Heat Transfer Pune University MCQs

 This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Modes Of Heat Transfer”.

1. The literature of heat transfer generally recognizes distinct modes of heat transfer. How many modes are there?

a) One

b) Two

c) Three

d) Four

Answer: c

Explanation: There are three modes of heat transfer i.e. radiation, convection and conduction.

2. Consider system A at uniform temperature t and system B at another uniform temperature T . Let the two systems be brought into contact and be thermally insulated from their surroundings but not from each other. Energy will flow from system A to system B because of

heat-transfer-questions-answers-modes-heat-transfer-q2

a) Temperature difference

b) Energy difference

c) Mass difference

d) Volumetric difference

Answer: a

Explanation: Greater the temperature imbalance the higher would be the rate of energy transfer.

3. An oil cooler in a high performance engine has an outside surface area 0.12 m 2 and a surface temperature of 65 degree Celsius. At any intermediate time air moves over the surface of the cooler at a temperature of 30 degree Celsius and gives rise to a surface coefficient equal to 45.4 W/ m 2 K. Find out the heat transfer rate?

a) 238.43 W

b) 190.68 W

c) 543.67 W

d) 675.98 W

Answer: b

Explanation: Q = (T 2 – T 1 ) A h = 0.12  45.4 = 190.68 W.

4. Unit of the rate of heat transfer is

a) Joule

b) Newton

c) Pascal

d) Watt

Answer: d

Explanation: Unit of heat transfer is Joule but the rate of heat transfer is joule per second i.e. watt.

5. Convective heat transfer coefficient doesn’t depend on

a) Surface area

b) Space

c) Time

d) Orientation of solid surface

Answer: a

Explanation: It is denoted by h and is dependent on space, time, geometry, orientation of solid surface.

6. The rate equation used to describe the mechanism of convection is called Newton’s law of cooling. So rate of heat flow by convection doesn’t depend on

a) Convective heat transfer coefficient

b) Surface area through which heat flows

c) Time

d) Temperature potential difference

Answer: c

Explanation: It is directly proportional to all of above except time.

7. How many types of convection process are there?

a) One

b) Three

c) Four

d) Two

Answer: b

Explanation: Forced, natural and mixed convection.

8. Thermal conductivity is maximum for which substance

a) Silver

b) Ice

c) Aluminum

d) Diamond

Answer: d

Explanation: Thermal conductivity of diamond is 2300 W/m K.

9. A radiator in a domestic heating system operates at a surface temperature of 60 degree Celsius. Calculate the heat flux at the surface of the radiator if it behaves as a black body

a) 697.2 W/m 2

b) 786.9 W/m 2

c) 324.7 W/m 2

d) 592.1 W/m 2

Answer: a

Explanation: As, q = Q/A = 5.67 * 10 -8  4 = 697.2.

10. Which of the following is an example of forced convection?

a) Chilling effect of cold wind on a warm body

b) Flow of water in condenser tubes

c) Cooling of billets in the atmosphere

d) Heat exchange on cold and warm pipes

Answer: b

Explanation: In forced convection, the flow of fluid is caused by a pump, fan or by atmospheric winds.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Steady And Unsteady Heat Transfer”.

1. Regarding one dimensional heat transfer, choose the correct statement.

a) Steady – f , Unsteady – f 

b) Steady – f , Unsteady – f 

c) Steady – f , Unsteady – f 

d) Steady – f , Unsteady – f 

Answer: a

Explanation: In case of one dimensional heat flow steady state is a function of x coordinate only while unsteady state is a function of x coordinate and time only.

2. Which statement is true regarding steady state condition?

a) There is a variation in temperature in the course of time

b) Heat exchange is constant

c) It is a function of space and time coordinates

d) Internal energy of the system changes

Answer: b

Explanation: Heat influx is always equal to heat efflux. It is a function of space coordinates only.

3. Which of the following is an example of steady state heat transfer?

a) Boilers and turbines

b) Cooling of I.C engine

c) Chilling effect of cold wind on a warm body

d) Electric bulb cools down by the surrounding atmosphere

Answer: d

Explanation: System is a perfect black body.

4. Heat transfer in a long, hollow cylinder which is maintained at uniform but different temperatures on its inner and outer surfaces may be assumed to be taking place in which direction?

a) Axial only

b) Unpredictable

c) Radial only

d) No heat transfer takes place

Answer: c

Explanation: Ambient temperature is uniform on the periphery of cylinder and temperature is uniform. So it takes place in the radial direction only.

5. Heat transfer takes place according to which law?

a) Newton’s law of cooling

b) Second law of thermodynamics

c) Newton’s second law of motion

d) First law of thermodynamics

Answer: b

Explanation: Second law states about heat transfer between source and sink.

6. Heat transfer takes place in liquids and gases is essentially due to

a) Radiation

b) Conduction

c) Convection

d) Conduction as well as convection

Answer: c

Explanation: Convection is a process by which thermal energy is transferred between solid and fluid flowing through it.

7. The appropriate rate equation for convective heat transfer between a surface and adjacent fluid is prescribed by

a) Newton’s first law

b) Wein’s displacement law

c) Kirchhoff’s law

d) Newton’s law of cooling

Answer: d

Explanation: The rate equation used to describe the mechanism of convection is called Newton’s law of cooling when the solid surface is cooled by the fluid.

8. Identify the wrong statement

a) The process of heat transfer is an irreversible process

b) For heat exchange, a temperature gradient must exist

c) A material medium is not necessary for heat transmission

d) Heat flow doesn’t depend on temperature

Answer: d

Explanation: Heat flows from higher to lower temperature.

9. During a cold winter season, a person prefers to sit near a fire. Which of the following modes of heat transfer provides him the maximum heat?

a) Conduction from the fire

b) If it is near the fire, convection sounds good

c) Convection and radiation together

d) Radiation will provide quick warmth

Answer: d

Explanation: Heat transfer by radiation can occur between two bodies even when they are separated by a medium colder than both of them.

10. Most unsteady heat flow occurs

a) Through the walls of the refrigerator

b) During annealing of castings

c) Through the walls of the furnace

d) Through lagged pipe carrying steam

Answer: b

Explanation: Under steady state condition, with time there is a change in temperature i.e. temperature field is a function of space and time.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Fourier Equation”.

1. The famous Fourier series is named after

a) Diller and Ryan

b) J.B. Joseph Fourier

c) Stefan- Boltzmann

d) Wein’s

Answer: b

Explanation: He gave this theory in 1824.

2. Fourier law of heat conduction is best represented by

a) Q = -k A d t /d x

b) Q = k A d x/d t

c) Q = -k A

d) Q = k d t/d x

Answer: a

Explanation: According to Fourier law of heat conduction, Q = -k A d t /d x.

3. Here are some assumptions that are made for Fourier law. Identify the wrong one

a) No internal heat generation

b) Steady state heat conduction

c) Non- linear temperature profile

d) Isotropic and homogenous material

Answer: c

Explanation: It has constant temperature gradient and a linear temperature profile.

4. Consider the following statements:

The Fourier heat conduction equation

Q = -k A d t /d x

Presumes

i) Steady state conditions

ii) Constant value of thermal conductivity

iii) Uniform temperature at the wall surface

iv) One dimensional heat flow

Which of these statements are correct?

a) i, ii and iii

b) i, ii and iv

c) i, iii and iv

d) i, iii and iv

Answer: d

Explanation: Thermal conductivity is different for different materials.

5. The diagram shows heat conduction through a plane wall. The surface temperature is 475 K and it radiates heat to the surroundings at 335 K. If thermal conductivity of the material is 12.5 W/m degree, find the temperature gradient. Let convective coefficient be 80 W/m 2 degree and radiation factor is 0.9

heat-transfer-questions-answers-fourier-equation-q5

a) – 1052.4 degree celsius

b) – 2052.4 degree celsius

c) – 3052.4 degree celsius

d) – 4052.4 degree celsius

Answer: a

Explanation: Heat conducted through the plate = convection heat losses + radiation heat losses. So, d t /d x = – 13155/12.5 = – 1052.4 degree Celsius.

6. Negative sign in Fourier heat conduction equation indicates

a) Heat always flow is in the direction of positive temperature gradient

b) Heat always flow in the direction of negative temperature gradient

c) No heat flow is there

d) Data is insufficient

Answer: b

Explanation: The ratio d t/d x represents the change in temperature per unit thickness i.e. the temperature gradient. So it represents heat flow in the direction of negative temperature gradient.

7. Transmission of heat i.e. molecular is smallest in case of

a) Gases

b) Liquids

c) Alloys

d) Solids

Answer: a

Explanation: In gases, atoms are arranged loosely, there is less molecular transmission of heat as compared to solids, liquids and alloys.

8. Which one is not the unit of thermal conductivity?

a) kcal/m hr K

b) KJ/m hr K

c) W/m s K

d) Cal/cm s K

Answer: c

Explanation: The unit kcal/m hr K could also be specified as J/m s K or W/m s K and this is actually done while quoting the numerical values of thermal conductivity.

9. “Thermal conductivity represents the amount of heat conducted across the unit area when a temperature difference of one kelvin”.

a) True

b) False

Answer: b

Explanation: It is across unit area and through unit distance.

10. Which of the following is the unit of thermal resistance?

a) degree/kcal

b) hour degree

c) s degree/kcal

d) degree/W

Answer: d

Explanation: Thermal resistance is expressed in the unit’s hr degree/kcal, degree/W and s degree/J.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Thermal Conductivity Of Materials”.

1. Thermal conductivity is defined as the heat flow per unit time

a) When the temperature gradient is unity

b) Across the wall with no temperature

c) Through a unit thickness of the wall

d) Across unit area where the temperature gradient is unity

Answer: d

Explanation: Thermal conductivity of a material is because of migration of free electrons and lattice vibrational waves.

2. Mark the matter with least value of thermal conductivity

a) Air

b) Water

c) Ash

d) Window glass

Answer: a

Explanation: For air, it is .024 W/ m degree i.e. lowest.

3. Which one of the following forms of water have the highest value of thermal conductivity?

a) Boiling water

b) Steam

c) Solid ice

d) Melting ice

Answer: c

Explanation: For ice, it is 2.25 W/m degree i.e. maximum.

4. The average thermal conductivities of water and air conform to the ratio

a) 50:1

b) 25:1

c) 5:1

d) 15:1

Answer: b

Explanation: For water, it is 0.55-0.7 W/m degree and for air it is .024 W/m degree.

5. Identify the very good insulator

a) Saw dust

b) Cork

c) Asbestos sheet

d) Glass wool

Answer: d

Explanation: Glass wool has a lowest thermal conductivity of 0.03 W/m degree amongst given option.

6. Most metals are good conductor of heat because of

a) Transport of energy

b) Free electrons and frequent collision of atoms

c) Lattice defects

d) Capacity to absorb energy

Answer: b

Explanation: For good conductors, there must be electrons that are free to move.

7. Heat conduction in gases is due to

a) Elastic impact of molecules

b) Movement of electrons

c) EM Waves

d) Mixing of gases

Answer: a

Explanation: If there is elastic collision then after sometime molecules regain its natural position.

8. The heat energy propagation due to conduction heat transfer will be minimum for

a) Lead

b) Water

c) Air

d) Copper

Answer: c

Explanation: It is because air has lowest value of thermal conductivity amongst given options.

9. Cork is a good insulator because

a) It is flexible

b) It can be powdered

c) Low density

d) It is porous

Answer: d

Explanation: Cork has thermal conductivity in the range of 0.05-0.10 which is very low so it can be porous.

10. Choose the false statement

a) For pure metal thermal conductivity is more

b) Thermal conductivity decreases with increase in the density of the substance

c) Thermal conductivity of dry material is lower than that of damp material

d) Heat treatment causes variation in thermal conductivity

Answer: b

Explanation: Thermal conductivity increase with increase in the density of a substance.

This set of Heat Transfer Interview Questions & Answers focuses on “General Heat Conduction Equation”.

1. From the concept of kinetic theory, mean travel velocity of the gas molecules is prescribed by the relation

a) V =  1/2

b) V =  1/2

c) V =  1/2

d) V =  1/2

Answer: b

Explanation: Heat transfer by conduction in gases occurs through transport of the kinetic energy of molecular motion resulting from the random movement and collisions of the molecules.

Where, G = Universal gas constant

M = Molecular weight of the gas

T = Absolute temperature.

2. Low temperature insulation are used when the enclosure is at a temperature lower than the ambient temperature. Which one of the following is not a low temperature insulation?

a) Asbestos

b) Cork

c) Cattle hair

d) Slag wool

Answer: a

Explanation: Asbestos is a high temperature insulation which is used when it is desired to prevent an enclosure at a temperature higher than the ambient from losing heat to its surroundings.

3. The value of Lorenz number in 10 -8 W ohms/K 2 is

a) 2.02

b) 2.35

c) 2.56

d) 2.45

Answer: d

Explanation: Lorenz no. = k/α T

4. For liquids, thermal conductivity is governed by the relation

a) k = A c p p 7/3 /M 1/3

b) k = 2A c p p 4/3 /M 2/3

c) k = A c p p 4/3 /M 1/3

d) k = A c p p 8/3 /M 1/3

Answer: c

Explanation: Area doesn’t depends on the nature of liquid but on the quantity (Ac p ) is nearly constant for all liquids. Where,

C P = Specific heat at constant pressure

M = Molecular weight of the gas

p = Mass density

A = Area.

5. Consider the following parameters

 Composition

 Density

 Porosity

 Structure

Then, thermal conductivity of glass wool varies from sample to sample because of variation is

a) i and ii

b) i, ii, iii and iv

c) i and iii

d) i, ii and iii

Answer: b

Explanation: There is a variation due to all the above options.

6. The thermal conductivity and the electrical conductivity of a metal at absolute temperature are related as

a) k/σ T

b) k/σ

c) k σ/T

d) k/T

Answer: a

Explanation: It is defined as the ratio of thermal conductivity to the product of electrical conductivity and temperature.

7. The relation Ϫ 2 t =0 is referred to as

a) Poisson’s equation

b) Fourier heat conduction equation

c) Solution for transient conduction

d) Laplace equation

Answer: d

Explanation: In the absence of internal heat generation or release of energy within the body, equation reduces to Laplace equation.

8. The unit of thermal diffusivity is

a) m 2 /hr-K

b) kcal/m 2 -hr

c) m 2 /hr

d) m/hr-K

Answer: c

Explanation: The quantity α = k/pc is called thermal diffusivity.

9. To effect a bond between two metal plates, 2.5 cm and 15 cm thick, heat is uniformly applied through the thinner plate by a radiant heat source. The bonding must be held at 320 K for a short time. When the heat source is adjusted to have a steady value of 43.5 k W/m 2 , a thermocouple installed on the side of the thinner plate next to source indicates a temperature of 345 K. Calculate the temperature gradient for heat conduction through thinner plate. In the diagram, the upper plate is 2.5 cm thick while the lower is 15 cm thick.

heat-transfer-interview-questions-answers-q9

a) – 1000 degree Celsius/m

b) – 2000 degree Celsius/m

c) – 3000 degree Celsius/m

d) – 4000 degree Celsius/m

Answer: a

Explanation: Temperature gradient = d t/d x = – 1000 degree Celsius/m. Also, q/A = – k (t 2 – t 1 )/δ.

10. The diffusion equations

Ɏ 2 t + q g =  

Governs the temperature distribution under unsteady heat flow through a homogeneous and isotropic material. The Fourier equation follows from this expression when

a) Temperature doesn’t depends on time

b) There is no internal heat generation

c) Steady state conditions prevail

d) There is no internal heat generation but unsteady state condition prevails

Answer: d

Explanation: In unsteady state condition, there is no internal heat generation.

This set of tough Heat Transfer Questions focuses on “Thermal Conductivity Of Different Materials”.

1. What is the thermal conductivity of asphalt in W/m K?

a) 0.064

b) 0.063

c) 0.062

d) 0.061

Answer: c

Explanation: Asphalt concrete has three main structural elements i.e. aggregates, binder and the contact layer between them. It is soft at high temperatures but brittle at low temperatures.

2. Which one is having the highest thermal conductivity?

a) Silver

b) Copper

c) Glass

d) Earth

Answer: a

Explanation: Thermal conductivity of silver is 410 W/m degree while that of copper, glass and earth are 385 W/m degree, 0.75 W/m degree and 0.138 respectively.

3. What is the thermal conductivity of magnetite in W/m K?

a) 3.7

b) 3.8

c) 3.9

d) 4.0

Answer: b

Explanation: This is having low thermal conductivity because of high intramolecular interactions between their outer edge molecules.

4. Which one is having the lowest thermal conductivity?

a) Furnace

b) Wool

c) Saw dust

d) Glass wool

Answer: d

Explanation: Thermal conductivity of glass wool is 0.03 W/m degree while that of a furnace, wool and saw dust are 0.30 W/m degree, 0.052 W/m degree and 0.07 respectively.

5. What is the thermal conductivity of coal in W/m K?

a) 0.25

b) 0.26

c) 0.27

d) 0.28

Answer: b

Explanation: Coal is composed primarily of carbon along with variable quantities of other elements such as hydrogen, sulfur, oxygen and nitrogen.

6. What is the thermal conductivity of cotton in W/m K?

a) 0.03

b) 0.04

c) 0.05

d) 0.06

Answer: d

Explanation: It is a soft material that grows in a protective capsule, around the seeds of cotton plants.

7. Which one is having the highest thermal conductivity?

a) Brass

b) Aluminum

c) Cast iron

d) Steel

Answer: b

Explanation: Thermal conductivity of aluminum is 225 W/m degree while that of brass, cast iron and steel are 107 W/m degree, 55 W/m degree and 20 respectively.

8. What is the thermal conductivity of apple in W/m K?

a) 0.513

b) 0.613

c) 0.713

d) 0.813

Answer: a

Explanation: It is a sweet fruit that has very low thermal conductivity in order to prevent it from damage.

9. What is the thermal conductivity of ice in W/m K?

a) 1.55

b) 1.66

c) 1.77

d) 1.88

Answer: d

Explanation: Frozen water in the form of an ordinary household ice cube. The white zone in the center is the result of tiny air bubbles.

