

a) One

b) Two

c) Three

d) Five

Answer: b

Explanation: In intercooling the compression process is divided into two stages. Air is compressed in first stage is cooled then further compressed in second stage.

12. Reheat and regeneration in combination used to increases ________

a) Net work output

b) Efficiency

c) Both net work output and efficiency

d) Doesn’t change net work output and efficiency

Answer: c

Explanation: Reheating increases the net work output. Reheating necessarily decreases the efficiency. Regeneration in the brayton cycle increases efficiency by recovery of waste heat. It doesn’t change the net work output.

13. Intercooling and regeneration in combination used to increases _________

a) Net work output

b) Efficiency

c) Doesn’t change net work output and efficiency

d) Both net work output and efficiency

Answer: d

Explanation: Intercooling mainly increases the net work output by decreasing compression work. Reheating necessarily decreases the efficiency. Regeneration in the brayton cycle increases efficiency by recovery of waste heat. It doesn’t change the net work output.

This set of Thermodynamics Multiple Choice Questions & Answers  focuses on “Rotary Compressor”.

1. Rotary compressors are used where ____ quantities of gas are needed at relatively ____ pressure.

a) large, high

b) large, low

c) small, high

d) small, low

Answer: b

Explanation: This is where rotary compressors are used.

2. Rotary compressor can be classified as

a) displacement compressor

b) steady-flow compressor

c) both of the mentioned

d) none of the mentioned

Answer: c

Explanation: These are the two types of rotary compressor.

3. In steady-flow compressor, compression occurs by

a) transfer of kinetic energy

b) transfer of potential energy

c) trapping air

d) all of the mentioned

Answer: a

Explanation: The transfer of kinetic energy occurs from a rotor.

4. In displacement compressor, compression occurs by

a) transfer of kinetic energy

b) transfer of potential energy

c) trapping air

d) all of the mentioned

Answer: c

Explanation: Here air is compressed by trapping it in reducing space.

5. The rotary positive displacement machines are ____ and compression is ____

a) cooled, isothermal

b) uncooled, isothermal

c) cooled, adiabatic

d) uncooled, adiabatic

Answer: d

Explanation: These are uncooled and adiabatic compression takes place.

6. The Roots blower and vane-type compressor are the types of

a) displacement compressor

b) steady-flow compressor

c) both of the mentioned

d) none of the mentioned

Answer: a

Explanation: These are the two types of rotary positive displacement machines.

7. For a Root blower, as pressure ratio increases, efficiency ____

a) increases

b) decreases

c) remains constant

d) none of the mentioned

Answer: b

Explanation: This can be seen by taking pressure ratios and calculating efficiencies for them.

8. The vane type compressor requires ____ the Roots blower.

a) equal work input

b) more work input

c) less work input

d) none of the mentioned

Answer: c

Explanation: This is true for given air flow and pressure ratio.

9. The centrifugal and axial flow compressor are the types of

a) displacement compressor

b) steady-flow compressor

c) both of the mentioned

d) none of the mentioned

Answer: b

Explanation: These are the two types of steady-flow compressors.

10. Which of the following is true for a centrifugal compressor?

a) rotation of impeller compresses the air

b) diffuser converts part of KE into internal energy

c) typical pressure ratio is around 1.4 to 1

d) all of the mentioned

Answer: d

Explanation: This is the working of a centrifugal compressor.

11. Which of the following is true for an axial-flow compressor?

a) blades are arranged in same manner as in reaction turbine

b) flow of air is along the axis of compressor

c) velocity of air changes when it passes through the blades

d) all of the mentioned

Answer: d

Explanation: This is the working of an axial-flow compressor.

12. For uncooled rotary compressor, compression process is ____ while ideal process is ____

a) isothermal, adiabatic

b) isentropic, adiabatic

c) adiabatic, isentropic

d) adiabatic, isothermal

Answer: c

Explanation: The isentropic process is reversible and adiabatic.

Answer: a

Explanation: This increases the temperature and enthalpy of gas.

This set of Steam and Gas Turbines Multiple Choice Questions & Answers  focuses on “Centrifugal Compressor”.

1. Centrifugal compressors are also known as ____________

a) turbo compressors

b) radial compressors

c) turbo & radial compressors

d) none of the mentioned

Answer: b

Explanation: Centrifugal compressors sometimes called as radial compressors.

2. In idealized turbo machinery, pressure rise is done by adding _______

a) Temperature

b) Mass

c) Kinetic energy

d) None of the mentioned

Answer: c

Explanation: In idealized turbo machinery the pressure rise is done by adding kinetic energy.

3. The pressure rise in impeller is _________ the rise in diffuser.

a) equal to

b) greater than

c) equal to or greater than

d) none of the mentioned

Answer: d

Explanation: Not in all cases pressure rise in diffuser equals to pressure rise in impeller in some cases pressure rise in impeller will be more than that of diffuser.

4. In centrifugal compressor velocity of flow leaving the impeller is equals to _______ in many cases.

a) Speed of sound

b) Double the speed of sound

c) Triple the speed of sound

d) None of the mentioned

Answer: a

Explanation: Initially the flow enters straightly into the impeller and the impeller pushes the flow inwards to spin faster and finally when the flow is leaving impeller it has almost speed of sound.

5. ________ are used as stationary compressors in steam turbines.

a) Moving vanes

b) Guide vanes

c) Moving & Guide vanes

d) None of the mentioned

Answer: b

Explanation: Guide vanes are used as stationary compressors where guide vanes reduce the velocity and according to Bernoulli’s theorem when velocity is reduced its pressure is increased.

6. ________________ has large change in inlet to exit radius when compared to centrifugal compressors.

a) Axial compressors

b) Axial & Centrifugal compressors

c) Centrifugal compressors

d) None of the mentioned

Answer: c

Explanation: Centrifugal compressors has large change in inlet to exit temperature when compared to axial compressor.

7. The first part in both centrifugal compressor and axial compressor are similar.

a) True

b) False

Answer: a

Explanation: The first in both centrifugal compressor and axial compressor are similar.

8. Centrifugal compressor __________ the energy by increasing the radius where as centrifugal fan decreases the energy.

a) increases

b) decreases

c) both follow the same

d) none of the mentioned

Answer: b

Explanation: Both centrifugal compressor and centrifugal fan are similar as both increases the energy by reducing the radius.

9. Centrifugal compressors increases the density of fluid more than ________

a) 1%

b) 3%

c) 5%

d) 7%

Answer: c

Explanation: Centrifugal compressors increases the density of fluid more than 5 %.

10. Centrifugal fan has relative fluid velocities ______

a) Mach number more than 0.3

b) Mach number less than 0.1

c) Mach number less than 0.3

d) Mach number more than 0.5

Answer: c

Explanation: Centrifugal fans often operate at fluid velocities less than mach number 0.3.

This set of Refrigeration Multiple Choice Questions & Answers  focuses on “Mechanism of Simple VCR System”.

1. A vapor compression refrigeration system is an improved type of air refrigeration system in which a suitable working substance, termed as refrigerant is used.

a) True

b) False

Answer: a

Explanation: A vapor compression refrigeration system is an improved type of air refrigeration system in which a suitable working substance, termed as refrigerant is used. Refrigerants such as ammonia, CO 2 and Sulphur dioxide etc. are used as refrigerants in a vapor compression refrigeration system.

2. The vapor compression refrigeration system is similar to a ________

a) latent heat pump

b) latent heat engine

c) generator

d) evaporator

Answer: a

Explanation: The vapor compression refrigeration system is a latent heat pump, as it pumps its latent heat from the brine and delivers it to the cooler.

3. The first vapor compression refrigeration system was developed by ________

a) L. G. Hamilton

b) Jacob Perkins

c) Stephen Rossi

d) Lenard Sandwik

Answer: b

Explanation: The first vapor compression refrigeration system was developed in 1834 by Jacob Perkins using hand operations.

4. The COP of vapor compression refrigeration compared to simple air refrigeration system is ________

a) high

b) low

c) same

d) can’t say

Answer: a

Explanation: COP of vapor compression refrigeration compared to simple air refrigeration system is high due to the use of refrigerant and as it can be employed over large range of temperatures.

5. Which of the following is not an advantage of vapor compression refrigeration system over air refrigeration system?

a) Smaller size for a given capacity of refrigeration

b) It has less running cost

c) It can be employed over wide range of temperature

d) There is no problem of leakage in vapor compression refrigeration system

Answer: d

Explanation: In vapor compression refrigeration system the prevention of leakage of the refrigerant is a major problem. The advantages of it are smaller size for a given capacity of refrigeration, it has less running cost and it can be employed over wide range of temperature.

6. The low pressure and temperature vapor refrigerant enters the ______ of the vapor compression system.

a) compressor

b) condenser

c) receiver

d) evaporator

Answer: a

Explanation: In the compressor the low pressure and temperature vapor refrigerant from the evaporator is drawn through the suction valve, where it is compressed to high pressure and temperature.

7. The high pressure and temperature vapor refrigerant enters the ______ of the vapor compression system.

a) compressor

b) condenser

c) receiver

d) evaporator

Answer: b

Explanation: In the condenser the high pressure and temperature vapor refrigerant from the compressor is condensed, during which the refrigerant gives up its latent heat.

8. In any compression refrigeration system there are how many pressure conditions?

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: In any compression refrigeration system there are 2 pressure conditions, namely high pressure and low pressure side.

9. Which of the following is not an advantage of vapor compression cycle over reversed Carnot cycle?

a) It’ COP and refrigeration effect can’t be increased

b) Use of expansion valve reduces the size and cost of plant

c) It is a practical cycle on which plant can be built

d) Wet compression of Carnot cycle is avoided

Answer: a

Explanation: It’ COP and refrigeration effect can be increased by use of superheated vapor at entry of compressor. The advantages are use of expansion valve reduces the size and cost of plant, it is a practical cycle on which plant can be built and wet compression of Carnot cycle is avoided.

10. The pipe line emanating from compressor up to the condenser is called ________

a) suction line

b) pipe line

c) liquid line

d) delivery line

Answer: d

Explanation: The pipe line emanating from compressor up to the condenser is called delivery line or discharge line or hot gas line. Whereas suction line is the low pressure vapor line from the exit end evaporator leading to the compressor suction side and liquid line carries the liquid from receiver to the control valve.

11. The coefficient of performance of a vapour compression refrigeration system is quite high as compared to the air refrigeration system.

a) True

b) False

Answer: a

Explanation: VCRS uses a refrigerant having specific properties. The most important properties include heat absorbing and rejecting capacities. Due to the occurrence of phase change of refrigerant in the Vapor compression refrigeration system and high-temperature range applicability, it has higher COP. VCRS used in applications like refrigerators for domestic and industrial applications.

12. In a refrigeration cycle, in which of the following heat absorption takes place?

a) Evaporator

b) Condenser

c) Expansion valve

d) Compressor

Answer: a

Explanation: Evaporator helps to raise the temperature of the refrigerant and convert that liquid form into gaseous form. For that, it absorbs heat energy from the items placed in the system. Utilizing that heat to make the phase change.

13. After which process during the VCR cycle, the highest temperature is achieved?

a) Evaporation

b) Expansion

c) Condensation

d) Compression

Answer: d

Explanation: The compression process is used to increase the pressure. So, according to Gay-Lussac’s law,

P ∝ T

As the highest pressure obtained after compression, the temperature obtained is also the highest.

14. The expansion valve in a refrigerator does not control the flow of refrigerant.

a) False

b) True

Answer: a

Explanation: Expansion valve acts as a diffuser, where pressure reduction is carried out using an orifice. By controlling the flow of refrigerant flowing through an orifice, the desired pressure drops obtained.

15. In a domestic vapour compression refrigerator, which of the following refrigerants commonly used?

a) CO 2

b) NH 3

c) Freon

d) Air

Answer: c

Explanation: FREON is a trading name for several gases called refrigerants. Commonly used refrigerants are hydrofluorocarbons and hydrocarbons. Though ammonia and dichlorodifluoromethane have the same boiling point, it is not toxic like ammonia. Hence, it is safe to use Freon than ammonia.

This set of Refrigeration Multiple Choice Questions & Answers  focuses on “Ideal VAR System”.

1. Which of the following is correct about VARS and VCRS?

a) VARS use mechanical energy, and VCRS use heat energy

b) VARS use heat energy, and VCRS use mechanical energy

c) Both use mechanical energy

d) Both use heat energy

Answer: b

Explanation: VCRS use mechanical energy i.e., VCRS uses compressor which withdraws energy from the evaporator and drawn into condenser after compression. VARS use heat energy i.e., heat exchanger or heat generators are used and by which desired effect is achieved.

2. The compressor from VCRS is replaced by which of the following in the VARS?

a) Absorber, Pump

b) Generator, Pressure reducing valve

c) Absorber, Pump, Generator, and Pressure reducing valve

d) Absorber, Rectifier, Generator, and Pressure reducing valve

Answer: c

Explanation: Compressor from the VCRS is replaced by an absorber, a pump, a generator, and a pressure reducing valve. These all components together do the same work as a compressor but by using the heat energy.

3. What is the purpose of using an absorber?

refrigeration-questions-answers-ideal-var-system-q3

a) Heat absorption

b) Heat rejection

c) Pressure reduction

d) Work done

Answer: b

Explanation: Vapour refrigerant from the evaporator enters the absorber. Cooling water is circulated to transfer the heat, and refrigerant’s strong solution is discharged to the generator.

4. What is the purpose of using a pump?

refrigeration-questions-answers-ideal-var-system-q3

a) Heat absorption

b) Heat rejection

c) Pressure reduction

d) Pressure increment

Answer: d

Explanation: The strong solution discharging from the absorber is pumped to the generator using a liquid pump. The liquid pump increases the pressure up to 10 bar.

5. What is the purpose of using a generator?

refrigeration-questions-answers-ideal-var-system-q3

a) Heat supplied

b) Heat rejection

c) Pressure reduction

d) Pressure increment

Answer: a

Explanation: Generator generates heat. From the generator, heat is supplied to the refrigerant, and the strong refrigerant is heated further by using some external source like gas or steam.

6. What is the purpose of using a pressure reducing valve?

refrigeration-questions-answers-ideal-var-system-q3

a) Heat supplied

b) Heat rejection

c) Pressure reduction

d) Pressure increment

Answer: c

Explanation: As the name suggests, pressure reducing valve reduces the pressure of a weak solution of refrigerant, and a weak solution is discharged into an absorber to make again it a strong solution.

7. What is the purpose of using an analyzer?

refrigeration-questions-answers-ideal-var-system-q7

a) Heat supplied

b) Removal of unwanted water particles

c) Pressure reduction

d) Removal of unwanted ammonia particles

Answer: b

Explanation: Analyzer is used to remove unwanted particles from entering the condenser. If the unwanted water particles enter the condenser, then these particles will enter the expansion valve too. These particles will freeze if passed through an expansion valve and will choke the pipeline.

8. What is the purpose of using a rectifier?

refrigeration-questions-answers-ideal-var-system-q7

a) Removal of all water particles by cooling

b) Heat supplied

c) Pressure reduction

d) Removal of unwanted ammonia particles

Answer: a

Explanation: Water vapors are not entirely removed in an analyzer, to condensate pure ammonia refrigerant, rectifier or dehydrator is used. Its function is to cool refrigerant coming out of an analyzer and condensate from rectifier returns to the analyzer by using the drip pipe.

9. What is the purpose of using a heat exchanger between the pump and the generator?

refrigeration-questions-answers-ideal-var-system-q7

a) Cool the strong hot solution

b) Heat the strong hot solution

c) Heat the weak hot solution

d) Cool the weak hot solution

Answer: d

Explanation: Heat exchanger between the pump and the generator is used to cool the weak hot solution of ammonia and water. The heat removed from the weak solution increases the temperature of the strong solution discharging the pump and entering the generator. Use of heat exchanger increases the economy of the plant.

10. What is the purpose of using a heat exchanger between condenser and evaporator?

refrigeration-questions-answers-ideal-var-system-q7

a) Heat the water particles

b) Heat the strong hot solution

c) Heat the weak hot solution

d) Sub-cooling of the refrigerant

Answer: d

Explanation: Heat exchanger between the condenser and the evaporator is used to sub-cool the refrigerant. Heat exchanger acts as a liquid sub-cooler. Refrigerant discharging from the condenser is sub-cooled by the low-temperature ammonia vapour in the evaporator.

This set of Refrigeration Multiple Choice Questions & Answers  focuses on “Vapour Refrigeration Cycles – Heat Pump”.

1. For a standard system with temperatures T 1 and T 2 , where T 1 < T a < T 2 (T a – Atmospheric Temperature). Q 1 is the heat extracted from a body at temperature T 1 , and Q 2 is heat delivered to the body at temperature T 2 . What is the C.O.P. of the heat pump for given conditions?

a) Q 2 / (Q 2 − Q 1 )

b) (Q 2 − Q 1 ) / Q 1

c) (Q 2 − Q 1 ) / Q 2

d) Q 1 / (Q 2 − Q 1 )

Answer: a

Explanation: As, C.O.P. = Desired effect / Work done

Here, work-done = Q 2 − Q 1

The desired temperature is T 2 . So, the heat delivered to achieve the desired temperature is Q 2 .

C.O.P. of the heat pump = Q 2 / (Q 2 − Q 1 ).

2. What is the difference between Heat Pump and Refrigerator?

a) Heat Pump Gives efficiency and refrigerator gives C.O.P.

b) Both are similar

c) Both are almost similar, just the desired effect is different

d) Work is output in refrigerator and work is input in heat pump

Answer: c

Explanation: Heat Pump and Refrigerator work on the same principle. Work needs to be given to get the desired effect. The characteristic which differentiates both of them is the temperature of the desired effect, heat pump desires for higher temperature whereas Refrigerator desires for lower temperature than atmospheric temperature.

3. What is the equation between efficiency of Heat engine and C.O.P. of heat pump?

a) η E =  P

b) η E = 1 /  P

c) η E /  P = 1

d) η E x  P = 0

Answer: b

Explanation: η E = W / Q hence for Carnot engine it is equal to (T 2 – T 1 ) / T 2 .

 P for Carnot cycle is equal to T 2 / (T 2 – T 1 ) .

So, these terms are related reciprocally.

4. How is the Relative coefficient of performance represented?

a) Theoretical C.O.P. / Actual C.O.P.

b) Actual C.O.P. / Theoretical C.O.P.

c) Theoretical C.O.P. x Actual C.O.P.

d) 1 / Theoretical C.O.P. x Actual C.O.P.

Answer: b

Explanation: Relative C.O.P. is the ratio of an actual to the theoretical coefficient of performance. It is used to show the deviation of C.O.P. due to the ideal state and real state conditions.

5. C.O.P. of the heat pump is always _____

a) one

b) less than One

c) greater than One

d) zero

Answer: c

Explanation: The second law of Thermodynamics states that a 100% conversion of heat into work is not possible without ideal conditions. So, efficiency will be less than 1. As C.O.P. is the reciprocal of efficiency, it tends to be more than 1.

6. For the systems working on reversed Carnot cycle, what is the relation between C.O.P. of Refrigerator i.e.  R and Heat Pump i.e.  P ?

a)  R +  P = 1

b)  R =  P

c)  R =  P – 1

d)  R +  P + 1 = 0

Answer: c

Explanation: If we put the values of C.O.P. for standard system i.e.  R = T 1 / (T 2 − T 1 ) and

 P = T 2 / (T 2 − T 1 ),

 P −  R = 1.

{T 2 / (T 2 − T 1 )} − {T 1 / (T 2 − T 1 )} = 1.

7. If the reversed Carnot cycle operating as a heat pump between temperature limits of 364 K and 294 K, then what is the value of C.O.P?

a) 4.2

b) 0.19

c) 5.2

d) 0.23

Answer: c

Explanation: C.O.P. of reversed Carnot cycle is given by,

C.O.P. = T 1 / (T 2 – T 1 )

= 364 / 

= 5.2.

8. A reversed Carnot cycle is operating between temperature limits of  33°C and  27°C. If it acts as a heat engine gives an efficiency of 20%. What is the value of C.O.P. of a heat pump operating under the same conditions?

a) 6.5

b) 8

c) 5

d) 2.5

Answer: c

Explanation: Temperature limits are given in the question so, we can calculate C.O.P. using the formula

C.O.P. = T 1 / (T 2 – T 1 )

But as the efficiency of the heat engine is given so directly by the relation, we can find out the C.O.P.

C.O.P. = 1 / η E

= 1 /  = 5.

9. If the coefficient of performance of the refrigerator is 4.67, then what is the value of the coefficient of performance of the heat pump operating under the same conditions?

a) 3.67

b) 5.67

c) 0.214

d) 9.34

Answer: b

Explanation: As we know, the equation between the coefficient of performance of the Refrigerator and heat pump:

 R =  P – 1

Hence, C.O.P. of heat pump = C.O.P. of Refrigerator + 1

= 4.67 + 1

= 5.67.

10. A heat pump is used to maintain a hall at 30°C when the atmospheric temperature is 15°C. The heat loss from the hall is 1200 kJ/min. Calculate the power required to run the heat pump if its C.O.P. is 40% of the Carnot machine working between the same temperature limits.

a) 0.495

b) 4.04

c) 0.247

d) 8.08

Answer: c

Explanation: Given data: T 1 = 30°C = 30 + 273 = 303 K

T 2 = 15°C = 15 + 273 = 288 K

Q 1 = 1200 kJ/min = 1200/60 = 2 kW

Calculations: C.O.P. of heat pump working on Carnot cycle,

Ideal C.O.P. = T 1 / (T 1 − T 2 )

= 303 / 

= 20.2

Actual C.O.P = 0.4 x Ideal C.O.P.

= 0.4 x 20.2 = 8.08

C.O.P. = Q 1 / W

Hence, W = Q 1 / C.O.P.

W = 2 / 8.08

W = 0.247 kW.

11. A heat pump which runs  rd of time removes on an average 2400 kJ/hr of heat. If power consumed is 0.25 kW, what is the value of the C.O.P.?

a) 4

b) 2

c) 8

d) 6

Answer: c

Explanation: Q 1 = 2400 kJ/hr

= 2400 / 

= 2 kW

C.O.P. = Q 1 / W

= 2 / 0.25

= 8.

12. C.O.P. of the refrigerator is always __________ the C.O.P. of the heat pump when both are working between the same temperature limits.

a) less than

b) greater than

c) equal to

d) inverse of

Answer: a

Explanation: C.O.P. = Desired effect / Work

As the desired effect for the heat pump is higher than the refrigerator. So, numerator value is higher for heat pump keeping denominator constant.

Can also be proved by this equation,

 R =  P – 1.

This set of Thermodynamics Multiple Choice Questions & Answers  focuses on “Refrigerator and Heat Pump”.

1. Which device maintains a body at a temperature lower than the temperature of the surroundings?

a) PMM1

b) PMM2

c) refrigerator

d) heat pump

Answer: c

Explanation: This is the main function of a refrigerator.

2. What does a refrigerant do?

a) absorbs the heat leakage into body from surroundings

b) evaporates in the evaporator

c) absorbs latent heat of vaporization form the body which is cooled

d) all of the mentioned

Answer: d

Explanation: Refrigerant is required for the proper functioning of a refrigerator.

3. Coefficient of performance is defined as

a) heat leakage/work input

b) work input/heat leakage

c) latent heat of condensation/work input

d) work input/latent heat of condensation

Answer: a

Explanation: Coefficient of performance is the performance parameter used in a refrigerator cycle.

4. Which device maintains a body at a temperature higher than the temperature of the surroundings?

a) PMM1

b) PMM2

c) refrigerator

d) heat pump

Answer: d

Explanation: This is the main function of a heat pump.

5. In a heat pump, there is heat leakage from the body to the surroundings.

a) true

b) false

Answer: a

Explanation: This is just opposite to a refrigerator.

6. What is the relation between COP of heat pump and refrigerator?

a) COP of pump=COP of refrigerator – 1

b) COP of pump=COP of refrigerator + 1

c) COP of pump=COP of refrigerator – 2

d) COP of pump=COP of refrigerator + 2

Answer: b

Explanation: This relation comes from the COP of pump and refrigerator.

7. Heat leakage from a heat pump to surroundings is always greater than work done on pump.

a) true

b) false

Answer: a

Explanation: =*.

8. Which of the following statements are true?

a) a heat pump provides a thermodynamic advantage over direct heating

b) COP for both refrigerator and pump cannot be infinity

c) work input for both refrigerator and pump is greater than zero

d) all of the mentioned

Answer: d

Explanation: W is the electrical energy used to drive the pump or refrigerator which cannot be zero.

9. Kelvin-Planck’s and Clausius’ statements are

a) not connected to each other

b) virtually two parallel statements of second law

c) violation of one doesn’t violate the other

d) none of the mentioned

Answer: b

Explanation: Kelvin-Planck’s and Clausius’ statements are equivalent in all aspects.

Answer: a

Explanation: This shows the equivalence of Kelvin-Planck’s and Clausius’ statements.

This set of Refrigeration Multiple Choice Questions & Answers  focuses on “Coefficient of Performance of Refrigeration – 1”.

1. What is the term C.O.P. referred in terms of refrigeration?

a) Capacity of Performance

b) Co-efficient of Plant

c) Co-efficient of Performance

d) Cooling for Performance

Answer: c

Explanation: Co-efficient of Performance is generally referred as C.O.P. for Refrigeration, which is used to measure the capacity or level up to which the refrigeration will occur.

2. C.O.P. can be expressed by which equation?

a) \

 

 \

 

 \

 

 \(\frac{Heat \,Transfer}{Work \,Done}\)

Answer: b

Explanation: Co-efficient of Performance is the ratio of the Refrigeration effect produced to the work done or work supplied to produce the effect.

Whereas the ratio- Work done to Heat transfer is called the efficiency.

3. What is the term relative C.O.P. referred in terms of refrigeration?

a) \

 

 \

 

 \

 

 \(\frac{Average \,C.O.P.}{Theoretical \,C.O.P.}\)

Answer: a

Explanation: Relative Co-efficient of Performance is generally referred as the ratio of Actual C.O.P. measured to the Theoretical C.O.P. assumed before calculations. It is in relation with the theoretical C.O.P. and generally expressed in “%” value.

4. Find the C.O.P. of a refrigeration system if the work input is 40 KJ/kg and refrigeration effect produced is 130 KJ/kg of refrigerant flowing.

a) 3.00

b) 2.25

c) 3.75

d) 3.25

Answer: d

Explanation: Here, refrigeration effect = 130 KJ/kg, work input = 40 KJ/kg

C.O.P. = \

 

 

.

5. Find the Relative C.O.P. of a refrigeration system if the work input is 60 KJ/kg and refrigeration effect produced is 130 KJ/kg of refrigerant flowing. Also Theoretical C.O.P. is 3.

a) 0.65

b) 0.79

c) 0.72

d) 0.89

Answer: c

Explanation: Given, work input = 60 KJ/kg, refrigeration effect = 130 KJ/kg and Theoretical C.O.P. = 3

Actual C.O.P. = \

 

 

Relative C.O.P. = \(\frac{Actual \,C.O.P.}{Theoretical \,C.O.P.}\)

= 2.167/3 = 0.722 i.e. = 72.2 %.

6. Find the C.O.P. of a refrigeration system if the work input is 30 KJ/kg and refrigeration effect produced is 120 KJ/kg of refrigerant flowing.

a) 3.00

b) 4.00

c) 0.75

d) 0.25

Answer: b

Explanation: Given, work input = 30 KJ/kg and refrigeration effect = 120 KJ/kg

C.O.P. = \

 

 

.

7. Which equation represents efficiency in general?

a) \

 

 \

 

 \

 

 \(\frac{Refrigeration \,effect}{Work \,Done}\)

Answer: c

Explanation: Work done to Heat transfer is called the efficiency in general sense, which can be applicable to Pumps, Refrigerators, and Turbines etc.

Whereas Co-efficient of Performance is the ratio of Refrigeration effect produced to the work done or work supplied to produce the effect.

8. The Co-efficient of Performance is always __________

a) greater than 1

b) less than 1

c) equal to 1

d) zero

Answer: a

Explanation: C.O.P. is always greater than 1, because in the ratio of C.O.P. the Refrigeration effect is always greater than the work done.

9. In a refrigerating machine, if the lower temperature is fixed, then the C.O.P. of machine can be increased by?

a) Increasing the higher temperature

b) Decreasing the higher temperature

c) Operating the machine at lower speed

d) Operating the machine at higher speed

Answer: b

Explanation: By decreasing the higher temperature, the change in temperatures can be reduced. This directly reduces the word done i.e. the denominator in C.O.P. Hence the C.O.P increases.

10. The reverse Carnot cycle C.O.P. can be expressed as _________ .

a) \

 

 \

 

 \

 

 \(\frac{t2-t1}{t1}\)

Answer: d

Explanation: Reverse Carnot C.O.P is the ratio of change in temperature i.e. t2-t1 to the lower temperature i.e. t1 here.

11. If a condenser and evaporator temperatures are 120 K and 60 K respectively, then reverse Carnot C.O.P is _________

a) 0.5

b) 1

c) 3

d) 2

Answer: b

Explanation: Reverse Carnot C.O.P. = \(\frac{t2-t1}{t1}\)

= \(\frac{120-60}{60}\)

= 1.

12. The C.O.P. of reverse Carnot cycle is most strongly dependent on which of the following?

a) Evaporator temperature

b) Condenser temperature

c) Specific heat

d) Refrigerant

Answer: a

Explanation: The C.O.P in the reverse Carnot cycle generally depends on the higher of the two temperatures, as it is located in the numerator of C.O.P equation and responsible directly for increase or decrease in C.O.P.

13. If a condenser and evaporator temperatures are 312 K and 273 K respectively, then reverse Carnot C.O.P is _________

a) 5

b) 6

c) 7

d) 8

Answer: c

Explanation: Reverse Carnot C.O.P. = \(\frac{t2-t1}{t1}\)

= \(\frac{312-271}{273}\)

= 7.

14. The C.O.P for reverse Carnot refrigerator is 2. The ratio of lowest temperature to highest temperature will be _____

a) twice

b) half

c) four times

d) three times

Answer: d

Explanation: Reverse Carnot C.O.P. = \(\frac{t2-t1}{t1}\)

2 = \(\frac{X-Y}{Y}\)

2 = \(\frac{X}{Y}\) – 1

Thus, \(\frac{X}{Y}\) = 3 i.e. X=3Y i.e. Higher temperature = 3 times Lower temperature.

15. If a condenser and evaporator temperatures are 250 K and 100 K respectively, then reverse Carnot C.O.P is _________

a) 5.5

b) 1.5

c) 2.5

d) 3.0

Answer: c

Explanation: Reverse Carnot C.O.P. = \(\frac{t2-t1}{t1}\)

= \(\frac{250-100}{100}\)

= 2.5.