10. Which one is having the lowest thermal conductivity?

a) Ash

b) Ice

c) Freon

d) Water

Answer: c

Explanation: Thermal conductivity of freon is 0.0083 W/m degree while that of ash, ice and water are 0.12 W/m degree, 2.25 W/m degree and 0.55 respectively.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Conduction Through A Plane Wall”.

1. In Cartesian coordinates the heat conduction equation is given by

a) d 2 t/dx 2 + d 2 t/dy 2 + d 2 t/dz 2 + q g =  

b) 2d 2 t/dx 2 + d 2 t/dy 2 + d 2 t/dz 2 + 34q g = 

c) d 2 t/dx 2 + 3d 2 t/dy 2 + d 2 t/dz 2 =  

d) 4d 2 t/dx 2 + d 2 t/dy 2 + d 2 t/dz 2 + 1/2q g =  

Answer: a

Explanation: This is one dimensional heat conduction through a homogenous, isotropic wall with constant thermal conductivity.

2. The temperature distribution in a large thin plate with uniform surface temperature will be

a) Logarithmic

b) Hyperbolic

c) Parabolic

d) Linear

Answer: d

Explanation: The temperature increases with increasing value of x. Temperature gradient will be positive i.e. linear.

3. Let us assume two walls of same thickness and cross-sectional area having thermal conductivities in the ratio 1/2. Let us say there is same temperature difference across the wall faces, the ratio of heat flow will be

a) 1

b) 1/2

c) 2

d) 4

Answer: b

Explanation: Q 1 = k 1 A 1 d t 1 /δ 1 and Q 2 = k 2 A 2 d t 2 /δ 2

Now, δ 1 = δ 2 and A 1 = A 2 and d t 1 = d t 2

So, Q 1 /Q 2 = ½.

4. The interior of an oven is maintained at a temperature of 850 degree Celsius by means of a suitable control apparatus. The oven walls are 500 mm thick and are fabricated from a material of thermal conductivity 0.3 W/m degree. For an outside wall temperature of 250 degree Celsius, workout the resistance to heat flow

a) 0.667 degree/W

b) 1.667 degree/W

c) 2.667 degree/W

d) 3.667 degree/W

Answer: b

Explanation: R t = 0.5/0.3 = 1.667 degree/W.

5. A plane slab of thickness 60 cm is made of a material of thermal conductivity k = 17.45 W/m K. Let us assume that one side of the slab absorbs a net amount of radiant energy at the rate q = 530.5 watt/m 2 . If the other face of the slab is at a constant temperature t 2 = 38 degree Celsius. Comment on the temperature with respect to the slab?

a) 87.5 degree Celsius

b) 32 degree Celsius

c) 47.08 degree Celsius

d) 32.87 degree Celsius

Answer: c

Explanation: Heat flux, q = k (t s – t f ) / Thickness. So, t s = 56.17 degree Celsius. Now, t = t s + (t f – t s ) x/Thickness.

6. The rate of heat transfer for a plane wall of homogenous material with constant thermal conductivity is given by

a) Q = kA (t 1 -t 2 )/δ

b) Q = 2kAx/ δ

c) Q = 2kAδx

d) Q = 2k/δ x

Answer: a

Explanation: Computations for heat flow can be made by substituting the value of temperature gradient into the general equation. The heat flow somehow doesn’t depend on x.

7. In case of homogeneous plane wall, there is a linear temperature distribution given by

a) t = t 1 + (t 2 -t 1 ) δ/x

b) t = t 2 – (t 2 -t 1 ) x/ δ

c) t = t 1 + (t 2 -t 1 ) x

d) t = t 1 + (t 2 -t 1 ) x/ δ

Answer: d

Explanation: The expression for steady state temperature distribution can be set up by integrating the Fourier rate equation.

8. The rate of convective heat transfer between a solid boundary and adjacent fluid is given by

a) Q = h A (t s – t f )

b) Q = h A

c) Q = (t s – t f )

d) Q = h (t s – t f )

Answer: a

Explanation: Here, h is heat transfer coefficient i.e. convective.

9. A homogeneous wall of area A and thickness δ has left and right hand surface temperatures of 0 degree Celsius and 40 degree Celsius. Determine the temperature at the center of the wall

a) 10 degree Celsius

b) 20 degree Celsius

c) 30 degree Celsius

d) 40 degree Celsius

Answer: b

Explanation: At the midpoint x = δ/2. So, temperature = 40 + /2 = 20 degree Celsius.

10. A rod of 3 cm diameter and 20 cm length is maintained at 100 degree Celsius at one end and 10 degree Celsius at the other end. These temperature conditions are attained when there is heat flow rate of 6 W. If cylindrical surface of the rod is completely insulated, determine the thermal conductivity of the rod material

a) 21.87 W/m degree

b) 20.87 W/m degree

c) 19.87 W/m degree

d) 18.87 W/m degree

Answer: d

Explanation: Q = k A C (t 1 – t 2 )/δ = 0.318 k.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Conduction Through A Composite Wall”.

1. A composite wall generally consists of

a) One homogenous layer

b) Multiple heterogeneous layers

c) One heterogeneous layer

d) Multiple homogenous layers

Answer: b

Explanation: Walls of houses where bricks are given a layer of plaster on either side.

2. Three metal walls of the same thickness and cross sectional area have thermal conductivities k, 2k and 3k respectively. The temperature drop across the walls  will be in the ratio

a) 3:2:1

b) 1:1:1

c) 1:2:3

d) Given data is insufficient

Answer: a

Explanation: As, δ 1 = δ 2 = δ 3 and cross sectional areas are same i.e. temperature drop varies inversely with thermal conductivity.

3. A composite wall is made of two layers of thickness δ 1 and δ 2 having thermal conductivities k and 2k and equal surface area normal to the direction of heat flow. The outer surface of composite wall are at 100 degree Celsius and 200 degree Celsius. The minimum surface temperature at the junction is 150 degree Celsius. What will be the ratio of wall thickness?

a) 1:1

b) 2:1

c) 1:2

d) 2:3

Answer: c

Explanation: Q = k 1 A 1 d t 1 / δ 1 = k 2 A 2 d t 2 / δ 2 Also areas are same.

4. Let us say thermal conductivity of a wall is governed by the relation k = k 0 . In that case the temperature at the mid-plane of the heat conducting wall would be

a) Av. of the temperature at the wall faces

b) More than average of the temperature at the wall faces

c) Less than average of the temperature at the wall faces

d) Depends upon the temperature difference between the wall faces

Answer: b

Explanation: k 0 is thermal conductivity at 0 degree Celsius. Here β is positive so it is more than average of the temperature at the wall faces.

5. Heat is transferred from a hot fluid to a cold one through a plane wall of thickness , surface area  and thermal conductivity . The thermal resistance is

a) 1/A (1/h 1 + δ/k + 1/h 2 )

b) A (1/h 1 + δ/k + 1/h 2 )

c) 1/A (h 1 + δ/k + h 2 )

d) A (h 1 + δ/k + 1/h 2 )

Answer: a

Explanation: Net thermal resistance will be summation of resistance through plane wall and from left side and right side of the wall.

6. Find the heat flow rate through the composite wall as shown in figure. Assume one dimensional flow and take

heat-transfer-questions-answers-conduction-composite-wall-q6

k 1 = 150 W/m degree

k 2 = 30 W/m degree

k 3 = 65 W/m degree

k 4 = 50 W/m degree

AB = 3 cm, BC = 8 cm and CD = 5 cm. The distance between middle horizontal line from the top is 3 cm and from the bottom is 7 cm

a) 1173.88 W

b) 1273.88 W

c) 1373.88 W

d) 1473.88 W

Answer: b

Explanation: Q = d t/ R T . R T = R 1 + R e q + R 2 = 0.02 + 0.01469 + 0.1 = 0.2669 degree/W.

7. A pipe carrying steam at 215.75 degree Celsius enters a room and some heat is gained by surrounding at 27.95 degree Celsius. The major effect of heat loss to surroundings will be due to

a) Conduction

b) Convection

c) Radiation

d) Both conduction and convection

Answer: c

Explanation: As there is temperature difference so radiation suits well.

8. “Radiation cannot be affected through vacuum or space devoid of any matter”. True or false

a) True

b) False

Answer: b

Explanation: It can be affected only by air between molecules and vacuum of any matter.

9. A composite slab has two layers having thermal conductivities in the ratio of 1:2. If the thickness is the same for each layer then the equivalent thermal conductivity of the slab would be

a) 1/3

b) 2/3

c) 2

d) 4/3

Answer: d

Explanation: 2 /1+2 = 4/3.

10. A composite wall of a furnace has two layers of equal thickness having thermal conductivities in the ratio 2:3. What is the ratio of the temperature drop across the two layers?

a) 2:3

b) 3:2

c) 1:2

d) log e 2 : log e 3

Answer: b

Explanation: We know that temperature is inversely proportional to thermal conductivity, so the ratio is 2:3.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Conduction Through A Cylindrical Wall”.

1. Typical examples of heat conduction through cylindrical tubes are not found in

a) Power plants

b) Oil refineries

c) Most process industries

d) Aircrafts

Answer: d

Explanation: Boilers have tubes in them, the condenser consist of a bank of tubes.

2. The rate of heat conduction through a cylindrical tube is usually expressed as

a) Per unit length

b) Per unit area

c) Only length

d) Only area

Answer: a

Explanation: It is expressed as per unit length rather than per unit area as done for plane walls.

3. A steel pipe of 20 mm inner diameter and 2 mm thickness is covered with 20 mm thick of fiber glass insulation . If the inside and outside convective coefficients are 10 W/m 2 degree and 5 W/m 2 degree, calculate the overall heat transfer coefficient based on inside diameter of pipe. In the diagram, the diameter of small circle is 20 mm

heat-transfer-questions-answers-conduction-cylindrical-wall-q3

a) 1.789 W/m 2 degree

b) 2.789 W/m 2 degree

c) 3.789 W/m 2 degree

d) 4.789 W/m 2 degree

Answer: b

Explanation: Q = 2 π l (t i – t 0 )/ [(1/h i r i ) + log e (r 3 /r 2 ) (1/k 2 ) + (1/h 0 r 3 )].

4. Logarithmic mean area of the cylindrical tube is given as

a) 2πr m

b) πr m l

c) 2πr m l

d) 2r m l

Answer: c

Explanation: It is known as equivalent area and r m = r 2 -r 1 /log e (r 2 /r 1 ).

5. A hot fluid is being conveyed through a long pipe of 4 cm outer diameter and covered with 2 cm thick insulation. It is proposed to reduce the conduction heat loss to the surroundings to one-third of the present rate by further covering with same insulation. Calculate the additional thickness of insulation

a) 11 cm

b) 12 cm

c) 13 cm

d) 14 cm

Answer: b

Explanation: Heat loss with existing insulation = 2 π k l (t 1 – t 2 )/log e (r 2 /r 1 ) and heat loss with additional insulation = 2 π k l (t 1 – t 2 )/log e (r 2 + x/r 1 ).

6. The heat flow equation through a cylinder of inner radius r 1 and outer radius r 2 is desired to be written in the same form as that for heat flow through a plane wall. For wall thickness (r 2 -r 1 ) the area will be

a) A 1 + A 2 /2

b) A 1 + A 2

c) A 2 – A 1 / log e (A 2 /A 1 )

d) A 1 + A 2 /2 log e (A 2 /A 1 )

Answer: a

Explanation: Here A 1 and A 2 are the inner and outer surface areas of tubes. The net area is A M .

7. A cylinder of radius r and made of material of thermal conductivity k 1 is surrounded by a cylindrical shell of inner radius r and outer radius 2r. This outer shell is made of a material of thermal conductivity k 2 . Net conductivity would be

a) k 1 + 3 k 2 /4

b) k 1 + k 2 /4

c) k 1 + 3k 2

d) k 1 + k 2

Answer: a

Explanation: Heat flowing per second is given by = k 1 (πr 2 ) (t 1 -t 2 ) δ. Shell heat is k 2 π [ 2 – r 2 ] (t 1 – t 2 )/ δ.

8. For steady state and a constant value of thermal conductivity, the temperature distribution associated with radial convection through a cylinder is

a) Linear

b) Parabolic

c) Logarithmic

d) Exponential

Answer: c

Explanation: As thermal conductivity is constant so we get a profile that is logarithmic in nature.

9. A cylindrical cement tube of radii 0.05 cm and 1.0 cm has a wire embedded into it along its axis. To maintain a steady temperature difference of 120 degree Celsius between the inner and outer surfaces, a current of 5 ampere is made to flow in the wire. Find the amount of heat generated per meter length. Take resistance of wire equal to 0.1 ohm per cm of length

a) 150 W/m length

b) 250 W/m length

c) 350 W/m length

d) 450 W/m length

Answer: b

Explanation: Resistance of wire = 10 ohm per m length. Heat generated =  2 10 = 250 W/m length.

10. A stainless steel tube with inner diameter 12 mm, thickness 0.2 mm and length 50n cm is heated electrically. The entire 15 k W of heat energy generated in the tube is transferred through its outer surface. Find the intensity of the current flow

a) 52 amps

b) 62 amps

c) 72 amps

d) 82 amps

Answer: a

Explanation: Power generated = 15 k W = 15000 W. Therefore, intensity of current flow =  ½ = 52 amps.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Conduction Through A Sphere”.

1. The temperature distribution associated with radial conduction through a sphere is represented by

a) Parabola

b) Hyperbola

c) Linear

d) Ellipse

Answer: b

Explanation: As conduction is radial i.e. in outward direction, so it follows the hyperbola equation..

2. The thermal resistance for heat conduction through a spherical wall is

a) (r 2 -r 1 )/2πkr 1 r 2

b) (r 2 -r 1 )/3πkr 1 r 2

c) (r 2 -r 1 )/πkr 1 r 2

d) (r 2 -r 1 )/4πkr 1 r 2

Answer: d

Explanation: We get this on integrating the equation Q = -k A d t/ d r from limits r 1 to r 2 and T 1 to T 2 .

3. The rate of conduction heat flow in case of a composite sphere is given by

a) Q = t 1 – t 2 / (r 2 – r 1 )/4πk 1 r 1 r 2 + (r 3 – r 2 )/4πk 2 r 2 r 3

b) Q = t 1 – t 2 / (r 2 – r 1 )/4πk 1 r 1 r 2 + (r 3 – r 2 )/4πk 2 r 2 r 3

c) Q = t 1 – t 2 / (r 2 – r 1 )/4πk 1 r 1 r 2 + (r 3 – r 2 )/4πk 2 r 2 r 3

d) Q = t 1 – t 2 / (r 2 – r 1 )/4πk 1 r 1 r 2 + (r 3 – r 2 )/4πk 2 r 2 r 3

Answer: c

Explanation: Here, convective film coefficient at the inner and outer surfaces are also considered.

4. The thermal resistance for heat conduction through a hollow sphere of inner radius r 1 and outer radius r 2 is

a) r 2 – r 1 /4πk r 1 r 2

b) r 2 /4πk r 1 r 2

c) r 1 /4πk r 1 r 2

d) 4πk r 1 r 2

Answer: a

Explanation: As Q = d t/ R T . Here R T is thermal resistance.

5. A spherical vessel of 0.5 m outside diameter is insulated with 0.2 m thickness of insulation of thermal conductivity 0.04 W/m degree. The surface temperature of the vessel is – 195 degree Celsius and outside air is at 10 degree Celsius. Determine heat flow per m 2 based on inside area

a) – 63.79 W/m 2

b) – 73.79 W/m 2

c) – 83.79 W/m 2

d) – 93.79 W/m 2

Answer: b

Explanation: Heat flow based on inside area = Q/4 π r 2 = – 73.79 W/m 2 .

6. The quantity d t/Q for conduction of heat through a body i.e. spherical in shape is

a) ln (r 2 /r 1 )/2πLk

b) ln (r 2 /r 1 )/πLk

c) ln (r 2 /r 1 )/2Lk

d) ln (r 2 /r 1 )/2πk

Answer: a

Explanation: We get this on integrating the equation Q = -k A d t/ d r from limits r 1 to r 2 and T 1 to T 2 .

7. A spherical vessel of 0.5 m outside diameter is insulated with 0.2 m thickness of insulation of thermal conductivity 0.04 W/m degree. The surface temperature of the vessel is – 195 degree Celsius and outside air is at 10 degree Celsius. Determine heat flow

a) – 47.93 W

b) – 57.93 W

c) – 67.93 W

d) – 77.93 W

Answer: b

Explanation: Q = 4 π k r 1 r 2 (t 1 – t 2 )/r 2 – r 1 = -57.93 W.

8. If we increase the thickness of insulation of a circular rod, heat loss to surrounding due to

a) Convection and conduction increases

b) Convection and conduction decreases

c) Convection decreases while that due to conduction increases

d) Convection increases while that due to conduction decreases

Answer: d

Explanation: In convection energy is transferred between solid and fluid but in conduction from T 1 to T 2 .

9. The following data pertains to a hollow cylinder and a hollow sphere made of same material and having the same temperature drop over the wall thickness

Inside radius = 0.1 m and outside surface area = 1 square meter

If the outside radius for both the geometrics is same, calculate the ratio of heat flow in the cylinder to that of sphere?

a) 0.056

b) 2.345

c) 1.756

d) 3.543

Answer: c

Explanation: For sphere r 2 =  1/2 = 0.282 m, for cylinder, l = A 2 /2 r 2 π = 0.565 m.

10. The oven of an electric store, of total outside surface area 2.9 m 2 dissipates electric energy at the rate of 600 W. The surrounding room air is at 20 degree Celsius and the surface coefficient of heat transfer between the room air and the surface of the oven is estimated to be 11.35 W/m 2 degree. Determine the average steady state temperature of the outside surface of the store

heat-transfer-questions-answers-conduction-sphere-q10

a) 38.22 degree Celsius

b) 48.22 degree Celsius

c) 58.22 degree Celsius

d) 68.22 degree Celsius

Answer: a

Explanation: Q = h A (t 0 – t a ).

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Shape Factor”.

1. Which of the following is a wrong statement according to the shape factor is equal to one?

a) For any surface completely enclosed by another surface

b) For infinite parallel planes radiating only to each other

c) For a flat or convex surface with respect to itself

d) Inside cylinder to outer cylinder of a long co-axial cylinder

Answer: d

Explanation: For a flat or convex surface, the shape factor with respect to itself is zero. This aspect stems from the fact that for any part of flat or convex surface, one cannot see any other part of the same surface.

2. Establish a relation for the shape factor of cylindrical cavity with respect to itself of depth h and diameter d. The cavity is closed on its outer surface with a flat plate

heat-transfer-questions-answers-shape-factor-2-q2

a) 4 h/4 h + d

b) 4 h + d/h

c) 4 h + d/4 h

d) h/4 h + d

Answer: a

Explanation: F 11 + F 12 = 1 and F 21 + F 22 = 1. But F 22 = 0, so F 11 = 1 – A 2 / A 1 .

3. The reciprocity theorem states that

a) F 12 = F 21

b) A 1 F 12 = A 2 F 21

c) A 1 F 21 = A 2 F 12

d) A 2 F 21 = A 1 F 12

Answer: b

Explanation: It indicates the net radiation exchange can be calculated by computing one way configuration factor from either surface to the other.

4. Two radiating surface A 1 = 6 m 2 and A 2 = 4 m 2 have shape factor F 12 = 0.1. Then the shape factor F 21 will be

a) 0.12

b) 0.18

c) 0.15

d) 0.10

Answer: c

Explanation: A 1 F 12 = A 2 F 21 .

5. What is the value of shape factor for two infinite parallel surfaces separated by a distance x?

a) 0

b) 1

c) x

d) Infinity

Answer: b

Explanation: All the radiation emitted by one falls on the other so shape factor is unity.

6. What is the shape factor of a sphere of diameter d inside a cubical box of length l = d?

heat-transfer-questions-answers-shape-factor-2-q6

a) π/6

b) 2 π/6

c) π/3

d) π/4

Answer: a

Explanation: F 11 + F 12 = 1 or F 12 = 1. So, F 21 = π/6.

7. What is the shape factor of hemispherical surface closed by a plane surface of diameter d?

a) 0

b) 1.5

c) 1

d) 0.5

Answer: d

Explanation: F 11 + F 12 = 1 and F 21 = 1. Therefore, F 11 = 0.5.

8. Establish a relation for shape factor for a conical cavity with respect to itself of depth h and diameter d. The cavity is closed on its outer surface with a flat plate

heat-transfer-questions-answers-shape-factor-2-q8

a) 1 – d/ (4 h 2 + d) 1/2

b) 1 – d/ (4 h 2 + d 2 ) 1/2

c) 1 – d/ (4 h + d 2 ) 1/2

d) 1 – d/ (h 2 + d 2 ) 1/2

Answer: b

Explanation: F 11 = 1 – A 2 / A 1 = 1 – 2 sin α = 1 – d/ (4 h 2 + d 2 ) ½ .

9. What is the shape factor for a hemispherical bowl with respect to itself of diameter d? The cavity is closed on its outer surface with a flat plate

a) 1.5

b) 1

c) 0.5

d) 2.5

Answer: c

Explanation: F 11 = 1 – A 2 / A 1 = 0.5.

10. Consider a system of concentric spheres of radius r 1 and r 2 (r 2 is greater than r 1 ). If r 1 = 5 cm, determine the radius r 2 if it is desired to have the value of shape factor F 21 equal to 0.6

heat-transfer-questions-answers-shape-factor-2-q10

a) 6.45 cm

b) 7.45 cm

c) 8.45 cm

d) 9.45 cm

Answer: a

Explanation: From reciprocity theorem, A 1 F 12 = A 2 F 21 .

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Effect Of Variable Conductivity”.

1. With variable thermal conductivity, Fourier law of heat conduction through a plane wall can be expressed as

a) Q = -k 0  A d t/d x

b) Q = k 0  A d t/d x

c) Q = –  A d t/d x

d) Q =  A d t/d x

Answer: a

Explanation: Here k 0 is thermal conductivity at zero degree Celsius.

2. The inner and outer surfaces of a furnace wall, 25 cm thick, are at 300 degree Celsius and 30 degree Celsius. Here thermal conductivity is given by the relation

K = (1.45 + 0.5 * 10 -5 t 2 ) KJ/m hr deg

Where, t is the temperature in degree centigrade. Calculate the heat loss per square meter of the wall surface area?

a) 1355.3 kJ/m 2 hr

b) 2345.8 kJ/m 2 hr

c) 1745.8 kJ/m 2 hr

d) 7895.9 kJ/m 2 hr

Answer: c

Explanation: Q = -k A d t/d x, Q d x = – k A d t = – (1.45 + 0.5 * 10 -5 t 2 ) A d t. Integrating over the wall thickness δ, we get Q = 436.45/0.25 = 1745.8 kJ/m 2 hr.

3. A plane wall of thickness δ has its surfaces maintained at temperatures T 1 and T 2 . The wall is made of a material whose thermal conductivity varies with temperature according to the relation k = k 0 T 2 . Find the expression to work out the steady state heat conduction through the wall?

a) Q = 2A k 0 (T 1 3 – T 2 3 )/3 δ

b) Q = A k 0 (T 1 3 – T 2 3 )/3 δ

c) Q = A k 0 (T 1 2 – T 2 2 )/3 δ

d) Q = A k 0 (T 1 – T 2 )/3 δ

Answer: b

Explanation: Q = -k A d t/d x = k 0 T 2 A d t/d x. Separating the variables and integrating within the prescribed boundary conditions, we get Q = A k 0 (T 1 3 – T 2 3 )/3 δ.

4. The mean thermal conductivity evaluated at the arithmetic mean temperature is represented by

a) k m = k 0 [1 + β (t 1 – t 2 )/2].

b) k m = k 0 [1 + (t 1 + t 2 )/2].

c) k m = k 0 [1 + β (t 1 + t 2 )/3].

d) k m = k 0 [1 + β (t 1 + t 2 )/2].

Answer: d

Explanation: At arithmetic mean temperatures i.e. (t 1 + t 2 )/2.

5. With respect to the equation k = k 0  which is true if we put β = 0?

a) Slope of temperature curve is constant

b) Slope of temperature curve does not change

c) Slope of temperature curve increases

d) Slope of temperature curve is decreases

Answer: a

Explanation: As temperature profile is linear so it is constant.

6. The accompanying sketch shows the schematic arrangement for measuring the thermal conductivity by the guarded hot plate method. Two similar 1 cm thick specimens receive heat from a 6.5 cm by 6.5 cm guard heater. When the power dissipation by the wattmeter was 15 W, the thermocouples inserted at the hot and cold surfaces indicated temperatures as 325 K and 300 K. What is the thermal conductivity of the test specimen material?

heat-transfer-questions-answers-effect-variable-conductivity-q6

a) 0.81 W/m K

b) 0.71 W/m k

c) 0.61 W/m K

d) 0.51 W/m K

Answer: b

Explanation: Q = k A (t 1 – t 2 )/δ. So, k = 0.71 W/m K.

7. If β is greater than zero, then choose the correct statement with respect to given relation

k = k 0 

a) k doesn’t depend on temperature

b) k depends on temperature

c) k is directly proportional to t

d) Data is insufficient

Answer: c

Explanation: k increases with increases temperature.

8. The unit of thermal conductivity doesn’t contain which parameter?

a) Watt

b) Pascal

c) Meter

d) Kelvin

Answer: b

Explanation: Its unit is W/m K.

9. The temperatures on the two sides of a plane wall are t 1 and t 2 and thermal conductivity of the wall material is prescribed by the relation

K = k 0 e 

Where, k 0 is constant and δ is the wall thickness. Find the relation for temperature distribution in the wall?

a) t 1 – t x / t 1 – t 2 = x

b) t 1 – t x / t 1 – t 2 = δ

c) t 1 – t x / t 1 – t 2 = δ/x

d) t 1 – t x / t 1 – t 2 = x/δ

Answer: d

Explanation: Q = -k A d t/d x = -k 0 e  d t/d x. Separating the variables and upon integration, we get Q/k 0 A = (t 1 – t 2 )/ δ . Therefore heat transfer through the wall, Q = k 0 A (t 1 – t 2 )/ δ . At x = x and t = t x we get the answer.

10. “If β is less than zero, then with respect to the relation k = k 0 , conductivity depends on surface area”.

a) True

b) False

Answer: b

Explanation: k decreases with increasing temperature.

This set of Heat Transfer Questions & Answers for freshers focuses on “Critical Thickness Of Insulation”.

1. A cable of 10 mm outside is to be laid in an atmosphere of 25 degree Celsius (h = 12.5 W/m 2 degree) and its surface temperature is likely to be 75 degree Celsius due to heat generated within it. How would the heat flow from the cable be affected if it is insulated with rubber having thermal conductivity k = 0.15 W/m degree?

a) 43.80 W per meter length

b) 53.80 W per meter length

c) 63.80 W per meter length

d) 73.80 W per meter length

Answer: b

Explanation: Q = 2 π d t/  log e (r c /r 0 ) = 53.80 W per meter length.

2. Chose the correct one with respect to the critical radius of insulation

a) There is more heat loss i.e. conductive

b) There occurs a decrease in heat flux

c) Heat loss increases with addition of insulation

d) Heat loss decreases with addition of insulation

Answer: c

Explanation: For a pipe heat loss is more at the critical radius.

3. A heat exchanger shell of outside radius 15 cm is to be insulated with glass wool of thermal conductivity 0.0825 W/m degree. The temperature at the surface is 280 degree Celsius and it can be assumed to remain constant after the layer of insulation has been applied to the shell. The convective film coefficient between the outside surface of glass wool and the surrounding air is estimated to be 8 W/m 2 degree. What is the value of a critical radius?

a) 9.31 mm

b) 10.31 mm

c) 11.31 mm

d) 12.31 mm

Answer: b

Explanation: Critical radius of insulation = k/h = 0.0825/8 = 0.01031 m = 10.31 mm.

4. For an object i.e. spherical the value of critical radius would be

a) 2k/3h

b) 3k/h

c) 2k/h

d) k/h

Answer: c

Explanation: It depends on the variation of angle with layers of insulation.

5. Maximum value of critical radius is

a) 0.01 m

b) 0.04 m

c) 0.06 m

d) 0.0001 m

Answer: a

Explanation: K for common insulating material is 0.05 W/ m degree.

6. An electric cable of aluminum  is to be insulated with rubber . If the cable is in air . Find the critical radius?

a) 80 mm

b) 160 mm

c) 40 mm

d) 25 mm

Answer: d

Explanation: Critical radius = 0.15/6 = 0.025 m = 25 mm.

7. The value of critical radius in case of a cylindrical hollow object is

a) 2k/h

b) 2h/k

c) k/h

d) h/k

Answer: c

Explanation: Unit is meter.

8. A wire of radius 3 mm and 1.25 m length is to be maintained at 60 degree Celsius by insulating it by a material of thermal conductivity 0.175 W/m K. The temperature of surrounding is 20 degree Celsius with heat transfer coefficient 8.5 W/ m 2 K. Find percentage increase in heat loss due to insulation?

a) 134.46 %

b) 124.23 %

c) 100.00 %

d) 12.55 %

Answer: a

Explanation: Q = 8.5   = 8.01 W. % increase =   = 134.46 %.

9. A pipe of outside diameter 20 mm is to be insulated with asbestos which has a mean thermal conductivity of 0.1 W/m degree. The local coefficient of convective heat to the surroundings is 5 W/square meter degree. Find the critical radius of insulation for optimum heat transfer from a pipe?

a) 10 mm

b) 20 mm

c) 30 mm

d) 40 mm

Answer: b

Explanation: Critical radius of insulation = k/h 0 = 0.1/5 = 0.02 m = 20 mm.

10. For insulation to be properly effective in restricting heat transmission, the pipe radius r 0 will be

a) Greater than critical radius

b) Less than critical radius

c) Equal to critical radius

d) Greater than or equal to critical radius

Answer: d

Explanation: Addition of insulating material doesn’t always decrease in the heat transfer rate.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Heat Generation Through Plane Wall”.

1. In case of heat conduction through the plane wall, which one of the following is not a correct assumption?

a) Steady state

b) Three dimensional heat flow

c) Volumetric heat generation must be constant

d) K must be constant

Answer: b

Explanation: There should be two dimensional heat flow. The differential equation describing the temperature distribution can be set up by making an energy balance.

2. If Q X is heat generated in at distance ‘x’, then heat conducted out at a distance  will be

a) Q X + 3d (Q X ) d x /d x

b) 2Q X + d (Q X ) d x /d x

c) d (Q X ) d x /d x

d) Q X + d (Q X ) d x /d x

Answer: d

Explanation: Q X + Q g = Q X + d X .

3. Notable example of uniform generation of heat within the conducting medium are

 Energy of a nuclear reactor

 Liberation of energy due to some exothermic chemical reactions

 Resistance heating in electrical appliances

Which of the statements made above are correct?

a) i, ii and iii

b) i and ii

c) i and iii

d) Only ii

Answer: a

Explanation: All the statements are correct with respect to plane wall heat conduction.

4. For a plane wall of thickness l with uniformly distributed heat generation q g per unit volume, the temperature t 0 at mid plane is given by

a) t 0 = q g l 2 /2k +t w

b) t 0 = q g l 2 /4k +t w

c) t 0 = q g l 2 /8k +t w

d) t 0 = q g l 2 /16k +t w

Answer: c

Explanation: t = q g /2k  x + t w . At mid plane i.e. x = l/2 we get t 0 = q g l 2 /8k +t w .

5. The temperature drop in a plane wall with uniformly distributed heat generation can be decreased by reducing

a) Wall thickness

b) Heat generation rate

c) Thermal conductivity

d) Surface area

Answer: a

Explanation: On decreasing wall thickness, generally temperature drop decreases.

6. Consider a slab of thickness δ with one side  insulated and other side  maintained at constant temperature. The rate of uniform heat generation within the slab is q g W/m 3 . Presuming that the heat conduction is in steady state and one dimensional along x direction, the maximum temperature in the slab would occur at x equal

a) δ/2

b) Zero

c) δ/4

d) δ

Answer: b

Explanation: Maximum temperature occurs at the insulated face of the wall where x = 0.

7. There occurs heat conduction and internal heat generation at uniform rate within the conduction medium itself in the following cases

 Drying of concrete

 Chemical processes

 Fuel elements in a nuclear reaction

Choose the correct option

a) i only

b) ii only

c) i and iii

d) i, ii and iii

Answer: d

Explanation: All are correct as there is heat conduction in all above cases. The rate of heat generation has to be controlled one, otherwise the resulting temperature growth might result in the failure of the medium.

8. The rear window of an automobile is made of thick glass i.e. AB = 5 mm and thermal conductivity is 0.8 W/m degree. To defrost this window, a thin transparent film type heating element has been fixed to its inner surface. For the conditions given below, determine the electric power that must be provided per unit area of window if a temperature 5 degree Celsius is maintained at its outer surface. Interior air temperature and the corresponding surface coefficient are 20 degree Celsius and 12 W/m 2 degree. Surrounding air temperature and the corresponding surface coefficient are – 15 degree Celsius and 70 W/m 2 degree. Electric heater provides uniform heat flux

heat-transfer-questions-answers-heat-generation-plane-wall-q8

a) 232.5 /m 2

b) 1232.5 /m 2

c) 2232.5 /m 2

d) 3232.5 /m 2

Answer: b

Explanation: (t I – t f )/(1/h i + δ/k) + q g = h 0 (t s – t 0 ).

9. Suppose heat is conducted due to electrons

Where, i = I/A and p is the resistivity, then

a) q g = 2i 2 p

b) q g = 3i 2 p

c) q g = i 2 p

d) q g = 4i 2 p

Answer: c

Explanation: It should be i 2 p. Here i is current density.

10. In case when both the surfaces of plane wall are at different temperature, we get an expression i.e.

T MAX – T W2 /T W1 – T W2 =  2 /4B

What is the value of B?

a) (q g )  2 /2k (T W1 – T W2 )

b) (q g )  3 /3k (T W1 – T W2 )

c) (q g )  4 /4k (T W1 – T W2 )

d) (q g )  5 /5k (T W1 – T W2 )

Answer: a

Explanation: T – T W2 /T W1 – T W2 = [1 – x/ δ] [B x/ δ +1].

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Dielectric Heating”.

1. Which one of the following materials are quickly heated by applying high frequency?

a) Textiles

b) Engines

c) Rubber

d) Coal

Answer: a

Explanation: They can be heated at high voltage alternating current to the plated of the condenser.

2. Generally heat generated depends on some parameters. It is directly proportional to

a) Time

b) Conductivity

c) Voltage

d) Distance between plates

Answer: c

Explanation: It generally depends on the voltage as directly proportional.

3. Consider a 1.2 m thick slab of poured concrete  with both of side surfaces maintained at a temperature of 20 degree Celsius. During its curing, chemical energy is released at the rate of 80 W/m 3 . Workout the maximum temperature of concrete

a) 30.73 degree celsius

b) 29.73 degree celsius

c) 28.73 degree celsius

d) 27.73 degree celsius

Answer: b

Explanation: t = q g  x/2 k + t w = 29.73 degree celsius.

4. The insulating material used in dielectric heating is

a) Coal

b) Silver

c) Coin

d) Wool

Answer: d

Explanation: Wool is good for heat conduction and from a dielectric heating point of view.

5. A composite slab consists of 5 cm thick layer of steel  on the left side and a 6 cm thick layer of brass  on the right hand side. The outer surfaces of the steel and brass are maintained at 100 degree Celsius and 50 degree Celsius. The contact between the two slabs is perfect and heat is generated at the rate of 4.2 * 10 5 k J/m 2 hr at the plane of contact. The heat thus generated is dissipated from both sides of composite slab for steady state conditions. Calculate the temperature at the interface

heat-transfer-questions-answers-dielectric-heating-q5

a) 115.26 degree celsius

b) 125.26 degree celsius

c) 135.26 degree celsius

d) 145.26 degree celsius

Answer: b

Explanation: Q 1 + Q 2 = Q g . Q 1 = k 1 A 1 t i – t 1 )/δ 1 and Q 2 = k 2 A 2 t i – t 2 )/δ 2 .

6. Unit of specific resistance is

a) Ohm mm 2 /m

b) Ohm mm

c) Ohm/m

d) Ohm mm/m

Answer: a

Explanation: Specific resistance is resistance per unit length.