This set of Refrigeration Questions and Answers for Freshers focuses on “Coefficient of Performance of Refrigeration – 2”.

1. Efficiency of the Refrigerator is _________ to the C.O.P of refrigerator.

a) inversely proportional

b) equal

c) independent

d) directly proportional

Answer: c

Explanation: Efficiency is the ratio of work done to heat supplied, whereas C.O.P is the ratio of Refrigeration effect to work done. Hence it is totally independent quantity.

2. What is the C.O.P. of a refrigeration system if the work input is 40 KJ/kg and work output is 80 KJ/kg and refrigeration effect produced is 130 KJ/kg of refrigerant flowing.

a) 3.00

b) 3.25

c) 2.25

d) 3.75

Answer: b

Explanation: C.O.P. = \

 

 

.

3. Find the Relative C.O.P. of a refrigeration system if the work input is 100 KJ/kg and refrigeration effect produced is 250 KJ/kg of refrigerant flowing. Also Theoretical C.O.P. is 3.

a) 0.65

b) 0.80

c) 0.83

d) 0.91

Answer: c

Explanation: Actual C.O.P. = \

 

 

Relative C.O.P. = \(\frac{Actual \,C.O.P.}{Theoretical \,C.O.P.}\)

= \(\frac{2.5}{3}\) = 0.833 i.e. = 83.3 %.

4. If a condenser and evaporator temperatures are 225 K and 100 K respectively, then reverse Carnot C.O.P is _________

a) 0.5

b) 1.5

c) 1.25

d) 1.75

Answer: c

Explanation: Reverse Carnot C.O.P. = \(\frac{t2-t1}{t1}\)

= \(\frac{225-100}{100}\)

= 1.25.

5. If a condenser and evaporator temperatures are 312 K and X K respectively, and C.O.P. is given as 5 then find the value of X.

a) 52

b) 65

c) 78

d) 82

Answer: a

Explanation: Reverse Carnot C.O.P. = \(\frac{t2-t1}{t1}\)

5 = \(\frac{312-X}{X}\)

6X = 312

X = 52 K.

6. The C.O.P for reverse Carnot refrigerator is 6. The ratio of lowest temperature to highest temperature will be _____

a) twice

b) three times

c) four times

d) seven times

Answer: d

Explanation: Reverse Carnot C.O.P. = \(\frac{t2-t1}{t1}\)

6 = \(\frac{X-Y}{Y}\)

6 = \(\frac{X}{Y}\) – 1

Thus, \(\frac{X}{Y}\) = 7 i.e. X=7Y i.e. Higher temperature = 7 times Lower temperature.

7. In general the ratio of lowest to highest temperature with respect to C.O.P. can be denoted by _________

a) C.O.P + 1 = Ratio of temperature

b) C.O.P/2 = Ratio of temperature

c) C.O.P + 4 = Ratio of temperature

d) C.O.P + 2 = Ratio of temperature

Answer: a

Explanation: In general, the ratio of lowest to highest temperature, say \(\frac{X-Y}{Y}\) = C.O.P. Hence, \(\frac{X}{Y}\) – 1 = C.O.P. i.e. \(\frac{X}{Y}\) = C.O.P + 1.

8. The C.O.P for reverse Carnot refrigerator is 2. The ratio of highest temperature to lowest temperature will be _____

a) 4 times

b) 3 times

c) 1/2 times

d) 1/3 times

Answer: d

Explanation: Reverse Carnot C.O.P. = \(\frac{t2-t1}{t1}\)

2 = \(\frac{X-Y}{Y}\)

2 = \(\frac{X}{Y}\) – 1

Thus, \(\frac{X}{Y}\) = 3 i.e. X=3Y i.e. Lowest temperature = \(\frac{1}{3}\) times Highest temperature.

9. The C.O.P of a reverse Carnot cycle doesn’t depend on which of the following?

a) Moisture

b) Evaporator temperature

c) Condenser temperature

d) Work done

Answer: a

Explanation: C.O.P for a reverse Carnot depends directly on the difference of Evaporator and Condenser temperature, and inversely on the Work done.

10. If a condenser and evaporator temperatures are ‘X’ K and 100 K respectively, and reverse Carnot C.O.P is 2.5 then find out the ‘X’.

a) 100 K

b) 150 K

c) 350 K

d) 200 K

Answer: c

Explanation: Reverse Carnot C.O.P. = \(\frac{t2-t1}{t1}\)

2.5 = \(\frac{X-100}{100}\)

X = 350 K.

This set of Refrigeration Multiple Choice Questions & Answers  focuses on “Classification of Refrigerants”.

1. Which of the following is not the type of refrigerant?

a) Organic refrigerants

b) Inorganic refrigerants

c) Azeotrope refrigerants

d) Halo-helium refrigerants

Answer: d

Explanation: Primary refrigerants are classified as Organic, i.e., Halo-Carbon refrigerants, Inorganic refrigerants, Azeotrope refrigerants, and Hydro-carbon refrigerants. Primary refrigerants are the one which takes a direct part in the refrigeration system.

2. When refrigerants take a direct part in the refrigeration system, then these types of refrigerants are called?

a) Primary

b) Secondary

c) Tertiary

d) Mixed

Answer: a

Explanation: Primary refrigerants are the one which takes a direct part in the refrigeration system. By absorbing the latent heat of vaporization effective refrigeration in produced.

3. The refrigerants which are first cooled by _________ refrigerants and then used for cooling purposes are known as ___________ refrigerants.

a) secondary, primary

b) primary, secondary

c) tertiary, primary

d) secondary, tertiary

Answer: b

Explanation Primary refrigerants are the one which takes a direct part in the refrigeration system. By absorbing the latent heat of vaporization effective refrigeration in produced. Whereas, refrigerants used for cooling purpose but firstly cooled by primary refrigerants are secondary refrigerants.

4. What is the number of halo-carbon compound refrigerants present as per ASHRAE?

a) 40

b) 41

c) 42

d) 43

Answer: c

Explanation: As per ASHRAE, i.e., American Society of Heating, Refrigeration and Air-conditioning Engineers, 42 Halo-carbon compounds are identified, but only a few are commonly used.

5. What does azeotrope mean?

a) Type of molecule

b) Type of bond

c) Stable mixture

d) Unstable mixture

Answer: c

Explanation: Azeotrope means a stable mixture of refrigerants; whose phases retain identical compositions over a large range of temperatures.

6. Which one of the following inorganic refrigerants is often used?

a) NH 3

b) CO 2

c) H 2 O

d) SO 2

Answer: a

Explanation: Ammonia, i.e., NH 3 , is widely used for the refrigeration system. Though it possesses high toxicity due to desirable properties to achieve maximum refrigeration effect is present, it is used frequently.

7. What is the refrigerant number of water?

a) R – 717

b) R – 744

c) R – 118

d) R – 100

Answer: c

Explanation: Water is having refrigerant number as R-118. Whereas, R-717, 744, 100 corresponds to Ammonia, Carbon dioxide, and ethyl chloride.

8. What is the number of halo-carbon compound refrigerants commonly used as per ASHRAE?

a) 42

b) 14

c) 41

d) 15

Answer: b

Explanation: As per ASHRAE, i.e., American Society of Heating, Refrigeration, and Air-conditioning Engineers, 42 Halo-carbon compounds are identified, but only a few are commonly used. The number of commonly used refrigerants is 14. Ex. R-11, 12, 13, 14, 21, 22, 30, 40, 100, 113, 114, 115, 123, 124, 134a, 152a.

9. Which of the following refrigerants are in the Freon group?

a) Organic refrigerants

b) Inorganic refrigerants

c) Azeotrope refrigerants

d) Hydrocarbon refrigerants

Answer: a

Explanation: Freon group is in the Organic or halo-carbon type of refrigerants. Halo-carbon compounds are produced and developed by freon family of refrigerants. Freon is the trademark of E.I. Du Pont de Nemours and Co., America.

10. Hydro-carbon refrigerants are commonly used nowadays.

a) True

b) False

Answer: b

Explanation: Hydro-carbons possess desired properties for refrigeration, but due to high flammability and explosive nature, hydrocarbons are replaced by another less explosive type of refrigerant.

This set of Refrigeration Multiple Choice Questions & Answers  focuses on “Compound VCR System – Cascade Systems – 1”.

1. Why is the cascade system used?

a) To attain low C.O.P.

b) To decrease the refrigeration effect

c) To produce low temperatures

d) To increase leakage loss

Answer: c

Explanation: If the vapor compression system is to be used for the production of temperatures, the common choice to stage type compression is a cascade system.

2. What is Cryogenics?

a) C.O.P. reduction

b) Low-temperature refrigeration

c) Work enhancement

d) High-temperature refrigeration

Answer: b

Explanation: Cryogenics means low-temperature refrigeration. It is applied to very low-temperature refrigeration applications. This word is derived Greek word Cryos meaning cold or frost.

3. What is the principal advantage of the cascade system?

a) Increase of work done

b) Decrease the refrigeration effect

c) Use of only one refrigerant

d) Use of multiple refrigerants

Answer: d

Explanation: Multiple refrigerants can be used in this type of system. It helps to attain lower temperatures, and the refrigeration effect can be enormously increased.

4. How is a cascade system structured?

a) VCR system in parallel combination

b) VCR system is not used

c) VCR system in a series combination

d) VAR system is used

Answer: c

Explanation: The cascade refrigeration system is structured by connecting two or more vapor compression refrigeration systems in series which use different refrigerants.

5. Low-temperature cascade system use refrigerants with ______ boiling temperature and high temperature cascade system use refrigerants with _________ boiling temperature.

a) low, high

b) high, low

c) low, low

d) high, high

Answer: a

Explanation: The high-temperature cascade system uses a refrigerant with high boiling temperatures such as R-12 or R-22. The low-temperature cascade system uses a refrigerant with low boiling temperatures such as R-13 or R-13 BI. This is used to achieve a higher coefficient of performance by increasing the refrigeration effect.

6. What does low boiling refrigerants ensure?

a) Lower C.O.P.

b) Higher C.O.P.

c) Lower pressure

d) Larger compressor displacement

Answer: b

Explanation: Low boiling temperature refrigerant has extremely high pressure. Due to this smaller compressor displacement is ensured. So, smaller displacement decreases overall work and increases the C.O.P. of the system.

7. Which intermediate refrigerants were used for the first time in the cascade system?

a) CO 2 and CO

b) CO and SO 2

c) NH 3 and CO 2

d) CO 2 and SO 2

Answer: d

Explanation: Firstly, the cascade system was used by Pietet in 1877. Cascade system was used for the liquefaction of oxygen using the Sulphur dioxide (SO 2 ) and Carbon dioxide (CO 2 ) as intermediate refrigerants.

8. The difference between low-temperature cascade condenser temperature and high-temperature cascade evaporator temperature is called temperature ______

a) subtraction

b) overview

c) overlap

d) gradient

Answer: c

Explanation: Temperature overlap is the difference between low-temperature cascade condenser temperature and high-temperature cascade evaporator temperature. Temperature overlap is a crucial phenomenon to attain higher C.O.P. because it shows the heat transfer in the system.

9. What is Intermediate temperature?

a) Temperature overlap is zero

b) Temperature overlap is infinity

c) The temperature of the low-temperature cascade system

d) The temperature of the high-temperature cascade system

Answer: a

Explanation: Intermediate temperature is achieved when the temperature overlap is zero i.e., the difference between low-temperature cascade condenser temperature and high-temperature cascade evaporator temperature is zero. Intermediate temperature indicates that both temperatures are equal. Intermediate temperature affects the heat transfer in the system.

This set of Refrigeration Questions and Answers for Campus interviews focuses on “Compound VCR System – Cascade Systems – 2”.

1. Multiple refrigerants can be used in the cascade refrigeration system.

a) True

b) False

Answer: a

Explanation: Multiple refrigerants can be used in the cascade refrigeration system. It helps to attain lower temperatures, and the refrigeration effect can be enormously increased.

2. How is the cascade system achieved?

a) VCR system in a parallel combination

b) VAR system in a series combination

c) VAR system in a parallel combination

d) VCR system in a series combination

Answer: d

Explanation: The cascade refrigeration system is structured by connecting two or more vapor compression refrigeration systems in series which use different refrigerants.

3. Cascade refrigeration system reduces the C.O.P.

a) True

b) False

Answer: b

Explanation: Due to the usage of multiple refrigerants and temperature overlap phenomenon, the refrigeration effect is increased, and overall work is reduced, leading to enhance the C.O.P. of the system.

4. What is the value of m1 in the following diagram?

refrigeration-questions-answers-cascade-systems-2-q4

a) 210 Q / (h 1 – h 7 ) kg/min

b) 210 Q / (h 1 – h 4 ) kg/s

c) 210 Q / (h 1 – h 4 ) kg/min

d) 210 Q / (h 1 – h 7 ) kg/s

Answer: c

Explanation: As m 1 flows through the low-temperature cascade system. So, when it passes through the evaporator i.e., between point 1 and point 4, there is a change in enthalpy, which produces the refrigeration effect.

Hence, R.E. = m x delta h

210 Q = m 1 x (h 1 – h 4 )

m 1 = 210 Q / (h 1 – h 1 ) kg/min.

5. What is the value of m 2 / m 1 in the following diagram?

refrigeration-questions-answers-cascade-systems-2-q4

a) (h 2 – h 4 ) / (h 5 – h 8 )

b) (h 6 – h 4 ) / (h 7 – h 8 )

c) (h 1 – h 4 ) / (h 6 – h 8 )

d) (h 2 – h 1 ) / (h 1 – h 8 )

Answer: a

Explanation: The cascade condenser is an intermediate between low temperature and high temperature cascade system. The heat between both systems is balanced hence, taking the thermal equilibrium of condenser.

Heat absorbed in the high temperature cascade system = Heat rejected in the low temperature cascade system

m 2 h 5 + m 1 h 4 = m 1 h 2 + m 2 h 8

m 1 (h 2 – h 4 ) = m 2 (h 5 – h 8 )

m 2 / m 1 = (h 2 – h 4 ) / (h 5 – h 8 ).

6. What is the value of work done in the following diagram?

refrigeration-questions-answers-cascade-systems-2-q4

a) m 1 (h 2 – h 4 ) + m 2 (h 5 – h 6 )

b) m 1 (h 2 – h 1 ) + m 2 (h 6 – h 5 )

c) m 1 (h 2 – h 8 ) + m 2 (h 5 – h 8 )

d) m 1 (h 2 – h 5 ) + m 2 (h 4 – h 8 )

Answer: b

Explanation: Work done is across the compressors.

Work = mass of refrigerant flowing through the compressor x change of enthalpy

Work done by low-temperature cascade system = m 1 (h 2 – h 1 )

Work done by high-temperature cascade system = m 2 (h 6 – h 5 )

Total work = Work done by low-temperature cascade system + Work done by high-temperature cascade system

Total work done = m 1 (h 2 – h 1 ) + m 2 (h 6 – h 5 ).

7. What is the value of C.O.P. in the following diagram?

refrigeration-questions-answers-cascade-systems-2-q4

a) 210 Q / m 1 (h 2 – h 4 ) + m 2 (h 5 – h 6 )

b) 210 Q / m 1 (h 2 – h 7 ) + m 2 (h 6 – h 8 )

c) 210 Q / m 1 (h 2 – h 8 ) + m 2 (h 5 – h 8 )

d) 210 Q / m 1 (h 2 – h 1 ) + m 2 (h 6 – h 5 )

Answer: d

Explanation: As m 1 flows through the low-temperature cascade system. So, when it passes through the evaporator i.e. between point 1 and point 4, there is change in enthalpy which produces the refrigeration effect.

Hence, R.E. = m x delta h

= 210 Q where, Q is the load on the low-temperature cascade system in tones of refrigeration

Work = mass of refrigerant flowing through the compressor x change of enthalpy

Work done by low-temperature cascade system = m 1 (h 2 – h 1 )

Work done by high-temperature cascade system = m 2 (h 6 – h 5 )

Total work = Work done by low-temperature cascade system + Work done by high-temperature cascade system

Total work done = m 1 (h 2 – h 1 ) + m 2 (h 6 – h 5 )

As, C.O.P. = Refrigeration effect / Total work done

= 210 Q / m 1 (h 2 – h 1 ) + m 2 (h 6 – h 5 ).

8. What is the value of power required to drive the systems in the following diagram?

refrigeration-questions-answers-cascade-systems-2-q4

a) m 1 (h 2 – h 4 ) + m 2 (h 5 – h 6 ) kW

b) m 1 (h 2 – h 1 ) + m 2 (h 8 – h 1 ) kJ/s

c) m 1 (h 2 – h 1 ) + m 2 (h 6 – h 5 ) kW

d) m 1 (h 2 – h 1 ) + m 2 (h 6 – h 5 ) kJ/min

Answer: c

Explanation: Power required to drive the system is the power required to do the work using compressors.

Work = mass of refrigerant flowing through the compressor x change of enthalpy

Work done by low-temperature cascade system = m 1 (h 2 – h 1 )

Work done by high-temperature cascade system = m 2 (h 6 – h 5 )

Total work = Work done by low-temperature cascade system + Work done by high-temperature cascade system

Total work done = m 1 (h 2 – h 1 ) + m 2 (h 6 – h 5 ) kJ/min

Power required = Work done / 60

= m 1 (h 2 – h 1 ) + m 2 (h 6 – h 5 ) kW.

9. What is the value of C.O.P. of low-temperature cascade system in terms of intermediate temperature in the following diagram?

refrigeration-questions-answers-cascade-systems-2-q4

If T EL and T CL = Evaporator and condenser temperatures for low-temperature cascade system

T EH and T CH = Evaporator and condenser temperatures for high-temperature cascade system

a) T EL / T CL – T CL

b) T CL / T CL – T EL

c) T CL / T I – T CL

d) T EL / T I – T CL

Answer: d

Explanation: Temperature of the condenser is higher than the temperature of the evaporator. The desired effect is obtained in the evaporator. So, by using the Carnot’s theorem to low-temperature cascade system,

C.O.P. = Temperature of the evaporator / Temperature of the condenser – Temperature of the evaporator

= T EL / T CL – T EL

As, when low-temperature cascade condenser temperature is equal to high-temperature cascade evaporator temperature.

T CL = T EH = T I

Hence, C.O.P. = T EL / T I – T CL .

10. What is the value of C.O.P. of high-temperature cascade system in terms of intermediate temperature in the following diagram?

refrigeration-questions-answers-cascade-systems-2-q4

If T EL and T CL = Evaporator and condenser temperatures for low-temperature cascade system

T EH and T CH = Evaporator and condenser temperatures for high-temperature cascade system

a) T EH / T CH – T CH

b) T I / T CH – T I

c) T CH / T I – T CH

d) T EH / T I – T CH

Answer: b

Explanation: Temperature of the condenser is higher than the temperature of the evaporator. The desired effect is obtained in the evaporator. So, by using the Carnot’s theorem to high-temperature cascade system,

C.O.P. = Temperature of the evaporator / Temperature of the condenser – Temperature of the evaporator

= T EH / T CH – T EH

As, when low-temperature cascade condenser temperature is equal to high-temperature cascade evaporator temperature.

T CL = T EH = T I

Hence, C.O.P. = T I / T CH – T I .

11. What is the value of pressure ratio if the pressure at point 1 and 2 is 1.809 and 3.467 bar, respectively?

refrigeration-questions-answers-cascade-systems-2-q4

a) 1.916

b) 1.916 bar

c) 0.521

d) 0.521 bar

Answer: a

Explanation: Pressure ratio is the ratio of pressure after the compression and pressure before the compression. As compression is completed at point 2 from the diagram.

So, pressure ratio = P 2 / P 1

= 3.467 / 1.809

= 1.916.

Pressure ratio is the ratio of two same quantities, so this value does not have any unit.

This set of Refrigeration Multiple Choice Questions & Answers  focuses on “Refrigerant Compressors – Multi – Stage Compression”.

1. Why is multi-stage compression used?

a) To improve C.O.P.

b) To decrease the refrigeration effect

c) To increase work

d) To increase leakage loss

Answer: a

Explanation: By expanding the refrigerant very close to the saturated liquid state refrigeration effect can be improved. And if the refrigerant is compressed very near to the saturated vapour line by using compression in more stages, overall work can be reduced. The relative decrease in the overall work done increases the C.O.P. of the system.

2. What is the expression for optimum intercooler or intermediate pressure P 2 if the cooling ratio is fixed in a compound compression refrigeration system with intercooling?

a) P 2 = P 1 / P 3

b) P 2 = P 3 / P 1

c) P 2 = √P 1 x P 3

d) P 2 = ∛P 1 x P 3

Answer: c

Explanation: Putting  = k and dW / dP 2 = 0 we get,

T 1 P 2 k – 1 / P 1 k = T 3 P 3 k / P 2 k + 1

For the given conditions T 1 = T 3 ,

P 2 x P 2 = P 1 x P 3

P 2 = √P 1 x P 3 .

3. Multistage compression provides effective lubrication.

a) False

b) True

Answer: b

Explanation: Due to the use of high-pressure compression by a single-cylinder, there are chances of obtaining very high discharge temperatures which might heat the cylinder head or burn the lubricating oil. So, with multistage burning of lubricating oil can be avoided as the discharge temperature will be lower. Hence, Multistage compression provides effective lubrication.

4. What is the value of optimum intermediate pressure is the cooling ratio is fixed in a compound VCR with intercooling having suction and discharge pressure as 2 and 8 bar respectively?

a) 2 bar

b) 4 bar

c) 6 bar

d) 8 bar

Answer: b

Explanation: Putting  = k and dW / dP 2 = 0 we get,

T 1 P 2 k – 1 / P 1 k = T 3 P 3 k / P 2 k + 1

For the given conditions T 1 = T 3 ,

P 2 x P 2 = P 1 x P 3

P 2 = √P 1 x P 3

= √2 x 8

= 4 bar.

5. Which of the following is not true about single-stage compression?

a) Size of the cylinder is very large

b) Difficult to reject a good amount of heat in a lesser time of compression

c) Volumetric efficiency will be low

d) Safe at a higher temperature

Answer: d

Explanation: As a single cylinder is used so for effective results large size cylinder will be used. Less time of compression and relatively volumetric efficiency will be lower. Due to the use of high-pressure compression by a single-cylinder, there are chances of obtaining very high discharge temperatures which might heat the cylinder head or burn the lubricating oil. Hence, it might not be safe in every situation.

This set of Refrigeration Multiple Choice Questions & Answers  focuses on “Actual VCR Cycle”.

1. Which of the following is not the difference between theoretical and actual VCR cycle?

a) Pressure drops in the evaporator

b) Pressure drops in the condenser

c) Compression of refrigerant is never polytropic

d) Compression of refrigerant is always isentropic

Answer: d

Explanation: Entropy is the randomness of molecules. Change in entropy cannot be kept constant. As any transfer of energy will provide energy to the particles and increasing the entropy. The entropy of the Universe tends to increase day by day. Hence, the compression process cannot be executed in an isentropic manner. Ideally, it can be assumed, but in the actual cycle, it is impossible to attain such a process.

2. What does process 2-3 represent?

refrigeration-questions-answers-actual-vcr-cycle-q2

a) Compression

b) Condensation

c) Evaporation with superheating

d) Expansion with subcooling

Answer: c

Explanation: Process 2-3 represents evaporation with superheating. The shift of exit of evaporation from point 2 to point 3 is due to

1) Automatic control of expansion valve

2) Absorbs a larger amount of heat

3) Refrigeration effect and work increases.

3. What does process 3-4 represent?

refrigeration-questions-answers-actual-vcr-cycle-q2

a) Pressure reduction

b) Compression

c) Condensation

d) Evaporation

Answer: a

Explanation: Due to the frictional resistance offered to the vapor refrigerant entering the compression system pressure drops from point 3 to point 4. Thus, actual suction pressure (P S ) lower than the evaporator pressure (P E ).

4. What does process 4-5 represent?

refrigeration-questions-answers-actual-vcr-cycle-q2

a) Temperature rise

b) Temperature fall

c) Compression

d) Expansion

Answer: a

Explanation: After pressure reduction and before compression begins, the temperature of cold refrigerant comes in contact with compression cylinder walls, which are at a higher temperature. Hence, due to heat transfer, the temperature of the refrigerant rises to point 5, which is the heating effect.

5. What does process 5-6 represent?

refrigeration-questions-answers-actual-vcr-cycle-q2

a) Expansion

b) Compression

c) Condensation

d) Evaporation

Answer: b

Explanation: Actual compression is denoted by process 5-6, which is neither isentropic nor polytropic, due to heat transfer between cylinder walls and vapor refrigerant. The temperature of cylinder walls is between cold suction vapor refrigerant and hot discharge vapor refrigerant. Pressure and temperature are increased in this process.

6. What does process 6-7 represent?

refrigeration-questions-answers-actual-vcr-cycle-q2

a) Temperature rise

b) Temperature fall

c) Compression

d) Expansion

Answer: b

Explanation: After compression and before Condensation begins, the temperature of hot refrigerant comes in contact with compression cylinder walls which are at a lower temperature. Hence, due to heat transfer, the temperature of the refrigerant falls to point 7, which is a cooling effect.

7. What does process 7-8 represent?

refrigeration-questions-answers-actual-vcr-cycle-q2

a) Condensation

b) Compression

c) Pressure reduction

d) Evaporation

Answer: c

Explanation: Due to the frictional resistance offered to the vapor refrigerant discharging the compression system pressure drops from point 7 to point 8. Thus, actual discharge pressure (P D ) higher than the condenser pressure (P C ).

8. What does process 8-9 represent?

refrigeration-questions-answers-actual-vcr-cycle-q2

a) Compression

b) Condensation

c) De-superheating

d) Expansion with subcooling

Answer: c

Explanation: Before condensing the refrigerant, to reject maximum heat to get stabilized state of refrigerant, the refrigerant must be at saturated state before condensing. Hence, de-superheating is carried out to get the refrigerant at a saturated level and then condensed.

9. What does process 9-10 represent?

refrigeration-questions-answers-actual-vcr-cycle-q2

a) Compression

b) Condensation

c) Evaporation

d) Expansion

Answer: b

Explanation: Process 9-10 shows the removal of latent heat where the dry saturated refrigerant is converted into a liquid refrigerant. This process is called Condensation.

10. What does process 10-11 represent?

refrigeration-questions-answers-actual-vcr-cycle-q2

a) Compression

b) Expansion

c) Evaporation with superheating

d) Subcooling

Answer: d

Explanation: The process 10-11 represents sub-cooling of the liquid refrigerant in the condenser before it is expanded. Sub-cooling is carried out as it increases the refrigerating effect per kg of the refrigerant flow. It does reduce the volume of refrigerant partially evaporated from liquid refrigerant while going through the expansion valve. The refrigerating effect can be increased by circulating a large amount of water, which is at a much lower temperature than condensing temperature.

11. What does process 11-1 represent?

refrigeration-questions-answers-actual-vcr-cycle-q2

a) Expansion

b) Condensation

c) Compression

d) Evaporation

Answer: a

Explanation: Process 11-1 represents expansion. The expansion of sub-cooled liquid refrigerant is carried out by throttling  from the condenser pressure to the evaporative pressure.

12. What does process 1-2 represent?

refrigeration-questions-answers-actual-vcr-cycle-q2

a) Expansion

b) Condensation

c) Compression

d) Evaporation

Answer: d

Explanation: Process 1-2 represents evaporation. In this process, absorption of heat is carried out and converting liquid refrigerant to saturated vapor condition. The evaporation process is known as Refrigeration effect.

13. What is the effect of a decrease in suction pressure on C.O.P.?

a) C.O.P. increases

b) C.O.P. decreases

c) C.O.P. remains the same

d) C.O.P. becomes zero

Answer: b

Explanation: As C.O.P. is the ratio of refrigeration effect to work done. Due to the decrease in the suction pressure, the refrigeration effect is decreases and work required for compression increases for the same amount of refrigerant flow. Hence, resulting in a reduction of the C.O.P. of the system and refrigeration cost also increases.

14. What is the effect of an increase in discharge pressure on C.O.P.?

a) C.O.P. decreases

b) C.O.P. increases

c) C.O.P. remains the same

d) C.O.P. becomes zero

Answer: a

Explanation: As C.O.P. is the ratio of refrigeration effect to work done. Due to the increase in the discharge pressure, the refrigeration effect decreases for the same amount of refrigerant flow. Hence, resulting in a small decrease in the C.O.P. of the system.

15. Which of the following factor of the actual VCR affects severely the coefficient of performance?

a) Increase in suction pressure

b) Decrease in discharge pressure

c) Decrease in suction pressure

d) Increase in discharge pressure

Answer: c

Explanation: As C.O.P. is the ratio of refrigeration effect to work done. Due to the increase in the discharge pressure, refrigeration effect decreases, but it is not that severe as is in the case of a decrease in suction pressure. Due to the decrease in the suction pressure, the refrigeration effect severely decreases, and work required for compression increases for the same amount of refrigerant flow. Hence, resulting in a severe decrease in the C.O.P. of the system and refrigeration cost also increases.

This set of Thermal Engineering Objective Questions & Answers focuses on “Heat Transfer by Conduction”.

1. If the radius of any current carrying conductor is less than the critical radius, then why the addition of electrical insulation will enable the wire to carry a higher current?

a) The heat loss from the wire would decrease

b) The heat loss from the wire would increase

c) The thermal resistance of the insulation is reduced

d) The thermal resistance of the conductor is increased

Answer: b

Explanation: If the radius of any current carrying conductor is less than the critical radius, then the addition of electrical insulation will enable the wire to carry a higher current because heat loss from the wire would increase. Radius of wire is inversely proportional to heat loss.

2. Which of the following substance has the minimum value of thermal conductivity?

a) Air

b) Water

c) Plastic

d) Rubber

Answer: a

Explanation: Among the given options air has the minimum value of thermal conductivity. Thermal conductivity of air is approximately K air =0.02.

3. In MLTθ system , what is the dimension of thermal conductivity?

a) ML -1 T -1 θ -3

b) MLT -1 θ -3

c) MLT -3 θ -1

d) MLT -2 θ -1

Answer: c

Explanation: Q=-KA

(ML 2 T -3 ) = k (L 2 ) 

K = ML 2 T -3 /Lθ , K=MLT -3 θ -1 .

4. For conduction through a spherical wall with constant thermal conductivity and with inner side temperature greater than outer wall temperature in 1-D heat transfer, what is the type of temperature

distribution?

a) Linear

b) Parabolic

c) Hyperbolic

d) Logarithmic

Answer: c

Explanation: For one dimensional steady heat flow-

i. Temperature distribution in slab is linear

ii. Temperature distribution in cylinder is logarithmic

iii. Temperature distribution in sphere is hyperbolic.