7. What maximum thickness of concrete can be poured without causing the temperature gradient to exceed 98.5 degree Celsius per meter anywhere in the slab? Consider a 1.2 m thick slab of poured concrete  with both of side surfaces maintained at a temperature of 20 degree Celsius. During its curing, chemical energy is released at the rate of 80 W/m 3 . Workout the maximum temperature of concrete

a) 2.64 m

b) 3.64 m

c) 4.64 m

d) 5.64 m

Answer: b

Explanation: d t/d x = q g /2 k. The temperature is largest at x = 0.

8. Dielectric heating apparatus consists of

a) 4 electrodes

b) Elemental strip

c) No Insulating material

d) 4 plates

Answer: b

Explanation: It consists of an elemental strip in the middle of the system.

9. The given expression can be used to solve the electrode temperature t w1 and t w2

q g δ = h 1 α 1 + h 2 α 2

Where, α 1 = A (t w 2 – t a ) and α 2 = (t w1 – t a ).

a) True

b) False

Answer: b

Explanation: α 1 = A (t w 1 – t a ) and α 2 = (t w2 – t a ).

10. A slab of insulating material of thickness 6 cm and thermal conductivity 1.4kJ/m hr deg is placed between and is in contact with two parallel electrodes, and is then subjected to high frequency dielectric heating at a uniform rate of 140,000kJ/m 3 hr. At steady state coefficients of combined radiation and convection are 42 and 48 kJ/m 2 hr deg. If atmospheric temperature is 25 degree Celsius, find surface temperatures?

a) 144.10 degree Celsius and 134.47 degree Celsius

b) 123.50 degree Celsius and 154.34 degree Celsius

c) 121.60 degree Celsius and 115.45 degree Celsius

d) 165.40 degree Celsius and 165.45 degree Celsius

Answer: c

Explanation: α = -q g x 2 /2k + h 1 α 1 /k + α 1 . At x =0.06 m and α = α 2 , α 2 = -180 + 2.8 α 1 . Also q g A δ = h 1 α 1 + h 2 α 2 .

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Heat Generation Through Cylinder”.

1. A 25 mm diameter egg roll  is roasted with the help of microwave heating. For good quality roasting, it is desired that temperature at the center of roll is maintained at 100 degree Celsius when the surrounding temperature is 25 degree Celsius. What should be the heating capacity in W/m 3 of the microwave if the heat transfer coefficient on the surface of egg roll is 20 W/m 2 degree?

a) 113.31 k W/m 3

b) 213.31 k W/m 3

c) 313.31 k W/m 3

d) 413.31 k W/m 3

Answer: b

Explanation: t max = t a + q g R/2h + q g R 2 /4k.

2. The constants of integration are to be determined from the relevant boundary conditions which are

 t = t w at r = R

 q g (π L R 2 ) = -k  

 Increasing temperature gradient

Choose the correct option?

a) Only i

b) Only ii

c) i and iii

d) i and ii

Answer: d

Explanation: Temperature gradient must be constant.

3. A concrete column used in bridge construction is cylindrical in shape with a diameter of 1 meter. The column is completely poured in a short interval of time and the hydration of concrete results in the equivalent of a uniform source strength of 0.7 W/kg. Determine the temperature at the center of the cylinder at a time when the outside surface temperature is 75 degree Celsius. The column is sufficiently long so that temperature variation along its length may be neglected. For concrete

Average thermal conductivity = 0.95 W/m K

Average density = 2300 kg/m 3

a) 190.92 degree Celsius

b) 180.92 degree Celsius

c) 170.92 degree Celsius

d) 160.92 degree Celsius

Answer: b

Explanation: t max = t w + q g R 2 /4k. We get, q g = 0.7 W/kg.

4. The temperature distribution profile for a solid cylinder is

a) Parabolic

b) Linear

c) Ellipse

d) Hyperbolic

Answer: a

Explanation: It must be parabolic for maximum heat conduction.

5. For a solid cylinder, maximum temperature difference occurring at the center of the rod is given by

a) t W – q g R 2 /4K

b) q g R 2 /4K

c) t W + q g R 2 /4K

d) t W + q g R 2 /4KL

Answer: c

Explanation: As temperature distribution profile is parabolic so on integrating between boundary conditions we get the result.

6. A slab of 12 cm thickness and generating heat uniformly at 10 6 W/m 3 has thermal conductivity of 200 W/m degree. Both surfaces of the slab are maintained at 150 degree Celsius. Determine the heat flow rate at the quarter planes

heat-transfer-questions-answers-heat-generation-cylinder-q6

a) 30000 W/m 2

b) 40000 W/m 2

c) 50000 W/m 2

d) 60000 W/m 2

Answer: a

Explanation: q g (π R 2 L) = (t w – t a )  h.

7. Consider a convective heat flow to water at 75 degree Celsius from a cylindrical nuclear reactor fuel rod of 50 mm diameter. The rate of heat generatioN is 50000000 W/m 3 and convective heat transfer coefficient is I kW/m 2 K. The outer surface temperature of the fuel element would be

a) 625 degree Celsius

b) 700 degree Celsius

c) 550 degree Celsius

d) 400 degree Celsius

Answer: b

Explanation: t w = t a + q g R/ 2h.

8. For a cylindrical rod with uniformly distributed heat sources, the thermal gradient at half the radius location will be

a) Four times

b) Twice

c) One fourth

d) One half

Answer: d

Explanation: t = t w + q g (R 2 – r 2 )/4k.  r = R/2 = 1/2 r = R .

9. The maximum temperature for cylindrical coordinate occurring at r = 0 is

a) t max = t a +q g R/h + q g R 2 /4k

b) t max = t a +q g R/4h + q g R 2 /4k

c) t max = t a +q g R/2h + q g R 2 /4k

d) t max = t a +q g R/6h + q g R 2 /4k

Answer: c

Explanation: As, t = t a +q g R/6h + q g R 2 /4k (R 2 –r 2 ).

10. In case of solid cylinder of radius R, the temperature distribution is given as

a) t – t w /t max – t w = 1 –  2

b) t – t w /t max – t w = 1 – 

c) t – t w /t max – t w = 1 –  3

d) t – t w /t max – t w = 1 –  4

Answer: a

Explanation: As we know for a long cylinder of radius R, t = t w + q g (R 2 – r 2 )/4k. On integrating this we get the answer. Where, t w is outer surface temperature and t max is along cylinder axis.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Heat Generation Through Sphere”.

1. Consider heat conduction through a solid sphere of radius R. There are certain assumptions

 Unsteady state conditions

 One-dimensional radial conduction

 Constant thermal conductivity

Identify the correct statements

a) i and iii

b) ii and iii

c) i, ii and iii

d) i and ii

Answer: b

Explanation: Statement 1 should be steady state condition.

2. An 8 cm diameter orange, approximately spherical in shape, undergoes ripening process and generates 18000 k J/m 3 hr of energy. If external surface of the orange is at 6.5 degree Celsius, find out the temperature at the center of the orange. Take thermal conductivity = 0.8 k J/ m hr degree for the orange material

a) 13.5 degree Celsius

b) 12.5 degree Celsius

c) 11.5 degree Celsius

d) 10.5 degree Celsius

Answer: b

Explanation: q g = 5000 W/m 3 , k = 0.222 W/m K and t = t W + q g R 2 /6K = 12.5 degree Celsius.

3. Consider the above problem, calculate the heat flow from the outer surface of the orange

a) 4.82 k J/hr

b) 5.82 k J/hr

c) 6.82 k J/hr

d) 7.82 k J/hr

Answer: a

Explanation: Q = 4/3 (π R 3 q g ) = 1.34 J/s.

4. What is the heat flow for steady state conduction for sphere?

a) 4 Q R + Q G = Q R + d R

b) 3 Q R + Q G = Q R + d R

c) 2 Q R + Q G = Q R + d R

d) Q R + Q G = Q R + d R

Answer: d

Explanation: Q R + Q G = Q R + d (Q R ) d R/d R.

Where, Q R = Heat conducted in at radius R

Q G = Heat conducted in the element

Q R + d R = Heat conducted out at radius R + d R.

5. The general solution for temperature distribution in case of solid sphere is

a) t = t W + q g (R 2 – r 2 )/4 k

b) t = t W + q g (R 2 – r 2 )/8 k

c) t = t W + q g (R 2 – r 2 )/6 k

d) t = t W + q g (R 2 – r 2 )/2 k

Answer: c

Explanation: The temperature distribution is parabolic.

6. A solid sphere of 8 cm radius has a uniform heat generation 0f 4000000 W/m 3 . The outside surface is exposed to a fluid at 150 degree Celsius with convective heat transfer coefficient of 750 W/m 2 K. If thermal conductivity of the solid material is 30 W/m K, determine maximum temperature

a) 444.45 degree Celsius

b) 434.45 degree Celsius

c) 424.45 degree Celsius

d) 414.45 degree Celsius

Answer: b

Explanation: q g (4 π R 3 /3) = h 4 π R 2 (t W – t a ), t w = 292.22 degree Celsius

T MAX = t w + q g R 2 /6 k.

7. Consider the above problem, find the temperature at 5 cm radius

a) 348.9 degree Celsius

b) 358.9 degree Celsius

c) 368.9 degree Celsius

d) 378.9 degree Celsius

Answer: d

Explanation: t – t w /t MAX – t w = 1 –  ½ .

8. Identify the correct boundary condition for a hollow sphere with inside surface insulated

a) At r = r 1 , the conduction region is perfectly insulated

b) At r = r 1 , the conduction region is partially insulated

c) Heat flow is infinity

d) Heat flow is negative

Answer: a

Explanation: In this range, the conduction region must be perfectly insulated.

9. A hollow sphere  of inner radius 6 cm and outside radius 8 cm has a heat generation rate of 4000000 W/m 3 . The inside surface is insulated and heat is removed by convection over the outside surface by a fluid at 100 degree Celsius with surface conductance 300 W/m 2 K. Make calculations for the temperature at the outside surfaces of the sphere

a) 105.6 degree Celsius

b) 205.6 degree Celsius

c) 305.6 degree Celsius

d) 405.6 degree Celsius

Answer: c

Explanation: q g 4 π (R 3 – r 3 )/3 = h 0 4 π r 2 (t 2 – t a ).

10. Consider the above problem, also calculate the temperature at the inside surfaces of the sphere

a) 138.3 degree Celsius

b) 327.8 degree Celsius

c) 254.7 degree Celsius

d) 984.9 degree Celsius

Answer: b

Explanation: t = t 2 + q g (R 2 – r 2 )/6 k – q g r 3 /3 k.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Fins”.

1. A very long copper rod 20 mm in diameter extends horizontally from a plane heated wall maintained at 100 degree Celsius. The surface of the rod is exposed to an air environment at 20 degree Celsius with convective heat transfer coefficient of 8.5 W/m 2 degree. Workout the heat loss if the thermal conductivity of copper is 400 W/m degree

a) 10.71 W

b) 20.71 W

c) 30.71 W

d) 40.71 W

Answer: b

Explanation: P/A = 4/d and m =  ½ =  ½ = 2.061.per meter.

2. Common applications of finned surfaces are with

 Electrical motors

 Economizers for steam power plant

 Convectors for steam and cold water heating systems

 Cooling coils

Identify the correct option

a) i, ii and iv

b) i, ii and iii

c) i, ii, iii and iv

d) i and ii

Answer: a

Explanation: It should be for hot water heating systems.

3. The extended surface used for the enhancement of heat dissipation is

a) Convective coefficient

b) Fourier number

c) Fin

d) No finned surface

Answer: c

Explanation: The surface area exposed to the surroundings is frequently increased by the attachment of protrusions to the surfaces, and the arrangement provides a means by which heat transfer rate can be improved.

4. It is said that fins can take a variety of forms

 Longitudinal fins of rectangular cross section attached to a wall

 Cylindrical tubes with radial fins

 Conical rod protruding from a wall

Identify the correct statement

a) i only

b) i and ii

c) ii and iii

d) i, ii and iii

Answer: d

Explanation: Option b is also known as annular fins.

5. A steel rod  1 cm in diameter and 5 cm long protrudes from a wall which is maintained at 10 degree Celsius. The rod is insulated at its tip and is exposed to an environment with h = 50 W/m 2 degree and t a = 30 degree Celsius. Calculate the fin efficiency

a) 56.57%

b) 66.57%

c) 76.57%

d) 86.57%

Answer: b

Explanation: Fin efficiency = tan h ml/ml, where m =  ½ = 25.82 per meter.

6. If the fin is sufficiently thin, so heat flows pertain to

a) One dimensional heat conduction

b) Two dimensional heat conduction

c) Three dimensional heat conduction

d) No heat flow is there

Answer: a

Explanation: As, δ is less than b, so one dimensional heat conduction is there.

7. If heat dissipation for one fin is given by 377.45 k J/hour, then what is the heat dissipation for 12 fins?

a) 7529.4 k J/hour

b) 6529.4 k J/hour

c) 5529.4 k J/hour

d) 4529.4 k J/hour

Answer: d

Explanation: For 12 fins, the heat dissipation will be equal to 12  = 4529.4 k J/hour.

8. In order to achieve maximum heat dissipation, the fin should be designed in such a way that has a

a) Maximum lateral surface towards the tip side of fin

b) Minimum lateral surface near the center line

c) Maximum lateral surface at the root side of fin

d) Maximum lateral surface near the center of fin

Answer: c

Explanation: Fins are so designed that lateral surface at the root side of the fin is maximum. This aspect results into higher heat dissipation.

9. A steel rod  1 cm in diameter and 5 cm long protrudes from a wall which is maintained at 10 degree Celsius. The rod is insulated at its tip and is exposed to an environment with h = 50 W/m 2 degree and t a = 30 degree Celsius. Calculate the rate of heat dissipation

a) 2.658 W

b) 3.658 W

c) 4.658 W

d) 5.658 W

Answer: b

Explanation: Q = k A m tan h ml (t 0 – t a ) = 3.658 W.

10. On a heat transfer surface, fins are provided to

a) Increase turbulence in flow for enhancing heat transfer

b) Increase temperature gradient so as to enhance heat transfer

c) Pressure drop of the fluid should be minimized

d) Surface area is maximum to promote the rate of heat transfer

Answer: d

Explanation: Fins are provided to a heat exchanger surface to augment the heat transfer by increasing the surface area exposed to the surroundings.

This set of Heat Transfer Interview Questions & Answers focuses on “Steady Flow of Heat Along a Rod”.

1. Which one is true regarding rectangular fin?

a) A C = b δ and P = 2

b) A C = 2 b δ and P = 2

c) A C = 3 b δ and P = 2

d) A C = 4 b δ and P = 2

Answer: a

Explanation: For rectangle, A =  . Where, b = width and δ = thickness.

2. Analysis of heat flow from the finned surface is made with the following assumptions

 Uniform heat transfer coefficient, h over the entire fin surface

 No heat generation within the fin generation

 Homogenous material

Identify the correct option

a) i only

b) i and ii only

c) i, ii and iii

d) ii only

Answer: c

Explanation: The knowledge of temperature distribution is necessary for their optimum design with regard to size and weight.

3. If heat conducted into the element at plane x is Q X = – k A C  X . Then heat conducted out of the element at plane  is

a) – 2k A C d/d x )

b) – k A C d/d x )

c) – 3k A C d/d x )

d) – 4k A C d/d x )

Answer: b

Explanation: Heat conducted out of the element is – [k A C  X + d x ].

4. A heating unit is made in the form of a vertical tube of 50 mm outside diameter and 1.2 m height. The tube is fitted with 20 steel fins of rectangular section with height 40 mm and thickness 2.5 mm. The temperature at the base of fin is 75 degree Celsius, the surrounding air temperature is 20 degree Celsius and the heat transfer coefficient between the fin as well as the tube surface and the surrounding air is 9.5 W/m 2 K. If thermal conductivity of the fin material is 55 W/m K, find the amount of heat transferred from the tube without fin

a) 98.44 W

b) 88.44 W

c) 78.44 W

d) 68.44 W

Answer: a

Explanation: Q = h A d t = h (π d 0 H) (t 0 – t INFINITY ).

5. The general solution of linear and homogenous differential equation  is of the form

a) γ = C 1 e 2 m x + C 2 e – m x

b) γ = C 1 e 3m x + C 2 e – m x

c) γ = C 1 e 4 m x + C 2 e – m x

d) γ = C 1 e m x + C 2 e – m x

Answer: d

Explanation: It should contain m x and – m x term.

6. For steady flow of heat along a rod, the general equation is

d 2 α/dx 2 – m 2 α = 0

The value of constant m is

a) (h P/k A C )

b) (h P/k A C ) 3/2

c) (h P/k A C ) 1/2

d) (h P/k A C ) 2

Answer: c

Explanation: This provides a general form of the energy equation for one dimensional heat flow.

7. In convection from the tip, we introduced a factor known as

a) Fin length

b) Correction length

c) No fin length

d) Radial length

Answer: b

Explanation: Just for simplicity we replace fin length by correction length.

8. Find the value of corrected length for rectangular fin?

Where, b is width and t is length of the fin

a) L C = L + b t/2 

b) L C = L + b t/ 

c) L C = L + 2 

d) L C = L + b t

Answer: a

Explanation: For rectangle, area = t b.

9. Which one is true for the spine?

a) A C = π d 2 /4 and P = 4 π d

b) A C = π d 2 /4 and P = 3 π d

c) A C = π d 2 /4 and P = π d

d) A C = π d 2 /4 and P = 2 π d

Answer: c

Explanation: A spine is a pin fin.

10. In convection from the tip what is the value of correction length?

a) L C = A C /P

b) L C = L + A C

c) L C = L + P

d) L C = L + A C /P

Answer: d

Explanation: It should contain all the three terms i.e. L, A and P.

This set of Heat Transfer Questions & Answers for experienced focuses on “Heat Dissipation From An Infinitely Long Fin “.

1. In heat dissipation from an infinitely long fin, the boundary conditions are

a) t = t 0 at x = infinity and t = t a at x = 0

b) t = t 0 at x = 0 and t = t a at x = infinity

c) t = t 0 at x = 0 and t = t a at x = 0

d) t = t 0 at x = infinity and t = t a at x = infinity

Answer: b

Explanation: These conditions must be approached when ml is greater than 5.

2. The temperature distribution in case of infinitely long fin is

a) t – t a /t 0 – t a = mx

b) t – t a /t 0 – t a = -mx

c) t – t a /t 0 – t a = e -m x

d) t – t a /t 0 – t a = log 

Answer: c

Explanation: Exponential curve should be here.

3. The rate of heat transfer in case of infinitely long fin is given by

a)  1/2 (t 0 – t a )

b)  1/2 (t 0 – t a )

c)  1/2 (t 0 – t a )

d)  1/2 (t 0 – t a )

Answer: a

Explanation: It should contain all the terms i.e. h, p, k and A.

4. Let us say there are two rods having same dimensions, one made of brass  and the other of copper , having one of their ends inserted into a furnace. At a section 10.5 cm away from the furnace, the temperature of brass rod is 120 degree Celsius. Find the distance at which the same temperature would be reached in the copper rod? Both ends are exposed to the same environment

a) 12.54 cm

b) 45.87 cm

c) 12.34 cm

d) 22.05 cm

Answer: d

Explanation: For brass rod, 120 = t a + (t 0 – t a ) e –m l and for copper rod, 120 = t a + (t 0 – t a ) e –M L So L = L 0 (k 2 /k 1 ) = 22.05 cm.

5. Three rods, one made of silver , second made of aluminum  and the third made of iron  are coated with a uniform layer of wax all around. The rods are placed vertically in a boiling water bath with 250 mm length of each rod projecting outside. If all the rods are having following dimensions i.e. diameter = 15 mm and length = 300 mm and have identical surface coefficient 12.5W/ m 2 K, work out the ratio of lengths up to which wax will melt on each rod

a) 2.45:1:1.732

b) 1.732:1:2.45

c) 2.45:1.732:1

d) 1.732:1:2.45

Answer: c

Explanation: k 1 /l 1 2 = k 2 /l 2 2 = k 3 /l 3 2 .

6. Let us assume there are two pieces of copper wire 0.1625 cm in diameter with a device that melts it at 195 degree Celsius. The wires are positioned vertically in air at 24 degree Celsius and the heat transfer coefficient of the wire is 17 W/ square m K. Let us say k = 335W/m K i.e. of wire. Find out the energy input?

heat-transfer-questions-answers-experienced-q6

a) 1.234W

b) 2.652W

c) 4.562W

d) 9.435W

Answer: b

Explanation: A C = π D 2 /4 = 2.073 * 10 –6 square meter. P = π D = 0.0051 m. Q FIN = k A C m (t 2 – t 1 ) = 1.326W.

7. A rod of 10 mm square section and 160 mm length with thermal conductivity of 50W/m K protrudes from a furnace wall at 200 degree Celsius with convective coefficient 20 W/ square m K. Make calculations for the heat convective up to 80 mm length

a) 6.84W

b) 7.34W

c) 4.54W

d) 5.47W

Answer: a

Explanation: Q = k A C m (t 2 – t 1 ), m = (P h/k A C ) 1/2 = 12.649 /m, so Q = 10.75W. At x = 80 mm,  = 1.01192, so T – 30/200 – 30 = 0.3635. Therefore net heat is 10.75 – k A C m (t 0.08 – t a ) = 6.84W.

8. A fin protrudes from a surface which is held at a temperature higher than that of its environment. The heat transferred away from the fin is

a) Heat escaping from the tip of the fin

b) Heat conducted along the fin length

c) Convective heat transfer from the fin surface

d) Sum of heat conducted along the fin length and that convected from the surface

Answer: c

Explanation: As the temperature is higher, so it’s convective.

9. The value of correction length for equilateral fin is

a) L C = 2 L + a/4 1/2

b) L C = L + a/4 1/2

c) L C = 3 L + a/4 1/2

d) L C = 6 L + a/4 1/2

Answer: b

Explanation: Area of triangle i.e. equilateral is  1/2 /4.

10. The parameter m = (h P/k A C ) 1/2 has been stated to increase in a long fin. If all other parameters are constant, then

a) Profile of temperature will remain the same

b) Along the length temperature drop will be less

c) The parameter influences the heat flow only

d) The temperature drop along the length will be steeper

Answer: d

Explanation: For an infinitely long fin t – t a /t 0 – t a = e – m x . Dimensionless temperature falls more with increase in factor m.

This set of Heat Transfer Interview Questions & Answers focuses on “Heat Dissipation From A Fin Insulated At The Tip”.

1. The relevant boundary conditions in case of heat dissipation from a fin insulated at the tip are

a) t = t 0 at x = 0 and d t/d x = 0 at x = 0

b) t = t 0 at x = 0 and d t/d x = 0 at x = 1

c) t = t 0 at x = 1 and d t/d x = 0 at x = 1

d) t = t 0 at x = infinity and d t/d x = 0 at x = infinity

Answer: b

Explanation: It should be at x = 0 and x = 1 respectively.

2. The temperature distribution in case of fin insulated at the tip is given by

a) t – t 0 /t 0 – t a = cos h m /cos ml

b) t – t 0 /t 0 – t a = cos h m /sin h ml

c) t – t 0 /t 0 – t a = cos h m /cos h ml

d) t – t 0 /t 0 – t a = cos m /sin ml

Answer: c

Explanation: It should contain cos h term and  term.

3. The rate of heat transfer from the fin in case of fin insulated at the tip is

a)  1/2 (t 0 – t a ) tan h ml

b)  1/2 (t 0 – t a ) tan h ml

c)  1/2 (t 0 – t a ) tan h ml

d)  1/2 (t 0 – t a ) tan h ml

Answer: a

Explanation: It should contains all the terms i.e. h, A, P, k.

4. “Fin is insulated at the tip”. What does that mean?

a) Less heat is transferred from the tip

b) Heat will transferred from tip only

c) More heat is transferred from the tip

d) No heat is transferred from the tip

Answer: d

Explanation: The fin is of finite length with the tip insulated and so no heat is transferred from the tip.

5. Find the heat transfer rate from a hot surface for 6 fins of 10 cm length? The base temperature of the fin is maintained at 200 degree Celsius and the film is exposed to a convection environment at 15 degree Celsius with convective coefficient 25W/square m K. Each fin has cross-sectional area 2.5 square centimeter and is made of a material having thermal conductivity 250W/m K

a) 120.34W

b) 130.18W

c) 145.46W

d) 165.43W

Answer: b

Explanation: n = 6 and l = 10 cm = 0.1 m, ml = 0.4472, Q = 6[ (2.5 * 10 – 4 )   tan h  = 130.18W.

6. An array of 10 fins of anodized aluminum  is used to cool a transistor operating at a location where the ambient conditions correspond to temperature 35 degree Celsius and convective coefficient 12W/square m K. The distance AB is 3 mm, EF is 0.4 mm. The length of the fin is 5 mm and has its base at 60 degree Celsius. Find the power dissipated by the fin array?

heat-transfer-interview-questions-answers-experienced-q6

a) 8.673W

b) 1.432W

c) 0.786W

d) 0.128W

Answer: c

Explanation: P = 2 = 6.8 mm, A =   = 1.2 square meter, m = 19.44 per meter. So, Q = k A m (t 0 –t a ) tan h ml = 0.0786, therefore heat loss from the array of 10 fins =   = 0.786W.

7. An electronic semiconductor device generates 0.16 k J/hr of heat. To keep the surface temperature at the upper safe limit of 75 degree Celsius, it is desired that the heat generated should be dissipated to the surrounding environment which is at 30 degree Celsius. The task is accomplished by attaching aluminum fins, 0.5 square mm and 10 mm to the surface. Work out the number of fins if thermal conductivity of fin material is 690W/m K and the heat transfer coefficient is 45k J/square m hr K. Neglect the heat loss from the tip of the fin

a) 4

b) 3

c) 2

d) 1

Answer: a

Explanation: P = 2 = 2 mm, A =   = 0.25 square meter. m = 22.85 per meter, so Q = k A m (t 0 –t a ) tan h ml = 39.77 * 10 -3 k J/hr per fin. So number of fins = 0.16/39.77 * 10 -3 = 4.02.

8. A rod of 10 mm diameter and 80 mm length with thermal conductivity 16W/ m K protrudes from a surface at 160 degree Celsius. The rod is exposed to air at 30 degree Celsius with a convective coefficient of 25W/square m K. How does the heat flow from this rod get affected if the same material volume is used for two fins of the same length? Assume short fin with insulated end

a) 12.25 %

b) 25.6 %

c) 23.4 %

d) 21.2 %

Answer: d

Explanation: Case 1 – m 1 = 25 per meter, m 1 l = 25 * 0.08 = 2. Therefore, Q 1 = 3.935W

Case 2 – d =0.00707 m, m 2 = 29.73 per meter, m 2 l = 2.378. Therefore, Q 2 = 2.385W

% increase in heat flow = 4.77 – 3.935/3.935 = 0.21.

9. Two rods A and B of the same length and diameter protrude from a surface at 120 degree Celsius and are exposed at air at 25 degree Celsius. The temperatures measured at the end of the rods are 50 degree Celsius and 75 degree Celsius. If thermal conductivity of material A is 20W/ m K, calculate it for B

a) 31.13W/m K

b) 41.13W/m K

c) 51.13W/m K

d) 61.13W/m K

Answer: c

Explanation: α/ α 0 = t – t a /t 0 – t a = 1/cos h ml. For rod A, cos h m 1 l =3.8. Similarly for rod 2, cos h m 2 l = 1.9, m 1 /m 2 = 1.599. So k 2 = k 1  2 = 51.13W/m K.

10. A centrifugal pump which circulates a hot liquid metal at 500 degree Celsius is driven by a 3600 rpm electric motor. The motor is coupled to the pump impeller by a horizontal steel shaft of dia 25 mm. Let us assume the motor temperature as 60 degree Celsius with the ambient air at 25 degree Celsius, what length of shaft should be specified between the motor and the pump? It may be presumed that the thermal conductivity of the shaft material is 35W/m K and that the convective film coefficient between the steel shaft and the ambient air is 15.7W/square m K

a) 38.96 cm

b) 54.76 cm

c) 23.76 cm

d) 87.43 cm

Answer: a

Explanation: 60 – 25/500 – 25 = cos h m /cos h ml = 1/cos h ml, so ml = 3.3. For a circular shaft of diameter d, P/A = 4/d, m = 8.47 per meter. So, l = 3.3/8.47 = 38.96 cm.

This set of Heat Transfer test focuses on “Fin Performance”.

1. The utility of fin in dissipating a given quantity of heat is generally assessed on the basic of how many parameters?

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: It depends on two parameters i.e. efficiency of fin and effectiveness of fin.

2. A copper steel rod  has been attached to a plane wall which is maintained at a temperature of 350 degree Celsius. The rod is 8 cm long and has the cross-section of an equilateral triangle with each side 5 mm. Determine the heat dissipation from the rod if it is exposed to a convection environment at 25 degree Celsius with unit surface conductance 100 W/m 2 degree. Consider end surface loss to be negligible

a) 10.26 W

b) 9.26 W

c) 8.26 W

d) 7.26 W

Answer: b

Explanation: For a fin of triangular cross-section, P = 3a and m =  ½ = 50.19 per meter.

3. For an infinitely long fin, the efficiency of fin is given by

a) 1/ml

b) 2/ml

c) 3/ml

d) 4/ml

Answer: a

Explanation: On simplify (p h k A C ) 1/2 (t 0 – t a ) tan h /h  (t 0 – t a ), we get it as tan h /ml. And here for very long fin numerator should be equal to 1.

4. For a fin of finite length with an insulated end, the fin efficiency is given as

a) tan h 

b) tan /ml

c) tan h /ml

d) tan h /ml

Answer: d

Explanation: On simplify (h p k A C ) 1/2 (t 1 – t 2 ) tan h /h  (t 1 – t 2 ), we get it as tan h /ml.

5. For an infinitely long fin, the effectiveness of fin is given as

a) (P k/h A C )

b) (P k/h A C ) 3/2

c) (P k/h A C ) 1/2

d) (P k/h A C ) 2

Answer: c

Explanation: Q FIN = (h k P A C ) 1/2 (t 2 – t 1 ), so effectiveness is (h P k A C ) 1/2 (t 2 – t 1 )/h A(t 0 – t a ).

6. For a straight rectangular fin of thickness δ and width b, choose the correct option

a) P/A C = 1/ δ

b) P/A C = 2/ δ

c) P/A C = 3/ δ

d) P/A C = 4/ δ

Answer: b

Explanation: As fin effectiveness in case of straight rectangular fin is  1/2 . So, P/A = 2/b δ.

7. “Effectiveness of fin is the ratio of the fin heat dissipation with fin to that of no fin”.

a) True

b) False

Answer: a

Explanation: Fins are used to enhance heat transfer rate and the use of fin on a surface can’t be recommended unless the enhancement in heat transfer justifies the extra cost and complexity associated with the fins.

8. Three fins of equal length and diameter but made of aluminum, brass and cast iron is heated to 200 degree Celsius at one end. If the fins dissipate heat to the surrounding air at 25 degree Celsius, the temperature at the free end will be least in

a) Brass fin

b) Cast iron fin

c) Aluminum fin

d) Each fin will have the same temperature

Answer: c

Explanation: Thermal conductivity of aluminum is higher than that of others.

9. Two long rods A and B of the same diameter have thermal conductivities k and 4k and have one of their end inserted into a furnace at 400 K. At 9.5 m away section from the furnace, the temperature of rod B is120 degree Celsius. So find out at what end from the furnace end, the same temperature would be reached in the rod A?

a) 0.25 m

b) 0.75 m

c) 0.15 m

d) 0.50 m

Answer: a

Explanation: k A /l A 2 = k B /l B 2 , so it is 0.25 m.

10. The figure shows a 5 cm diameter rod, 90 cm long, which is having its lower face grinded smooth. The remainder of the rod is exposed to 32 degree Celsius room air and a surface coefficient heat transfer equal to 6.8 W/m 2 degree exists between the rod surface and the room air. The grinder dissipates mechanical energy at the rate of 35 W. If thermal conductivity of rod material is 41.5 W/m degree, find the temperature of the rod at the point where the grinding is taking place

heat-transfer-test-q10

a) 161.45 degree celsius

b) 151.45 degree celsius

c) 141.45 degree celsius

d) 131.45 degree celsius

Answer: b

Explanation: For a circular rod of diameter d, P/A = π/d and m =  1 ⁄ 2 = 3.62 per meter.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Design Considerations For Fins”.

1. The following factors need consideration for the optimum design of fins

 Cost

 Space considerations

 Weight considerations

Choose the correct option

a) i only

b) i and ii only

c) i, ii and iii

d) ii only

Answer: c

Explanation: The design will be considered optimum when the fins require minimum cost of manufacture are light in weight.

2. A heating unit is made in the form of a vertical tube of 50 mm outside diameter and 1.2 m height. The tube is fitted with 20 steel fins of rectangular section with height 40 mm and thickness 2.5 mm. The temperature at the base of fin is 75 degree Celsius, the surrounding air temperature is 20 degree Celsius and the heat transfer coefficient between the fin as well as the tube surface and the surrounding air is 9.5 W/m 2 K. If thermal conductivity of the fin material is 55 W/m K, find the amount of heat transferred from the tube with fin

a) 1234 .98 W

b) 1004.84 W

c) 6539.83 W

d) 3829.46 W

Answer: b

Explanation: Heat flow rate convicted from the base, Q b = h A b (t 0 – t INFINITY ) and heat flow rate convicted from the fins, Q f = n k A C m (t 0 – t a ).

3. The fins would be effective for heat conduction if the ratio P k/h A C is

a) Greater than 5

b) Less than 5

c) Equal to 5

d) Varies between 2 to 9

Answer: a

Explanation: The ratio perimeter divided by area multiply by length must be greater than 5.

4. Consider the following statements pertaining to heat transfer through fins

 They must be arranged at right angles to the direction of flow of working fluid

 The temperature along the fin is variable and accordingly heat transfer rate varies along the fin elements

 Fins are equally effective irrespective whether they are on the hot side or cold side of the fluid

 Fins are made of materials that have thermal conductivity higher than that of wall

Identify the correct statements

a) iii and iv

b) i and iv

c) ii and iii

d) i and ii

Answer: d

Explanation: The statements made at serial number 3 and 4 are wrong. Fins are located on the side where the convective coefficient has a low value.

5. An increase in fin effectiveness is caused by a high value of

 Convective coefficient

 Thermal conductivity

 Circumference

 Area

Identify the correct statement

a) i and iii

b) iii and iv

c) ii and iv

d) ii and iii

Answer: c

Explanation: Refer to the expression for fin effectiveness, an increase in fin effectiveness is caused by high value of circumference and thermal conductivity.

6. A steel strap is serving as a support for the steam pipe. The strap is welded to the pipe and bolted to the ceiling. The junction between the support strut and the ceiling is adiabatic, and the outside temperature of steam pipe is 105 degree Celsius. The strut AB is 60 cm high and AD = BC = 12.5 cm. It is 0.3 cm thick. Workout the rate at which heat is lost to the surrounding air by the support strut. It may be assumed that thermal conductivity for steel is 45 W/m degree, the total outside surface coefficient is 17 W/m 2 degree and the surrounding air is at 32 degree Celsius

heat-transfer-questions-answers-design-consideration-fins-q6

a) 178 W

b) 168 W

c) 158 W

d) 148 W

Answer: a

Explanation: α x /α 0 = t x – t a /t 0 – t a = cos m /cos ml.

7. Choose the correct option regarding fin efficiency and fin effectiveness

a) 2 Fin effectiveness = A FIN /A B 

b) 3 Fin effectiveness = A FIN /A B 

c) Fin effectiveness = A FIN /A B 

d) ½ Fin effectiveness = A FIN /A B 

Answer: c

Explanation: On simplify the equations of fin efficiency and fin effectiveness we get the result.

8. The handle of a saucepan, 30 cm long and 2 cm in diameter, is subjected to 100 degree Celsius temperature during a certain cooking operation. The average unit surface conductance over the handle surface is 7.35 W/m 2 degree in the kitchen air at 24 degree Celsius. The cook is likely to grasp the last 10 cm of the handle and hence the temperature in this region should not exceed 38 degree Celsius. What should be the thermal conductivity of the handle material to accomplish it? The handle may be treated as a fin insulated at the tip

a) 18.36 W/m degree

b) 17.36 W/m degree

c) 16.36 W/m degree

d) 15.36 W/m degree

Answer: b

Explanation: α x /α 0 = t x –t a /t 0 –t a = cos m /cos ml. Now, for a circular handle of diameter d, P/A = 4/d.