5. Which of the following expresses the thermal diffusivity of a substance in terms of thermal conductivity of a substance , mass density  and specific heat ?

a) k 2 ρc

b) 1/ρkc

c) k/ρc

d) ρc/k 2

Answer: c

Explanation: Thermal diffusivity indicates the ease at which energy get diffused in the volume of the substances. It is defined as the ratio of the thermal conductivity to the heat capacity of the substance.

Therefore, α = k/ρc.

6. A copper block and an air mass block having similar dimensions are subjected to symmetrical heat transfer from one face of the each block. The other face of the block will be reaching to the same temperature at a rate?

a) Faster in air block

b) Faster in copper block

c) Equal in air as well as copper block

d) Data in sufficient

Answer: b

Explanation: Thermal conductivity is minimum in gases and maximum in solids. Thermal conductivity of copper is higher than air. Hence heat flow will be in copper block rather than air block.

7. The outer surface of a long cylinder is maintained at constant temperature. The cylinder does not have any heat source. The temperatures in the cylinder will _________

a) Increase linearly with radius

b) Decrease linearly with radius

c) Be independent of radius

d) Vary logarithmically with radius

Answer: d

Explanation: The temperature distribution will vary logarithmically with radius for cylinder.

α = \(\frac{2πLk}{

 

}\).

8. A steam pipe is covered with two layers of insulating materials, with the better insulating material forming the outer. What is the effect on heat conducted if the two layers are interchanged?

a) Will increase

b) Will decrease

c) Will remain unaffected

d) May increase or decrease depending upon the thickness of each layer

Answer: d

Explanation: Let k 1 and k 2 be the thermal conductivity of the two insulators, k 1 being the better insulator.

Q = \(\frac{A∆T}{\frac{ln⁡

 

}{k1}+\frac{ln⁡

 

}{k2}}\)

If we interchange the insulators, the value of Q might increase or decrease depending upon the thickness of each layer.

9. A plane wall is 20cm thick with an area perpendicular to heat flow of 1m 2 and has a thermal conductivity of 0.5W/mK. A temperature difference of 100°C is imposed across it. What is the ratio of heat flow?

a) 0.10 kW

b) 0.15 kW

c) 0.20 kW

d) 0.25 kW

Answer: c

Explanation: Thermal resistance, R th = \(\frac{L}{KA}=\frac{0.2}{0.5×1}=0.5\frac{K}{W} \)

Heat transfer, Q = \(\frac{}{R_{th}}=\frac{100}{0.5}\)=200W=0.2kW.

10. An insulating material with a thermal conductivity k = 0.12W/mK is used for a pipe carrying steam. The local coefficient of heat transfer to the surrounding h = 4 W/m 2 K. In order to provide effective insulation, what should be the minimum outer diameter of the pipe?

a) 45mm

b) 60mm

c) 75mm

d) 90mm

Answer: b

Explanation: Minimum outer radius should be critical radius. The minimum critical radius is given by,

r c = \(\frac{k}{h}=\frac{0.12}{4}\)=0.03m=30mm

Hence, outer diameter = r c =  mm = 60mm.

11. In a long cylindrical rod of radius R and a surface heat flux of q 0 , what is the uniform internal heat generation rate?

a) 2q 0 /R

b) 2q 0

c) q 0 /R

d) 2q 0 /3R

Answer: a

Explanation: In steady state, Heat generated = Heat conducted at surface

q g × πR 2 L = q 0 × 2πRL

q g = 2q 0 /R.

12. A plane slab of 100mm thickness generates heat. It is observed that the temperature drop between the center and its surface to be 50°C. If the thickness is increased to 200mm the temperature difference will be ___________

a) 100°C

b) 200°C

c) 400°C

d) 600°C

Answer: b

Explanation: T max = T wall + q g L 2 /2k

(T max – T wall ) = q g L 2 /2k

(T max – T wall ) α L 2

\(\frac{2}{1}=\frac{L_2}{L_1} \)

(T max – T wall ) 2 /50 = \

 

\) 2

(T max – T wall )2 = 400°C.

13. As the temperature increases, the thermal conductivity of a gas ____________

a) Increases

b) Decreases

c) Remain constant

d) Increases up to a certain temperature and then decreases

Answer: a

Explanation: In gases heat conduction occurs by molecular momentum transfer when high velocity, high temperature molecules collide with the low velocity low temperature molecules. Thermal conductivity of gases increases with temperature because molecular momentum transfer increases with increase in temperature.

This set of Thermal Engineering online test focuses on “Heat Transfer by Convection”.

1. Hot air at 50°C. If the forced heat transfer convection is 75 W/m 2 K, the heat gain rate gain rate by the plate through an area of 2m 2 will be _____

a) 15KW

b) 22.5KJ/sec

c) 7.5KJ/sec

d) 25KJ/sec

Answer: a

Explanation: Rate of heat transfer

Q = Q = hA∆T

Q = 75×2×

Q = 15 kW.

2. Nusselt number for fully developed turbulent flow in a pipe is given by Nu=C Re a Pr b . the values of a and b are?

a) a=0.5 and b=0.33 for heating and cooling

b) a=0.5 and b=0.4 for heating and b=0.3 for cooling

c) a=0.8 and b=0.4 for heating and b=0.3 for cooling

d) a=0.8 and b=0.3 for heating and b=0.4 for cooling

Answer: c

Explanation: For fully developed turbulent flow in a pipe, Nusselt number

Nu=0.023 Re 0.8 Pr n

Where, n= 0.4 for heating

And n = 0.3 for cooling.

3. Which of the following non-dimensional numbers is used for transition from laminar flow to turbulent flow in the free convection?

a) Reynolds number

b) Grashoff number

c) Peclet number

d) Rayleigh number

Answer: b

Explanation: Grashoff’s number plays similar role to that of Reynolds number in forced convection. Its value indicates whether the fluid flow of natural convection is laminar or turbulent. The critical value of Grashoff number indicates that the transition from laminar to turbulent flow.

4. Which non-dimensional numbers relates the thermal boundary layer and hydrodynamic boundary layer?

a) Rayleigh number

b) Peclet number

c) Grashoff number

d) Prandtl number

Answer: d

Explanation: Prandtl number is a dimensionless number that relates the thermal boundary layer and hydrodynamic boundary layer.

\(\frac{δ}{t} = \frac{hydrodynamic \, boundary \, layer}{thermodynamic \, boundary \, layer}\) = Pr 1/3 .

5. The laminar boundary layer occurs when a cold fluid over a hot plate. In which of the following positions, the temperature gradient assumes zero value?

a) At bottom of boundary layer

b) In mid free stream of fluid

c) At top of boundary layer

d) At the junction of laminar and turbulent boundary layer

Answer: c

Explanation: The value of temperature gradient dT/dy decreases with the distance from plate and at the top of boundary layer it becomes zero as the temperature equals to free stream temperature of the fluid.

6. The characteristic length for computing Grashoff number of horizontal cylinder is ___________

a) The length of the cylinder

b) The diameter of the cylinder

c) The perimeter of the cylinder

d) The radius of the cylinder

Answer: b

Explanation: The diameter becomes the characteristics length in case of free convection. Whereas length is for forced convection.

7. In laminar developing flow through a pipe with constant wall temperature, the magnitude of the pipe wall inner surface convective heat transfer coefficient shall be maximum at the:

a) Middle length of flow

b) Beginning of flow

c) End of flow

d) One third of the length of flow

Answer: d

Explanation: In laminar flow through pipe nusselt number constant for constant wall temperature. Therefore wall inner surface convection heat transfer coefficient will be constant for the laminar flow in pipe.

8. For a fluid with Prandtl number Pr>1, momentum boundary layer thickness?

a) Decreases rapidly compared to the thermal boundary layer thickness

b) And thermal boundary layer thickness increase at the same rate

c) Increases rapidly as compared to thermal boundary layer thickness

d) And thermal boundary layer thickness decrease at the same rate

Answer: c

Explanation: \

 

 1/3

For Pr>1 , δ>t

Therefore momentum boundary layer thickness increases rapidly compared to the thermal boundary layer thickness.

9. Water  flows over a flat plate which is heated over the entire length. Which one of the following relationship between the hydrodynamic boundary layer thickness  and the thermal boundary layer thickness t is true?

a) t > δ

b) t < δ

c) t = δ

d) Cannot be predicted

Answer: b

Explanation: \

 

 1/3

Since Prandtl number Pr = 6 > 1, therefore from above equation,

t < δ.

10. If q w =2500x where x is in m and in the direction of flow , the bulk mean temperature of the water leaving the pipe in °C is ________

a) 71

b) 76

c) 79

d) 81

Answer: d

Explanation: q w ×πDL = mc p (t 2 -t 1 )

5000×π×0.05×3=0.01×4.18×10 3 (t 2 -20)

t 2 =76°C

At the outlet flux can be calculated by,

Q w = h(t s -t 1 )

5000=1000(t s -76)

t s =81°C.

This set of Thermal Engineering Questions & Answers for Exams focuses on “Heat Transfer by Radiation”.

1. What is the equivalent emissivity for radiant heat exchange between a small body  in a very large enclosure ?

a) 0.5

b) 0.4

c) 0.2

d) 0.1

Answer: b

Explanation: When body 1 is completely enclosed by body 2,body 1 is very small.

ε=\(\frac{1}{\frac{1}{ε1}+A1/A2

 

} \)

Since, A1/A2 is very less whole term is neglected.

ε = 0.4.

2. If a body is at 2000K, the wavelength at which the body emits maximum amount of radiation is __________

a) 1.45µm

b) 1.45cm

c) 0.345cm

d) 0.345µm

Answer: a

Explanation: From Wien’s displacement law,

λ max T = 2898µmk

λ max ×2000 = 2898µmk

λ max = 1.45µm.

3. Which of the following parameter is not responsible for loss of heat from a hot surface in room?

a) Temperature of the surface

b) Emissivity of the surface

c) Temperature of the air in the room

d) Dimensions of the room

Answer: d

Explanation: For radiation heat transfer

Q = ∈σ(T 1 4 -T 2 4 )

And A=πDL

Hence heat transfer will depend upon temperature of the surface and surrounding, surface Emissivity and length and diameter of the pipe.

4. Hot coffee in a cup is allowed to cool. Its cooling rate is measured and found to be greater than the value calculated by conduction, convection and radiation measurement. The difference is due to:

a) Properties of coffee changing with temperature

b) Currents of air flow in the room

c) Underestimation of the emissivity of coffee

d) Evaporation

Answer: d

Explanation: Hot coffee in a cup is allowed to cool, while measuring its cooling rate evaporation is considered so its value is greater than the value of calculated by conduction, convection and radiation.

5. Which of the following statement is incorrect?

a) For metals, the value of absorptivity is high

b) For non-conducting materials, reflectivity is low

c) For polished surfaces, reflectivity is high

d) For gases, reflectivity is very low

Answer: a

Explanation: The emissivity of metallic surface is generally small. Therefore absorptivity will be high. Emissivity of non conducting material is high so reflectivity low. For highly polished surface reflectivity is high.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Heat Exchangers”.

1. A periodic type heat exchanger is known as _____

a) Direct contact heat exchanger

b) Indirect contact heat exchanger

c) Recuperator

d) Regenerator

Answer: d

Explanation: A periodic type heat exchanger is known as regenerator. Recuperator are heat exchanger in which two fluids are separated by a wall.

2. Surface heat exchangers are also known as _______

a) Direct contact heat exchanger

b) Indirect contact heat exchanger

c) Recuperator

d) Regenerator

Answer: c

Explanation: Recuperator are heat exchanger in which two fluids are separated by a wall. Recuperator is also called surface heat exchangers. In direct flow heat exchanger two fluids exchange heat by coming in direct contact.

3. In which type of heat exchanger the same space is occupied by the hot and cold gases, between which heat is exchanged?

a) Recuperator

b) Regenerator

c) Direct contact heat exchanger

d) Indirect contact heat exchanger

Answer: b

Explanation: A periodic type heat exchanger is known as regenerator. Regenerator heat exchanger in which the same space is occupied by the hot and cold gases, between which heat is exchanged. Recuperator are heat exchanger in which two fluids are separated by a wall.

4. Which of the following is not an application of regenerator?

a) Jet condenser

b) Steam power plant

c) Oxygen producer

d) Blast furnace

Answer: a

Explanation: Regenerators find their applications in steam power plant, oxygen producer and blast furnace. The application of direct contact type heat exchanger are open feed water heaters, de-super heater and jet condensers.

5. What property of cold fluid remains constant in case of evaporator?

a) Volume

b) Temperature

c) Entropy

d) Enthalpy

Answer: b

Explanation: In case of evaporator the temperature remains constant as pressure does not change. The temperature of hot fluid decreases.

6. LMTD in case of condenser will ________ for counter and parallel flow heat exchanger.

a) Remain same

b) Not be equal

c) Cannot be determined

d) Will not change with time

Answer: a

Explanation: In case of condenser and evaporator ∆T1 and ∆T2 remains same in case of counter and parallel flow heat exchanger. Therefore LMTD in case of condenser will be same for counter and parallel flow heat exchangers.

7. How many categories heat exchanger can be classified based on nature of heat exchange process?

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: The classification of heat exchanger on the basis of nature of heat exchange:

Direct contact

Recuperator

Regenerator

8. Which type of flow arrangement is this?

a) Parallel

b) Counter

c) Regenerator

d) Shell and tube

Answer: b

Explanation: In this type of arrangement fluids enter and leaves from opposite side, also flow in opposite direction. Therefore in this arrangement type flow is counter flow.

9. Which type of flow arrangement is this?

a) Parallel

b) Counter

c) Regenerator

d) Shell and tube

Answer: a

Explanation: In this type of arrangement fluids enter and leaves from same side, also flow in same direction. Therefore in this arrangement type flow is parallel flow.

10. In which of the following recuperator heat exchanger is not used?

a) Evaporator

b) Chemical factories

c) Automobile radiators

d) Condensers

Answer: b

Explanation: Recuperator are heat exchanger in which two fluids are separated by a wall. Recuperator is not used in chemical factories. It is used in condenser, evaporator and automobile radiator.

11. Number of categories heat exchanger can be classified based on the direction of flow of fluids is ______

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: The classification of heat exchanger on the basis of the direction of flow of fluids are:

Counter flow heat exchanger

Parallel flow heat exchanger

Cross flow heat exchangers

12. Number of categories heat exchanger can be classified based on the mechanical design of heat exchanger surface is ______

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: The classification of heat exchanger on the basis of the mechanical design of heat exchanger surface:

Concentric tube

Shell and tube

Multiple shell

13. Number of categories heat exchanger can be classified based on the physical state of heat exchanging fluid is ____

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: The classification of heat exchanger on the basis of the physical state of heat exchanging fluid:

Condenser

Evaporator

14. In which of the following type of heat exchanger the heat exchange between the two fluids occur by their complete physical mixing?

a) Direct contact heat exchanger

b) Indirect contact heat exchanger

c) Recuperator

d) Regenerator

Answer: a

Explanation: In direct contact type of heat exchanger the heat exchange between the two fluids occur by their complete physical mixing. Recuperator are heat exchanger in which two fluids are separated by a wall. Regenerator heat exchanger in which the same space is occupied by the hot and cold gases, between which heat is exchanged.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “ Heat Transfer – Fins”.

1. A finned surface consists of root or base area of 1m 2 and fin surface area of 2m 2 . The average heat transfer coefficient for finned surface is 20W/m 2 K effectiveness of fins provided is 0.75. If finned surface with surface with root or base temperature of 50°C is transferring heat to fluid of 30°C, then the rate of heat transfer without using fin is _______

a) 400W

b) 800W

c) 1000W

d) 1200W

Answer: b

Explanation: (Q without ) = h A (T h – T infinity )

Q = 20×2×

Q = 800W.

2. Heat pipe is widely used nowadays because _________

a) It acts as an insulator

b) It acts as conductor and insulator

c) It acts as a superconductor

d) It acts as a fin

Answer: c

Explanation: Heat pipe is a heat transfer device. The thermal conductance of heat pipe may be several hundred 500times than that best available metal conductor, hence they act as super conductor.

3. Which the following statements pertaining to heat transfer through fins?

a) Fins are equally effective irrespective of whether they are on the hot side or cold side of the fluid

b) The temperature along the fin is variable and hence the rate of heat transfer varies along the elements of the fin

c) The fins may be made of materials that have a lower conductivity than the material of the wall

d) Fins must be arranged at right angles to the direction of the flow of the working fluid

Answer: b

Explanation: Effectiveness of fin = \(\sqrt{\frac{kP}{hA}}\)

Fins are made up of material as that of base metal is integral but higher thermal conductivity metal are preferable as temperature drop will be close to uniform temperature.

4. Addition of fin to the surface increases the heat transfer if \Missing open brace for subscript Equal to one

b) Greater than one

c) Less than one

d) Greater than one but less than two

Answer: c

Explanation: ε = \(\sqrt{\frac{kP}{hA}}\)

For increased heat transfer, εfin should be higher. Hence εfin should be greater than one \(\sqrt{\frac{kP}{hA}}\) should be less than one.

5. The insulated tip temperature of a rectangular longitudinal fin having an excess  root temperature of θ 0 is _____

a) θ 0 tanh

b) θ 0 /sinh

c) θ 0 tanh/

d) θ 0 /cosh

Answer: d

Explanation: Temperature distribution for insulated tip

\(\frac{θ}{θ_0} = \frac{coshm}{cosh⁡} \)

X=l, θ=\(\frac{θ_0}{cosh⁡}\).

6. A fin length l, protrudes from a surface held at temperature T 0 ; it being higher than the ambient temperature T a . The heat dissipation from the free end of the fin is stated to be negligibly small. What is the temperature gradient  x=l at the tip of the fin?

a) Zero

b) (T 0 -T l )/l

c) h(T 0 -T a )

d) (T l -T a )/(T 0 -T a )

Answer: a

Explanation: It is mentioned that heat dissipation for the free end of fin is stated to be negligibly small

 x=l = 0.

7. On heat transfer surface, fins are provided?

a) To increase temperature gradient so as to enhance heat transfer

b) To increase turbulence in flow for enhancing

c) To increase surface area to promote the rate of heat transfer

d) To decrease the pressure drop of the fluid

Answer: c

Explanation: Fins are the extended surface of a body. Fins are provided to increase the surface area to promote the rate of heat transfer.

8. Extended surface are used to increase the rate of heat transfer. When the convective heat transfer coefficient h=mk, the addition of extended surface will:

a) Increase the rate of heat transfer

b) Decrease the rate of heat transfer

c) Not increase the rate of heat transfer

d) Increase the rate of heat transfer when the length of the fin is very large

Answer: c

Explanation: For the fins, Effectiveness = \(\sqrt{\frac{kP}{hA}}=\frac{km}{h} \) And when h = mk effectiveness will become unity and by providing fin, heat transfer rate will be unity.

9. Three fins of equal length and diameter but made of aluminum, brass and cast iron are heated to 2000C at one end. If the fins dissipate heat to the surrounding air is 250C the temperature at the end will be least in case of ______

a) Aluminum fin

b) Brass fin

c) Cast-iron fin

d) Each fin have the same temperature at the free end

Answer: c

Explanation: Effectiveness = \(\sqrt{\frac{kP}{hA}}\)

Effectiveness is proportional to thermal conductivity. In the given material thermal conductivity is minimum for cast iron. Therefore of cast iron is least effective.

10. An increase in pin fin effectiveness is caused by high value of _________

a) Convective coefficient

b) Change in temperature

c) Thermal conductivity

d) Sectional Area

Answer: c

Explanation: Effectiveness α \

 

 should be high. Effectiveness is directly proportional to thermal conductivity.

11. Which one of the following configuration has the highest fin effectiveness?

a) Thin, closely spaced fins

b) Thin, widely spaced fins

c) Thick, widely spaced fins

d) Thick, closely spaced fins

Answer: a

Explanation: Effectiveness = \

 

 

 and with small cross sectional area . Hence instead of using 1fin with large cross section area we can use number of thin fins to keep the term \(\sqrt{\frac{P}{A}}\) high i.e. to keep high effectiveness.

12. A fin will be necessary and effective only when _____

a) k is small and h is large

b) k is large and h is also large

c) k is small and h is also small

d) k is large and h is small

Answer: d

Explanation: Effectiveness α \

 

 and inversely proportional to h .

This set of Power Plant Engineering Multiple Choice Questions & Answers  focuses on “Rankine cycle”.

1. What is the unit of Heat rate?

a) kJ/KW

b) KW/kJ

c) kJ

d) KW

Answer: a

Explanation: Heat rate is the rate of input required to produce unit shaft output.

2. Rankine cycle operating on low pressure limit of p1 and high pressure limit of p2 ___________

a) has higher thermal efficiency than the Carnot cycle operating between same pressure limits

b) has lower thermal efficiency than Carnot cycle operating between same pressure limits

c) has same thermal efficiency as Carnot cycle operating between same pressure limits

d) may be more or less depending upon the magnitudes of p1 and p2

Answer: a

Explanation: Area under P-V curve for Rankine will be more as compared to Carnot cycle.

3. Rankine efficiency of a Steam Power Plant ___________

a) improves in Summer as compared to that in Winter

b) improves in Winter as compared to that in Summer

c) is unaffected by climatic conditions

d) none of the mentioned

Answer: b

Explanation: In winters, the temperature of cooling water is low, which increases Condenser’s efficiency.

4. Rankine cycle comprises of ___________

a) two isentropic processes and two constant volume processes

b) two isentropic processes and two constant pressure processes

c) two isothermal processes and two constant pressure processes

d) none of the mentioned

Answer: b

Explanation: Rankine cycle is a reversible cycle which have two constant pressure and two constant temperature processes.

5. In Rankine cycle, the work output from the turbine is given by ___________

a) change of internal energy between inlet and outlet

b) change of enthalpy between inlet and outlet

c) change of entropy between inlet and outlet

d) change of temperature between inlet and outlet

Answer: b

Explanation: Work output = h1 – h2.

6. Which of the following contributes to the improvement of efficiency of Rankine cycle in a Thermal Power Plant?

a) reheating of steam at intermediate stage

b) regeneration use of steam for heating Boiler feed water

c) use of high pressures

d) all of the mentioned

Answer: d

Explanation: The regenerative features effectively raise the nominal cycle heat input temperature, by reducing the addition of heat from the Boiler/fuel source at the relatively low feedwater temperatures that would exist without regenerative feedwater heating.

7. Match the following:

i) Boiler                 A. reversible adiabatic expansion of steam

ii) turbine               B. constant pressure heat heat addition

iii) Condenser            C. reversible adiabatic compression

iv) pump                  D. constant pressure heat rejection

a) i-B ii-A iii-D iv-C

b) i-A ii-C iii-D iv-A

c) i-B ii-D iii-C iv-A

d) i-A ii-D iii-B iv-C

Answer: a

Explanation: Working fluid in Rankine cycle undergoes 4 processes, expansion in turbine, heat addition in Boiler, heat rejection in Condenser and compression in pump.

8. What is the actual turbine inlet temperature in Rankine cycle?

a) 700C

b) 800C

c) 550C

d) 1150C

Answer: c

Explanation: The TIT is of the range 500-570C.

9. Rankine cycle efficiency of a good Steam Power Plant may be in the range of?

a) 15 to 20%

b) 35 to 45%

c) 70 to 80%

d) 90 to 95%

Answer: b

Explanation: Efficiency of Rankine cycle in actual working condition is found to be between 35 to 45%.

10. A simple Rankine cycle operates the Boiler at 3 MPa with an outlet temperature of 350°C and the Condenser at 50 kPa. Assuming ideal operation and processes, what is the thermal efficiency of this cycle?

a) 7.7

b) 17.7

c) 27.7

d) 37.7

Answer: c

Explanation: Fixing the states; h1 = 340.5 kJ/kg, h2 = h1 + v1  = 343.5 kJ/kg, h3 = 3115.3 kJ/kg, s3 = 6.7428 kJ/kg – K, x4 = 0.869, and h4 = 2343.9 kJ/kg. Thus, η = 1 – Qout / Qin = 1 –  /  = 27.7%.

11. A simple Rankine cycle produces 40 MW of power, 50 MW of process heat and rejects 60 MW of heat to the surroundings. What is the utilization factor of this co generation cycle neglecting the pump work?

a) 50

b) 60

c) 70

d) 80

Answer: b

Explanation: Application of the first law to the entire cycle gives Qin = Qp + Qreject + W = 150 MW. The utilization factor is then =  / Qin = 60%.

This set of Thermodynamics Multiple Choice Questions & Answers  focuses on “Simple Steam Power Cycle and Rankine Cycle”.

1. A power cycle continuously converts ____ into ____

a) heat, heat

b) work, heat

c) heat, work

d) work, work

Answer: c

Explanation: Here heat is the energy released by burning of fuel and work is done as shaft work.

2. In the vapour power cycle, working fluid undergoes a change of phase.

a) true

b) false

Answer: a

Explanation: Here working fluid is water.

3. The path followed in a vapour power cycle is

a) boiler-condenser-turbine-pump

b) boiler-turbine-condenser-pump

c) boiler-turbine-pump-condenser

d) boiler-pump-turbine-condenser

Answer: b

Explanation: In the boiler, water takes heat then expands in turbine going into condenser where it condenses into water and then it is pumped back into boiler.

4. For a fluid undergoing cycle process,

a) there is no net change in its internal energy

b) energy transfer as heat is equal to the energy transfer as work

c) both of the mentioned

d) none of the mentioned

Answer: c

Explanation: When a fluid undergoes a cycle process, this changes take place.

5. For a vapour power cycle,

a) net heat input is converted into net work output

b) Q1-Q2 = Wt-Wp

c) efficiency = 1 – 

d) all of the mentioned

Answer: d

Explanation: Here Q1 is the heat transferred to the fluid and Q2 is the heat rejected, Wt is work transferred from fluid and Wp is work transferred into fluid.

6. In a Rankine cycle, all the processes are ideal.

a) true

b) false

Answer: a

Explanation: The Rankine cycle is an ideal cycle and also a reversible cycle.

7. For a Rankine cycle, which of the following is true?

a) a reversible constant pressure heating process happens in steam boiler

b) reversible adiabatic expansion of steam in turbine

c) reversible constant pressure heat rejection in condenser

d) all of the mentioned

Answer: d

Explanation: All the processes are ideal in Rankine cycle.

8. The liquid water handled by pump is

a) incompressible

b) with increase in pressure, there is a little change in density or specific volume

c) both of the mentioned

d) none of the mentioned

Answer: c

Explanation: In a pump, reversible adiabatic compression of liquid takes place.

9. The work ratio is defined as the ratio of

a) positive work output to net work output

b) net work output to positive work output

c) heat input to work output

d) none of the mentioned

Answer: b

Explanation: The work ratio = Wnet / Wt.

10. Steam rate is the rate of steam flow required to produce unit shaft output.

a) true

b) false

Answer: a

Explanation: It is the capacity of a steam plant and steam rate = 1/.

11. Heat rate is given by 

a) cycle efficiency

b) 3600 / cycle efficiency

c) cycle efficiency / 3600

d) cycle efficiency * 3600

Answer: b

Explanation: Heat rate is the rate input required to produce unit work output.

12. Which of the following statement is true?

a) during compression, specific volume of the fluid should be kept small

b) during expansion, specific volume of the fluid should be kept large

c) both of the mentioned

d) none of the mentioned

Answer: c

Explanation: The larger the specific volume, more is the work produced or consumed by the steady-flow device.

Answer: a

Explanation: This is the reason why steam power plants is so popular.

This set of Thermodynamics Multiple Choice Questions & Answers  focuses on “Ideal Regenerative Cycle and Regenerative Cycle”.

1. The mean temperature of heat addition can be increased by

a) increasing the amount of heat supplied at high temperatures

b) decreasing the amount of heat added at low temperatures

c) both of the mentioned

d) none of the mentioned

Answer: c

Explanation: These are the two ways of increasing mean temperature of heat addition.

2. In the ideal regenerative cycle, the condensate after leaving the pump circulates around the turbine casing.

a) true

b) false

Answer: a

Explanation: Through this heat transfer takes place between the vapour flowing through the turbine and liquid flowing around the turbine.

3. The efficiency of an ideal regenerative cycle is given by

a) 1 – 

b) 1 – 

c) 1 – 

d) none of the mentioned

Answer: b

Explanation: The efficiency of a cycle is given by 1 – .

4. The efficiency of an ideal regenerative cycle is ____ the Carnot cycle efficiency.

a) greater than

b) equal to

c) less than

d) none of the mentioned

Answer: b

Explanation: For both the cycles, efficiency is given by 1 – .

5. When compared with the Rankine cycle, the ideal regenerative cycle has

a) less net work output

b) more steam rate

c) more efficient

d) all of the mentioned

Answer: d

Explanation: These indiate that the ideal regenerative cycle is better than the Rankine cycle but it is not practicable.

6. The ideal regenerative cycle is not practicable because

a) reversible heat transfer can’t be obtained in finite time

b) heat exchanger in turbine is mechanically impracticable

c) there is high moisture content of steam in the turbine

d) all of the mentioned

Answer: d

Explanation: These reasons result in the ideal regenerative cycle being not used practically.

7. For a regenerative cycle, which of the following is true?

a) efficiency = /Q1

b) efficiency = /Q1

c) steam rate = 3600/

d) all of the mentioned

Answer: d

Explanation: These are the expressions for steam rate and cycle efficiency for a regenerative cycle.

8. The efficiency of regenerative cycle will be ____ the efficiency of the Rankine cycle.

a) greater than

b) equal to

c) less than

d) none of the mentioned

Answer: a

Explanation: The reason being, with regeneration, the mean temperature of heat addition increases.

9. Which of the following is an assumption for heaters?

a) they are adequately insulated

b) there is no heat gain from or heat loss to the surroundings

c) both of the mentioned

d) none of the mentioned

Answer: c

Explanation: This assumption is necessary for the heaters.

Answer: a

Explanation: This comes from the equation obtained for the regenerative cycle.

This set of tough Thermodynamics Questions focuses on “Mean Temperature of Heat Addition and Reheat Cycle”.

1. In the Rankine cycle, heat is added reversibly at

a) constant pressure and constant temperature

b) constant pressure and infinite temperature

c) infinite pressure and constant temperature

d) infinite pressure and infinite temperature

Answer: b

Explanation: This is a basic fact about Rankine cycle.

2. The efficiency of Rankine cycle is given by

a) 1 – 

b) 1 – 

c) 1 – 

d) none of the mentioned

Answer: c

Explanation: Here T2 is the temperature of heat rejection and Tmean is the mean temperature of heat addition.

3. Which of the following statement is true?

a) for given Tmean, lower is the T2, higher will be the efficiency of Rankine cycle

b) the lowest possible temperature of heat rejection is the surroundings temperature

c) higher is the mean temperature of heat addition, higher will be the efficiency

d) all of the mentioned

Answer: d

Explanation: The efficiency of the Rankine cycle = 1 – .