9. Let us assume a square section fin split longitudinally and used as two fins. This will result in

a) Increase or decrease in heat transfer depending on the material of fin

b) Heat flow remains constant

c) Decrease in heat transfer

d) Increase in heat transfer

Answer: d

Explanation: Heat transfer will definitely increases because it split into two fins i.e. more surface area.

10. Mark the false statement regarding effectiveness of fin

a) A high value of film coefficient adversely affects the fin effectiveness

b) Fin effectiveness is improved if the fin is made from a material of low conductivity

c) Fin effectiveness is improved by having thin fins

d) It can also be improved by having closely spaced fins

Answer: b

Explanation: It should be of high thermal conductivity.

This set of Heat Transfer quiz focuses on “Heat Flow Through Triangular And Parabolic Fins”.

1. Which fin yields the maximum heat flow per unit weight?

a) Straight triangular fin

b) Curved triangular fin

c) Parabolic fin

d) Hyperbolic fin

Answer: a

Explanation: In straight triangular fin, there is maximum heat flow.

2. Heat dissipation by every segment of the fin is

a) Sometimes same

b) Same

c) Not same

d) Sometimes same or sometimes not same

Answer: c

Explanation: It is always different as fins are not uniform with respect to cross-sectional area.

3. “If a fin of a constant cross section is used, there would be wastage of material”. Chose the correct option

a) True

b) False

Answer: a

Explanation: Cross section must vary to utilize the material.

4. Which one is true regarding parabolic fin?

a) It dissipates the minimum amount of heat at a minimum material cost

b) It dissipates the minimum amount of heat at a maximum material cost

c) It dissipates the maximum amount of heat at a maximum material cost

d) It dissipates the maximum amount of heat at a minimum material cost

Answer: d

Explanation: In this case, a parabolic fin is of great practical importance.

5. For parabolic fin, the curve follows which law?

a) y = C/x 2

b) y = C x 4

c) y = C x 2

d) y = C x 1/2

Answer: c

Explanation: Equation of parabola is y = 4 x 2 or x = 4 y 2 .

6. The correction length for cylindrical fin is

a) L C = L + d/4

b) L C = 2 L + d/4

c) L C = 3 L + d/4

d) L C = 4 L + d/4

Answer: a

Explanation: Area = π d 2 /4. Where, d is the diameter.

7. Provision of fins on a given heat transfer surface will be more effective if there is

a) Fewer but thick fins

b) Large number of thick fins

c) Fewer but thin fins

d) Large number of thin fins

Answer: d

Explanation: Increase in ratio of perimeter P to be cross sectional area A C brings about improvement in the effectiveness of fins.

8. The heat dissipation at any section of parabolic fin is given by

a) (t 2 – t 1 )  

b) k (t 2 – t 1 )  

c) k (t 2 – t 1 ) 

d) k (t 2 – t 1 ) 

Answer: b

Explanation: Q = q x (A X ) = k (t 2 – t 1 )  .

9. An air cooled cylindrical wall is to be fitted with triangular fins of 3 cm thickness at base and 12 cm in height. The fins are made from stainless steel with density 8000 kg/m 3 and thermal conductivity 17.5 W/m K. The wall temperature is 600 degree Celsius and the fin is exposed to an environment with t a = 30 degree Celsius and h = 20 W/m 2 K. What is the temperature distribution along the fin?

a) t = 10 + 250 I 0 [6.056  1/2 ].

b) t = 20 + 250 I 0 [6.056  1/2 ].

c) t = 30 + 250 I 0 [6.056  1/2 ].

d) t = 40 + 250 I 0 [6.056  1/2 ].

Answer: c

Explanation: α/α 0 = t – t 0 /t 0 – t a = I 0 [2 B  ½ ]/ I 0 [2 B  ½ ]. Here B =  ½ = 3.028.

10. Consider the above problem, make calculations for the rate of heat flow per unit mass of fin material used

a) 126.53 W/kg

b) 154.76 W/kg

c) 134.87 W/kg

d) 165.46 W/kg

Answer: a

Explanation: Q = b  ½ α 0 I 1 [2 B  ½ / I 0 [2 B  ½ = 1822 W. Mass of fin per meter width = 14.4 kg. Therefore rate of heat flow per unit mass = 1822/14.4 = 126.53 W/kg.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Thermometric Well”.

1. The convective coefficient in boilers and condenser is

a) Low

b) High

c) Moderate

d) Depends on time and temperature

Answer: b

Explanation: For low heat transfer rate it is high.

2. In thermometer, heat is dissipated from which end?

a) Back side

b) Front side

c) Middle

d) No heat is dissipated

Answer: a

Explanation: In thermometric well heat dissipation is from the end.

3. The thermometric well is treated as a fin of

a) It is not treated as a fin

b) Infinite length

c) Finite length

d) It depends on the amount of heat dissipation

Answer: c

Explanation: Generally for maximum heat dissipation it is treated as a fin of finite length.

4. Thermometric well apparatus doesn’t consist of

a) Thermometer

b) Pipe line

c) Oil

d) Ammeter

Answer: d

Explanation: It doesn’t consist of an ammeter.

5. For what purpose we use thermometric well?

a) To measure current

b) To measure resistance

c) To measure temperature of gas

d) To stop heat dissipation

Answer: c

Explanation: We use this apparatus for the measurement of a temperature of gas flowing through pipeline.

6. In thermometric well, the error can be minimized by

 The value of parameter ml slows down by some quantity.

 Lagging of the tube so that conduction of heat along its length is arrested.

a) True

b) False

Answer: b

Explanation: Here, we increase the value of parameter ml.

7. Suppose there are two reservoirs A and B. Let T A and T B be the temperatures of respective reservoirs. At any instant say T A = T B . In this case, maximum temperature occurs at

a) Middle

b) Left

c) Right

d) All over the rod

Answer: a

Explanation: Here temperatures are same so x = ½ i.e. middle of the rod.

8. The relevant boundary conditions in case of heat transfer from a bar connected to two heat sources at different temperatures are

a) α = α 1 at x = 0 and α = α 2 at x = infinity

b) α = α 1 at x = 1 and α = α 2 at x = 2L

c) α = α 1 at x = infinity and α = α 2 at x = 1

d) α = α 1 at x = 0 and α = α 2 at x = L

Answer: d

Explanation: It should be at x = o and x = L.

9. “The tube that is used in a thermometric well is considered as a hollow fin and the temperature distribution is obtained by using the relation applicable to infinitely long fin”.

a) True

b) False

Answer: b

Explanation: The relation applicable to fin with insulated tip is used here.

10. “The diameter of well does not have any effect on temperature measurement by the thermometer”.

a) True

b) False

Answer: a

Explanation: Here, there is no relation between well diameter and temperature.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Time Constant”.

1. The time constant of a thermocouple is the time taken to

a) Minimum time taken to record a temperature reading

b) Attain 50% of initial temperature difference

c) Attain the final value to be measured

d) Attain 63.2% of the value of the initial temperature difference

Answer: d

Explanation: The time constant of a thermocouple represents the time required to attain 63.2% value.

2. A thermocouple junction of spherical form is to be used to measure the temperature of the gas stream. The junction is at 20 degree Celsius and is placed in a gas stream which is at 200 degree Celsius. Make calculations for junction diameter needed for the thermocouple to have thermal time constant of one second. Assume the thermos-physical properties as given below

k = 20 W/ m K

h = 350 W/m 2 K

c = 0.4 k J/kg K

p = 8000 kg/m 3

a) 0.556 mm

b) 0.656 mm

c) 0.756 mm

d) 0.856 mm

Answer: b

Explanation: T = p V c/h A = p r c/3h. So, r = 3 h T/p c = 0.000328 m = 0.328 m.

3. A low value of time constant can be achieved for a thermocouple by

 Increasing the wire diameter

 Increasing the value heat transfer coefficient

 Use light metals of low density and low specific heat

a) ii and iii

b) i and iii

c) i and ii

d) i, ii and iii

Answer: a

Explanation: Diameter of wire should be less.

4. Which of the following has units of time constant? 

a) p V/h A

b) p c/h A

c) p V c/h A

d) V c/h A

Answer: c

Explanation: It has the unit of time and is time constant of the system.

5. “Thermal radiation suffers no attenuation in a vacuum”.

a) True

b) False

Answer: a

Explanation: It is gradual loss of intensity of any kind of flux.

6. How does the body temperature falls or rises with time?

a) Logarithmic

b) Parabolic

c) Linear

d) Exponentially

Answer: d

Explanation: The rate depends on the parameter h A/p V c.

7. The lumped parameter solution for transient conduction can be conveniently stated as

a) t – t a /t I – t a = 2 exponential (- B I F 0 )

b) t – t a /t I – t a = exponential (- B I F 0 )

c) t – t a /t I – t a = 3 exponential (- B I F 0 )

d) t – t a /t I – t a = 6 exponential (- B I F 0 )

Answer: b

Explanation: This is the general solution for lumped system parameter.

8. An iron billet  measuring 20 * 15 * 80 cm is exposed to a convective flow resulting in convection coefficient of 11.5 W/m 2 K. Determine the Biot number

a) 0.02376

b) 0.008974

c) 0.004563

d) 0.006846

Answer: d

Explanation: B = h L C /k = 0.006846.

9. A mercury thermometer with bulb idealized as a sphere of 1 mm radius is used for measuring the temperature of fluid whose temperature is varying at a fast rate. For mercury

k = 10 W/m K

α = 0.00005 m 2 /s

h = 10 W/m 2 K

If the time for the temperature change of the fluid is 3 second, what should be the radius of thermocouple to read the temperature of the fluid?

For the thermocouple material

k = 100 W/m K

α = 0.0012 m 2 /s

h = 18 W/m 2 K

a) .864 mm

b) .764 mm

c) .664 mm

d) .564 mm

Answer: a

Explanation: T = k l/h α. So, radius is 0.864 mm.

10. A thermocouple junction of spherical form is to be used to measure the temperature of the gas stream. The junction is at 20 degree Celsius and is placed in a gas stream which is at 200 degree Celsius. Make calculations for junction diameter needed for the thermocouple to have thermal time constant of one second. Assume the thermos-physical properties as given below

k = 20 W/ m K

h = 350 W/m 2 K

c = 0.4 k J/kg K

p = 8000 kg/m 3

a) 0.456 mm

b) 0.556 mm

c) 0.656 mm

d) 0.756 mm

Answer: c

Explanation: T = p V c/h A = p r c/3 h.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Response of a Thermocouple”.

1. “The response of a thermocouple is defined as the time required for the thermocouple to reach the surrounding temperature when it is exposed to it”.

a) True

b) False

Answer: b

Explanation: It is the source temperature.

2. The sensitivity of thermocouple is defined as the time required by thermocouple to reach how much percentage of its steady state values?

a) 43.3

b) 53.2

c) 63.3

d) 73.3

Answer: c

Explanation: The time constant of a thermocouple represents the time required to attain 63.2% value.

3. The response time for different sizes and materials of thermocouple wires usually lie between

a) 0.04 to 2.5 seconds

b) 0.06 to 1.2 seconds

c) 0.02 to 0.04 seconds

d) 2.4 to 9.4 seconds

Answer: a

Explanation: Depending upon the type of fluid used, the response time for different sizes and materials of thermocouple wires usually lie between o.o4 to 2.5 seconds.

4. A thermocouple junction of spherical form is to be used to measure the temperature of the gas stream. The junction is at 20 degree Celsius and is placed in a gas stream which is at 200 degree Celsius. Make calculations for time required by the thermocouple to reach 197 degree Celsius temperature. Assume the thermos-physical properties as given below

k = 20 W/ m K

h = 350 W/m 2 K

c = 0.4 k J/kg K

p = 8000 kg/m 3

a) 1.094 seconds

b) 2.094 seconds

c) 3.094 seconds

d) 4.094 seconds

Answer: d

Explanation: t – t a /t I – t a = exponential .

5. An egg with mean diameter of 4 cm and initially at 25 degree Celsius is placed in an open boiling water container for 4 minutes and found to be boiled at a particular level. For how long should a similar egg boil at the same level, when refrigerator temperature is 5 degree Celsius? Use lumped parameter theory and assume following properties of egg

k = 12 W/m K

h = 125 W/m 2 K

c = 2 k J/kg K

p = 1250 kg/m 3

a) 251.49 seconds

b) 261.49 seconds

c) 271.49 seconds

d) 281.49 seconds

Answer: c

Explanation: T t – T INFINITY /T i – T INFINITY = e – b T .

6. A person is found dead at 5 pm in a room whose temperature is 20 degree Celsius. The temperature of body is measured to be 25 degree Celsius, when found and heat transfer coefficient is estimated to be 8 W/m 2 K. Modeling the body as a 30 cm diameter, 1.7 cm long cylinder. Estimate the time of death of person.

a) 13.55 seconds

b) 12.55 seconds

c) 11.55 seconds

d) 10.55 seconds

Answer: d

Explanation: T t – T INFINITY /T i – T INFINITY = e – b T .

7. The following data pertains to the junction of a thermocouple wire used to measure the temperature of the gas stream.

Density = 8500 kg/m 3 , specific heat = 325 J/kg K, thermal conductivity = 40 W/m K and the heat transfer coefficient between the junction and gas = 215 W/m 2 K.

If thermocouple junction can be approximated as 1 mm diameter sphere, determine how long it will take for the thermocouple to read 99% of the initial temperature difference

a) 9.86 seconds

b) 8.86 seconds

c) 7.86 seconds

d) 6.86 seconds

Answer: a

Explanation: t – t a /t i – t a = exponential .

8. A thermocouple junction in the form of 4 mm radius sphere is to be used to measure the temperature of a gas stream. The junction is initially at 35 degree Celsius and is placed in a gas stream which is at 300 degree Celsius. The thermocouple is removed from the hot gas stream after 10 seconds and kept in still air at 25 degree Celsius with convective coefficient 10 W/m 2 K. Find out the time constant of the thermocouple. Assume the thermos-physical properties as given below

h = 37.5 W/m 2 K

p = 7500 kg/m 3

c = 400 J/kg K

a) 6.67 seconds

b) 106.67 seconds

c) 206.67 seconds

d) 306.67 seconds

Answer: b

Explanation: T = p V c/h A = p r c/3 h = 106.67 seconds.

9. What percentage of water an average human body can have?

a) 52%

b) 62%

c) 72%

d) 82%

Answer: c

Explanation: Average human body is 72% water by mass.

10. Heisler charts are valid if

a) Fourier number is equal to 0.2

b) Fourier number is less than 0.2

c) Fourier number is greater than 0.2

d) Fourier number is equal to 0.4

Answer: c

Explanation: The solution to the transient heat flow in infinite flat plates are available in the form of these charts.

This set of Heat Transfer MCQs focuses on “Transient Heat Conduction Solids With Infinite Thermal Conductivity”.

1. A flat wall of fire clay, 50 cm thick and initially at 25 degree Celsius, has one of its faces suddenly exposed to a hot gas at 950 degree Celsius. If the heat transfer coefficient on the hot side is 7.5 W/m 2 K and the other face of the wall is insulated so that no heat passes out of that face, determine the time necessary to raise the center of the wall to 350 degree Celsius. For fire clay brick

heat-transfer-mcqs-q1

Thermal conductivity = 1.12 W/m K

Thermal diffusivity = 5.16 * 10 -7 m 2 /s

a) 43.07 hours

b) 53.07 hours

c) 63.07 hours

d) 73.07 hours

Answer: a

Explanation: t – t a /t 0 – t a = 0.86. Also, α T/l 2 = 0.32.

2. Glass spheres of 2 mm radius and at 500 degree Celsius are to be cooled by exposing them to an air stream at 25 degree Celsius. Find maximum value of convective coefficient that is permissible. Assume the following property values

Density = 2250 kg/m 3

Specific heat = 850 J/kg K

Conductivity = 1.5 W/m K

a) 245 W/m 2 K

b) 235 W/m 2 K

c) 225 W/m 2 K

d) 215 W/m 2 K

Answer: c

Explanation: l = volume/surface area = r/3. So, h =   /r.

3. The transient response of a solid can be determined by the equation. 

a) – 4 p V c = h A (t – t 0 )

b) – 3 p V c = h A (t – t 0 )

c) – 2 p V c = h A (t – t 0 )

d) – p V c = h A (t – t 0 )

Answer: d

Explanation: It can be determined by relating rate of change of internal energy with conductive heat exchange at the surface.

4. A 2 cm thick steel slab heated to 525 degree Celsius is held in air stream having a mean temperature of 25 degree Celsius. Estimate the time interval when the slab temperature would not depart from the mean value of 25 degree Celsius by more than 0.5 degree Celsius at any point in the slab. The steel plate has the following thermal physical properties

Density = 7950 kg/m 3

C P = 455 J/kg K

K = 46 W/m K

a) 6548 s

b) 6941 s

c) 4876 s

d) 8760 s

Answer: b

Explanation: t – t a /t I – t a = exponential . Now A/V = 100 per meter.

5. An average convective heat transfer coefficient for flow of air over a sphere has been measured by observing the temperature-time history of a 12 mm diameter copper sphere (density = 9000 kg/m 3 and c = 0.4 k J/kg K) exposed to air at 30 degree Celsius. The temperature of the sphere was measured by two thermocouples one located at the center and the other near the surface. The initial temperature of the ball was 75 degree Celsius and it decreased by 10 degree Celsius in 1.2 minutes. Find the heat transfer coefficient

a) 27.46 W/m 2 K

b) 21.76 W/m 2 K

c) 29.37 W/m 2 K

d) 25.13 W/m 2 K

Answer: d

Explanation: t – t a /t I – t a = exponential . So, h = 25.13 W/m 2 K.

6. Transient condition means

a) Conduction when temperature at a point varies with time

b) Very little heat transfer

c) Heat transfer with a very little temperature difference

d) Heat transfer for a short time

Answer: a

Explanation: The term transient or unsteady state designates a phenomenon which is time dependent.

7. Which of the following is not correct in a transient flow process?

a) The state of matter inside the control volume varies with time

b) There can be work and heat interactions across the control volume

c) There is no accumulation of energy inside the control volume

d) The rate of inflow and outflow of mass are different

Answer: c

Explanation: In transient heat conduction there is accumulation of energy inside the control volume.