4. If we ____ the superheat at constant pressure then the cycle efficiency ____

a) decrease, increases

b) increase, decreases

c) increase, increases

d) decrease, decreases

Answer: c

Explanation: Increasing the superheat at constant pressure increases the mean temperature of heat addition and cycle efficiency also increases.

5. The maximum temperature of steam that can be used is not fixed.

a) true

b) false

Answer: b

Explanation: It is fixed from metallurgical considerations.

6. To prevent erosion of blades, quality should not fall below

a) 85%

b) 90%

c) 95%

d) 100%

Answer: a

Explanation: Thus the maximum moisture content which is allowed at the turbine exhaust is not to exceed 15%.

7. To fix the maximum steam pressure at the inlet of turbine we need to first fix

a) the maximum steam temperature at turbine inlet

b) minimum temperature of heat rejection

c) the minimum quality of steam at turbine exhaust

d) all of the mentioned

Answer: d

Explanation: These all are needed to be fixed to fix to maximum steam pressure at turbine inlet.

8. Which of the following is true about a reheat cycle?

a) used to limit the quality at turbine exhaust at 0.85 when steam pressure is higher than max

b) after partial expansion in turbine, steam is brought back to boiler

c) the steam is reheated by combustion gases

d) all of the mentioned

Answer: d

Explanation: This is the functioning of a reheat cycle.

9. The correct sequence of expansion in a reheat cycle is

a) HP turbine – LP turbine – constant pressure in boiler

b) HP turbine – constant pressure in boiler – LP turbine

c) LP turbine – constant pressure in boiler – HP turbine

d) LP turbine – HP turbine – constant pressure in boiler

Answer: b

Explanation: Here HP is the high pressure turbine and LP is the low pressure turbine.

10. Why is steam not allowed to to expand deep into two-phase region before being taken for reheating.

a) to protect the reheater tubes

b) to prevent solid deposits being left behind while evaporating

c) both of the mentioned

d) none of the mentioned

Answer: c

Explanation: These solid deposits are difficult to remove from the reheater tubes.

11. Why should the reheat pressure be optimized?

a) a low reheat pressure brings down the mean temperature of heat addition and hence the cycle efficiency

b) a high reheat pressure increases the moisture content at turbine exhaust

c) both of the mentioned

d) none of the mentioned

Answer: c

Explanation: These are the reasons why we need to optimize reheat pressure.

12. The optimum reheat pressure is ____ times that of the initial steam pressure.

a) 0.2

b) 0.23

c) 0.25

d) all of the mentioned

Answer: d

Explanation: It lies in the range 0.2-0.25 for most of the modern power plants.

13. With the use of reheat,

a) the net work output of the plant increases

b) there is only a marginal increase in cycle efficiency

c) the quality of steam at turbine exhaust is kept within a limit

d) all of the mentioned

Answer: d

Explanation: These can be considered the main advantages of using reheat.

Answer: b

Explanation: The maximum number of reheats used till now is two only.

This set of Power Plant Engineering Multiple Choice Questions & Answers  focuses on “Binary Vapour Cycle”.

1. Rankine cycle efficiency of a good steam power plant may be in the range of ___________

a) 15 to 20%

b) 35 to 45%

c) 70 to 80%

d) 90 to 95%

Answer: b

Explanation: The efficiency of a Rankine cycle lies in the range of 35-45 percent mostly.

2. Rankine cycle operating on low pressure limit of p 1 and high pressure limit of p 2 ___________

a) has higher thermal efficiency than the Carnot cycle operating between same pressure limits

b) has lower thermal efficiency than Carnot cycle operating between same pressure limits

c) has same thermal efficiency as Carnot cycle operating between same pressure limits

d) may be more or less depending upon the magnitudes of p 1 and p 2

Answer: a

Explanation: For two pressure limits, the thermal efficiency of a Rankine cycle is more than that of the Carnot cycle operating between the same pressure difference which is evident in the P-V diagrams of both the cycle.

3. Rankine efficiency of a steam power plant ___________

a) improves in summer as compared to that in winter

b) improves in winter as compared to that in summer

c) is unaffected by climatic conditions

d) none of the mentioned

Answer: b

Explanation: Rankine cycle efficiency depends upon condenser temperature.

4. Rankine cycle comprises of ___________

a) two isentropic processes and two constant volume processes

b) two isentropic processes and two constant pressure processes

c) two isothermal processes and two constant pressure processes

d) none of the mentioned

Answer: b

Explanation: A rankine cycle is a thermodynamic cycle which comprises of two isentropic and two isobaric processes which is a major thermodynamic cycle in various applications.

5. In Rankine cycle the work output from the turbine is given by ___________

a) change of internal energy between inlet and outlet

b) change of enthalpy between inlet and outlet

c) change of entropy between inlet and outlet

d) change of temperature between inlet and outlet

Answer: b

Explanation: The power output from the turbine is given by the change of internal energy between inlet and outlet.

6. Regenerative heating i.e., bleeding steam to reheat feed water to boiler ___________

a) decreases thermal efficiency of the cycle

b) increases thermal efficiency of the cycle

c) does not affect thermal efficiency of the cycle

d) may increase or decrease thermal efficiency of the cycle depending upon the point of extraction of steam

Answer: b

Explanation: Bleeding decreases the thermal efficiency of the cycle as it reheats feed water to the boiler.

7. Regenerative cycle thermal efficiency ___________

a) is always greater than simple Rankine thermal efficiency

b) is greater than simple Rankine cycle thermal efficiency only when steam is bled at particular pressure

c) is same as simple Rankine cycle thermal efficiency

d) is always less than simple Rankine cycle thermal efficiency

Answer: a

Explanation: A regeneration is done to increase the efficiency of a reheat cycle so naturally the efficiency of a regenerative cycle is much more than a Rankine cycle.

8. In a regenerative feed heating cycle, the optimum value of the fraction of steam extracted for feed heating ___________

a) decreases with increase in Rankine cycle efficiency

b) increases with increase in Rankine cycle efficiency

c) is unaffected by increase in Rankine cycle efficiency

d) none of the mentioned

Answer: b

Explanation: In case of regenerative feed heating cycle, the optimum value of the fraction of steam extracted for feed heating decreases with increase in Rankine cycle efficiency. As the efficiency of regenerative cycle is more than the Rankine cycle.

9. In a regenerative feed heating cycle, the greatest economy is affected ___________

a) when steam is extracted from only one suitable point of steam turbine

b) when steam is extracted from several places in different stages of steam turbine

c) when steam is extracted only from the last stage of steam turbine

d) when steam is extracted only from the first stage of steam turbine

Answer: b

Explanation: Steam extraction is an important factor in deciding the economy of a regenerative cycle. The greatest economy is affected when steam is extracted from several places in different stages of steam turbine.

10. What is the maximum percentage gain in Regenerative feed heating cycle thermal efficiency?

a) increases with number of feed heaters increasing

b) decreases with number of feed heaters increasing

c) remains same unaffected by number of feed heaters

d) none of the mentioned

Answer: a

Explanation: In case of regenerative feed heating cycle, the optimum value of the fraction of steam extracted for feed heating decreases with increase in Rankine cycle efficiency. As the efficiency of regenerative cycle is more than the Rankine cycle. Naturally, the maximum percentage gain in regenerative feed heating cycle thermal efficiency increases with number of feed heaters increasing.

This set of Power Plant Questions and Answers for Freshers focuses on “Binary Vapour Cycle – II”.

1. How can we differentiate Rankine cycle from Carnot cycle?

a) Heat addition process of Rankine cycle is reversible isothermal whereas heat addition process of Carnot cycle is reversible isobaric

b) Heat addition process of Rankine cycle is reversible isobaric whereas heat addition process of Carnot cycle is reversible isothermal

c) Heat addition process of Rankine cycle is reversible isentropic whereas heat addition process of Carnot cycle is reversible isothermal

d) Both cycles are identical except the working fluid used

Answer: b

Explanation: Heat addition process of Rankine cycle is reversible isobaric whereas heat addition process of Carnot cycle is reversible isothermal. This is one of the major differences in both the cycles.

2. What is the relation between efficiencies of Rankine cycle and Carnot cycle for the same pressure ratio?

a) (η r ) = (η c )

b) (η r ) > (η c )

c) (η r ) < (η c )

d) none of the mentioned

Answer: c

Explanation: The relation between efficiencies of Rankine cycle and Carnot cycle for the same pressure ratio is given by,

(η r ) < (η c ).

3. If T m be the mean temperature of heat addition in Rankine cycle as shown in diagram,

what will the formula for efficiency of Rankine cycle?

power-plant-questions-answers-freshers-q3

a) (η r ) = (T 3 / T m )

b) (η r ) = 1 – (T 3 / T m )

c) (η r ) = 1 – (T 2 / T m )

d) (η r ) = (T 2 / T m )

Answer: b

Explanation:The required formula becomes

(η r ) = 1 – (T 3 / T m ).

4. The maximum efficiency of Rankine cycle (η r ) is the function of ___________

a) the mean temperature of heat addition (T m ) only

b) the mean temperature of heat addition (T m ) and temperature of steam at the exit of the turbine

c) the mean temperature of heat addition (T m ) and temperature of steam at the entry of the turbine

d) the mean temperature of heat addition (T m ) and temperature of steam at exit of the condenser

Answer: a

Explanation: The maximum efficiency of Rankine cycle (η r ) is the function of

the mean temperature of heat addition (T m ) only.

5. What is the effect of superheated steam on efficiency of Rankine cycle?

a) efficiency of Rankine cycle decreases with increase in superheat of the steam

b) efficiency of Rankine cycle increases with increase in superheat of the steam

c) efficiency of Rankine cycle is not affected by change in superheat of the steam

d) none of the mentioned

Answer: b

Explanation: The effect of superheated steam on efficiency of Rankine cycle is that the efficiency of Rankine cycle increases with increase in superheat of the steam.

6. What is the effect of increase in pressure at which heat is added on the pump work in the Rankine cycle?

a) the pump work increases with increase in pressure of heat addition

b) the pump work decreases with increase in pressure of heat addition

c) the pump work does not change with increase in pressure of heat addition

d) the pump work either increases or decreases with increase in pressure of heat addition

Answer: a

Explanation: The effect of increase in pressure at which heat is added on the pump work in the Rankine cycle is that the pump work increases with increase in pressure of heat addition.

7. When the pressure at which heat is added in Rankine cycle increases, the moisture content at the turbine exhaust?

a) increases

b) decreases

c) remains same

d) cannot say

Answer: a

Explanation: With an increase in the pressure at which heat is added in Rankine cycle increases, the moisture content at the turbine exhaust increases.

8. What is the condition for increasing the chances of corrosion of blades of turbine?

a) decrease in the pressure difference between which the Rankine cycle operates

b) increase in the pressure difference between which the Rankine cycle operates

c) increases and decreases in the pressure difference between which the Rankine cycle operates

d) none of the mentioned

Answer: b

Explanation: The condition for increasing the chances of corrosion of blades of turbine is an increase in the pressure difference between which the Rankine cycle operates.

9. What is the maximum content of moisture allowed at the turbine exhaust in the steam power plant?

a) 50 %

b) 60 %

c) 30 %

d) 15 %

Answer: d

Explanation: The maximum content of moisture allowed at the turbine exhaust in the steam power plant is 15 %.

10. Which of these is a binary cycle?

a) Mercury-steam cycle

b) Mercury-water cycle

c) Mercury-Sodium cycle

d) None of the mentioned

Answer: a

Explanation: The binary cycle is mainly a Mercury-Steam cycle.

This set of Thermodynamics Multiple Choice Questions & Answers  focuses on “Binary Vapor Cycles”.

1. Which of the following fluid can be used in place of water?

a) diphenyl ether

b) aluminium bromide

c) mercury

d) all of the mentioned

Answer: d

Explanation: These fluids are better than water in high temperature range.

2. Which of the following statement is true?

a) only mercury has been used in place of water

b) diphenyl ether decomposes at high temperatures

c) both of the mentioned

d) none of the mentioned

Answer: c

Explanation: Also aluminium bromide is a possibility which can be considered.

3. Mercury is a better fluid in high temperature range.

a) true

b) false

Answer: a

Explanation: This is because its vaporization pressure is relatively low even at higher temperatures.

4. Why is mercury unsuitable at low temperatures?

a) its saturation pressure becomes very high

b) its specific volume is very low at such a high pressure

c) both of the mentioned

d) none of the mentioned

Answer: d

Explanation: Its saturation pressure becomes very low and specific volume is very large.

5. In a binary cycle, ____ cycles with ____ working fluid are coupled.

a) two, same

b) two, different

c) three, same

d) three, different

Answer: b

Explanation: In a binary cycle, heat rejected by one can be utilized by other.

6. To vaporize one kg of water, ____ kg of mercury must condense.

a) 5-6

b) 6-7

c) 7-8

d) 8-9

Answer: c

Explanation: This depends on the properties of mercury and water.

7. When mercury cycle is added to the steam cycle,

a) the mean temperature of heat addition increases

b) efficiency decreases

c) maximum pressure is high

d) all of the mentioned

Answer: a

Explanation: The increase in the mean temperature of heat addition increases the efficiency and the maximum pressure is also low.

8. Initially in a reciprocating steam engine,

a) a binary cycle was used

b) steam was used in the high temperature

c) ammonia or sulphur dioxide was used in the low temperature range

d) all of the mentioned

Answer: d

Explanation: Steam from engine at higher temperature and pressure was used to evaporate sulphur dioxide or ammonia which operated in another cycle.

9. In a mercury-steam cycle, mercury cycle is called ____ and steam cycle is called ____

a) bottoming cycle, topping cycle

b) topping cycle, bottoming cycle

c) both are called bottoming cycle

d) both are called topping cycle

Answer: b

Explanation: The mercury-steam cycle represents a two-fluid cycle.

Answer: a

Explanation: In this three-fluid cycle, sulphur dioxide cycle is added in the low temperature range.

This set of Thermodynamics Multiple Choice Questions & Answers  focuses on “Mass Balance and Energy Balance in a Simple Steady Flow Process”.

1. Equation of continuity comes from

a) conservation of energy

b) conservation of mass

c) conservation of work

d) conservation of heat

Answer: b

Explanation: w1=w2 i.e., we get 1=2 and this is called equation of continuityMisplaced &.

2. In a flow process, the work transfer may be of which type?

a) external work

b) flow work

c) all of the mentioned

d) none of the mentioned

Answer: c

Explanation: Flow work is the displacement work and external work mostly comprises of shaft work.

3. The total rate of flow of all energy streams entering the control volume must equal to that of leaving the control volume.

a) true

b) false

Answer: a

Explanation: Given statement is true by the conservation of energy.

4. Which of the following represents the steady flow energy equation?

a) Q+Wx=-/2+g

b) Q+Wx=+/2+g

c) Q-Wx=-/2+g

d) Q-Wx=+/2+g

Answer: d

Explanation: This equation is the general form of SFEE and it involves conservation of mass and energy.

5. When more than one fluid stream is in a control volume, which of the following is more convenient?

a) energy flow per unit time

b) energy flow per unit mass

c) all of the mentioned

d) none of the mentioned

Answer: a

Explanation: It makes calculations less difficult.

6. In the differential form, the SFEE becomes

a) dQ+dW=dh+VdV+gdZ

b) dQ-dW=dh+VdV+gdZ

c) dQ+dW=dh-VdV-gdZ

d) dQ-dW=dh-VdV+gdZ

Answer: b

Explanation: This equation is the differential form of SFEE.

7. The steady flow energy equation is applied to which of the following processes?

a) pipe line flows

b) heat transfer processes

c) combustion processes

d) all of the mentioned

Answer: d

Explanation: These are the applications of SFEE.

8. When more than one fluid stream enters or leaves the control volume, which type of balance is taken?

a) mass balance

b) energy balance

c) mass balance and energy balance

d) none of the mentioned

Answer: c

Explanation: Both energy and mass balance are considered here.

9. What are the different kinds of external work?

a) shear work

b) electrical work

c) all of the mentioned

d) none of the mentioned

Answer: c

Explanation: Given two kinds of external work are important.

Answer: c

Explanation: At inlet, flow work=-pvdm and at exit, flow work=pvdm.

This set of Thermodynamics Multiple Choice Questions & Answers  focuses on “Second Law of Thermodynamics”.

1. Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate of waste heat rejection to a nearby river is 50 MW, determine the net power output for this heat engine. thermodynamics-questions-answers-second-law-thermodynamics-q1

a) 30 MW

b) 40 MW

c) 50 MW

d) 60 MW

Answer: a

Explanation: Net power output = 80 – 50 MW = 30 MW.

2. Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate of waste heat rejection to a nearby river is 50 MW, determine the thermal efficiency for this heat engine. thermodynamics-questions-answers-second-law-thermodynamics-q1

a) 47.5 %

b) 27.5 %

c) 37.5 %

d) none of the mentioned

Answer: c

Explanation: The thermal efficiency of heat engine = net work output / heat input

= 30/80 = 0.375 = 37.5 %.

3. A car engine with a power output of 50 kW has a thermal efficiency of 24 percent. Determine the fuel consumption rate of this car if the fuel has a heating value of 44,000 kJ/kg . thermodynamics-questions-answers-second-law-thermodynamics-q3

a) 0.00273 kg/s

b) 0.00373 kg/s

c) 0.00473 kg/s

d) 0.00573 kg/s

Answer: c

Explanation: Q = 50/0.24 = 208.3 kW,

hence fuel consumption rate = 208.3 kW / 44000 kJ/kg = 0.00473 kg/s.

4. The food compartment of a refrigerator is maintained at 4°C by removing heat from it at a rate of 360 kJ/min. If the required power input to the refrigerator is 2kW, determine the coefficient of performance of the refrigerator.

thermodynamics-questions-answers-second-law-thermodynamics-q4

a) 4

b) 3

c) 2

d) 1

Answer: b

Explanation: COP =  = 3.

5. The food compartment of a refrigerator is maintained at 4°C by removing heat from it at a rate of 360 kJ/min. If the required power input to the refrigerator is 2kW, determine the rate of heat rejection to the room that houses the refrigerator. thermodynamics-questions-answers-second-law-thermodynamics-q4

a) 450 kJ/min

b) 460 kJ/min

c) 470 kJ/min

d) 480 kJ/min

Answer: d

Explanation: Q = 360 +  = 480 kJ/min.

6. A heat pump is used to meet the heating requirements of a house and maintain it at 20°C. On a day when the outdoor air temperature drops to 2°C, the house is estimated to lose heat at a rate of 80,000 kJ/h. If the heat pump under these conditions has a COP of 2.5, determine the power consumed by the heat pump. thermodynamics-questions-answers-second-law-thermodynamics-q6

a) 32000 kJ/h

b) 33000 kJ/h

c) 34000 kJ/h

d) 35000 kJ/h

Answer: a

Explanation: W = Q/COP = 80000 kJ/h / 2.5 = 32000 kJ/h.

7. A heat pump is used to meet the heating requirements of a house and maintain it at 20°C. On a day when the outdoor air temperature drops to 2°C, the house is estimated to lose heat at a rate of 80,000 kJ/h. If the heat pump under these conditions has a COP of 2.5, determine the rate at which heat is absorbed from the cold outdoor air. thermodynamics-questions-answers-second-law-thermodynamics-q6

a) 32000 kJ/h

b) 48000 kJ/h

c) 54000 kJ/h

d) 72000 kJ/h

Answer: b

Explanation: The rate at which heat is absorbed = 80000 – 32000 = 48000 kJ/h.

8. An air-conditioner provides 1 kg/s of air at 15°C cooled from outside atmospheric air at 35°C. Estimate the amount of power needed to operate the air-conditioner.

a) 1.09 kW

b) 1.19 kW

c) 1.29 kW

d) 1.39 kW

Answer: d

Explanation: Q = m*cp* = 20.08 kW

COP = / = 14.4

hence power needed = 20/14.4 = 1.39 kW.

9. A cyclic machine, as shown below, receives 325 kJ from a 1000 K energy reservoir. It rejects 125 kJ to a 400 K energy reservoir and the cycle produces 200kJ of work as output. Is this cycle reversible, irreversible, or impossible?

a) reversible

b) irreversible

c) impossible

d) none of the mentioned

Answer: c

Explanation: The Carnot efficiency = 1 –  = 0.6 and real efficiency =  = 0.615 which is greater than the Carnot efficiency hence cycle is impossible.

10. In a cryogenic experiment you need to keep a container at -125°C although it gains 100 W due to heat transfer. What is the smallest motor you would need for a heat pump absorbing heat from the container and rejecting heat to the room at 20°C?

a) 97.84 kW

b) 98.84 kW

c) 99.84 kW

d) 95.84 kW

Answer: a

Explanation: COP = 1.022 and thus power required = 100/1.022 = 97.84 kW.

Answer: b

Explanation: W = thermal efficiency * Q thus Q = 1/ = 0.952 kJ.

This set of Engineering Materials & Metallurgy Multiple Choice Questions & Answers  focuses on “Thermometry and the Types of Thermometers”.

1. Measurement of elevated temperatures is defined as ___________

a) Thermometry

b) Pyrometry

c) Metallography

d) Radiography

Answer: b

Explanation: Pyrometry deals with elevated temperatures, generally around 950 F. The apparatus that is used in this process is known as a pyrometer. Thermometry generally deals with the measurement of temperatures below 950 F.

2. What temperature does the dark red color generally deal with?

a) 950 F

b) 1150 F

c) 1175 F

d) 1300 F

Answer: b

Explanation: Temperature of metals can be estimated by simply looking at the color of the hot body. Dark red is assigned a temperature of 1150 F, whereas for faint red, dark cherry, and cherry red it is 950 F, 1175 F, and 1300 F in that order.

3. What temperature is the dark orange color associated with?

a) 1475 F

b) 1650 F

c) 1750 F

d) 1800 F

Answer: b

Explanation: Temperature of metals can be estimated by simply looking at the color of the hot body. Dark orange is associated with a temperature of about 1150 F, while for bright cherry, orange, and yellow it is 1475 F, 1750 F, and 1800 F correspondingly.

4. Bimetallic strips are employed in ________ thermometers.

a) Vapor-pressure

b) Liquid-expansion

c) Metal-expansion

d) Resistance

Answer: c

Explanation: Bimetallic strips made by bonding of high-expansion and low-expansion metals are used in the common thermostat. When used as an industrial temperature indicator, these can be bent into a coil.

5. Bimetallic strips contain _______ as a metal.

a) Muntz metal

b) Yellow brass

c) Bronze

d) Aluminum

Answer: b

Explanation: Bimetallic strips include invar as one metal and yellow brass as another. For higher temperatures, nickel alloy can be used. These can be used in temperatures ranging from -100 F to 1000 F.

6. Why is invar used in bimetallic strips?

a) Low density

b) Low coefficient of expansion

c) High-temperature resistance

d) High abrasion resistance

Answer: b

Explanation: Most bimetallic strips are composed of invar and yellow brass as metal. Invar has the advantage of low coefficient of expansion, whereas yellow brass has the ability to be used at low temperatures.

7. _______ is commonly used in liquid-expansion thermometers.

a) Bourdon tube

b) Spinning rotor gauge

c) McLeod gauge

d) Manometer

Answer: a

Explanation: Liquid-expansion thermometers consist of a bulb and an expansible device. The bulb is exposed to the temperature that needs to be measured and usually, a Bourdon tube is used as an expansion device. These are connected by capillary tubing and are filled with a medium.

8. Resistance thermometer generally makes use of ________ for the measurement of resistance.

a) Potentiometer

b) Adruino

c) Diode bridge

d) Wheatstone bridge

Answer: d

Explanation: Resistance thermometers are based on the principle of increase in electrical resistance with increasing temperature. It consists of a resistance coil mounted in a protecting tube which is connected to a resistance measuring instrument. Generally, Wheatstone bridge is used in this process.

9. Which of these materials is not used for resistance coils?

a) Nickel

b) Copper

c) Titanium

d) Platinum

Answer: c

Explanation: Resistance coils are generally made of nickel, copper, or platinum. Nickel and copper can be used in the temperature range of 150-500 F, whereas platinum can be used between -350 to 1100 F.

10. Liquid expansion thermometers are filled with ________

a) Mercury

b) Amalgam

c) Gallium

d) Cesium

Answer: a

Explanation: The liquid-expansion thermometer has the entire system filled with an organic liquid or mercury. Mercury is used at a temperature range of -35 to 950 F. Alcohol and creosote are used at -110 to 160 F, and 20 to 400 F respectively.

This set of Thermodynamics Multiple Choice Questions & Answers  focuses on “Pure Substance”.

1. Which of the following represents the specific volume during phase transition.

a) Vf-Vg

b) Vg-Vf

c) Vf+Vg

d) none of the mentioned

Answer: b

Explanation: Here Vg is the specific volume of the saturated vapour and Vf is the specific volume of the saturated liquid.

2. At critical point, value of Vg-Vf is

a) two

b) one

c) zero

d) infinity

Answer: c

Explanation: As pressure increases, there is a decrease in Vg-Vf and at critical point its value becomes zero.

3. Above the critical point, the isotherms are continuous curves.

a) true

b) false

Answer: a

Explanation: These continuous curves approach equilateral hyperbolas at large volumes and low pressures.

4. A rigid tank contains 50 kg of saturated liquid water at 90°C. Determine the pressure in the tank and the volume of the tank.

a) 0.0518 m 3

b) 0.0618 m 3

c) 0.0718 m 3

d) 0.0818 m 3

Answer: a

Explanation: P = [email protected] C = 70.183 kPa

v = [email protected] C = 0.001036 m 3 /kg

Total volume of the tank = mv = ( 0.001036 m 3 /kg)

= 0.0518 m 3 .

5. A piston –cylinder device contains 0.06m 3 of saturated water vapour at 350 kPa pressure. Determine the temperature and mass of the vapour inside the cylinder.

a) 0.104 kg

b) 0.124 kg

c) 0.134 kg

d) 0.114 kg

Answer: d

Explanation: T = [email protected] = 138.86°C

v = [email protected] = 0.52422 m 3 /kg

m = V/v = 0.06 m 3 /0.52422 m 3 /kg = 0.114 kg.

6. A rigid tank contains 10 kg of water at 90°C. If 8 kg of the water is in the liquid form and the rest is in the vapour form, determine the pressure in the tank.

a) 60.183 kPa

b) 70.183 kPa

c) 80.183 kPa

d) 90.183 kPa

Answer: b

Explanation: P = [email protected] °C = 70.183 kPa.

7. A rigid tank contains 10 kg of water at 90°C. If 8 kg of the water is in the liquid form and the rest is in the vapour form, determine the volume of the tank.

a) 1.73 m 3

b) 2.73 m 3

c) 3.73 m 3

d) 4.73 m 3

Answer: d

Explanation: P = [email protected] °C = 70.183 kPa

@ 90°C, vf = 0.001036 m 3 /kg and vg = 2.3593 m 3 /kg

V = Vf + Vg = mf vf + mg vg = 4.73 m 3 .

8. An 80 litre vessel contains 4 kg of R-134a at a pressure of 160 kPa. Determine the temperature.

a) -10.60°C

b) -13.60°C

c) -15.60°C

d) -19.60°C

Answer: c

Explanation: v = V/m = 0.080 m 3 /4 kg = 0.02 m 3 /kg

@ 160kPa, vf = 0.0007437 m 3 /kg; vg = 0.12348 m 3 /kg

vf < v < vg Therefore T = [email protected] = -15.60°C.

9. An 80 litre vessel contains 4 kg of R-134a at a pressure of 160 kPa. Determine the quality.

a) 0.127

b) 0.137

c) 0.147

d) 0.157

Answer: d

Explanation: v = V/m = 0.080 m 3 /4 kg = 0.02 m 3 /kg

@ 160kPa, vf = 0.0007437 m 3 /kg; vg = 0.12348 m 3 /kg.

vf < v < vg

x = / vfg = 0.157.

10. An 80 litre vessel contains 4 kg of R-134a at a pressure of 160 kPa. Determine the volume occupied by the vapour phase.

a) 0.0775 m 3

b) 0.0575 m 3

c) 0.0975 m 3

d) 0.0375 m 3

Answer: a

Explanation: v = V/m = 0.080 m 3 /4 kg = 0.02 m 3 /kg

@ 160kPa, vf = 0.0007437 m 3 /kg; vg = 0.12348 m 3 /kg

vf < v < vg

x = / vfg = 0.157

mg = x*m = 0.628kg

Vg = mg*vg = 0.0775 m 3 or 77.5 litre.

Answer: b

Explanation: v = RT/P = * /

= 0.026325 m 3 /kg.

This set of Fluid Mechanics Multiple Choice Questions & Answers  focuses on “Thermodynamic Relations”.

1. The symbol of Helmholtz free energy is_________

a) A

b) H

c) B

d) E

Answer: a

Explanation: Helmholtz free energy is defined as the thermodynamic potential that measures the useful work obtained from a closed thermodynamic system. It is done at a constant volume and temperature. IUPAC name is ‘A’.

2. Which among the following is the formula for Helmholtz free energy?

a) U+TS

b) U+TV

c) U-TS

d) UTV

Answer: c

Explanation: Helmholtz free energy is defined as the thermodynamic potential that measures the useful work obtained from a closed thermodynamic system. It is done at a constant volume and temperature. IUPAC name is ‘A’. .

3. What is the unit of Helmholtz free energy?

a) Kelvin

b) Joule

c) Kilowatt

d) Newton

Answer: b

Explanation: Helmholtz free energy is defined as the thermodynamic potential that measures the useful work obtained from a closed thermodynamic system. The SI unit of Helmholtz free energy is Joule.

4. What is the symbol for Gibbs free energy?

a) A

b) H

c) G

d) E

Answer: c

Explanation: Gibbs free energy is defined as the thermodynamic potential that is used to calculate the maximum amount of reversible work. It takes place at a constant pressure and temperature. The IUPAC name for Gibbs free energy is ‘G’.

5. Gibbs free energy is also known as______

a) Free energy

b) Free entropy

c) Free enthalpy

d) Free motion

Answer: c

Explanation: Gibbs free energy is defined as the thermodynamic potential that is used to calculate the maximum amount of reversible work. Gibbs free energy is also known as free enthalpy. It takes place at a constant pressure and temperature.

6. What is the unit of Gibbs free energy?

a) Kelvin

b) Joule

c) Kilowatt

d) Newton

Answer: b

Explanation: Gibbs free energy is defined as the thermodynamic potential that is used to calculate the maximum amount of reversible work. It takes place at a constant pressure and temperature. The SI unit is Joule.

7. Which among the following is the formula for Gibbs free energy?

a) H-T∆S

b) H-T

c) T∆S

d) H-S

Answer: a

Explanation: Gibbs free energy is defined as the thermodynamic potential that is used to calculate the maximum amount of reversible work. It takes place at constant temperature and pressure. It is given by H-T∆S.

8. All the energy relations satisfy the mathematical condition in thermodynamics.

a) True

b) False

Answer: a

Explanation: All the thermodynamic energy relations satisfy the mathematical condition of being a set of continuous variables. They are a function of state variables themselves.

9. The performance of a flow device is expressed in terms of _________

a) Adiabatic efficiency

b) Isentropic efficiency

c) Thermal efficiency

d) Mechanical efficiency

Answer: b

Explanation: The performance of a flow device is expressed in terms of its isentropic efficiency. The actual performance of the device is compared with that of its isentropic device. It happens at the same inlet and exit conditions.

10. Isentropic efficiency is defined as ________

a) The power output of actual turbine/ power output if the turbine were isentropic

b) The power input of actual turbine/ power output if the turbine were isentropic

c) The power output of actual turbine/ power output if the turbine were adiabatic

d) The power output of actual turbine/ power output if the turbine were polytropic

Answer: a

Explanation: The performance of a flow device is expressed in terms of its isentropic efficiency. The actual performance of the device is compared with that of its isentropic device. It happens at the same inlet and exit conditions.

This set of Thermodynamics Multiple Choice Questions & Answers  focuses on “Ideal Gas-1”.

1. An ideal gas is one which obeys the law pv=RT at all pressures and temperatures.

a) true

b) false

Answer: a

Explanation: Though such a gas is hypothetical.

2. The value of universal gas constant is

a) 8.2353

b) 8.3143

c) 8.5123

d) none of the mentioned

Answer: b

Explanation: This value comes from the Avogadro law when we put p=760 mm Hg = 1.013*10^5 N/m^2 , T=273.15 K , v=22.4 m^3/kgmol.

3. Which of the following statement is true?

a) characteristic gas constant is given by dividing the universal gas constant by the molecular weight

b) Avogadro’s number  = 6.023 * 10^26 molecules/kgmol

c) Boltzmann constant  = 1.38 * 10^-23 J/molecule K

d) all of the mentioned

Answer: d

Explanation: The Boltzmann constant is given by universal gas constant divided by Avogadro number.

4. The equation of state of an ideal gas is given by

a) pV=mRT, here R is characteristic gas constant

b) pV=nRT, here R is universal gas constant

c) pV=NKT

d) all of the mentioned

Answer: d

Explanation: Characteristic gas constant is given by dividing the universal gas constant by the molecular weight and Boltzmann constant is given by universal gas constant divided by Avogadro number.

5. Specific heats are constant for an ideal gas.

a) true

b) false

Answer: a

Explanation: For an ideal gas, specific heats are constant and it satisfies the equation of state.

6. For real gases,

a) specific heats vary appreciably with temperature

b) specific heats vary little with pressure

c) both of the mentioned

d) none of the mentioned

Answer: c

Explanation: This is in contrast to an ideal gas for which specific heats are constant.

7. At constant temperature, 

a) u change when v or p changes

b) u does not change when v or p changes

c) u does not change when t changes

d) u always remains constant

Answer: b

Explanation: Internal energy does not change with change in v or p but changes only when temperature changes.

8. For an ideal gas, internal energy is a function of temperature only.

a) true

b) false

Answer: a

Explanation: This is known as Joule’s law.

9. Which of the following statement is true? 

a) the equation du=c*dT holds good for an ideal gas for any process

b) for gases other than ideal ones, the equation holds true for a constant volume process only

c) for an ideal gas c is constant and hence Δu=c*ΔT

d) all of the mentioned

Answer: d

Explanation: All these statements are true and also internal energy is a function of temperature only.

10. Which of the following statement is correct for an ideal gas?

a) h=u+pv

b) h=u+RT

c) h=f

d) all of the mentioned

Answer: d

Explanation: For an ideal gas, pv=RT and hence h is a function of temperature only.

Answer: c

Explanation: Δh=ΔT = cp*ΔT.

This set of Thermodynamics Multiple Choice Questions & Answers  focuses on “Ideal Gas-2”.

1. The value of cp and cv depend on

a) temperature of the gas

b) ɣ and R

c) pressure of the gas

d) all of the mentioned

Answer: b

Explanation: The value of cp and cv depends on the number of atoms in a molecule and the molecular weight of the gas.

2. Which of the following statement is true?

a) value of ɣ for monoatomic gases is 5/3

b) value of ɣ for diatomic gases is 7/5

c) for polyatomic gases, the value of ɣ is approximately taken as 4/3

d) all of the mentioned

Answer: d

Explanation: These values of ɣ can be shown by the classical kinetic theory of gases.

3. The maximum and minimum values of ɣ is

a) 1.33, 1

b) 2.00, 1

c) 1.67, 1

d) 1.25, 1

Answer: c

Explanation: ɣ=1 when cp=cv and ɣ=1.67=5/3 for monoatomic gases.

4. Which of the following equation can be used to compute the entropy change between any two states of an ideal gas?

a) s2-s1 = cv*ln + R*ln

b) s2-s1 = cp*ln – R*ln

c) s2-s1 = cp*ln + cv*ln

d) all of the mentioned

Answer: d

Explanation: Any of the given three equations can be used.

5. For a reversible adiabatic change, ds=0.

a) true

b) false

Answer: a

Explanation: For a reversible adiabatic process, change in entropy is zero.

6. For a reversible adiabatic process,

a) p*

ɣ

 = constant

b) T*

ɣ) = constant

c) T*

ɣ/ɣ) = constant

d) all of the mentioned

Answer: d

Explanation: All these relations come from the pv=RT and p*

ɣ

 = constant.

7. Which of the following is true for a polytropic process?

a) p

 is used to describe the process

b) it is not adiabatic

c) it can be reversible

d) all of the mentioned

Answer: d

Explanation: These are the properties of an adiabatic process.

8. In the equation p

, n is calculated by

a)  / 

b)  / 

c)  / 

d) none of the mentioned

Answer: b

Explanation: It comes from the p1*

 = p2*

.

9. For entropy change in a polytropic process, which of the following statement is true?

a) when n=ɣ, the entropy change becomes zero

b) if p2>p1, for n<=ɣ, the entropy of the gas decreases

c) for n>ɣ, the entropy of the gas increases

d) all of the mentioned

Answer: d

Explanation: This comes from the relation s2-s1 = [ɣ/{nɣ}]*R*ln.

10. Polytropic specific heat is given by cn=cvɣ/ .

a) true

b) false

Answer: a

Explanation: The polytropic specific heat is used in the relation Qr=cn*ΔT.

Answer: c

Explanation: This comes from the relation Qr=cn*ΔT and cn=cvɣ/.

This set of Chemical Process Calculation Multiple Choice Questions & Answers  focuses on “Ideal Gas – I”.

1. ___________is any measurable characteristics of a substance that can be calculated or deduced.

a) Property

b) State

c) Phase

d) None of the mentioned

Answer: a

Explanation: Property is any measurable characteristics of a substance that can be calculated or deduced.

2. ________ of a system gives the condition of a system as specified by its properties.

a) Property

b) State

c) Phase

d) None of the mentioned

Answer: b

Explanation: State of a system gives the condition of a system as specified by its properties.

3. A ________ is completely homogeneous and uniform state of matter

a) Property

b) State

c) Phase

d) None of the mentioned

Answer: c

Explanation: State of a system gives the condition of a system as specified by its properties.

4. A compound may consists of _______ phase.

a) Only one

b) Only two

c) Not more than two

d) One or more than one

Answer: d

Explanation: A compound may consists of one or more than one phase.

5. Mercury and water  in a same container would represent two different phases.

The statement is

a) Correct

b) Incorrect

c) Can`t say

d) None of the mentioned

Answer: a

Explanation: A compound may consists of one or more than one phase.

6. Incorrect equation for an ideal is

a) PV = nRT

b) Pv = RT

c) PM = ρRT

d) None of the mentioned

where: P – Absolute Pressure of the gas

V – Total volume occupied by the gas

n – Number of moles of the gas

T – Absolute temperature

v – Specific molar volume

ρ – Density of the gas

Answer: d

Explanation: All three are correct as a & b are ideal gas law and c is in the form of density.

7. For an ideal gas, the specific molar volume of the gas is doubled then the pressure would be 

a) Same as before

b) Double

c) Half

d) None of the mentioned

Answer: c

Explanation: Pressure and specific volumes are inversely proportional.

8. Specific molar volume for an ideal gas is

a) Volume per mass

b) Volume per molecular weight

c) Volume per mole

d) None of the mentioned

Answer: c

Explanation: Specific molar volume for an ideal gas is Volume per mole.

9. The inverse of molar volume is

a) Molar density

b) Mole fraction

c) Molar specific volume

d) None of the mentioned

Answer: a

Explanation: The inverse of molar volume is Molar density .

Answer: a

Explanation: Gas constant = R = PV/nT.

This set of Chemical Process Calculation Multiple Choice Questions & Answers  focuses on “Ideal gas – II”.

1. At constant temperature the pressure of an ideal gas is doubled its density becomes

a) Half

b) Double

c) Same

d) None

Answer: b

Explanation: Density is proportional to pressure.

2. For an ideal gas, incorrect statement is

a) Molecules do not occupy any space

b) No attractive force exist between the molecules

c) The gas molecules move in random, straight line motion

d) None of the mentioned

Answer: d

Explanation: Conditions for a gas to behave as predicated by the ideal gas law.

3. Correct statement is

a) Gases at low pressure or high temperature behave as an ideal gas

b) Gases at high pressure or low temperature behave as an ideal gas

c) Gases at high density behave as an ideal gas

d) None of the mentioned

Answer: a

Explanation: Gases at low pressure or high temperature behave as an ideal gas.

4. What is the volume of 20 gms of Oxygen in Litre at standard conditions?

a) 10

b) 12

c) 14

d) 16

Answer: c

Explanation: At standard conditions, 1 mole of of gas occupy 22.4 litre volume.

5. What is the density of Carbon-di-oxide in kg/m3 at 27 degree celcius and 100 KPa?

a) 1.76

b) 2.76

c) 3.76

d) 4.76

Answer: a

Explanation: density = /.

6. Usually while measuring the specific gravity of a gas, the reference gas that is taken is air.

The above given statement is-

a) False

b) True

c) Depends on the gas

d) None of the mentioned

Answer: b

Explanation: Air is usually taken at the same temperature and pressure.

7. What is the volume of 2.5 gm moles Nitrogen gas at STP? At 100 KPa pressure and 300 K temperature, 2 gm moles of the gas occupy 20 cubic meter volume.

a) 11.23

b) 22.46

c) 33.69

d) 44.92

Answer: d

Explanation: P1V1/n1T1 = P2V2/n2T2.

8. Which one has the lowest density at room temperature?

a) Ne

b) N2

c) NH3

d) CO

Answer: c

Explanation: NH3 have minimum molecular weight.

9. 1 atm pressure and 0°C’ condition is known as

a) Room temperature and pressure

b) Standard temperature and pressure

c) Atmospheric temperature and pressure

d) None of the mentioned

Answer: b

Explanation: 1 atm pressure and 0°C condition is known as standard temperature and pressure .

Answer: a

Explanation: V = nRT/P.

This set of Chemical Process Calculation Multiple Choice Questions & Answers  focuses on “Real gases”.

1. The set of physical conditions at which the density and other properties of a liquid and vapour become identical, is called

a) Corresponding state

b) Critical state

c) Equation of state

d) None of the mentioned

Answer: b

Explanation: The set of physical conditions at which the density and other properties of a liquid and vapour become identical is known as Critical state.

2. Critical point occurs at

a) Highest temperature and pressure at which gas and liquid can coexist

b) Lowest temperature and pressure at which gas and liquid can coexist

c) Highest temperature and lowest pressure at which gas and liquid can coexist

d) Lowest temperature and highest pressure at which gas and liquid can coexist

Answer: a

Explanation: Critical point occurs at highest temperature and pressure.

3. Material in a state above its critical point is

a) Critical fluid

b) Supercritical fluid

c) Plasma fluid

d) None of the mentioned

Answer: b

Explanation: Material in a state above its critical point is called Supercritical fluid.

4. “Any compound should have the same reduced volume at the same reduced pressure and reduced temperature.” Is known as

a) Law of corresponding states

b) Law of critical state

c) Law of reduced state

d) None of the mentioned

Answer: a

Explanation: Any compound should have the same reduced volume at the same reduced pressure and reduced temperature law of corresponding states.

5. Compressibility factor z is

a) z = PV/RT

b) z = Pv/RT

c) z = PM/RT

d) z = PV/nT

Where: P – Absolute Pressure of the gas

V – Total volume occupied by the gas

n – Number of moles of the gas

T – Absolute temperature

v – Specific molar volume

Answer: b

Explanation: z = Pv/RT.

6. For an ideal gas, the compressibility factor is

a) 0.5

b) 1.0

c) 1.5

d) 2.0

Answer: b

Explanation: z = Pv/RT = 1, for an ideal gas.

7. A parameter that indicates the degree of nonsphericity of a molecule is called

a) Compressibility factor

b) Acentric factor

c) Holburn factor

d) None of the mentioned

Answer: b

Explanation: A parameter that indicates the degree of nonsphericity of a molecule is called Acentric factor.

8. The summation of each of the partial pressure of the component in a system equals to the total pressure. Is

a) Dalton`s law

b) Boyle`s law

c) Charles`s Law

d) None of the mentioned

Answer: a

Explanation: The summation of each of the partial pressure of the component in a system equals to the total pressure is Dalton`s law.

9. For the Van der Waals real gas equation

Statement I: The unit of constant a is atm.L2/mole2

Statement II: The unit of constant b is L

a) Both statements are false

b) Only one can be true at a time

c) Both statement are true

d) None of the mentioned

Answer: b

Explanation: In the first statement, Specific molar volume is used and in second one, total volume. So for the given equation both statement can not be true at a time.

Answer: a

Explanation: Corrected and normalized conditions of temperature, pressure and volume, normalized by their respective critical conditions is called Reduced variables.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Psychrometers”.

1. Wet bulb depression, under saturated ambient air conditions _____________

a) Is always positive

b) Is always negative

c) Is always zero

d) May have value depending upon the dew point temperature

Answer: c

Explanation: The wet bulb depression, under saturated ambient air conditions is always zero. At RH=100% DBT=WBT=DPT. Hence depression is equal to zero.

2. What is the minimum temperature to which water can be cooled in a cooling tower?

a) The dew point temperature of air

b) The dew bulb temperature of air

c) They dry bulb temperature of air

d) The ambient air temperature

Answer: b

Explanation: At saturation condition the dry bulb, wet bulb and dew point temperature are same with 100% relative humidity. The minimum temperature to which water can be cooled in cooling tower is equal to the wet bulb temperature of the air.

3. When a stream of moist air is passed over a cold and dry cooling coil such that no condensation takes place, then the air stream will get along the line of __________

a) Constant wet bulb temperature

b) Constant dew point temperature

c) Constant relative humidity

d) Constant enthalpy

Answer: b

Explanation: When a stream of moist air is passed over a cold and dry cooling coil such that no condensation takes place, then the cooling is sensible cooling. It is along dew point temperature as it remains constant during sensible cooling.

4. During adiabatic saturation process of air, wet bulb temperature ___________

a) Increases and dry bulb temperature remains constant

b) Remains constant and dry bulb temperature increases

c) Remains constant and dry bulb temperature decreases

d) Decreases and dry bulb temperature remains constant

Answer: c

Explanation: During adiabatic saturation process, enthalpy and wet bulb temperature remains constant and dry bulb temperature decreases. In this the dew point temperature increases.

5. The performance of an evaporation condenser largely depends on ________

a) Dry bulb temperature of air

b) Wet bulb temperature of air

c) Hot water temperature

d) Air conditional room temperature

Answer: a

Explanation: The evaporative cooler performance is directly related to evaporate water at a given relative humidity. The temperature of the water does not have a great effect upon the cooling produced through evaporation.

6. What happens when warm saturated air is cooled?

a) Excess moisture condenses

b) Excess moisture condenses but relative humidity remains unchanged

c) Excess moisture condenses and specific humidity increases but relative humidity remains unchanged

d) Specific humidity increases and relative humidity increases

Answer: b

Explanation: When warm saturated air is cooled the excess of moisture condenses. The relative humidity remains constant but specific humidity decreases due to which the excess of moisture condenses.

7. Air at 35 and 25 degree Celsius dew point temperature passes through the water shower whose temperature is maintained at 20 degree Celsius. What is the process involved?

a) Cooling and humidification

b) Sensible cooling

c) Cooling and dehumidification

d) Heating and humidification

Answer: c

Explanation: The process involved is cooling and dehumidification. Since the temperature of spray water is less than the dew point temperature cooling and dehumidification will take place.

8. Which one of the following is correct according to temperature?

a) Effective temperature is the index which correlates the combined effects of air dry bulb temperature, air humidity and air movement upon human comfort

b) The value of effective temperature in winter and summer should be same for human comfort

c) Effective temperature and wet bulb temperature are same

d) The valve of the effective temperature should be higher in winter than in summer for comfort

Answer: a

Explanation: The valve of effective temperature is low in winter and high in summer. In winter it is 19 degrees Celsius and in summer 22 degrees Celsius. Therefore the effective temperature is the index which correlates the combined effects of air dry bulb temperature, air humidity and air movement upon human comfort.

9. What is the relative humidity when the wet bulb depression is zero?

a) 100%

b) 60%

c) 40%

d) Zero

Answer: a

Explanation: Wet bulb depression is the difference between the dry bulb and wet bulb temperature. Hence, when DBT = WBT implies that relative humidity = 100%. Therefore when the wet bulb depression is zero the relative humidity is equal to 100%.

10. The latent heat load in an auditorium is 25% of sensible heat load. What is the value of sensible heat fact?

a) 0.3

b) 0.5

c) 0.8

d) 1.0

Answer: c

Explanation: Sensible heat factor is the ratio of sensible heat and the total heat is equal to latent heat plus sensible heat

SHF = SH/ , but LH = 0.25SH

Hence, SHF = 0.8.

11. What is the saturation temperature at the partial pressure of water vapor in the air water vapor mixture called?

a) Dry bulb temperature

b) Wet bulb temperature

c) Dew point temperature

d) Saturation temperature

Answer: c

Explanation: At 100 RH, WBT = DBT = DPT. The temperature corresponding to the saturation temperature at the partial pressure of water vapor is called dew point temperature.

This set of Chemical Process Calculation Multiple Choice Questions & Answers  focuses on “Psychrometric chart – I”.

1. A chart showing the humidity versus temperature along with all the other properties of moist air is called

a) Relative humidity chart

b) Psychrometric chart

c) Air vapour phase chart

d) None of the mentioned

Answer: b

Explanation: A chart showing the humidity versus temperature along with all the other properties of moist air is called psychrometric chart.

2. The representation on the humidity chart of the energy balance in which the heat transfer to water from the air is assumed to equal the enthalpy of vaporization of liquid water is Wet Bulb Line.

The above given statement is

a) True

b) False

c) Not defined well

d) None of the mentioned

Answer: a

Explanation: The representation on the humidity chart of the energy balance in which the heat transfer to water from the air is assumed to equal the enthalpy of vaporization of liquid water is Wet Bulb Line.

3. The humidity chart is formally known as the psychrometric chart. The statement is

a) True

b) False

c) Not defined well

d) None of the mentioned

Answer: a

Explanation: The humidity chart is formally known as the psychrometric chart.

4. Psychrometric chart is plotted between the parameters

a) Humidity and temperature

b) Humidity and pressure

c) Humidity and volume of the dry air

d) None of the mentioned

Answer: a

Explanation: Psychrometric chart is plotted between Humidity and temperature.

5. The idea of wet bulb temperature is based on the equilibrium between the rates of energy transfer to the ______ and the evaporation of ______

a) Water and vapour

b) Bulb, water

c) Bulb and vapour

d) None of the mentioned

Answer: b

Explanation: The idea of wet bulb temperature is based on the equilibrium between the rates of energy transfer to the bulb and the evaporation of water.

6. Humidity measurement is done with

a) Thermometer

b) Hygrometer

c) Barometer

d) None of the mentioned

Answer: b

Explanation: Humidity measurement is done with Hygrometer.

7. Absolute humidity is the ______ amount of water vapour that the air could hold at certain temperature.

a) Equal

b) Lowest

c) Greatest

d) None of the mentioned

Answer: c

Explanation: Absolute humidity is the greatest amount of water vapour that the air could hold at certain temperature.

8. For saturated air, relative humidity is

a) 100%

b) 50%

c) 0%

d) None of the mentioned

Answer: a

Explanation: For saturated air, relative humidity is 100%.

9. The degree of saturation is the ratio of the ______ humidity to the _______ humidity at the same __________

a) Actual specific, saturated specific, temperature

b) Saturated specific, actual specific, pressure

c) Actual specific, saturated specific, pressure

d) Saturated specific, actual specific, temperature

Answer: a

Explanation: The degree of saturation is the ratio of the actual specific humidity to the saturated specific humidity at the same temperature.

Answer: c

Explanation: The degree of saturation varies between 0 to 1.

This set of Chemical Process Calculation Multiple Choice Questions & Answers  focuses on “Psychrometric chart – II”.

chemical-process-calculation-questions-answers-psychrometric-chart-ii-q1

1-10. The above given psychrometric chart has been marked with several points from 1-9. Each point represent some specific point or curve or a line.

1. Point 1 represents?

a) % Relative humidity curve

b) 100% Relative humidity

c) Wet bulb line

d) None of the mentioned

Answer: a

chemical-process-calculation-questions-answers-psychrometric-chart-ii-q1a

Explanation: point 1 represents the % Relative humidity curve.

2. Point 2 represents?

a) Specific humid volume

b) Specific enthalpy values

c) Adiabatic cooling line

d) Relative humidity curve

Answer: c

Explanation: Point 2 represents the Adiabatic cooling line.

3. Point 3 represents?

a) Specific humid volume

b) Specific enthalpy values

c) Adiabatic cooling line

d) Relative humidity curve

Answer: b

Explanation: Point 3 represents the Specific enthalpy values.

4. Point 4 represents?

a) Specific humid volume

b) Specific enthalpy values

c) Adiabatic cooling line

d) Relative humidity curve

Answer: a

Explanation: Point 4 represents the Specific humid volume.

5. Point 5 represents?

a) Dew point temperature

b) Dry bulb temperature

c) Wet bulb temperature

d) None of the mentioned

Answer: b

Explanation: Point 5 represents the dry bulb temperature.

6. Point 6 represents?

a) Dew point temperature

b) Dry bulb temperature

c) Wet bulb temperature

d) None of the mentioned

Answer: c

Explanation: Point 6 represents the wet bulb temperature.

7. Point 7 represents?

a) Dew point temperature

b) Dry bulb temperature

c) Wet bulb temperature

d) None of the mentioned

Answer: a

Explanation: Point 7 represents the dew point temperature.

8. Point 8 represents?

a) % Relative humidity curve

b) 100% Relative humidity

c) Wet bulb line

d) None of the mentioned

Answer: b

Explanation: Point 8 represents 100% Relative humidity.

9. Point 9 represents?

a) Dry point

b) Wet point

c) Dew point

d) None of the mentioned

Answer: c

Explanation: Point 9 represents dew point.

Answer: a

Explanation: Adiabatic cooling lines which are the same  as the wet bulb or psychrometric line.

This set of Air-Conditioning Multiple Choice Questions & Answers  focuses on “Psychrometric Chart”.

1. What do vertical and uniformly spaced lines indicate on the psychrometric chart?

a) DPT

b) WBT

c) DBT

d) Specific humidity

Answer: c

Explanation: The psychrometric chart has DBT on the x-axis. So, the vertical and uniformly spaced lines denote dry bulb temperature and spaced by 5°C.

2. Which of the following is represented by curved lines on the psychrometric chart?

a) Specific humidity

b) Relative humidity

c) WBT

d) DPT

Answer: b

Explanation: The psychrometric chart has DBT on the x-axis and specific humidity on the y-axis. The curved line on the grid shows the relative humidity. From 0 to 100 %, these lines are drawn at an interval of 10%.

3. What is represented by inclined straight lines but non-uniformly spaced on the psychrometric chart?

a) Specific humidity

b) Relative humidity

c) WBT

d) DPT

Answer: c

Explanation: The psychrometric chart has DBT on the x-axis and specific humidity on the y-axis. The inclined lines from the saturation curve and on the x-axis denote wet-bulb temperature.

4. What is represented by horizontal lines but non-uniformly spaced on the psychrometric chart?

a) Specific humidity

b) Relative humidity

c) WBT

d) DPT

Answer: d

Explanation: The psychrometric chart has DBT on the x-axis and specific humidity on the y-axis. The horizontal lines from the saturation curve and parallel to the abscissa denote dew point temperature.

5. What is represented by inclined straight lines but uniformly spaced on the psychrometric chart?

a) Enthalpy

b) Relative humidity

c) WBT

d) DPT

Answer: a

Explanation: The psychrometric chart has DBT on the x-axis and specific humidity on the y-axis. The inclined lines from the saturation curve and parallel to the wet-bulb temperature lines denote Enthalpy.

6. What is represented by obliquely inclined straight lines but uniformly spaced on the psychrometric chart?

a) Enthalpy

b) Relative humidity

c) Specific volume

d) DPT

Answer: c

Explanation: The psychrometric chart has DBT on the x-axis and specific humidity on the y-axis. The obliquely inclined lines from the saturation curve and on the dry-bulb temperature lines denote Specific volume.

7. Specific humidity lines are also known as_________

a) moisture content lines

b) relative humidity

c) specific volume

d) moist lines

Answer: a

Explanation: Specific humidity is the moisture present in the air mixture. So, specific humidity denotes moisture content. Hence, these lines are also known as moisture content lines.

8. Relative humidity on saturation curve has value of ________% at various dry bulb temperatures.

a) 0

b) 50

c) 10

d) 100

Answer: d

Explanation: Relative humidity is the ratio of actual mass of water vapor in the given volume of moist air to the mass of water vapor in the same amount of saturated air. So, on the saturation curve, for every dry bulb temperature, the value of relative humidity remains 100%.

9. Enthalpy lines and specific volume lines are the same.

a) False

b) True

Answer: a

Explanation: Though both of the lines representing given properties are inclined straight lines and uniformly spaced, but the slope of both the lines are different. The specific volume lines are obliquely inclined.

10. A psychrometric chart is a graphical representation of various physical properties of dry air.

a) True

b) False

Answer: b

Explanation: Psychrometric chart is a graphical representation of the various thermodynamic properties of moist air. This is used to find out properties of air in the field of air conditioning.

11. Which of the following represents sensible cooling on the psychrometric chart?

a) Horizontal line

b) Vertical line

c) Inclined line

d) Curve

Answer: a

Explanation: Sensible cooling is when DBT decreases. And as per the psychrometric chart, horizontal lines show the change in just DBT and represent sensible cooling.

12. Which of the following represents sensible heating on the psychrometric chart?

a) Horizontal line

b) Vertical line

c) Inclined line

d) Curve

Answer: a

Explanation: Sensible cooling is when DBT rises. And as per the psychrometric chart, horizontal lines show the change in just DBT and represent sensible heating. It is in the opposite direction of sensible cooling.

13. What is the range of DBT on the psychrometric chart usually used?

a) 0 to 30°C

b) 0 to 45°C

c) -6 to 45°C

d) -2 to 45°C

Answer: c

Explanation: Generally, the temperature range of dry bulb temperature lines on the psychrometric chart is from -6 to 45°C. These lines are drawn with a difference of every 5°C of DBT.

14. Which of the following is not correct to fill the blank: The psychrometric chart is normally drawn for standard atmospheric pressure of ________

a) 760 mm of Hg

b) 76 cm of Hg

c) 1.01325 bar

d) 1.01325 torr

Answer: d

Explanation: The psychrometric chart is drawn on the standard values. The standard atmospheric pressure has a value 1.01325 bar. If we convert into other units, then 760 mm of Hg and 76 cm of Hg are also appropriate values of standard pressure, but 1.01325 torr is not. So, this cannot be filled in the blank space.

15. On the Psychrometric chart, the coordinates of any point on it will have abscissa as specific humidity value and ordinate as a dry bulb temperature value.

a) False

b) True

Answer: a

Explanation: The psychrometric chart has DBT and Sp. Humidity on the x-axis and y-axis, respectively. So, any point on the chart will have specific humidity as ordinate and dry bulb temperature as abscissa.

This set of Thermodynamics Multiple Choice Questions & Answers  focuses on “Psychrometric Chart and Process”.

1. Which of the following statement is true?

a) the chart is plotted for pressure equal to 760mm Hg

b) the constant wbt line represents adiabatic saturation process

c) the constant wbt line coincides with constant enthalpy line

d) all of the mentioned

Answer: d

Explanation: All these come from the psychrometric chart.

2. In sensible heating or cooling,

a) work done remains constant

b) dry bulb temperature or air remains constant

c) both of the mentioned

d) none of the mentioned

Answer: a

Explanation: The dry bulb temperature of air changes.

3. When humidity ratio of air ____ air is said to be dehumidified.

a) increases

b) decreases

c) remains constant

d) none of the mentioned

Answer: b

Explanation: when it increases, air is said to be humidified.

4. Air can be cooled and dehumidified by

a) circulating chilled water in tube across air flow

b) placing evaporator coil across air flow

c) spraying chilled water to air

d) all of the mentioned

Answer: d

Explanation: These are the ways of cooling and dehumidifying air.

5. Cooling and dehumidification of air is done in summer air conditioning.

a) true

b) false

Answer: a

Explanation: This is a common process in summer air conditioning.

6. Heating and humidification is done in

a) summer air conditioning

b) winter air conditioning

c) both of the mentioned

d) none of the mentioned

Answer: b

Explanation: This is opposite to summer air conditioning.

7. Which of the following is an absorbent?

a) silica gel

b) activated alumina

c) both of the mentioned

d) none of the mentioned

Answer: c

Explanation: Both of these are examples of absorbents.

8. When air passes through silica gel,

a) it absorbs water vapour molecules

b) latent heat of condensation is released

c) dbt of air increases

d) all of the mentioned

Answer: d

Explanation: This process is called chemical dehumidification.

9. In adiabatic evaporative cooling, heat transfer between chamber and surroundings is ____

a) zero

b) high

c) low

d) none of the mentioned

Answer: a

Explanation: No heat transfer takes place between chamber and surroundings in adiabatic evaporative cooling.

10. The cooling tower uses the phenomenon of evaporative cooling to cool warm water above the dbt of air.

a) true

b) false

Answer: b

Explanation: The cooling tower uses the phenomenon of evaporative cooling to cool warm water below the dbt of air.

Answer: c

Explanation: These are the two factors considered.

This set of Air-Conditioning Multiple Choice Questions & Answers  focuses on “Psychrometry – Heating and Cooling Factor – 1”.