8. A cylindrical stainless steel  ingot, 10 cm in diameter and 25 cm long, passes through a heat treatment furnace which is 5 meter in length. The initial ingot temperature is 90 degree Celsius, the furnace gas is at 1260 degree Celsius and the combined radiant and convective surface coefficient is 100 W/m 2 K. Determine the maximum speed with which the ingot moves through the furnace if it must attain 830 degree Celsius temperature. Take thermal diffusivity as 0.45 * 10 -5 m 2 /s

a) . 000116 m/s

b) .000216 m/s

c) . 000316 m/s

d) . 000416 m/s

Answer: b

Explanation: t – t a /t I – t a = exponential . Now, A/V = 2/r L = 0.48 per cm. Also, T = 1158.53 second so required velocity is 0.25/1158.53.

9. The curve for unsteady state cooling or heating of bodies is

a) Hyperbolic curve asymptotic both to time and temperature axis

b) Exponential curve asymptotic both to time and temperature axis

c) Parabolic curve asymptotic to time axis

d) Exponential curve asymptotic to time axis

Answer: d

Explanation: α/α 0 = exponential [- h A T/p c V], which represents an exponential curve.

10. What is the wavelength band for TV rays?

a) 1 * 10 3 to 34 * 10 10 micron meter

b) 1 * 10 3 to 2 * 10 10 micron meter

c) 1 * 10 3 to 3 * 10 10 micron meter

d) 1 * 10 3 to 56 * 10 10 micron meter

Answer: b

Explanation: This is the maximum and minimum wavelength for TV rays.

This set of Heat Transfer Multiple Choice Questions & Answers focuses on “Transient Heat Conduction In Solids With Finite Conduction”.

1. Diagram shows transient heat conduction in an infinite plane wall. Identify the correct boundary condition in transient heat conduction in solids with finite conduction

heat-transfer-multiple-choice-questions-answers-q1

a) t = t i at T = 0

b) d t /d x = 1 at x = 0

c) d t /d x = infinity at x = 1

d) d t / d x = infinity at x = 0

Answer: a

Explanation: d t / d x = 0 at x = 0. The solution of controlling differential equation in conjunction with initial boundary conditions would give an expression for temperature variation both with time and position.

2. Let there is some conduction resistance, then temperature becomes a function of

 Biot number

Fourier number

 Dimensionless parameter

a) i and ii

b) ii and iii

c) i and iii

d) i, ii and iii

Answer: d

Explanation: It should be the function of all of the above i.e. Biot number, Fourier number and dimensionless parameters which includes all the dimensionless numbers.

3. The value of Biot number and Fourier number, as used in the Heisler charts, are evacuated on the basis of a characteristics parameter s which is the thickness in case of plates and the surface radius in case of cylinders and spheres.

a) True

b) False

Answer: b

Explanation: It must be semi thickness instead of thickness.

4. A large steel plate 50 mm thick is initially at a uniform temperature of 425 degree Celsius. It is suddenly exposed on both sides to an environment with convective coefficient 285 W/m 2 K and temperature 65 degree Celsius. Determine the center line temperature.

For steel, thermal conductivity = 42.5 W/m K and thermal diffusivity = 0.043 m 2 /hr

a) 261 degree Celsius

b) 271 degree Celsius

c) 281 degree Celsius

d) 291 degree Celsius

Answer: c

Explanation: t 0 – t a /t I – t a = 0.6.

5. With respect to above problem, determine the temperature inside the plate 12.5 mm from the mid plane after 3 minutes

a) 272.36 degree Celsius

b) 262.36 degree Celsius

c) 252.36 degree Celsius

d) 22.35 degree Celsius

Answer: a

Explanation: x/l = 0.5 and t 0 – t a /t I – t a = 0.96.

6. When Biot number exceeds 0.1 but is less than 100, use is made of Heislers charts for the solution of transient heat conduction.

a) True

b) False

Answer: a

Explanation: The Heisler charts are extensively used to determine the temperature distribution and heat flow rate when both conduction and convection resistance are of equal importance.

7. In transient heat conduction, the two significant dimensionless parameters are

a) Reynolds number and Fourier number

b) Reynolds number and Biot number

c) Reynolds number and Prandtl number

d) Biot number and Fourier number

Answer: d

Explanation: These two are dimensionless numbers. Biot number is given by the ratio of internal conduction resistance to the surface convection resistance whereas Fourier number signifies the degree of penetration of heating or cooling effect through a solid.

8. A 12 cm diameter cylindrical bar, initially at a uniform temperature of 40 degree Celsius, is placed in a medium at 650 degree Celsius with convective coefficient of 22 W/m 2 K. Determine the time required for the center to reach 255 degree Celsius. For the material of the bar:

Thermal conductivity = 20 W/m K

Density = 580 kg/m 3

Specific heat = 1050 J/kg K

a) 1234.5 seconds

b) 1973.16 seconds

c) 3487.3 seconds

d) 2896.4 seconds

Answer: b

Explanation: 1/B I = k/h R = 0.1515, t – t a /t I – t a = 0.647. X / l = 0 . Therefore, α T/R 2 = 0.18.

9. Consider the above problem, calculate the temperature of the surface at this instant

a) 476.4 degree Celsius

b) 453.5 degree Celsius

c) 578.9 degree Celsius

d) 548.6 degree Celsius

Answer: c

Explanation: r/R = 1 and 1/B I = 0.1515. t 0 – t a /t I – t a = 0.18.

10. A solid which extend itself infinitely in all directions of space is termed as finite solid.

a) True

b) False

Answer: b

Explanation: It is known as infinite solid. This type of solid can extend itself in x-direction, y-direction and z-direction.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Biot Number”.

1. A gold ring  measuring 15 * 10 * 60 cm is exposed to a surface where h = 11.5 W/m 2 K. Find the value of biot number

a) 0.68

b) 0.58

c) 0.48

d) 0.38

Answer: c

Explanation: Biot number = h l/k = 0.48.

2. In the lumped system parameter model, the variation of temperature with time is

a) Linear

b) Exponential

c) Sinusoidal

d) Cube

Answer: b

Explanation: t – t a /t I – t a = exponential [-h A T/p V c].

3. Which of the following dimensionless number gives an indication of the ratio of internal  resistance to the surface  resistance?

a) Biot number

b) Fourier number

c) Stanton number

d) Nusselt number

Answer: a

Explanation: It is the ratio of conduction resistance to that of convective resistance.

4. Lumped parameter analysis for transient heat conduction is essentially valid for

a) B I < 0.1

b) 1 < B I < 10

c) 0.1 < B I < 0.5

d) It tends to infinity

Answer: a

Explanation: It is generally accepted that lump system analysis is applicable if Biot number is less than 0.1.

5. In the non-dimensional Biot number, the characteristics length is the ratio of

a) Perimeter to surface area of solid

b) Surface area to perimeter of solid

c) Surface area to volume of solid

d) Volume of solid to its surface area

Answer: d

Explanation: We introduced characteristics length for lump system analysis.

6. During heat treatment, cylindrical pieces of 25 mm diameter, 30 mm height and at 30 degree Celsius are placed in a furnace at 750 degree Celsius with convective coefficient 80 W/m 2 degree. Calculate the time required to heat the pieces to 600 degree Celsius. Assume the following property values

Density = 7850 kg /m 3

Specific heat = 480 J/kg K

Conductivity = 40 W/m degree

a) 226 sec

b) 326 sec

c) 426 sec

d) 526 sec

Answer: b

Explanation: t – t a /t I – t a = exponential .

7. The quantity h L C /k is known as

a) Biot number

b) Fourier number

c) Stanton number

d) Nusselt number

Answer: a

Explanation: Biot number = conduction resistance/convection resistance.

8. For a plat plate  the heat exchange occurs from both the sides. The characteristics length is equal to

a) δ/4

b) δ/3

c) δ/2

d) δ

Answer: c

Explanation: l = δ b h/2 b h = δ/2.

9. Fourier number is given by

a) α T/L C 2

b) 2 α T/L C 2

c) 3 α T/L C 2

d) 4 α T/L C 2

Answer: a

Explanation: It signifies the degree of penetration of heating or cooling effect through a solid. Where, α is thermal diffusivity, T is time constant and L C is characteristics length

10. Identify the correct relation between Biot number and Fourier number

a) 4 b T = B I F 0

b) 2 b T = B I F 0

c) 3 b T = B I F 0

d) b T = B I F 0

Answer: d

Explanation: B I = h L C /k and F 0 = α T/L C 2 .

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Lump System Analysis”.

1. According to lumped system analysis, solid possesses thermal conductivity that is

a) Infinitely large

b) Infinitely small

c) Moderate

d) 50% small

Answer: a

Explanation: Solutions to the many of the transient heat flow problems are obtained by the lumped system parameter analysis.

2. The temperature and rate of heat conduction are undoubtedly dependent on

a) Time coordinates

b) Space coordinates

c) Mass coordinates

d) Both time and space coordinates

Answer: d

Explanation: It should depend on both time coordinates and space coordinates.

3. Glass spheres of 2 mm radius and at 500 degree Celsius are to be cooled by exposing them to an air stream at 25 degree Celsius. Find the minimum time required for cooling to a temperature of 60 degree Celsius. Assume the following property values

Density = 2250 kg/m 3

Specific heat = 850 J/kg K

Conductivity = 1.5 W/m K

a) 13.78 seconds

b) 14.78 seconds

c) 15.78 seconds

d) 16.78 seconds

Answer: b

Explanation: t – t a /t I – t a = exponential .

4. Which is true regarding lumped system analysis?

 Conductive resistance = 0

 Convective resistance = 0

 Thermal conductivity = 0

 Thermal conductivity = infinity

Identify the correct statements

a) i and ii

b) i, ii and iv

c) i and iv

d) ii and iv

Answer: c

Explanation: Solids have infinite thermal conductivities. It implies that internal conductance resistance is very low.

5. Which of the following is an example of lump system analysis?

a) Heating or cooling of fine thermocouple wire due to change in ambient temperature

b) Heating of an ingot in an furnace

c) Cooling of bars

d) Cooling of metal billets in steel works

Answer: a

Explanation: Others are the examples of non-periodic variation.

6. What is the criterion for the applicability of lump system analysis?

a) Mean length

b) Normal length

c) Characteristics length

d) Mass no

Answer: c

Explanation: The first set in establishing criteria for the applicability of lump system analysis is to define a characteristics length.

7. What is the value of characteristics length for cylinder?

a) R/5

b) R/4

c) R/3

d) R/2

Answer: d

Explanation: π R 2 L/2 π R L = R/2.

8. During heat treatment, cylindrical pieces of 25 mm diameter, 30 mm height and at 30 degree Celsius are placed in a furnace at 750 degree Celsius with convective coefficient 80 W/m 2 degree. Find the value of biot number if thermal conductivity is 40 W/m degree

a) 0.0082

b) 0.0072

c) 0.0062

d) 0.0052

Answer: a

Explanation: For a cylindrical piece, the characteristic linear dimension is, l = volume/surface area = .00441 m. So, biot number = hl/k = -.00882.

9. What is the value of characteristics length for sphere?

a) R/2

b) R/3

c) R/4

d) R/5

Answer: b

Explanation: 4/3 π R 3 /4 π R 2 = R/3.

10. What is the value of characteristics length for cube?

a) L/3

b) L/4

c) L/5

d) L/6

Answer: d

Explanation: L 3 /6 L 2 = L/6.

This set of Heat Transfer online test focuses on “Transient Heat Conduction In Infinite Thick Solids”.

1. “An infinite solid is one which extends itself infinitely in all directions of space”. Identify the correct option

a) True

b) False

Answer: a

Explanation: If the infinite solid is split in the middle by the plane, each half is known as semi-infinite solid.

2. The boundary conditions in case of transient heat conduction in infinite thick solids are

 t  = t i

 t  = t a for T greater than zero

 t  = t i for T greater than zero

Identify the correct statements

a) i and ii

b) i and iii

c) ii and iii

d) i, ii and iii

Answer: d

Explanation: Boundary conditions are those by which we could find out the values of constant.

3. The perturbation time varies as

a) d 2 /α

b) 2 d 2 /α

c) 3 d 2 /α

d) 4 d 2 /α

Answer: a

Explanation: At penetration depth d, there will be 1% perturbation.

4. The temperature perturbation at all the surface has penetrated to the depth

a) 1.6  1/2

b) 2.6  1/2

c) 3.6  1/2

d) 4.6  ½

Answer: c

Explanation: At penetration depth d, there will be 1% perturbation at a time t.

5. A water line is buried underground in dry soil that has an assumed initial temperature of 4.5 degree Celsius. The pipe may have no flow through it for long period of time, yet it will not be drained in order that no freezing occurs, the pipe must be kept at a temperature not lower than 0 degree Celsius. The pipe is to be designed for a 30 hour period at the beginning of which the soil surface temperature suddenly drops to – 17 degree Celsius. Workout the minimum earth covering needed above the water pipe so as to prevent the possibility of freezing during 36 hour cold spell. The soil in which the pipe is buried has the following properties

heat-transfer-online-test-q5

Density = 640 kg/m 3

Specific heat = 1843J/kg degree

Thermal conductivity = 0.345 W/m degree

a) 0.25 m

b) 0.35 m

c) 0.45 m

d) 0.55 m

Answer: b

Explanation: t – t a /t i – t a = erf [x/2  ½ ].

6. At the penetration depth d, there will be 1% perturbation at a time given by

a) 4 d 2 /13 α

b) 3 d 2 /13 α

c) 2 d 2 /13 α

d) d 2 /13 α

Answer: d

Explanation: d/2  ½ = 1.8.

7. A large steel ingot, which has been uniformly heated to 750 degree Celsius, is hardened by quenching it in an oil bath that is maintained at 25 degree Celsius. What length of time is required for the temperature to reach 600 degree Celsius at a depth of 1 cm? Thermal diffusivity for the steel ingot is 1.21 * 10 -5 m 2 /s. The ingot may be approximated as a flat plate

a) 4.55 seconds

b) 3.55 seconds

c) 2.55 seconds

d) 1.55 seconds

Answer: c

Explanation: t – t a /t I – t a = erf [x/2  ½ ].

8. A mild steel plate 5 cm thick and initially at 40 degree Celsius temperature is suddenly exposed on one side to a fluid which causes the surface temperature to increase to and remain at 90 degree Celsius. Calculate maximum time that the slab be treated as a semi-infinitely body

For steel, thermal diffusivity = 1.25 * 10 – 5 m 2 /s

a) 100 seconds

b) 200 seconds

c) 300 seconds

d) 400 seconds

Answer: b

Explanation: T MAX = δ 2 /4 α  2 = 200 seconds.

9. Consider the above problem, find the temperature at the center of the slab one minute after the change in surface temperature

a) 66 degree Celsius

b) 76 degree Celsius

c) 86 degree Celsius

d) 96 degree Celsius

Answer: a

Explanation: t – t a /t I – t a = erf [x/2  ½ ]. Therefore, temperature at the center of the slab is t a + 0.48 (t I – t a ).

10. Water pipes are to be buried underground in a wet soil (thermal diffusivity = 2.78 * 10 -3 m 2 /s) which is initially at 4.5 degree Celsius. The soil surface temperature suddenly drops to -5 degree Celsius and remains at this value for 10 hours. Calculate the minimum depth at which the pipe be laid if the surrounding soil temperature is to remain above 0 degree Celsius. The soil may be considered as semi-infinite solid

a) 0.467 m

b) 0.367 m

c) 0.267 m

d) 0.167 m

Answer: d

Explanation: t – t a /t I – t a = erf [x/2  ½ ]. Thus x = 2   ½ = 0.167 m.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Periodic Variation”.

1. When the surface temperature variation inside a solid are periodic in nature, the profile of temperature variation with time may assume

a) Triangular

b) Linear

c) Parabolic

d) Hyperbolic

Answer: a

Explanation: Any type of waveform can be analyzed and resolved into an infinite number of sine and cosine waves.

2. The surface temperature oscillates about the mean temperature level in accordance with the relation

a) α S,T – α S,A = 2 sin 

b) α S,T – α S,A = 5 sin 

c) α S,T – α S,A = sin 

d) α S,T – α S,A = 3 sin 

Answer: c

Explanation: α S,T = t S,T – t M .

3. The temperature variation of a thick brick wall during periodic heating or cooling follows a sinusoidal waveform. During a period of 24 hours, the surface temperature ranges from 25 degree Celsius to 75 degree Celsius. Workout the time lag of the temperature wave corresponding to a point located at 25 cm from the wall surface. Thermo-physical properties of the wall material are; thermal conductivity = 0.62 W/m K; specific heat = 450 J/kg K and density = 1620 kg/m 3

a) 3.980 hour

b) 6.245 hour

c) 2.648 hour

d) 3.850 hour

Answer: b

Explanation: d T = x/2  ½ where x = 0.25 m and n = frequency.

4. A single cylinder 2-stroke engine operates at 1500 rpm. Calculate the depth where the temperature wave due to variation in cylinder is damped to 1% of its surface value. For the cylinder material, thermal diffusivity = 0.042 m 2 /hr

a) 0.1996 cm

b) 0.3887 cm

c) 0.2774 cm

d) 0.1775 cm

Answer: d

Explanation: α X,A = α S,A exponential [-x  ½ ] where frequency = 1500 * 60.

5. The temperature distribution at a certain time instant through a 50 cm thick wall is prescribed by the relation

T = 300 – 500 x – 100 x 2 + 140 x 3

Where temperature t is in degree Celsius and the distance x in meters has been measured from the hot surface. If thermal conductivity of the wall material is 20 k J/m hr degree, calculate the heat energy stored per unit area of the wall

a) 4100 k J/hr

b) 4200 k J/hr

c) 4300 k J/hr

d) 4400 k J/hr

Answer: a

Explanation: d t/d x = -500 + 200 x + 420 x 2 . Now heat storage rate = Q IN – Q OUT = 10000 – 5900 = 4100 k J/hr.

6. A large plane wall, 40 cm thick and 8 m 2 area, is heated from one side and temperature distribution at a certain time instant is approximately prescribed by the relation

T = 80 – 60 x +12 x 2 + 25 x 3 – 20 x 4

Where temperature t is in degree Celsius and the distance x in meters. Make calculations for heat energy stored in the wall in unit time.