1. What is the By-pass factor for heating coil, if td 1 = temperature at entry, td 2 = temperature at exit and td 3 = coil temperature?

a) td 3 – td 1 / td 3 – td 1

b) td 3 – td 2 / td 3 – td 1

c) td 3 – td 2 / td 2 – td 1

d) td 3 – td 2 / td 1 – td 2

Answer: b

Explanation: By-pass factor is the amount of air by-passed in the process. So, for heating coil,

BPF = Temperature difference between coil and exit / Temperature difference between coil and entry

= td 3 – td 2 / td 3 – td 1 .

2. What is the By-pass factor for cooling coil, if td 1 = temperature at entry, td 2 = temperature at exit and td 3 = coil temperature?

a) td 3 – td 1 / td 3 – td 1

b) td 3 – td 2 / td 3 – td 1

c) td 3 – td 2 / td 2 – td 1

d) td 2 – td 3 / td 1 – td 3

Answer: d

Explanation: By-pass factor is the amount of air by-passed in the process. So, for cooling coil,

BPF = Temperature difference between exit and coil / Temperature difference between entry and coil

= td 2 – td 3 / td 1 – td 3 .

3. If the value of BPF for one row of a coil is y then what is the value of BPF for n similar rows?

a) n / y

b) n + y

c)  n

d) n x y

Answer: c

Explanation: BPF for multiple similar rows of a coil is the power of one BPF to the number of rows. i.e.  n .

4. What is the value of sensible heat given out by the coil?

a) U A c t m

b) U A c

c) U t m

d) U A c 2 t m

Answer: a

Explanation: Sensible heat given out by the coil is the product of overall heat transfer coefficient, the surface area of the coil and logarithmic mean temperature difference.

So, Sensible heat = U A c t m .

5. What is the formula of logarithmic mean temperature in terms of By-pass factor?

a) T m = td 2 – td 1 / log e [1/BPF]

b) T m = td 2 – td 1 / log 10 [BPF]

c) T m = td 2 – td 1 / log e [BPF]

d) T m = td 2 – td 3 / log e [BPF]

Answer: c

Explanation: Logarithmic mean temperature difference for the given arrangement is,

T m = td 2 – td 1 / log e [td 3 – td 1 / td 3 – td 2 ]

As, BPF for the coil = [td 3 – td 1 / td 3 – td 2 ]

T m = td 2 – td 1 / log e [BPF].

6. What is the efficiency of the coil?

a) 1 + BPF

b) 1 / BPF

c) BPF

d) 1 – BPF

Answer: d

Explanation: As the by-pass factor is the inefficiency, so the contact factor or efficiency of the coil is given by 1 – BPF.

7. What is the contact factor for heating coil, if td 1 = temperature at entry, t 2 = temperature at exit and td 3 = coil temperature?

a) td 3 – td 1 / td 3 – td 1

b) td 2 – td 1 / td 3 – td 1

c) td 3 – td 2 / td 2 – td 1

d) td 3 – td 2 / td 1 – td 2

Answer: b

Explanation: Contact factor or efficiency of the coil is 1 – By-pass factor. So, for heating coil,

η H = 1 – BPF

= 1 – [td 3 – td 2 / td 3 – td 1 ]

= td 2 – td 1 / td 3 – td 1 .

8. What is the contact factor for cooling coil, if td 1 = temperature at entry, td 2 = temperature at exit and td 3 = coil temperature?

a) td 1 – td 2 / td 1 – td 3

b) td 2 – td 1 / td 3 – td 1

c) td 3 – td 2 / td 2 – td 1

d) td 3 – td 2 / td 1 – td 2

Answer: a

Explanation: Contact factor or efficiency of the coil is 1 – By-pass factor. So, for cooling coil,

η C = 1 – BPF

= 1 – [td 2 – td 3 / td 1 – td 3 ]

= td 1 – td 2 / td 1 – td 3 .

9. A coil with low BPF has better performance.

a) True

b) False

Answer: a

Explanation: As BPF is the amount of air bypassed in the process and is the inefficiency of the coil. So, lower the value of BPF then better is the performance and vice-versa.

10. The value of BPF for the heating and the cooling coil is different under the same temperature conditions.

a) True

b) False

Answer: b

Explanation: BPF for heating coil is, BPF = td 3 – td 2 / td 3 – td 1

BPF for cooling coil is, BPF = td 2 – td 3 / td 1 – td 3

As the numerator and denominator are reversed in the cooling coil than the heating coil. So, the value we get for either of the coils is the same.

This set of Air-Conditioning online test focuses on “Psychrometry – Heating and Cooling Factor – 2”.

1. What is the By-pass factor for heating coil, if td 1 = 19°C, td 2 = 25°C and td 3 = 37°C?

a) 0.75

b) 0.4

c) 0.7

d) 0.1

Answer: c

Explanation: By-pass factor is the amount of air by-passed in the process. So, for heating coil,

BPF = Temperature difference between coil and exit / Temperature difference between coil and entry

= td 3 – td 2 / td 3 – td 1

= 37 – 25 / 37 – 19

= 0.70588 = 0.7.

2. What is the By-pass factor for cooling coil, if td 1 = 49°C, td 2 = 36°C and td 3 = 30°C?

a) 0.315

b) 0.31

c) 0.320

d) 0.3

Answer: a

Explanation: By-pass factor is the amount of air by-passed in the process. So, for the cooling coil,

BPF = Temperature difference between exit and coil / Temperature difference between entry and coil

= td 2 – td 3 / td 1 – td 3

= 36 – 30 / 49 – 30

= 0.315.

3. If the value of BPF for one row of the coil is one, then what is the value of BPF for 3872 similar rows?

a) 3872

b) 3871

c) 3873

d) 1

Answer: d

Explanation: BPF for multiple similar rows of the coil is the power of one BPF to the number of rows. i.e.  n

So,  3872 = 1, for any value of n, the value comes one only.

4. If the value of BPF for one row of the coil is 0.4, then what is the value of BPF for 12 similar rows?

a) 0.00000167

b) 0.0000167

c) 0.00167

d) 0.000167

Answer: b

Explanation: BPF for multiple similar rows of the coil is the power of one BPF to the number of rows. i.e.  n

So,  12 = 0.0000167.

5. What is the value of sensible heat given out by the coil, if U = 198 W/m 2 K, A c = 11 m 2 and t m = 0°C?

a) 1

b) 0

c) 594594

d) 543636

Answer: b

Explanation: Sensible heat given out by the coil is the product of the overall heat transfer coefficient, the surface area of the coil, and logarithmic mean temperature difference.

So, Sensible heat = U A c t m = 198 x 11 x  = 594594.

6. What is the value of BPF, if U = 1.9 W/m 2 K, A c = 1.1 m 2 and m a = 0.97 kg?

a) 0.14

b) 0.12

c) 0.13

d) 0.15

Answer: b

Explanation: BPF = e -[U A c / 1.022 m a ] = e -[1.9 x 1.1/ 1.022 x 0.97] = e -[2.1082] = 0.12.

7. What is the value of logarithmic mean temperature if td 1 = 300 K, td 2 = 400 K and BPF = 0.75?

a) 70°C

b) 79°C

c) 85°C

d) 75°C

Answer: d

Explanation: Logarithmic mean temperature difference for the given arrangement is,

T m = td 2 – td 1 / log e [td 3 – td 1 / td 3 – td 2 ]

As, BPF for the coil = [td 3 – td 1 / td 3 – td 2 ]

T m = td 2 – td 1 / log e [BPF]

= 400 – 300 / ln  = 100 / 0.287 = 348.43 K = 75.43°C = 75°C.

8. What is the efficiency of the coil, if BPF = 0.6913?

a) 1.6913

b) 0.4467

c) 0.3087

d) 1.4467

Answer: c

Explanation: As the by-pass factor is the inefficiency, so the contact factor or efficiency of the coil is given by 1 – BPF = 1 – 0.6913 = 0.3087.

9. What is the contact factor for heating coil, if td 1 = 11°C, td 2 = 32°C and td 3 = 47°C?

a) 0.5833

b) 0.4833

c) 0.7833

d) 0.1833

Answer: a

Explanation: By-pass factor is the amount of air by-passed in the process. So, for heating coil,

BPF = Temperature difference between coil and exit / Temperature difference between coil and entry

= td 3 – td 2 / td 3 – td 1

= 47 – 32 / 47 – 11

= 0.4167

η H = 1 – BPF = 1 – 0.4167 = 0.5833.

10. What is the contact factor for cooling coil, if td 1 = 44°C, td 2 = 26°C and td 3 = 19°C?

a) 0.62

b) 0.72

c) 0.78

d) 0.18

Answer: b

Explanation: By-pass factor is the amount of air by-passed in the process. So, for the cooling coil,

BPF = Temperature difference between coil and exit / Temperature difference between coil and entry

= td 2 – td 3 / td 1 – td 3

= 26 – 19 / 44 – 19

= 0.28

η C = 1 – BPF = 1 – 0.28 = 0.72.

11. For given conditions, td 1 = 288 K, td 2 = 301 K and td 3 = 314 K, the value of by-pass factor and contact factor is equal.

a) True

b) False

Answer: a

Explanation: BPF = 314 – 301 / 314 – 288 = 13 / 26 = 0.5 and η = 1 – BPF = 1 – 0.5 = 0.5

So, for given set of values the value of contact factor and by-pass factor is equal.

12. Higher the value of contact factor, poor is the performance of the coil.

a) True

b) False

Answer: b

Explanation: The contact factor is the efficiency of the coil. So, if the value of it is higher, then better is the performance and vice-versa. It is the opposite of BPF.

This set of Air-Conditioning Multiple Choice Questions & Answers  focuses on “Psychrometry – Humidification and Dehumidification”.

1. Which of the following denotes Humidification process on the psychrometric chart?

a) A horizontal line with an arrow towards right

b) Vertical line with arrow upwards

c) Horizontal line with an arrow towards right

d) Vertical line with arrow downwards

Answer: b

Explanation: Psychrometric chart has DBT on the x-axis and sp. Humidity on the y-axis. So, if a process is going up on the chart, then there is an increase in relative as well as specific humidity. So, the vertical line with an arrow upwards denotes humidification process.

2. Which of the following denotes dehumidification process on the psychrometric chart?

a) A horizontal line with an arrow towards the right

b) Vertical line with arrow upwards

c) A horizontal line with an arrow towards the right

d) Vertical line with arrow downwards

Answer: d

Explanation: Psychrometric chart has DBT on the x-axis and sp. Humidity on the y-axis. So, if a process is going down on the chart, then there is a decrease in relative as well as specific humidity. So, the vertical line with an arrow downwards denotes dehumidification process.

3. Which of the following is denoted by LH = (h 2 – h 1 ) = h fg (W 2 – W 1 )?

a) Sensible cooling

b) Sensible heating

c) Humidification

d) Dehumidification

Answer: c

Explanation: Latent heat transfer i.e., LH = (h 2 – h 1 ) = h fg (W 2 – W 1 ) denotes Humidification. As in the humidification process, there is an increase in specific humidity, so W 2 > W 1 and also increase in enthalpy, h 2 > h 1 .

4. Which of the following is denoted by LH = (h 1 – h 2 ) = h fg (W 1 – W 2 )?

a) Sensible cooling

b) Sensible heating

c) Humidification

d) Dehumidification

Answer: d

Explanation: Latent heat transfer i.e., LH = (h 1 – h 2 ) = h fg (W 1 – W 2 ) denotes dehumidification. As in the dehumidification process, there is a decrease in specific humidity, so W 1 > W 2 and also decrease in enthalpy, h 1 > h 2 .

5. What is the relation between DBT and WBT if the relative humidity is 100%?

a) DBT = WBT

b) DBT > WBT

c) DBT ≫ WBT

d) DBT < WBT

Answer: a

Explanation: When the dry bulb temperature is equal to the wet-bulb temperature, then the relative humidity tends to be 100%.

6. The minimum temperature to which moist air can be cooled under ideal conditions in a spray washer is ___________

a) DPT of inlet air

b) Water inlet temperature

c) WBT of inlet air

d) Water outlet temperature

Answer: c

Explanation: Wet-bulb temperature is the minimum temperature to which moist air can be cooled under ideal conditions in a spray washer. It is an integral part of the process.

7. Which of the following process is used in winter air conditioning?

a) Humidification

b) Dehumidification

c) Heating and Humidification

d) Cooling and Dehumidification

Answer: c

Explanation: In Winter, the weather is dry and cold, so the process of heating and humidification is done to get the desired comfort conditions.

8. Which of the following process is used in summer air conditioning?

a) Humidification

b) Dehumidification

c) Heating and Humidification

d) Cooling and Dehumidification

Answer: d

Explanation: In Summer, the weather is humid and hot, so the process cooling and dehumidification is done to get the desired comfort conditions.

9. What is the value of Latent heat transfer in the process of humidification having enthalpies as 20 and 40 kJ / kg of dry air respectively?

a) 60

b) 40

c) -20

d) 20

Answer: d

Explanation: Latent heat transfer i.e., LH = (h 2 – h 1 ) = h fg (W 2 – W 1 ) denotes Humidification. As in the humidification process, there is an increase in specific humidity, so W2 > W1 and also increase in enthalpy, h 2 > h 1 . So, as per given data, h 1 = 40 and h 2 = 20 kJ / kg of dry air.

So, latent heat transfer = h 2 – h 1 = 40 – 20 = 20 kJ/ kg of dry air.

10. What is the value of Latent heat transfer in the process of dehumidification having enthalpies as 19.45 and 13.67 kJ/ kg of dry air respectively?

a) 5.78

b) -5.78

c) 33.12

d) -33.12

Answer: a

Explanation: Latent heat transfer i.e., LH = (h 1 – h 2 ) = h fg (W 1 – W 2 ) denotes dehumidification. As in the dehumidification process, there is a decrease in specific humidity, so W 1 > W 2 and also decrease in enthalpy, h 1 > h 2 . So, as per given data, h 1 = 19.45 and h 2 = 13.67 kJ / kg of dry air.

So, latent heat transfer = h 1 – h 2 = 19.45 – 13.67 = 20 kJ / kg of dry air.

11. What is the value of Latent heat transfer in a process of humidification having h fg = 10 kJ / kg of dry air, W 1 = 0.0071 kJ / kg of dry air and W 2 = 0.0017 kJ / kg of dry air?

a) 0.054

b) 0.0054

c) 0.54

d) 5.4

Answer: a

Explanation: Latent heat transfer i.e., LH = (h 2 – h 1 ) = h fg (W 2 – W 1 ) denotes Humidification. As, in the humidification process there is an increase in specific humidity so W 2 > W 1 and also increase in enthalpy, h 2 > h 1 .

So, latent heat transfer = h fg (W 2 – W 1 ) = 10  = 0.054 kJ / kg of dry air.

12. During humidification process the dry bulb temperature ___________

a) increases

b) decreases

c) tends to zero

d) remains the same

Answer: d

Explanation: During the humidification process, DBT remains the same as the increase is in specific humidity and relative humidity. As the process is vertical on chart keeping DBT constant.

13. Which of the following is not an indirect method of air-washer humidification?

a) Using vaporization after cooling

b) Using heated spray water

c) Using re-circulated spray water without prior heating of air

d) By pre-heating the air and then washing it with re-circulated air

Answer: a

Explanation: There are direct and indirect methods of obtaining humidification and dehumidification. There are three types of Indirect methods which are as follows:

i.Using heated spray water

ii.Using re-circulated spray water without prior heating of air

iii.By pre-heating the air and then washing it with re-circulated air.

14. During the sensible cooling process, specific humidity decreases.

a) True

b) False

Answer: b

Explanation: Sensible cooling process is the one where dry bulb temperature decreases. It is a horizontal process on the psychrometric chart. Hence, the specific humidity neither increases nor decreases, but there might be some change in relative humidity.

15. In order to cool and dehumidify a stream of moist air, it must be passed over the coil at a temperature between DBT and WBT.

a) True

b) False

Answer: b

Explanation: In order to cool and dehumidify a stream of moist air, it must be passed over the coil at a temperature which is lower than the dew point temperature of the incoming stream to get desire state of the air.

This set of Energy Engineering Multiple Choice Questions & Answers  focuses on “Types of Biogas Plants – 1”.

1. Which types of plant are fed and emptied regularly?

a) Batch type plants

b) Continuous type plants

c) Dome type plants

d) Drum type plants

Answer: b

Explanation: Continuous plants are fed and emptied continuously. They empty automatically through the overflow whenever new material is filled in. therefore, the substance must be fluid and homogeneous.

2. Gas production of continuous plant is higher than which of the following plant?

a) Batch plant

b) Dome plant

c) Drum plant

d) Flexible gas biogas plant

Answer: a

Explanation: Continuous plants are suitable for rural households as the necessary work fits well into daily routine. Gas production is constant and higher than in batch plants. Today, nearly all biogas plants are operating on a continuous mode.

3. What type of plant is a floating gas holder plant?

a) Batch plant

b) Continuous plant

c) Semi-batch plant

d) Semi-continuous plant

Answer: d

Explanation: This is one of the common designs in India, coming under category of semi-continuous feed plant. It has a cylindrical floating biogas holder on top of the well shaped digester. This digester is vividly used in rural sides.

4. For how much percent is biogas normally designed to hold?

a) 35%

b) 20%

c) 50%

d) 80%

Answer: c

Explanation: As the biogas is produced in the digester, it rises vertically and gets accumulated and stored in the biogas holder at a constant pressure of 8-10 cm of water column. The biogas holder is normally designed to store 50% of the daily gas production.

5. Where was fixed dome concept plant developed?

a) China

b) America

c) India

d) Japan

Answer: c

Explanation: The plant based on fixed dome concept was developed in India in the middle 1970s. The Chinese fixed plants use a seasonal crop wastes as the major feed stock for feeding, therefore, their design is based on principle of semi-batch-feed digester.

6. Indian fixed dome digesters are designed for holding what pressure capacity?

a) 0 – 90 cm of water column

b) 70 – 90 cm of water column

c) 50 – 65 cm of water column

d) 10 – 25 cm of water column

Answer: a

Explanation: The Indian fixed dome plant design use the principle of displacement of slurry inside the digester for storage of biogas in the fixed gas storage chamber. Indian fixed dome biogas digesters are designed for pressure inside the plant varying from 0 to 90 cm of water column.

7. Fixed dome biogas plant is best suitable for which type of plant?

a) Continuous type

b) Batch type

c) Semi-batch type

d) Semi continuous type

Answer: b

Explanation: The discharge opening is located on water wall surface of the outlet displacement chamber and it spontaneously controls the maximum pressure. It is best suited for batch process especially when daily feeding is adopted in small quantities.

8. Which plants digester is fabricated by using rubber?

a) Flexible Bag biogas plant

b) Fixed dome biogas plant

c) Floating drum biogas plant

d) Khadi and village industries type biogas plant

Answer: a

Explanation: Main unit of the plant including the digester is fabricated by using rubber, high strength plastic, neoprene or red mud plastic. The inlet and outlet are made of heavy duty PVC tube. A small pipe of same PVC tube is fixed on top of the plant as gas outlet pipe.

9. Which type of plant is portable?

a) Flexible Bag biogas plant

b) Fixed dome biogas plant

c) Floating drum biogas plant

d) Khadi and village industries type biogas plant

Answer: a

Explanation: Flexible bag biogas plant is portable and can easily be erected. It requires support from outside, upto the slurry level, to maintain the shape as per its design configuration, which is done by placing the bag inside a pit dug on site.

10. In which type of plant the weight needs to be added on top to build the desired pressure?

a) Flexible Bag biogas plant

b) Fixed dome biogas plant

c) Floating drum biogas plant

d) Khadi and village industries type biogas plant

Answer: a

Explanation: The depth of the pit should be in proportion to height of the digester so that the mark of initial slurry level is in line with the ground level. The outlet pipe is fixed in such a way that its outlet opening is also in line with the ground level. Some weight has to be added on the top of the bag to build the desired pressure to convey the generated gas to the point of utilization.

11. What is the main advantage of flexible bag plant?

a) The fabrication can be centralized for mass production

b) It has highest gas storage capacity

c) It is portable from place to place

d) Easy to understand and work on it

Answer: a

Explanation: Advantage of flexible bag plant is that the fabrication can be centralized for mass production. Individuals or agencies having land and basic infrastructure can take up fabrication of the biogas plant with small investment.

12. Khadi village industries type biogas plant is example of _________

a) flexible Bag biogas plant

b) fixed dome biogas plant

c) floating drum biogas plant

d) semi-batch type

Answer: c

Explanation: Khadi and village industries type biogas plant  is an example of Floating drum biogas plant. It is a semi continuous plant. It is one of the most common types of biogas plant used in rural area of India.

13. Vertical type KVIC is more commonly used than horizontal type.

a) True

b) False

Answer: a

Explanation: The KVIC  model is a floating holder semi continuous fed bio gas plant and has two types,  Vertical and  Horizontal. The vertical type is more commonly used and the horizontal type is used only in the high water table region.

14. Which part of the KVIC is made of mixture of cement concrete and brick ballast?

a) Foundation

b) Digester

c) Gas holder

d) Inlet and outlet pipe

Answer: a

Explanation: Foundation of a KVIC is compact base made of a mixture of cement concrete and brick ballast. The foundation is well compacted using wooden ram and then top surface is cemented to prevent any percolation & seepage.

This set of Tough Energy Engineering Questions and Answers focuses on “Types of Biogas Plants – 2”.

1. The ____________ is made of bricks and cement mortar and it’s inside wall are plastered with a mixture of cement and sand.

a) Foundation

b) Digester

c) Gas holder

d) Inlet and outlet pipe

Answer: b

Explanation: Digester  of KVIC is a cylindrical shaped well like structure, constructed using the foundation as its base. The digester is made of bricks and cement mortar and it inside walls are plastered with a mixture of cement and sand.

2. From what material is biogas holder drum made?

a) Mild steel

b) Cast iron

c) Aluminum alloys

d) Rot iron

Answer: a

Explanation: The biogas holder drum of KVIC model is normally made of mild steel sheets. The biogas holder rests on a ledge constructed inside the walls of the digester well. If the KVIC model is made with water jacket on top of the digester wall, no ledge is made and the drum of the biogas holder is placed inside the water jacket.

3. What is the weight of biogas holder?

a) 8-10 kg/m 2

b) 10-15 kg/m 2

c) 8-18 kg/m 2

d) 25-30 kg/m 2

Answer: a

Explanation: The weight of the biogas holder is 8-10 kg/m 2 . The biogas holder of KVIC  is also fabricated out of fiber glass reinforced plastic , high density polyethylene  or Ferroconcrete .

4. What type of movement does a biogas holder of KVIC has?

a) Linear Movement

b) Transverse movement

c) Rotary movement

d) Circular movement

Answer: c

Explanation: The biogas holder of a KVIC moves up and down on a guide pipe situated in the centre of the digester. The biogas holder has a rotary movement that helps in breaking the scum-mat formed on the top surface of the slurry. The weight of the biogas holder is 8-10 kg/m 2 . So that it can store biogas at a constant pressure of 8-10 cm of water column.

5. How is the inlet pipe of KVIC made?

a) Cement mortar

b) Asbestos cement concrete

c) White cement

d) PVC

Answer: b

Explanation: The inlet pipe is made out of cement concrete  or Asbestos cement concrete  or pipe. The one end of the inlet pipe is concrete to the mixing tank and the other end goes inside the digester on the inlet side of the partition wall and rests on a support made of bricks of about 1 feet height.

6. How is the outlet pipe of KVIC made?

a) Cement mortar

b) Asbestos cement concrete

c) White cement

d) PVC

Answer: b

Explanation: The outlet pipe is made out of cement concrete  or asbestos cement concrete  or pipe. The one end of the outlet pipe is connected to the outlet tank and the other end goes inside the digester, on the outlet side of the partition wall and rests on a support made of about bricks of about 1 feet height.

7. How is the Biogas outlet pipe of KVIC made?

a) Cement

b) PVC

c) Asbestos

d) GI pipe

Answer: d

Explanation: The biogas outlet pipe is fixed on the top middle portion of the biogas holder, which is made of a small of GI pipe fitted with socket and a gat valve. The biogas generated in the plant and stored in the biogas holder is taken through the gas outlet pipe via pipeline to the place of utilization.

8. What is the life time of Floating drum plant?

a) Short

b) Large

c) Life time

d) Average

Answer: a

Explanation: Even though the floating drum plant is built with all reinforced constituents and hard materials the life of floating drum plant is short which is upto 15 years; And in tropical coastal regions its just about 5 years.

9. Which plant has no movable parts?

a) KVIC

b) JANATA model

c) Pragati design plant

d) Ferro – cement plant

Answer: b

Explanation: The janata model consists of a digester and fixed biogas holder as gas storage chamber covered by a dome shaped enclosed roof structure. The entire plant is made of bricks and cement masonry and constructed underground. Unlike the KVIC, the janata model has no movable part.

10. Which part of the janata bio gas plant comprises of the fermentation chamber and the gas storage chamber?

a) Foundation

b) Digester

c) Inlet chamber

d) Outlet chamber

Answer: b

Explanation: The digester is a cylindrical tank resting on the foundation. The top surface of the foundation serves as the bottom of the digester. The digester of janata biogas plant comprises of the fermentation chamber and the gas storage chamber.

11. The Gas storage chamber of the digester of janata model is designed to store what percentage og daily gas?

a) 5%

b) 19%

c) 33%

d) 58%

Answer: c

Explanation: The gas storage chamber is also cylindrical in shape and is the integral part of the digester and located just above the fermentation chamber. The GSC is designed to store 33% of the daily gas production from the plant.

12. Which plant has fixed hemi-spherical roof?

a) Fixed dome plant

b) Floating drum plant

c) Chinese plant

d) Flexible bag biogas plant

Answer: a

Explanation: Fixed dome plant has a fixed hemi-spherical roof which forms the cover of the digester and constructed with brick and cement concrete mixture, after which it is plastered with cement mortar. The dome is only an enclosed roof designed in such a way to avoid steel reinforcement.

13. The upper portion of the ______________ is in the shape of bell mouth and constructed using bricks and cements mortar in fixed dome plant.

a) gas storage chamber

b) inlet chamber

c) outlet chamber

d) digester

Answer: b

Explanation: The upper portion of the inlet chamber is in the shape of bell mouth and constructed using bricks and cements mortar in fixed dome plant. It s outer wall is kept inclined to the cylindrical wall of the digester so that the feed material can flow easily into the digester by gravity.

14. The outer chamber upper is called as ___________

a) outlet displacement chamber

b) outlet discharge chamber

c) outlet density chamber

d) outlet dwell chamber

Answer: a

Explanation: Outer chamber is a rectangular shaped located just on the opposite side of the inlet chamber. The bottom opening of the outlet chamber is connected to the gate and the upper portion is much wider and is known as Outlet displacement chamber .

This set of Thermal Engineering Questions and Answers for Aptitude test focuses on “Availability and Irreversibility”.

1. Availability function  is given by ______

a) U-p 0 V-T 0 S

b) U+p 0 V-T 0 S

c) –U+p 0 V-T 0 S

d) p 0 V-T 0 S

Answer: b

Explanation: W max = ∅1-∅ 0 is known as availability of closed system at state 1. Where ∅1 availability function at state 1 and ∅ 0 is the availability function of dead state. The availability function is defined as ∅ = U+p 0 V-T 0 S

2. What is the lowest temperature at which heat rejection can takes place?

a) 100°C

b) 0K

c) 273K

d) Temperature equal to surrounding

Answer: d

Explanation: The lowest temperature at which heat rejection can takes place is equal to the temperature of surrounding. Efficiency will be maximum if heat rejection takes place at temperature of surrounding.

3. Gibbs function is given by _____

a) G = H – TS

b) G = pV – TS

c) G = U – pV

d) G = U + pV

Answer: a

Explanation: Gibbs function is given by G = H – TS. Also it is given by G =U +pV -TS

4. What is the part of heat energy which is converted into mechanical work which reduce system in state of equilibrium?

a) Available energy

b) Electrical energy

c) Unavailable energy

d) Hydraulic energy

Answer: a

Explanation: Available energy is the part of heat energy which is converted into mechanical work. It is converted by ideal processes which reduce system in state of equilibrium.

5. The _________ obtainable from a certain heat input in a cyclic heat engine is called _________

a) Maximum work, unavailable energy

b) Maximum work, available energy

c) Minimum work, available energy

d) Minimum work, unavailable energy

Answer: b

Explanation: The maximum work obtainable from a certain heat input in a cyclic heat engine is called available energy. Available energy is the part of heat energy which is converted into mechanical work.

6. What is the part of heat energy which cannot be converted into mechanical work which reduce system in state of equilibrium?

a) Available energy

b) Electrical energy

c) Unavailable energy

d) Hydraulic energy

Answer: c

Explanation: Unavailable energy is the part of heat energy which cannot be converted into mechanical work. It cannot be converted even by ideal processes which reduce system in state of equilibrium.

7. The _________ has to be rejected by any cyclic heat engine is called _________

a) Maximum energy, unavailable energy

b) Maximum energy, available energy

c) Minimum energy, available energy

d) Minimum energy, unavailable energy

Answer: d

Explanation: The minimum energy that has to be rejected by any cyclic heat engine is called unavailable energy. Unavailable energy is the part of heat energy which cannot be converted into mechanical work.

8. The difference between idealized reversible work and actual work done by the system is called _____

a) Reversibility

b) Irreversibility

c) Gain in available energy

d) Decrease in unavailable energy

Answer: b

Explanation: The difference between idealized reversible work and actual work done by the system is called irreversibility of the process. Irreversibility = loss in available energy = increase in unavailable energy.

9. Availability is also known as __________

a) Exergy

b) Ronthalpy

c) Enthalpy

d) Entropy

Answer: a

Explanation: The maximum work obtainable from a certain heat input in a cyclic heat engine is called available energy. Availability is also known as exergy.

10. Which of the following statements applicable to a perfect gas will also be true for an irreversible process?

a) dQ = dU + pdV

b) dQ = du – pdV

c) Tds = dU + pdV

d) Tds = du – pdV

Answer: c

Explanation: Tds = dU + pdV

All are properties thus it is applicable for reversible as well as irreversible process.

11. Which of the following statement is correct?

a) Available energy is generally conserved

b) Available energy is maximum theoretical work obtainable

c) Available energy can either be negative or positive

d) Available energy is minimum theoretical work obtainable

Answer: b

Explanation: The maximum work obtainable from a certain heat input in a cyclic heat engine is called available energy. Available energy is the part of heat energy which is converted into mechanical work.

12. When system undergoes a process such that ∫dQ/T = 0 and ∆S>0, the process is ______

a) Irreversible adiabatic

b) Reversible adiabatic

c) Isothermal

d) Isobaric

Answer: a

Explanation: ∆S more than zero which implies irreversible process. ∫dQ/T = 0 is adiabatic.

This set of Energy Engineering Multiple Choice Questions & Answers  focuses on “Fuels and Combustion”.

1. In presence of which gas is the fuel burnt to generate energy in the form of heat?

a) Oxygen

b) Hydrogen

c) Methane

d) Nitrogen

Answer: a

Explanation: The fuel is burnt in the presence of oxygen to generate energy in the form of heat. This heat energy can be used for electrical power generation in steam power plants and for propelling ships, automobiles and locomotives, etc.

2. Which are the main constituents of fuel from given options?

a) Carbon and Nitrogen

b) Oxygen and Hydrogen

c) Carbon and Hydrogen

d) Helium and Oxygen

Answer: c

Explanation: Carbon and Hydrogen are the main constituents of a fuel. In addition to them, fuel also contains sulfur, oxygen and nitrogen in minimal quantities. Depending on the fuel, the percentage of carbon ranges from 50-95%, hydrogen (H 2 ) 2-6%, oxygen (O 2 ) 2-4%, sulfur (S 2 ) 0.5-3% and Nitrogen  5-7%. A solid fuel may also contain 2-30% ash. In a fuel carbon, hydrogen and sulfur are combustible elements whereas nitrogen and ash are incombustible elements. Since the major combustible elements in fuel are carbon and hydrogen it can also be called as Hydrocarbon fuel.

3. Which fuel is used widely in steam power plants?

a) Oil

b) Gas

c) Coal

d) Petroleum

Answer: c

Explanation: Coal is the oldest form of fuel and is still used in a large scale throughout the world by steam power plants as well as all power generation plants. Coal is a heterogeneous compound and its constituents are always carbon, hydrogen, oxygen, sulfur, nitrogen and certain mineral non combustibles.

4. What is the phenomenon of the formation of coal called?

a) Metamorphism

b) Diagenis’

c) Photosynthesis

d) Protolith

Answer: a

Explanation: The phenomenon by which the buried vegetation consisting wood, grass, shrubs etc, transformed into coal is known as metamorphism. The nature of coal will depend upon the type of vegetation buried and nature and duration of metamorphism. There are different types of metamorphism namely contact metamorphism, regional metamorphism, cataclastic metamorphism, hydrothermal metamorphism, burial metamorphism and shock metamorphism.

5. On what basis is the coal classified?

a) Period of formation

b) Depending on capacity to burn

c) Region/area where is it formed

d) Physical and chemical composition

Answer: d

Explanation: The coal is classified on the basis of its physical and chemical composition. The proximate and ultimate analyses are the common tests which are used to find the commercial value of the coal. The proximate analysis gives characteristics of coal such as percentage of moisture, ash and volatile matter. Analysis of coal gives an indication about fusion temperature and heating value of the coal.

6. What is the use of electrostatic precipitations in steam power plant?

a) To remove the steam

b) To draw the coal powder into boiler

c) To remove the feed water

d) To remove fly ash

Answer: d

Explanation: The electrostatic precipitators are extensively used in the steam power plant for removal of fly ash from the electric utility boiler emission. Since the water inside the boiler is converted into steam, the fire for converting is fuelled by burning the coal which produces high amount of fly ash, so that the ash’ can be reused. The use of electrostatic precipitators is growing rapidly because of the new strict air code and environmental laws. An electrostatic can be designed to operate at any desired efficiency.

7. Why is ‘make-up water’ added to drum continuously?

a) To remove the impurities in tube

b) To replace the water that has been converted into steam

c) To keep the system cool externally

d) To compensate for water loss trough blow down

Answer: d

Explanation: Make-up water is added to compensate for the losses of water incurred by blow downs or leakages occurring in boiler, and also to maintain desired water level in boiler steam drum. Blow down and leakages are common since there is continuous flow of condensate from condenser to boiler.

8. What causes failure of boiler tube?

a) Heating the tubes, when desired water level is not maintained

b) Induced pressure in the water

c) Over use of boiler

d) Hardness of water

Answer: d

Explanation: Hardness of water causes failure of boiler tube. Hard water consists of calcium and magnesium salts. Hardness in water will form deposits on the tube water surfaces which will lead to overheating and failure of tubes. Thus the salts have to be removed from the water.

This set of Power Systems Multiple Choice Questions & Answers  focuses on “Fuel Combustion and Combustion Equipments”.

1. Which of the following is not an advantage of stoker firing?

a) Zero capital investment

b) Greater combustion capacity

c) Uniform feeding of coal into furnance

d) Saving in labour cost

Answer: a

Explanation: Stokers are the mechanical arrangement which is used to feed coals into the furnace. Due to the mechanical equipment combustion capacity increases with uniform feeding of coal of into furnace, also labour cost is reduced. Stoker firing requires installation of mechanical stoker hence, capital investment requires.

2. In case of underfeed stokers fuel is supplied in the furnace _________

a) below the point of air admission

b) above the point of air admission

c) beside the point of air admission

d) by shovels

Answer: a

Explanation: Mechanical stokers are of two types namely underfeed stokers and overfeed stokers. The two differ in manner of feeding of coal below or above the level at which primary air is admitted in the furnace. In case of underfeed stokers fuel is supplied into the furnace below the point of air admission.

3. Which of the following is most advantageous and most widely method of solid fuel firing?

a) Stoker firing

b) Underfeed firing

c) Spreader stoker firing

d) Pulverised fuel firing

Answer: d

Explanation: The advantages of using pulverised fuel outweigh the disadvantages and all the modern power plants uses pulverised fuel firing method. In pulverised fuel firing better combustion is achieved. Pulverization is a means of exposing a large surface area to the action of oxygen and consequently helping the combustion.

4. What is the main disadvantage of spreader stoker?

a) Poor control

b) Difficult operation

c) More unwanted fly ash

d) Dangerous operation

Answer: c

Explanation: In spreader stoker the coal is not fed into the furnace by means of grate. Coal is fed by means of hopper, and function of grade is only to collect and remove the ash out of furnace. Spreader stokers can burn any type of coal from lignite to samianthracite, whether they are free burning or cocking. Their disadvantage is that fly ash is more.

5. Which do you mean by pulverization?

a) Burning of crushed coal

b) Burning of uncrushed coal

c) Crushing of coal into small partials

d) Breaking coal into small particsl

Answer: c

Explanation: Pulverization is crushing of coal into fine particles. This exposes large surface area to the action of oxygen and consequently helps the combustion. Solids fuels can be used in a powderd and burn like oil and gas.

6. What is the amount of air required to burn the pulverised coal?

a) Less than air required to burn lumped coal

b) More than air required to burn lumped coal

c) More than air required to burn lumped coal

d) No air required to burn the lapped coal

Answer: a

Explanation: The surface area of pulverised coal is increased in almost the ratio of 400:1, therefore high rates of combustion are possible. Moreover a smaller quantity of air is required than when the fuel is burnt in lumped form.

7. Capital cost of pulverised fuel firing is less than hand firing.

a) True

b) False

Answer: b

Explanation: Hand firing does not require any capital investment. For pulverised fuel firing separate coal preparation plant is required which makes the installation expensive.

8. What is the main requirement of good coal burner?

a) Minimum operating and maintenance cost

b) Production of uniform flame with complete combustion

c) The burner should be easy to handle and control

d) Attainment of proper designed

Answer: b

Explanation: Main requirement of good coal burner is complete combustion with production of uniform flame. Minimum operating and maintenance cost, simplicity in operation, attainment of proper design all are requirements of complete combustion.

9. Coal burners are employed to fire the pulverised coal along with primary air in the furnace.

a) True

b) False

Answer: a

Explanation: Coal burner’s fire pulverised coal with stream of primary air into the furnace. Secondary air is then admitted separately below the burner, around the burner or elsewhere.

10. Spread stockers can be employed for boiler capacitors of _________

a) 1000 kg/hr – 14000 kg/hr

b) 586 kg/hr – 1086 kg/hr

c) 1500 kg/hr – 40000 kg/hr

d) 70,000 kg/hr – 140,000kg/hr

Answer: d

Explanation: In spreader stoker coal is fed from the coal hopper into the path of the rotor by means of a conveyor and rotor throws the coal into the furnace. Grate collects the ash and moves it out of the furnace.

This set of Engineering Chemistry Multiple Choice Questions & Answers  focuses on “Analysis of Coal”.

1. In analysis of coal, determination of _______ is done by Kjeldahl method.

a) Volatile matter

b) Nitrogen

c) Ash

d) Oxygen

Answer: b

Explanation: In ultimate analysis of coal, Nitrogen is determined by Kjeldahl method. The method involves 3 steps which are digestion, distillation and titration. In digestion, sample is digested in boiling concentrated sulphuric acid in presence of catalyst and ammonium sulphate is obtained. In distillation, excess NaOH solution is added and ammonia is obtained. In titration, ammonia is determined by volumetric acid solution. The result can be expressed in terms of %N.

2. In determination of carbon and hydrogen by ultimate analysis, increase in weight of calcium chloride bulb represents ______

a) weight of water formed

b) weight of carbon dioxide formed

c) carbon

d) hydrogen

Answer: a

Explanation: The gaseous products of combustion are passed through two bulbs. One containing weighed amount of anhydrous Calcium chloride which absorbs water other containing weighed amount of potassium hydroxide which absorbs carbon dioxide.

3. Hydrogen available for combustion is lesser than the actual one.

a) true

b) false

Answer: a

Explanation: Oxygen is present in combined form with hydrgen in the coal. It is very difficult to break hydrogen-oxygen bond and hence, that combined form of hydrogen is not available for combustion. Thus, free hydrogen available for combustion is lesser than the actual one.

4. Good quality of coal should have _____ % of oxygen.

a) low

b) high

c) 100

d) 0

Answer: a

Explanation: Good quality of coal should have less % of oxygen because increase in 1% oxygen content decreases calorific value approximately by 1.7%. Ideal fuel should have high calorific value and hence, less %oxygen.

5. Percentage of ash by analysis of coal is given by _____

a)  × 100

b)  × 100

c) /weight of sample × 100

d)  × 100

Answer: a

Explanation: Ash is weight of residue obtained after complete combustion of 1 g of coal at 700‐750°C. Ash in % is given by, %A =  × 100.

6. In coal, Sulphur is usually present in the extent of __________

a) 0.5-3%

b) 90-95%

c) 70-75%

d) 80-85%

Answer: a

Explanation: Sulphur is generally present in the extent of 0.5 to 3% and is usually derived from ores like gypsum, iron pyrites, etc. High percentage of sulphur in coal is undesirable to be used for making coke in the iron industry since it affects the quality and properties of steel.

7. In ultimate analysis, %hydrogen is given by __________

a) (Increase in weight of CaCl 2 tube×2×100 / weight of coal sample taken×18)

b) 

c) (Increase in weight of CaCl 2 tube×4×100 / weight of coal sample taken×18)

d) 

Answer: a

Explanation: %hydrogen=  × 100

In ultimate analysis of coal,

%H = (Increase in weight of CaCl 2 tube×2×100 / weight of coal sample taken×18).

8. In ultimate analysis, %carbon is given by __________

a) 

b) 

c) (Increase in weight of CaCl 2 tube×12×100 / weight of coal sample taken×44)

d) (Increase in weight of CaCl 2 tube×2×100 / weight of coal sample taken×18)

Answer: a

Explanation: %C =  ×100

In the ultimate analysis of coal,

%C = .

9. __________ is the weight of residue obtained after burning a weighed amount of dry coal.

a) Ash

b) Volatile matter

c) Moisture

d) Carbon

Answer: a

Explanation: Ash is an unwanted matter and non-combustible which is the residue left after all combustible substances are burnt off. %Ash=  × 100.

10. In proximate analysis of coal, %fixed carbon = __________

a) 100 – % 

b) 100 – % 

c) 100 – %  + %volatile matter

d) 100 – % 

Answer: a

Explanation: In proximate analysis of coal, firstly percentage of ash, volatile matter and moisture is found out. Percentage of fixed carbon can be found out by subtracting the above percentages from 100 since coal consists of ash, moisture, volatile matter and fixed carbon.

This set of Fluid Mechanics Multiple Choice Questions & Answers  focuses on “Model Analysis”.

1. What is a model analysis?

a) A small-scale replica

b) Actual structure

c) Theory structure.

d) Adopted structure

Answer: a

Explanation: Model analysis is defined as a small-scale replica of the actual structure. Model analysis need not be smaller all the time. They can even be larger than the prototype.

2. What is a prototype?

a) A small-scale replica

b) Actual structure

c) Theory structure.

d) Adopted structure

Answer: b

Explanation: Prototype is the actual structure that needs to be constructed. For a better understanding of the model, we prepare a model analysis. They can even be larger than the prototype.

3. Advantage of a model analysis is_________

a) Performance cannot be predicted

b) The relationships between the variable cannot be obtained

c) Shear stress to thermal diffusivity

d) Alternative designs can be predicted

Answer: d

Explanation: One of the major advantages of the model analysis is that we can predict the alternative designs. It can also predict the performance of the machine in advance.

4. Why do we need a model analysis?

a) For determining the dimensions

b) To provide a safe design

c) To check the shear stress

d) To check the thermal diffusivity

Answer: b

Explanation: One of the major advantages of the model analysis is that we can predict the alternative designs. It provides the safest design in the most economical way.

5. The similarity between the motion of model and prototype is_________

a) Dynamic similarity

b) Potential similarity

c) Kinematic similarity

d) Design similarity

Answer: c

Explanation: Kinematic similarity is defined as the similarity between motion of the model and the prototype. It exists in between the model and prototype. The points in the model and prototype are of the same magnitude.

6. The similarity between the forces of model and prototype is ________

a) Dynamic similarity

b) Potential similarity

c) Kinematic similarity

d) Design similarity

Answer: a

Explanation: Dynamic similarity is defined as the similarity between forces of a model and the prototype. It exists in between the model and prototype. The points in the model and prototype are of the same magnitude.

7. Which among these forces does not act in a moving fluid?

a) Inertial force

b) Viscous force

c) Gravity force

d) Drag

Answer: d

Explanation: Drag does not take place in moving fluids. Moving fluids are restricted by a viscous force and move along an inertial force. The gravitational force tends to act perpendicular to the fluid surface.

8. What is the formula for elastic force?

a) Elastic stress/area

b) Elastic strain/area

c) Elastic stress*area

d) Elastics stress* Elastic strain

Answer: c

Explanation: Elastic force is a force that is developed by a material to retain to its original position. It regains its shape after a period of time. When an elastic material is compressed or stretched, it develops an elastic force.

9. For a dynamic similarity between a model and a prototype, the ratio of their forces in the model and the prototype must be equal.

a) True

b) False

Answer: a

Explanation: For a dynamic similarity between a model and a prototype, the ratio of their forces in the model and the prototype must be equal. It means that the dynamic similarity between a model and a prototype must be the same.

This set of Thermal Engineering Multiple Choice Questions & Answers  focuses on “Centrifugal & Reciprocating Pumps”.

1. Which hydraulic machine convert mechanical energy into pressure energy?

a) Pump

b) Refrigerator

c) Heat engine

d) Turbine

Answer: a

Explanation: Pump is a hydraulic machine which convert mechanical energy into hydraulic energy or pressure energy. Pump is a device which transfer the fluid from suction head to delivery head.

2. Which of the following is a positive displacement pump?

a) Centrifugal pump

b) Diffuser pump

c) Axial pump

d) Reciprocating pump

Answer: d

Explanation: Reciprocating pump and rotary pumps are positive displacement pumps. Centrifugal pump, diffuser pump and axial pump are Dynamic pumps.

3. Which of the following is a dynamic pump?

a) Centrifugal pump

b) Rotary pump

c) Multiple rotor

d) Reciprocating pump

Answer: a

Explanation: Reciprocating pump and rotary pumps are positive displacement pumps. Centrifugal pump, diffuser pump and axial pump are Dynamic pumps.

4. Dynamic pumps are also known as ______

a) Positive displacement pumps

b) Non-positive displacement pumps

c) Reciprocating pump

d) Rotary pump

Answer: b

Explanation: Dynamic pumps are also known as non-positive displacement pumps or roto-dynamic pumps. Reciprocating pump and rotary pumps are positive displacement pumps.

5. Hydraulic machine convert mechanical energy into hydraulic energy by means of centrifugal force acting on the fluid ______

a) Centrifugal pump

b) Rotary pump

c) Multiple rotor

d) Reciprocating pump

Answer: a

Explanation: Pump is a hydraulic machine which convert mechanical energy into hydraulic energy or pressure energy. In centrifugal pump mechanical energy is converted into hydraulic energy by means of centrifugal force.

6. The rise in pressure head at any point of the impeller is proportional to ______

a) \

 

 \

 

 \

 

 \(\frac{u^2}{g} \)

Answer: b

Explanation: From Bernoulli equation \(\frac{∆p}{ρ}+\frac{∆u^2}{2g}+∆z=0 \)

Here, u 1 = 0 and ∆z = 0

\(\frac{∆p}{ρ} = \frac{u^2}{2g} \)

7. Which of the following is not an advantage of centrifugal pump?

a) Low initial cost

b) High efficiency

c) Difficult installation

d) Easy maintenance

Answer: c

Explanation: Advantages of centrifugal pump are:

Low initial cost

Easy maintenance

Easy installation

High efficiency

Uniform and continuous discharge

8. When the speed of the centrifugal pump is doubled, the power required to drive the pump will be _____

a) Increase 8 times

b) Increase 4 times

c) Double

d) Remain the same

Answer: a

Explanation: PαN 3 so P 2 =8P 1

9. Why an air vessel is used in a reciprocating pump?

a) To obtain a continuous supply of water at uniform rate

b) To reduce suction head

c) To increase the delivery head

d) To reduce cavitation

Answer: a

Explanation: Air vessel are fitted to delivery side of reciprocating pump. An air vessel is used in a reciprocating pump to obtain a continuous supply of water at uniform rate and to reduce power consumption.

10. Negative slip occurs in reciprocating pumps, when delivery pipe is ______

a) Long and suction pipe is short and pump is running at low speed

b) Long and suction pipe is short and pump is running at high speed

c) Short and suction pipe is long and pump is running at low speed

d) Short and suction pipe is long and pump is running at high speed

Answer: a

Explanation: Negative slip occurs in reciprocating pumps, when delivery pipe is long and suction pipe is short. Also, pump should be running at low speed.

11. Why is a minimum of net positive suction head  required for a hydraulic pump?

a) To prevent cavitation

b) To increase discharge

c) To increase suction head

d) To increase efficiency

Answer: a

Explanation: NPSH is the difference of absolute pressure head at the inlet to the pump and the vapour pressure head corresponding to temperature of the liquid pumped plus the velocity head. The NPSH should always be positive to keep away the cavitation.

12. For a given centrifugal pump _____

a) The discharge varies directly as the speed

b) The head varies inversely as the speed

c) The power varies as the square of the speed

d) The discharge varies as the square of the speed

Answer: a

Explanation: The dimensionless number for centrifugal pump is \(\frac{Q}{ND^3}\)

Hence discharge is directly proportional to speed.

This set of Heat Transfer Operations Multiple Choice Questions & Answers  focuses on “Types of Condensers”.

1. What is the name of the following condenser setup?

heat-transfer-operations-questions-answers-types-condensers-q1

a) Spray Condenser

b) Packed Column

c) Baffled Column

d) Fluidised column

Answer: b

Explanation: The bed is stable and at a low height, hence the given figure is packed column. If it had been in free state, it would have been a fluidised column.

2. What is the name of the following condenser setup?

heat-transfer-operations-questions-answers-types-condensers-q2

a) Spray Condenser

b) Tube side Condenser

c) Baffled Column

d) Fluidised column

Answer: b

Explanation: The tube side condenser is a special condenser in which the coolant flows in the shell and the condensation in the tubes, hence the name tube side condenser. It cannot usually handle high pressure drops.

3. What is the name of the following condenser setup?

heat-transfer-operations-questions-answers-types-condensers-q3

a) Spray Condenser

b) Tube side Condenser

c) Baffled Column

d) Sparged tube column

Answer: d

Explanation: The sparged tube condenser consists of a perforated tube/ pipe with holes for injecting bubbles of vapour into a pool of liquid. The condensate being immiscible float at the top of the coolant, or maybe the coolant is the same material as the vapour.

4. What is the name of the following condenser setup?

heat-transfer-operations-questions-answers-types-condensers-q4

a) Spray Condenser

b) Tube side Condenser

c) Baffled Column

d) Sparged tube column

Answer: a

Explanation: The spray type condenser is actually a direct contact condenser where in which the coolant and gas come in direct contact as the coolant is sprayed with the use of nozzles in a vessel filled with the vapour.

5. What is the name of the following condenser setup?

heat-transfer-operations-questions-answers-types-condensers-q5

a) Spray Condenser

b) Tube side Condenser

c) Baffled Column

d) Sparged tube column

Answer: c

Explanation: The baffled column condenser is similar to the spray condenser, with the difference that the coolant is directed to flow over a set of trays in a column. The vapour is supplied from the bottom as shown in the diagram.

6. What is the name of the following condenser setup?

heat-transfer-operations-questions-answers-types-condensers-q6

a) Tray Type

b) Jet Condenser

c) Sparged Tube

d) Air-cooled

Answer: d

Explanation: The air cooled condensers are a separate category of condensers which use air as the cooling medium, it may be natural/forced convection air cooled condensers. As an example for this being the commonly used indoor refrigerators and air conditioners.

7. What is the name of the following condenser setup?

heat-transfer-operations-questions-answers-types-condensers-q7

a) Tray Type

b) Jet Condenser

c) Sparge Tube

d) Air-cooled

Answer: b

Explanation: The jet condenser is a device in which a jet of coolant liquid forced into a hot vapour stream, de-superheating the vapour and hence condensing it. Here a jet of liquid is injected into a vessel carrying vapour via a plate/pipe and a nozzle located at the centre axis.

8. What is the name of the following condenser setup?

heat-transfer-operations-questions-answers-types-condensers-q8

a) Tray Type

b) Jet Condenser

c) Horizontal shell side

d) Air-cooled

Answer: c

Explanation: The shell side condenser is a special condenser in which the coolant flows in the tubes and the condensation happens in the shell, hence the name shell side condenser. This condenser can handle large pressure drops.

This set of Heat Transfer Operations Multiple Choice Questions & Answers  focuses on “Condensers – Heat Transfer Coefficients – 1”.

1. Which one of the following is not an assumption of condensation heat regime taken to calculate the heat transfer coefficient?

a) Presence of linear temperature profile

b) Absence of high pressure

c) Absence of viscous shear of the vapour

d) Thickness of the film is too small to create a temperature difference

Answer: b

Explanation: The three assumptions of the of condensation heat regime taken to calculate the heat transfer coefficient are –

Absence of viscous shear of the vapour

Thickness of the film is too small to create a temperature difference

Presence of linear temperature profile

2. What is the expression for the flow velocity of the falling film in a vertical condenser?

a) U=\

 

 

\)

b) U=\

 

\)

c) U=\

 

\)

d) U=\

 

 

\)

Answer: d

Explanation: U=\

 

 

\) is the correct equation obtained by deriving the Mass balance equation from the Nusselt’s theory of condensation.

Weight of fluid = Bouyancy + Shear

g=\

 

dy=∫μ du

U = \

 

 

\)

When the vapour density is negligible, \

 

 

\)

3. What is the term y in the expression for velocity of condensate flow?

U=\

 

 

\)

a) Film thickness

b) Film thickness at y

c) Distance from the wall at x

d) Film thickness at x

Answer: d

Explanation: The term y in the expression for velocity of condensate flow is the distance from the cooling wall inside the film at a distance x from the top. It would be clear from the diagram –

4. What is the expression for mass flow rate of condensate in a condenser?

a) M°=\

 

 

\)

b) M°=\

 

 

\)

c) M°=\

 

 

\)

d) M°=\

 

 

\)

Answer: d

Explanation: Weight of fluid = Bouyancy + Shear

g=\

 

dy=∫μ du

U = \

 

 

\)

Integrating the equation with φ = 0, M° = \

 

 

 = \frac{\rho^2}{\mu}g

 

\) we get the required expression.

5. What is the term delta in the expression for velocity of condensate flow?

M°=\

 

 

\)

a) Final stable film thickness

b) Film thickness at y

c) Average film thickness

d) Film thickness at x

Answer: d

Explanation: The term delta in the expression for velocity of condensate flow is the film thickness at a distance x from the top.

6. What is the expression for flow velocity of the condensate if the density of the vapour is not zero?

a) U=\

 

 

\)

b) U=\

 

\)

c) U=\

 

\)

d) U=\

 

 

\)

Answer: d

Explanation: U=\

 

 

\)

The derivation shown is –

Weight of fluid = Bouyancy + Shear

g=\

 

dy=∫μ du

U = \

 

 

\)

7. What is the expression for mass flow rate of condensate in a condenser?

a) M°=\

 

 

\)

b) M°=\

 

 

\)

c) M°=\

 

 

\)

d) M°=\

 

 

\)

Answer: d

Explanation: Integrating the equation M°=\

 

 

dy = \frac{\rho}{\mu}g

 

\)

8. What is the expression for the laminar film thickness of the condensate at a distance of x from the top of the condenser?

a) δ=[4K(T sat – T L )μ x)/(ρ gh fg )] 1/2

b) δ=[4K(T sat – T L )μ x)/(ρ gh fg )] 1/4

c) δ=[4K(T sat – T L )μ x)/(ρ gh fg )] 1/8

d) δ=[K(T sat – T L )μ x)/(ρ gh fg )] 1/4

Answer: b

Explanation: The derivation considering vapour density is –

\

 

 

\)

∴ equating

\(\delta^3\frac{ds}{dx} = \frac{K

\mu}{\rho

g\lambda}\)

\(\int\delta^3d\delta = \int \frac{K

\mu dx}{\rho

g\lambda}\)

\(\frac{\delta^4}{4} = \int \frac{K

\mu x}{\rho

g\lambda}\)

\(\delta = [\frac{4K

\mu x}{\rho

g\lambda}]^{1/4}\)

9. What is the term T L in the Nusselt theory of condensation equation for film thickness?

\

 

 Liquid temperature

b) Gas temperature

c) Wall temperature

d) Bulk temperature

Answer: a

Explanation: The term T L is the liquid temperature of the film-

10. What is the expression for Averaged convective heat transfer coefficient for a vertical condenser?

a) h VER =0.943 \

 

 h VER =0.943 \

 

 h VER =0.943 \

 

 h VER =0.943 \([\frac{K^3 p^2 g h_{fg}}{\mu L

}]^{25}\)

Answer: a

Explanation: The derivation for the averaged heat transfer coefficient can be derived as –

\(\delta = [\frac{4K

\mu x}{\rho

g\lambda}]^{1/4}\)

h x = \(\frac{K}{\delta}\)

h avg = \(\frac{4}{3} \frac{K}{\delta}\)

h VER =0.943 \([\frac{K^3 p^2 g \lambda}{\mu L

}]^{\frac{1}{4}} \)

h VER =0.943 \([\frac{K^3 p^2 g h_{fg}}{\mu L

}]^{0.25} \)

11. What is the relation between the averaged heat transfer coefficient over the entire condenser length vs its value at a point x?

a) \

 

 \

 

 

 \

 

 

 \(\frac{h_{avg}}{h_x} =\frac{3}{4} \)

Answer: c

Explanation: h avg =\(\frac{4}{3}\) K/δ and h x = K/δ hence dividing the two expressions we get \(\frac{h_{avg}}{h_x} = \frac{4}{3} \).

12. What is the relation between the averaged heat transfer coefficient over the entire condenser length vs its film thickness value at a point x?

a) h avg = \

 

 h avg = \

 

 h avg = \

 

 h avg = \(\frac{4}{3} \delta/K \)

Answer: a

Explanation: The correct representation of the average value is 4K/3Δ because the given statement is a dimensional and calculation result obtained after the integration of h X as the basic equation.

13. What is the relation between the heat transfer coefficient vs its film thickness value at a point x?

a) h x = K/δ

b) h avg = \

 

 h avg = δ/K

d) h avg = \(\frac{4}{3} \delta/K \)

Answer: a

Explanation: The term h x = K/δ is the dimensionally obtained term by minutely examining the dimensions of each of the given terms hx, k and δ, the diagram describing delta is –

14. What is the value of the constant C when the condenser is placed vertically?

h VER =C \

 

 0.0943

b) 0.943

c) 0.725

d) 0.633

Answer: b

Explanation: The derivation can be presented as-

\(\delta = [\frac{4K

\mu x}{\rho

g\lambda}]^{1/4}\)

h x = \(\frac{K}{\delta}\)

h avg = \(\frac{4}{3} \frac{K}{\delta}\)

h VER =0.943 \([\frac{K^3 p^2 g \lambda}{\mu L

}]^{\frac{1}{4}} \)

h VER =0.943 \([\frac{K^3 p^2 g h_{fg}}{\mu L

}]^{0.25}\)

Here the term is 0.943

15. What is the value of the constant C when the condenser is placed horizontally?

h HOR =C \

 

 0.942

b) 0.725

c) 0.325

d) 0.027

Answer: b

Explanation: The expression derived from the Equation, –

\(\delta = [\frac{4K

\mu x}{\rho

g\lambda}]^{1/4}\)

h x = \(\frac{K}{\delta}\)

h avg = \(\frac{4}{3} \frac{K}{\delta}\)

h VER =0.943 \([\frac{K^3 p^2 g \lambda}{\mu L

}]^{\frac{1}{4}} \)

h VER =0.943 \([\frac{K^3 p^2 g h_{fg})}{\mu L

}]^{0.25} \)

This equation is for vertical wall, for horizontal wall, we multiply with the factor of sin, we get the value 0.725.

This set of Heat Transfer Operations Problems focuses on “Condensers – Heat Transfer Coefficients – 2”.

1. For the given equation of heat transfer coefficient, which one of the following condition should hold true?

h VER =0.943 \

 

 Re < 1800

b) Re > 1800

c) Re < 1000

d) Re < 18000

Answer:a

Explanation: The expression is valid only when the Reynolds number is less than the required 1800 value otherwise the constant term in the given equation would have to be changed according to the new value of forces and mass balances.

2. If the Reynolds Number is more than 1800, then what is the equation for the heat transfer coefficient of the condenser?

a) h=0.0077 Re 0.4 [K 3 ρ 2 g ⁄ μ 2 ] 1/3

b) h=0.0077 Re 0.4 [K 3 ρ 2 g ⁄ μ 2 ] 1/4

c) h=0.0077 Re 0.4 [K 4 ρ 2 g ⁄ μ 2 ] 1/3

d) h=0.0077 Re 0.4 [K 3 ρ 3 g ⁄ μ 2 ] 1/3

Answer:a

Explanation: The expression is valid only when the Reynolds number is less than the required 1800 value otherwise the constant term in the given equation would have to be changed according to the new value of forces and mass balances. If the Reynolds Number increases to a higher value, then we use the Kirkbride formula to derive the given equation:

h=0.0077 Re 0.4 [K 3 ρ 2 g ⁄ μ 2 ] 1/3

3. When do we apply the Kirkbride equation for determination of heat transfer coefficient in a condenser?

h=0.0077 Re 0.4 [K 3 ρ 2 g ⁄ μ 2 ] 1/3

a) Re < 1800

b) Re > 1800

c) Re < 1000

d) Re < 18000

Answer:b

Explanation: The expression is valid only when the Reynolds number is less than the required 1800 value otherwise the constant term in the given equation would have to be changed according to the new value of forces and mass balances. If the Reynolds Number increases to a higher value, then we use the Kirkbride formula. For below 1800 value, Re < 1800, the expression is –

h HOR =0.943 \([\frac{K^3 p^2 g h_{fg}}{\mu L

}]^{0.25}\)

And thus for answer when Re > 1800

h=0.0077 Re 0.4 [K 3 ρ 2 g ⁄ μ 2 ] 1/3

4. A 0.5m square plate is to be used to condense dry saturated steam at 0.08bar, if the wall surface is to be maintained at 18.5℃, calculate the film thickness at a distance of 25cm.

Given T sat = 41.5℃, density = 995kg/m 3 , h fg = 2402 KJ/m hr K , K = 2.22 KJ/mK, μ = 8.01x 10 –4 Ns/m 2

a) 0.0364 m

b) 0.024 m

c) 0.0046 m

d) 0.0056 m

Answer:a

Explanation: The equation for Film thickness is given by Nusselt theory of condensation as

\(\delta = [\frac{4K

\mu x}{\rho gh_{fg}}]^{1/4}\)

Here T sat = 41.5℃, T = 18.5℃, density = 995kg/m 3 , h fg = 2.4 KJ/m 2 K, K = 2.22 KJ/mK, μ = 8.01x 10 –4 N/m 2 s. Hence we get δ=0.0364

δ=[(4×2.22×10 3 x x8.01x 10 -4 x 0.25)/995×9.8×2.4×10 3 ] 1/4 =0.036m.

5. A 0.5m square plate is to be used to condense dry saturated steam at 0.08bar, if the wall surface is to be maintained at 18.5℃, calculate the Convective heat transfer coefficient h at a distance of 25cm.

Given T sat = 41.5℃, density = 995kg/m 3 , h fg = 2402 KJ/m 2 K, K = 2.22 KJ/mK, μ = 8.01 x 10 –4 N/m 2 s

a) 160×10 5 J/m 2 K

b) 3.5×10 5 J/m 2 K

c) 450×10 5 J/m 2 K

d) 150×10 5 J/m 2 K

Answer:b

Explanation: We can apply the equation/relation for the value to be h =\(\frac{k}{\Delta}\) = 2.22×10 3 /0.0064 = 3.5×105 J/m 2 K.

The equation for Film thickness is given by Nusselt theory of condensation as

\(\delta = [\frac{4K

\mu x}{\rho gh_{fg}}]^{1/4}\)

δ=[4×2.22×10 3 xx8.01 x 10 -4 x0.25 ⁄ 995×9.8x2402x10 3 ] 1/4 =0.0064m

Given T sat = 41.5℃, density = 995kg/m 3 , h fg = 2402 KJ/m hr K, K = 2.22 KJ/mK, μ = 8.01x 10 –4 Ns/m 2 Hence we get δ=0.0064m.

6. What is the value of h avg of a condenser at a distance of 25cm if a 0.5m square plate is to be used to condense dry saturated steam at 0.08bar, if the wall surface is to be maintained at 18.5℃?

Given T sat = 41.5℃, density = 995kg/m 3 , h fg = 2402 KJ/m hr K, K = 2.22 KJ/mK, μ = 8.01x 10 –4 Ns/m 2

a) 50×105 J/m 2 K

b) 10×105 J/m 2 K

c) 350×105 J/m 2 K

d) 26 x105 J/m 2 K

Answer:d

Explanation: We know the relation between them to be

h VER =0.943\([\frac{K^3 p^2 g h_{fg}}{\mu x

}]^{0.25}\)

h VER =0.943\([\frac{^3 x 995^2 x 9.8x2402x1000}{8.01×0.25}]^{0.25}\)=2572000=2572000

Here T sat = 41.5℃, T = 18.5℃, density = 995kg/m 3 , h fg = 2402 KJ/m 2 K, K = 2.22 KJ/mK, μ = 8.01 x 10 –4 N/m 2 s. Hence we get δ=0.000148 and h = 200 x10 5 J/m 2 K.

7. A 0.5m square plate is to be used to condense dry saturated steam at 0.08bar, if the wall surface is to be maintained at 18.5℃, calculate the mean flow velocity at a distance of 25cm.

Given T sat = 41.5℃, density = 995kg/m 3 , h fg = 2402 KJ/m hr K, K = 2.22 J/mK, μ = 8.01x 10 –4 Ns/m 2

a) 0.09 m/s

b) 0.9 m/s

c) 0.6m/s

d) 169.6m/s

Answer:a

Explanation: v = \

 

 

 

x8.01x 10 -4 x0.25)⁄ 995×9.8x2402x10 3 ] 1/4 =0.000148m

Given T sat = 41.5℃, density = 995kg/m 3 , h fg = 2402 KJ/m hr K, K = 2.22 J/mK, μ= 8.01x 10 – 4 Ns/m 2 Hence we get δ=1.48×10 -4 m.

8. A 0.5m square plate is to be used to condense dry saturated steam at 0.08bar, if the wall surface is to be maintained at 18.5℃, calculate the mean mass flow at a distance of 25cm.

Given T sat = 41.5℃, density = 995kg/m 3 , h fg = 2402 KJ/m hr K, K = 2.22 J/mK, μ = 8.01x 10 –4 Ns/m 2

a) 0.135 Kg/hr

b) 22.39 Kg/hr

c) 0.135 kg/sec

d) 22.39 kg/s

Answer:d

Explanation: M = VρA = 0.09x995x0.25 = 22.39 kg/s where

v = ρgδ 2 / 3μ = (995x10x0.000148 2 )/(3×8.01×10 -4 )=0.09m/s.

9. What is the expression for mean flow velocity of the film fluid in a vertical condenser?

a) v = ρgδ / 3μ

b) v = ρgδ 2 / μ

c) v = ρgδ 2 / 3

d) v = ρgδ 2 / 3μ

Answer:d

Explanation: The mean flow velocity is the velocity with which the film of condensate moves down the wall of the condenser, hence the derived formula is, v = ρgδ 2 / 3μ.

10. What is the expression for total heat transferred in a condenser of length L and diameter D?

a) h avg A Δ T sat-L

b) h avg A Δ T LMTD

c) h X A ΔT sat-L

d) h X A ΔT LMTD

Answer:a

Explanation: The total heat transferred is the measurement done for the whole condenser setup and cannot be used for a particular point. Hence we use the h avg as the heat transfer coefficient and LMTD is not applicable here.

11. What is the expression for total steam condensate rate?

a) M = V / h fg

b) M = h fg /V

c) M = V / h avg

d) M = VϸA

Answer:d

Explanation: m = VϸA is the dimensionally sane equation among all the others which do not match the dimension of M = MT -1 and hence VϸA is the correct option.

12. What is the expression for Reynolds number for the falling film in a condsenser?

a) Re= \

 

 Re= \

 

 Re= \

 

 Re= \(\frac{\ddot{M}}{\rho u}\)

Answer:a

Explanation: Re=\(\frac{4\ddot{M}}{\rho u}\) is the dimensionally sane equation among all the others which do not match the dimension of Reynolds number which is dimensionless and hence Re=\(\frac{4\ddot{M}}{\rho u}\) is the correct option.

This set of Fluid Mechanics Multiple Choice Questions & Answers  focuses on “Fluid Properties – 1”.

1. Which one of the following is the unit of mass density?

a) kg = m 3

b) kg = m 2

c) kg = m

d) kg = ms

Answer: a

Explanation: Mass Density is defined as the mass per unit volume, i.e., p = m ⁄v

Thus, the unit of p is kg = m 3 .

2. The specific gravity of a liquid has

a) the same unit as that of mass density

b) the same unit as that of weight density

c) the same unit as that of specific volume

d) no unit

Answer: d

Explanation: The specific gravity of a liquid is the ratio of two similar quantities  which makes it unitless.

3. The specific volume of a liquid is the reciprocal of

a) weight density

b) mass density

c) specific weight

d) specific volume

Answer: b

Explanation: Specific volume is defined as the volume per unit mass.

v = v⁄m = 1 / m⁄v = 1⁄p

where p is the mass density.

4. Which one of the following is the unit of specific weight?

a) N = m 3

b) N = m 2

c) N = m

d) N = ms

Answer: a

Explanation: Specific weight is defined as the weight per unit volume, i.e.,

γ = w / v

Thus, unit of is N = m 3 .

5. Which one of the following is the dimension of mass density?

a) [M 1 L -3 T 0 ].

b) [M 1 L 3 T 0 ].

c) [M 0 L -3 T 0 ].

d) [M 0 L 3 T 0 ].

Answer: a

Explanation: Mass Density is defined as the mass per unit volume, i.e.,

[p] = [m]/[v] = [m] /[L 3 ] = [ML -3 ].

6. Which one of the following is the dimension of specific gravity of a liquid?

a) [M 1 L -3 T 0 ].

b) [M 1 L 0 T 0 ].

c) [M 0 L -3 T 0 ].

d) [M 0 L 0 T 0 ].

Answer: d

Explanation: The specific gravity of a liquid is the ratio of two similar quantities  which makes it dimensionless.

7. Which one of the following is the dimension of specific volume of a liquid?

a) [M 1 L -3 T 0 ].

b) [M -1 L 3 T 0 ].

c) [M -1 L -3 T 0 ].

d) [M 0 L 3 T 0 ].

Answer: b

Explanation: Specific volume is defined as the volume per unit mass. Thus,

[v] = [V]/[m] = [L 3 ]/[M] = [M -1 L 3 ].

8. Which one of the following is the dimension of specific weight of a liquid?

a) [ML -3 T -2 ].

b) [ML 3 T -2 ].

c) [ML -2 T -2 ].

d) [ML 2 T -2 ].

Answer: c

Explanation: Specific weight is defined as the weight per unit volume, i.e.,

fluid-mechanics-questions-answers-fluid-properties-q8

9. Two fluids 1 and 2 have mass densities of p1 and p2 respectively. If p1 > p2, which one of the following expressions will represent the relation between their specific volumes v1 and v2?

a) v1 > v2

b) v1 < v2

c) v1 = v2

d) Cannot be determined due to insufficient information.

Answer: b

Explanation: Specific volume is defined as the volume per unit mass.

v = v⁄m = 1 / m⁄v = 1⁄p

where p is the mass density. Thus, if p1 > p2, the relation between the specific volumes v1 and v2

will be represented by v1 < v2.

10. A beaker is filled with a liquid up to the mark of one litre and weighed. The weight of the liquid is found to be 6.5 N. The specific weight of the liquid will be

a) 6:5 kN = m 3

b) 6:6 kN = m 3

c) 6:7 kN = m 3

d) 6:8 kN = m 3

Answer: a

Explanation: Specific weight is defined as the weight per unit volume, i.e.,

γ = w⁄V

Thus, γ = 6:5 ⁄10 -3 N ⁄ m 3 = 6:5 kN/m 3 .

11. A beaker is filled with a liquid up to the mark of one litre and weighed. The weight of the liquid is found to be 6.5 N. The specific gravity of the liquid will be

a) 0.65

b) 0.66

c) 0.67

d) 0.68

Answer: b

Explanation: Specific gravity of a liquid is defined as the ratio of the density of the liquid to that of water.

fluid-mechanics-questions-answers-fluid-properties-q11

Thus, S = 0:66.

12. A beaker is filled with a liquid up to the mark of one litre and weighed. The weight of the liquid is found to be 6.5 N. The specific volume of the liquid will be

a) 1 l =kg

b) 1:5 l =kg

c) 2 l =kg

d) 2:5 l =kg

Answer: b

Explanation: Specific volume is defined as the volume per unit mass. Thus,

fluid-mechanics-questions-answers-fluid-properties-q12

This set of Fluid Mechanics Interview Questions and Answers focuses on “Fluid Properties – 2”.

1. Calculate the specific weight and weight of 20dm 3 of petrol of specific gravity 0.6.

a) 5886,117.2

b) 5886,234.2

c) 11772,117.2

d) None of the mentioned

Answer: a

Explanation: Specific weight = density*acceleration due to gravity

=.6*1000*9.81=5886N/m 3

Weight=volume*specific weight

=5886*0.02=117.2N.

2. If 200m 3 of fluid has a weight of 1060N measured on the planet having acceleration due to gravity 6.625m/s2, what will be it’s specific volume?

a) 0.8

b) 0.7

c) 0.9

d) 0.5

Answer: a

Explanation: Specific weight=Weight/volume

= /volume

=density*acceleration due to gravity

=1/

Specific volume=1060/.

3. For an incompressible fluid does density vary with temperature and pressure?

a) It varies for all temperature and pressure range

b) It remains constant

c) It varies only for lower values of temperature and pressure

d) It varies only for higher values of temperature and pressure

Answer: b

Explanation: For an incompressible fluid, the change in density is negligible. Thus it does not change with temperature and pressure.

4. Specific gravity is what kind of property?

a) Intensive

b) Extensive

c) None of the mentioned

d) It depends on external conditions

Answer: a

Explanation: It is independent of quantity of matter present.

5. If there is bucket full of oil and bucket full of water and you are asked to lift them, which one of the two will require more effort given that volume of buckets remains same?

a) Oil bucket

b) Water bucket

c) Equal effort will be required to lift both of them

d) None of the mentioned

Answer: b

Explanation: Density of water is more that oil. Hence, its weight for same volume of oil will also be higher. Therefore, more effort will be required.

6. If the fluid has specific weight of 10N/m 3 for a volume of 100dm 3 on a planet which is having acceleration due to gravity 20m/s2 , what will be its specific weight on a planet having acceleration due to gravity 4m/s2?

a) 5 N/m 3

b) 50 N/m 3

c) 2 N/m 3

d) 10 N/m 3

Answer: c

Explanation: For same volume, specific weight is directly proportional to acceleration due to gravity

Specific weight=4*10/20=2.

7. Should Specific Wieght of incompressible fluid only be taken at STP?

a) Yes, as specific weight may show large variation with temperature and pressure

b) No, it can be taken for any temperature and pressure

c) It should be taken at standard temperature but pressure may be any value

d) It should be taken at standard pressure but temperature may be any value

Answer: b

Explanation: Specific weight is inversely proportional to volume. For incompressible fluid , variation of volume with temperature and pressure is negligible for practical consideration. Therefore, specific weight remains constant.

8. An instrument with air as fluid was involved in some experiment which was conducted during day in desert. Due to some reason experiment couldn’t be conducted during day and had to be conducted during night. However there were considerable errors in obtained values. What might be the reason of these errors?

a) It was human error

b) It was instrumental error

c) Error was due to the fact that experiment was conducted at night

d) None of the mentioned

Answer: c

Explanation: In Desert areas, temperature at night is considerably lower than at day. Due to this air contracts at night. Hence, it’s specific volume changes. As specific volume was characteristic property utilized, results obtained showed error due to change in specific volume.

9. A stone weighed 177 N on earth. It was dropped in to oil of specific gravity 0.8 on a planet whose acceleration due to gravity is 5m/s2. It displaced oil having weight of 100N. What was the volume of oil displaced by the stone?

a) 25 Litres

b) 15 Litres

c) 25 m 3

d) None of the mentioned

Answer: a

Explanation: Volume displaced=oil displaced/=100/ .

10. An compressible fluid’s specific gravity was measured on earth, on a planet having acceleration due to gravity 5.5 times that of earth, and in space at STP. Where will it be having highest value?

a) on the earth

b) on the planet

c) in the space

d) it will be constant everywhere

Answer: d

Explanation: Specific gravity is characteristic property of fluid and is independent of external conditions.

This set of Heat Transfer Operations Multiple Choice Questions & Answers  focuses on “Types of Evaporators”.

1. Which one of the following is not a type of evaporator?

a) Forced Circulation

b) Natural Circulation

c) Nucleate Boiling

d) Gasketed evaporators

Answer: d

Explanation: Evaporators can be classified on numerous parameters, such as the circulation type- Natural or Forced, Boiling type- Nucleate and Non-Nucleate, etc. Gasketed Evaporators is not a class of evaporators.

2. Refrigerators use liquid coolants which evaporate in an evaporator installed in a closed chamber.

a) True

b) False

Answer: a

Explanation: In an air-conditioning system or a refrigeration system, cooling occurs by passing a compressed coolant, for example R-22  or R-410A, to evaporate/vaporize to gas inside the setup/refrigerator while taking in heat from the surrounding target area, for example a refrigerator cage or rooms, in the process.

3. Which one of the following is not a suitable application of evaporators?

a) Refrigeration

b) Cooling

c) Heating

d) Crystallisation

Answer: c

Explanation: Heating is not a possible application of evaporators, whereas it is usually used for concentrating solutions as a feed to crystallization, re-evaporation of compressed liquids, refrigeration applications, and generation of vapours for process applications.

4. What is the driving force for evaporation to take place?

a) Difference in partial pressure

b) Difference in pressure

c) Difference in Concentration

d) Difference in temperature

Answer: a

Explanation: The driving force for evaporation to take place is the partial pressure difference of the vapour from the evaporating surface and the surrounding that promotes mass transfer to take place. It is not concentration/pressure differences as they hold for liquid and bulk respectively.

5. Crystallizers are one of the most important setups in industries nowadays. It is solely used to dry a solution to an extent to obtain the crystals of the solute. Which one of the following is the most suitable operation to carry out this process?

a) Forced Circulation

b) Natural Circulation

c) Nucleate Boiling

d) Non-nucleate Boiling

Answer: a

Explanation: Forced circulation is observed when boiling is suppressed at the heating surface due to the hydrostatic heads operating on it, hence falling film evaporators which generally use crystallizers, viscous fluids, waste streams, and other fouling fluids, nucleate boiling would naturally increase their fouling rate and is hence is not suited.

6. Why don’t we use Natural convection for evaporating waste streams, crystallizers, and viscous fluids?

a) Slow process

b) Sedimentation problem

c) Prevents fouling at the heating surface

d) Causes overheating

Answer: c

Explanation: Forced circulation is observed when boiling is suppressed at the heating surface due to the hydrostatic heads operating on it, hence falling film evaporators which generally use crystallizers, viscous fluids, waste streams, and other fouling fluids, nucleate boiling would naturally increase their fouling rate and is hence is not suited.

7. Falling film evaporators are those in which evaporation takes place from the film interface with nucleate boiling at the wall.

a) True

b) False

Answer: b

Explanation: Falling film evaporators are those in which evaporation takes place from the film interface with no nucleate boiling at the wall because of the extensive fouling that is created by it.

8. How are the tube surfaces in falling film evaporators heated to enhance evaporation?

a) Heaters at about 200℃

b) Heaters at above 200℃

c) Heaters at just above 100℃

d) Steam condensing at the outer wall

Answer: d

Explanation: Falling film evaporators are those in which evaporation takes place from the film interface with no nucleate boiling at the wall because of the extensive fouling that is created by it. Hence to avoid nucleate boiling, we usually use steam as the heating agent.

9. Which one of the following is not a subset of Nucleate boiling evaporators?

a) Climbing Film Evaporator

b) Rising Film Evaporator

c) Short-tube Vertical Evaporator

d) Falling Film Evaporator

Answer: d

Explanation: Falling film evaporators are those in which evaporation takes place from the film interface with no nucleate boiling at the wall because of the extensive fouling that is created by it. Hence to avoid nucleate boiling, we usually use steam as the heating agent.

10. Recognize the following evaporator.

heat-transfer-operations-questions-answers-types-evaporators-q10

a) Falling Film Evaporator

b) Short-tube Vertical Evaporator

c) Climbing Film Evaporator

d) Basket-type Evaporator

Answer: c

Explanation: The direction of flow of fluid in the tubes clearly indicates the that the fluid is rising up, hence it is known as rising film/ climbing film evaporator.

11. Recognize the following evaporator.

heat-transfer-operations-questions-answers-types-evaporators-q11

a) Falling Film Evaporator

b) Short-tube Vertical Evaporator

c) Climbing Film Evaporator

d) Basket-type Evaporator

Answer: b

Explanation: The flow direction is downward and the heating is done by a gas that has temperature high enough to cause nucleate boiling. As the tubes are short and there is a basket, it is known as a short tube vertical evaporator.

12. Recognize the following evaporator.

heat-transfer-operations-questions-answers-types-evaporators-q12

a) Falling Film Evaporator

b) Short-tube Vertical Evaporator

c) Climbing Film Evaporator

d) Basket-type Evaporator

Answer: a

Explanation: The flow direction is downward and the heating is done by a gas that has temperature not high enough to cause nucleate boiling. As the tubes are short and there is a basket, it is known as a Falling Film evaporator.

This set of Refrigeration online quiz focuses on “Evaporators Types, Factors Affecting it and Defrosting Methods”.

1. Why is the evaporator used?

a) To improve C.O.P.

b) To decrease the refrigeration effect

c) To absorb heat

d) To reject heat

Answer: c

Explanation: Evaporator is used to absorb the heat from the medium being cooled and get the refrigeration effect. Multiple evaporators are used to enhance the refrigeration effect and eventually the C.O.P. too.

2. What is the pressure side of a refrigerating system in which evaporator is used?

a) Low pressure

b) Zero pressure

c) Negative pressure

d) High pressure

Answer: a

Explanation: Evaporator is used to convert liquid refrigerant into vapor refrigerant by absorbing heat. The liquid coming from the expansion valve is converted into vapor and forwarded to the compressor for compression. Evaporator gives the refrigeration effect. So, it is used on the lower pressure side to provide input for the compressor.

3. Evaporator is also called as Freezing, cooling or chilling coil.

a) False

b) True

Answer: b

Explanation: Evaporator is used to convert liquid refrigerant into vapor refrigerant by absorbing heat. The liquid coming from the expansion valve is converted into vapor and forwarded to the compressor for compression. Evaporator gives the refrigeration effect. So, it is also called a Freezing, cooling, or chilling coil.

4. What evaporator does to the refrigerant coming from expansion valve in terms of state?

a) high-pressure liquid refrigerant

b) low-pressure liquid and vapor refrigerant

c) low-pressure vapor refrigerant

d) high-pressure vapor refrigerant

Answer: b

Explanation: The liquid coming from the expansion valve is converted into vapor and forwarded to the compressor for compression. So, the evaporator absorbs the heat, and phase change occurs. There is a slight change in pressure but negligible. Thus, the state of refrigerant changes from low-pressure liquid to low-pressure vapor.

5. What is the formula for the capacity of evaporator?

a) Q = U A 2 (T 2 – T 1 )

b) Q = U A (T 2 – T 1 )

c) Q = A (T 2 – T 1 )

d) Q = U (T 2 – T 1 )

Answer: b

Explanation: Capacity of the evaporator is the amount of heat absorbed by it over a given period of time. So, the heat absorbed or capacity is the heat convection, can be given as,

Q = U A (T 2 – T 1 ), where U is overall heat transfer coefficient, A is the surface area of the evaporator and T 1 & T 2 are the saturation temperature of refrigerant and temperature of the medium to be cooled respectively.

6. Which of the following is not a factor affecting the capacity of evaporator?

a) Velocity of refrigerant

b) The thickness of the evaporator coil wall

c) Material

d) Evaporator pressure

Answer: d

Explanation: Capacity of the evaporator is the amount of heat absorbed by it over a given period of time. Increase in the velocity can increase the heat transfer up to a certain extent. The thickness of the coil wall affects the convection, and hence capacity can be changed. The material should be a good conductor to get optimum heat transfer. Hence, evaporator pressure does not affect the capacity, but the temperature does have.

7. What is the value of the fluid side heat transfer coefficient when liquid flows through the evaporator shell?

a) C x m

b) C x  0.5

c) C x  2

d) C x  3

Answer: b

Explanation: Fluid side heat transfer coefficient is denoted as h f and has a formula,

h f = [  0.8 (C p μ / K) 0.4 ] W / m 2 K

So, for liquid flowing through the shell, the value is h f = C x  0.5 = C x √m.

8. What is the value of fluid side heat transfer coefficient when air flows over finned coil by forced convection?

a) C x  0.1

b) C x  0.2

c) C x  0.3

d) C x  0.4

Answer: d

Explanation: Fluid side heat transfer coefficient is denoted as h f and has a formula,

h f = [  0.8 (C p μ / K) 0.4 ] W / m 2 K

So, for air flowing over finned coil by forced convection, the value is h f = C x  0.4

9. What is the value of the fluid side heat transfer coefficient when air flows over cold pipes by natural convection?

a) 0.2 (T a – T m / D) 1/4

b) 0.2 (T a – T m / D)

c) 0.2 (T a – T m / D) 1/3

d) 0.2 (T a – T m / D) 1/2

Answer: a

Explanation: Fluid side heat transfer coefficient is denoted as h f and has a formula,

h f = [  0.8 (C p μ / K) 0.4 ] W / m 2 K

So, for air flowing over cold pipes by natural convection, the value is h f = 0.2 (T a – T m / D) 1/4

10. The bare tube evaporators are called as _______ evaporators.

a) extended surface

b) bare surface

c) pipe surface

d) prime surface

Answer: d

Explanation: The bare tube evaporators are also known as prime surface evaporators due to the simple construction of it. The bare tube coil is easy to clean and defrost; hence, it is called prime surface evaporator.

11. Which type of evaporator is generally used in home freezers, ice cream cabinets, etc?

a) Shell and coil evaporator

b) Finned evaporator

c) Shell and tube evaporator

d) Plate evaporator

Answer: d

Explanation: Plate evaporators are generally used in home freezers, ice cream cabinets, beverage coolers, and locker plants, etc. In this type of evaporator, coils are either welded on one side of the plate or between two plates which are welded together at the edges.

12. Which type of evaporator is generally used for wine cooling and chilling oil in the petroleum industry?

a) Tube-in-tube evaporator

b) Finned evaporator

c) Shell and tube evaporator

d) Plate evaporator

Answer: a

Explanation: Tube-in-tube evaporators are used for wine cooling and chilling oil in the petroleum industry. In this type of evaporator, one tube is inside another tube. It provides a high heat transfer rate.

13. Which type of evaporator is used in household refrigerators?

a) frosting evaporator

b) defrosting evaporator

c) non-frosting evaporator

d) non-corrosive evaporator

Answer: a

Explanation: Evaporators used in household refrigerators, low-temperature evaporators, and bare pipe coils in storage boxes come under frosting type. These operate at temperature always below 0°C. Coils frost continuously and need to be defrosted at regular interval of time.

14. Which type of evaporator is generally used to cool drinking water?

a) Shell and tube evaporator

b) Finned evaporator

c) Shell and coil evaporator

d) Plate evaporator

Answer: c

Explanation: The shell and coil evaporators are the dry expansion evaporators used to chill water. The shell may be sealed or open. If the shell is sealed, then it is used for cooling drinking water.

15. Finned evaporators are also known as prime surface evaporators.

a) False

b) True

Answer: a

Explanation: Finned evaporators are the ones which use a number of fins to increase the contact surface for heat transfer. The specifications of fins depend on the required heat transfer rate. As fins are the extended material used for heat transfer enhancement so, finned evaporators are also called as extended surface evaporators whereas, bare tube coil evaporators are called as prime surface evaporators.

This set of Refrigeration Multiple Choice Questions & Answers  focuses on “Expansion Devices”.

1. The device which divides the high pressure side and the low pressure side of a refrigerating system is known as _____________

a) condenser device

b) evaporator device

c) receiver device

d) expansion device

Answer: d

Explanation: The expansion device is an important device that divides the high pressure side and the low pressure side of a refrigerating system. It is also known as metering device or throttling device.

2. Which of the following function is not performed by the expansion device?

a) It reduces the high pressure liquid refrigerant to low pressure liquid refrigerant

b) It maintains the desired pressure difference between the high and low pressure sides of the system

c) It controls the flow of refrigerant according to the load on the evaporator

d) It compresses the vapor

Answer: d

Explanation: The expansion device performs the following functions, it reduces the high pressure liquid refrigerant to low pressure liquid refrigerant, it maintains the desired pressure difference between the high and low pressure sides of the system and It controls the flow of refrigerant according to the load on the evaporator.

3. The expansion device used with flooded evaporators is known as expansion valves.

a) True

b) False

Answer: b

Explanation: The expansion device used with flooded evaporators is known as float valves and the expansion device used with dry expansion evaporators are called expansion valves.

4. The expansion device is placed between which two components?

a) Condenser and evaporator

b) Compressor and condenser

c) Evaporator and compressor

d) Receiver and evaporator

Answer: d

Explanation: The expansion device is placed between receiver  and evaporator .

5. Which of the following is not an advantage of capillary tube?

a) The cost of capillary tube is less

b) A high starting motor is not required

c) No receiver is needed

d) A capillary tube designed for a specific condition will also work efficiently for other conditions

Answer: d

Explanation: A capillary tube designed for a specific condition will not work efficiently for other conditions as the length is directly and inner diameter is indirectly proportional to the frictional resistance. The longer the tube and smaller the diameter, greater is the pressure drop created in the refrigerant flow.

6. Capillary tube, as an expansion device is used in?

a) Water coolers

b) Domestic refrigerators

c) Room air conditioners

d) All of the mentioned

Answer: d

Explanation: The capillary tube as an expansion device is used in small capacity hermetic sealed refrigeration units such as in water coolers, domestic refrigerators and room air conditioners, etc.

7. Which one of the following is also known as a constant superheat valve?

a) Capillary tube

b) Hand-operated expansion valve

c) Thermostatic Expansion valve

d) Low side float valve

Answer: d

Explanation: Thermostatic expansion valve is also called a constant superheat valve because it maintains a constant superheat of the vapor refrigerant at the end of the evaporator coil, by controlling the flow of liquid refrigerant through the evaporator.

8. Thermostatic expansion valves are usually set for a superheat of?

a) 10°C

b) 5°C

c) 8°C

d) 15°C

Answer: b

Explanation: Thermostatic expansion valves are usually set for a superheat of 5°C for better efficiency of the refrigeration cycle.

9. The low-side float valve is located between the condenser and evaporator.

a) True

b) False

Answer: b

Explanation: The low-side float valve is located in the low pressure side i.e. between the evaporator and the compressor suction line. Whereas the high side float valve is located in the high pressure side i.e. between the condenser and evaporator.

10. The thermostatic expansion valve operates on the changes in the ___________

a) degree of superheat at exit from the evaporator

b) temperature of the evaporator

c) pressure in the condenser

d) pressure in the evaporator

Answer: a

Explanation: The thermostatic expansion valve operates on the changes in the degree of superheat at exit from the evaporator. Thermostatic expansion valves are usually set for a superheat of 5° C for better efficiency of the refrigeration cycle.