For wall material:

Thermal conductivity = 6 W/m K and thermal diffusivity = 0.02 m 2 /hr.

a) 870.4 W

b) 345.6 W

c) 791.04 W

d) 238.5 W

Answer: c

Explanation: Q IN = – k A  X = 0 = 2880 W and Q OUT = – k A  X = 0.4 = 2088.96 W.

7. Consider the above problem, calculate rate of temperature change at 20 cm distance from the side being heated

a) 0.777 degree Celsius/hour

b) 0.888 degree Celsius/hour

c) 0.999 degree Celsius/hour

d) 0.666 degree Celsius/hour

Answer: b

Explanation: d t/d T = α d 2 t/d x 2 = 0.888 degree Celsius/hour.

8. At a certain time instant, the temperature distribution in a long cylindrical fire tube can be represented approximately by the relation

T = 650 + 800 r – 4250 r 2

Where temperature t is in degree Celsius and radius r is in meter. Find the rate of heat flow such that the tube measures: inside radius 25 cm, outside radius 40 cm and length 1.5 m.

For the tube material

K = 5.5 W/m K

α = 0.004 m 2 /hr

a) 3.672 * 10 8 W

b) 3.672 * 10 2 W

c) 3.672 * 10 5 W

d) – 3.672 * 10 5 W

Answer: d

Explanation: Q = – k A , Rate of heat storage = Q IN – Q OUT = – 3.672 * 10 5 W.

9. Consider he above problem, find the rate of change of temperature at the inside surface of the tube

a) – 35 degree Celsius/hour

b) – 45 degree Celsius/hour

c) – 55 degree Celsius/hour

d) – 65 degree Celsius/hour

Answer: c

Explanation: d t/d T = α [d 2 t/d r 2 + d t/r d r] = – 55 degree Celsius/hour.

10. Time lag is given by the formula

a) x/2 [1/  ½ ].

b) x/3 [1/  ½ ].

c) x/4 [1/  ½ ].

d) x/5 [1/  ½ ].

Answer: a

Explanation: The time interval between the two instants is called the time lag.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Transmissivity”.

1. The value of transmissivity may vary from

a) 0-1

b) 1-2

c) 3-4

d) 4-5

Answer: a

Explanation: This is a very small quantity. The response of the body to incident radiations is, however, completely independent of and unaffected by the simultaneous emission from the body.

2. Of the radiant energy 350W/m 2 incident upon a surface 250W/m 2 is absorbed, 60W/m 2 is reflected and the remainder is transmitted through the surface. Workout the value for absorptivity for the surface material

a) 0.113

b) 0.114

c) 0.115

d) 0.116

Answer: c

Explanation: Transmissivity = fraction of total energy transmitted by the body = 350 – /350 = 0.115.

3. Transmissivity can also be defined as a ratio of transmitted radiation to that of incident energy flow.

a) True

b) False

Answer: b

Explanation: It is the ratio of incident energy flow to that of transmitted radiations.

4. A cavity with a small hole will always behave as a

a) White body

b) Transparent body

c) Black body

d) Opaque body

Answer: c

Explanation: No radiation passing in and out will have any effect upon the equilibrium of radiation inside the cavity.

5. What is the wavelength band of gamma rays?

a) 1 * 10 -7 to 4 * 10 -4 micron meter

b) 2 * 10 -7 to 2.4 * 10 -4 micron meter

c) 3 * 10 -7 to 3.4 * 10 -4 micron meter

d) 4 * 10 -7 to 1.4 * 10 -4 micron meter

Answer: d

Explanation: This is the maximum and minimum wavelength for gamma rays.

6. Transmissivity is defined as

a) Fraction of total energy transmitted by the body

b) Fraction of total energy reflected by the body

c) Fraction of total energy absorbed by the body

d) Fraction of total energy absorbed and radiated by the body

Answer: a

Explanation: It is a fraction of total energy transmitted by the body.

7. A polished metal pipe 5 cm outside diameter and 370 K temperature at the outer surface is exposed to ambient conditions at 295 K temperature. The emissivity of the surface is 0.2 and the convection coefficient of heat transfer is 11.35 W/m 2 degree. Calculate the overall coefficient of heat transfer by the combined mode of convection and radiation

a) 11.04 W/m 2 degree

b) 12.04 W/m 2 degree

c) 13.04 W/m 2 degree

d) 14.04 W/m 2 degree

Answer: c

Explanation: The total heat exchange can be expressed as, Q = U A d t where U is the overall coefficient of heat transfer.

8. Radiant energy with an intensity of 800 W/m 2 strikes a flat plate normally. The absorptivity is thrice the reflectivity and twice the transmissivity. Determine the rate of transmission

a) 155.47 W/m 2

b) 145.47 W/m 2

c) 135.47 W/m 2

d) 125.47 W/m 2

Answer: b

Explanation: Rate of transmission = T Q 0 = 145.47 W/m 2 .

9. With respect to incident radiation, transmissivity varies with

a) Time

b) Temperature

c) Surface area

d) Wavelength

Answer: d

Explanation: For transparent body, transmissivity is equal to unity.

10. A body through which all the incident radiations passes, is called

a) Opaque body

b) Black body

c) Transparent body

d) White body

Answer: c

Explanation: For transparent body, transmissivity is unity.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Reflectivity”.

1. Reflectivity is defined as

a) Fraction of total energy transmitted by the body

b) Fraction of energy reflected by the body

c) Fraction of total energy absorbed by the body

d) Fraction of total energy absorbed and radiated by the body

Answer: b

Explanation: It is total energy reflected by the body. Its value depends upon the nature of the surface of the body, its temperature and wavelength of incident radiations.

2. Of the radiant energy 350W/m 2 incident upon a surface 250W/m 2 is absorbed, 60W/m 2 is reflected and the remainder is transmitted through the surface. Workout the value for reflectivity for the surface material

a) 0.181

b) 0.171

c) 0.161

d) 0.151

Answer: b

Explanation: Reflectivity = fraction of total energy reflected by the body = 60/350 = 0.171.

3. A thin metal plate of 4 cm diameter is suspended in atmospheric air whose temperature is 290 K. This plate attains a temperature of 295 K when one of its face receives radiant energy from a heat source at the rate of 2 W. If heat transfer coefficient on both surfaces of the plate is stated to be 87.5 W/m 2 K, workout the energy lost by reflection

a) 0.8 W

b) 0.3 W

c) 0.9 W

d) 0.10 W

Answer: c

Explanation: Heat loss by convection from both sides of the plates = 2 h A d t = 1.1 W. Energy lost by reflection = 2.0 – 1.1 = 0.9 W.

4. The value of reflectivity depends upon

 Body temperature

 Wavelength of reflected radiation

 Nature of surface of body

Identify the correct statement

a) i and ii

b) ii and iii

c) ii only

d) i and iii

Answer: d

Explanation: It should be wavelength of incident radiation. It is the fraction of total energy reflected by the body.

5. When a surface absorbs a certain fixed percentage of striking radiation the surface is called

a) Grey body

b) Black body

c) White body

d) Opaque body

Answer: a

Explanation: It is absorbing a certain fixed amount of radiations.

6. Which one is true regarding gases?

a) Reflectivity is unity

b) Absorptivity is zero

c) Reflectivity is zero

d) Transmissivity is zero

Answer: c

Explanation: Gases are known to reflect very little of the radiation incident on their surface.

7. Thin glass plate is an example of

a) Transparent body

b) Opaque body

c) Black body

d) White body

Answer: a

Explanation: For a transparent body, transmissivity is zero.

8. Radiant energy with an intensity of 800 W/m 2 strikes a flat plate normally. The absorptivity is thrice the reflectivity and twice the transmissivity. Determine the rate of reflection

a) 518.20 W/m 2

b) 418.20 W/m 2

c) 318.20 W/m 2

d) 218.20 W/m 2

Answer: d

Explanation: Rate of reflection = p Q 0 = 218.20 W/m 2 .

9. The total radiant energy A impinging upon a body would be partially or totally absorbed, reflected from its surface or transmitted through it. Identify the correct statement

heat-transfer-questions-answers-reflectivity-q9

a) B is the energy that is absorbed by the body

b) B is the energy that is reflected by the body

c) C is the energy that is absorbed by the body

d) D is the energy that is transmitted by the body

Answer: b

Explanation: A = B + C + D.

10. Which one is true for an opaque body?

a) Transmissivity is zero

b) Reflectivity is zero

c) Absorptivity is zero

d) Reflectivity is unity

Answer: a

Explanation: For the opaque body, sum of absorptivity and reflectivity is unity.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Absorptivity”.

1. Energy strikes a vertical hemispherical plate with an intensity of 640 W/m 2 . The absorptivity is thrice the transmissivity and twice the reflectivity. Determine the rate of transmission

a) 236.74 W/m 2

b) 116.37 W/m 2

c) 187.87 W/m 2

d) 456.09 W/m 2

Answer: b

Explanation: Rate os transmission =  /3 = 116.37 W/m 2 .

2. The absorptivity of a surface depends upon

 Direction of reflected radiation

 Temperature of the surface

 Composition

Identify the correct statements

a) i, ii and iii

b) i and iii

c) ii only

d) ii and iii

Answer: d

Explanation: It depend upon the direction of incident radiation. Absorptivity is the fraction of total energy absorbed by the body.

3. What is the absorptivity of the gray body?

a) Below unity

b) Unity

c) 2

d) 2.5

Answer: a

Explanation: Gray body absorbs a certain fixed percentage of impinging radiations.

4. A body that reflects all the incident thermal radiations is called a

a) Opaque body

b) Gases

c) Black body

d) Specular body

Answer: d

Explanation: In this case reflection is regular. For such bodies reflectivity is unity and transmissivity is equal to zero.

5. Energy strikes a vertical hemispherical plate with an intensity of 640 W/m 2 . The absorptivity is thrice the transmissivity and twice the reflectivity. Determine the rate of absorption

a) 449.12 W/m 2

b) 349.12 W/m 2

c) 249.12 W/m 2

d) 149.12 W/m 2

Answer: b

Explanation: Rate of absorption =   = 349.12 W/m 2 .

6. Let 220 W/m 2 of radiant energy is absorbed by a convex surface, 90 W/m 2 is reflected and 40 W/m 2 is transmitted through it. What is the value of absorptivity?

a) 0.72

b) 0.62

c) 0.52

d) 0.42

Answer: b

Explanation: It is the ratio of incident energy flow to that of absorbed radiations. Absorptivity = 220/ = 0.62.

7. For a grey surface

a) Reflectivity equals emissivity

b) Emissivity equals transmissivity

c) Absorptivity equals reflectivity

d) Emissivity is constant

Answer: d

Explanation: Emissivity is constant only for grey surfaces because grey surface radiates much more than any of the other surface.

8. Which quantity can be neglected for gases?

a) Transmissivity

b) Reflectivity

c) Absorptivity

d) None can be neglected

Answer: b

Explanation: Gases are known to reflect very little of the radiation incident on their surface.

9. Absorptivity of a body is equal to its emissivity

a) Under thermal equilibrium conditions

b) For a polished surface

c) At one particular temperature

d) At shorter wavelengths

Answer: a

Explanation: In most case absorptivity of a body is equal to its emissivity under thermal equilibrium conditions.

10. Electromagnetic waves strikes a hot body maintained at 900 degree Celsius which has a reflectivity of 0.66 and a transmissivity of 0.022. Let the absorbed flux be 60 W/m 2 . Determine the rate of incident flux

a) 168.67 W/m 2

b) 178.67 W/m 2

c) 188.67 W/m 2

d) 198.67 W/m 2

Answer: c

Explanation: Absorptivity + Reflectivity + Transmissivity = 1. So, incident flux = 60/0.318 = 188.67 W/m 2 .

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Black Body”.

1. Radiation heat transfer is characterized by

a) Due to bulk fluid motion, there is a transport of energy

b) Thermal energy transfer as vibrational energy in the lattice structure of the material

c) Movement of discrete packets of energy as electromagnetic waves

d) There is circulation of fluid by buoyancy effects

Answer: c

Explanation: EM waves is characterized by radiant heat transfer.

2. Which is true regarding radiation?

a) Radiation travels only in medium

b) Radiation travels without any medium

c) Radiation travels in medium but sometimes without medium

d) Radiation travels in medium or without medium

Answer: d

Explanation: It is true that radiation travels in medium or without medium.

3. Radiation exchange occurs in

a) Solid

b) Vacuum

c) Liquid

d) Gas

Answer: b

Explanation: It occurs more effectively only in vacuum.

4. Energy released by a radiating surface is not continuous but is in the form of successive and separate packets of energy called

a) Photons

b) Protons

c) Electrons

d) Neutrons

Answer: a

Explanation: The photons are propagated through space as rays.

5. The electromagnetic waves are emitted as a result of

 Vibrational movement

 Rotational movement

Atomic or sub-atomic particles comprising the matter

Identify the correct statement

a) i and ii

b) ii and iii

c) i and iii

d) i, ii and iii

Answer: d

Explanation: The emission occurs when the body is excited by an oscillating electrical signal.

6. Thermal radiations occur in the portion of electromagnetic spectrum between the wavelengths

a) 10 -2 to 10 -4 micron

b) 10 -1 to 10 -2 micron

c) 0.1 to 10 2 micron

d) 10 -2 micron onwards

Answer: c

Explanation: The radiation waves propagates with the speed of light, and their wavelength ranges from 0.1 to 100 micron meter.

7. A perfectly black body

a) Absorbs all the incident radiation

b) Allow all the incident radiation to pass through it

c) Reflects all the incident radiation

d) Has its surface coated with lamp black or graphite

Answer: a

Explanation: A black body is a perfect emitter and absorber.

8. What is the wavelength band for cosmic rays?

a) Up to 45 * 10 -7 micron meter

b) Up to 23 * 10 -7 micron meter

c) Up to 19 * 10 -7 micron meter

d) Up to 4 * 10 -7 micron meter

Answer: d

Explanation: This is its maximum wavelength.

9. For a prescribed wavelength a black body radiates how much energy at the temperature of body?

a) Maximum

b) Minimum

c) 20%

d) 50%

Answer: a

Explanation: A black body neither reflects nor transmits any amount of incident radiations.

10. Radiation emitted by a black surface is a function of wavelength and temperature but is independent of direction.

a) True

b) False

Answer: a

Explanation: The black body is a diffused emitter.

This set of Heat Transfer online quiz focuses on “Spectral And Spatial Energy Distribution”.

1. The distribution of radiant energy is non uniform with respect to

a) Wavelength

b) Direction

c) Both wavelength and direction

d) Time

Answer: c

Explanation: It should be non-uniform with respect to both wavelength and direction.

2. Spatial distribution is also known as directional distribution.

a) True

b) False

Answer: a

Explanation: Spatial or directional both are same.

3. Of the radiant energy 350W/m 2 incident upon a surface 250W/m 2 is absorbed, 60W/m 2 is reflected and the remainder is transmitted through the surface. Workout the value for absorptivity for the surface material

a) 0.123

b) 0.714

c) 0.684

d) 0.386

Answer: b

Explanation: Absorptivity = 250/350 = 0.714.

4. Thermal radiation strikes a surface which has a reflectivity of 0.55 and transmissivity of 0.032. A quantity known as flux is found out to be 95 W/m 2 . Determine the rate of incident flux.

a) 123.34 W/m 2

b) 333.37 W/m 2

c) 122.27 W/m 2

d) 227.27 W/m 2

Answer: d

Explanation: Incident flux = 95/1 – 0.032 – 0.55 = 227.27 W/m 2 .

5. What is the wavelength for visible light?

a) 3.9 * 10 -1 to 7.8 * 10 -1 micron meter

b) 4.9 * 10 -1 to 7.8 * 10 -1 micron meter

c) 5.9 * 10 -1 to 7.8 * 10 -1 micron meter

d) 6.9 * 10 -1 to 7.8 * 10 -1 micron meter

Answer: a

Explanation: This is the maximum and minimum wavelength for visible light.

6. Radiant energy with an intensity of 800 W/m 2 strikes a flat plate normally. The absorptivity is thrice the reflectivity and twice the transmissivity. Determine the rate of absorption

a) 236.40 W/m 2

b) 336.40 W/m 2

c) 436.40 W/m 2

d) 536.40 W/m 2

Answer: c

Explanation: Rate of absorption = 0.5455 * 800 = 436.40 W/m 2 .

7. A thin metal plate of 4 cm diameter is suspended in atmospheric air whose temperature is 290 K. This plate attains a temperature of 295 K when one of its face receives radiant energy from a heat source at the rate of 2 W. If heat transfer coefficient on both surfaces of the plate is stated to be 87.5 W/m 2 K, workout the reflectivity of the plate

a) 0.35

b) 0.45

c) 0.55

d) 0.65

Answer: b

Explanation: Heat loss by convection from both sides of the plates = 2 h A d t = 1.1 W. Energy lost by reflection = 2.0 – 1.1 = 0.9 W.

8. On a clear night there is radiation from earth’s surface to space. On such a night, the water particles on the plant leaves radiate to the sky whose temperature may be taken as 200 K. The water particles receive heat by convection from the surrounding air, the convection heat transfer coefficient has a value of 30 W/m 2 K. If the water should not freeze, make calculations for the air temperature

a) 280.474 K

b) 345.645 K

c) 123.456 K

d) 874.387 K

Answer: a

Explanation: Heat radiated to sky = Heat received by convection. So, temperature of air = 224.22/30 + 273 = 280.474 K.

9. Isothermal furnaces, with small apertures, approximate a black body and are frequently used to calibrate heat flux gauges.

a) True

b) False

Answer: a

Explanation: A small hole leading into a cavity thus acts very nearly as a black body.

10. A surface element emits radiation in all directions, the intensity of radiation is however same in different directions.

a) True

b) False

Answer: b

Explanation: It must be different in different directions.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Planck’s Law”.

1. The energy emitted by a black surface should not vary in accordance with

a) Wavelength

b) Temperature

c) Surface characteristics

d) Time

Answer: d

Explanation: It is time independent. For a prescribed wavelength, the body radiates much more energy at elevated temperatures.

2. In the given diagram let r be the length of the line of propagation between the radiating and the incident surfaces. What is the value of solid angle W?

heat-transfer-questions-answers-plancks-law-q2

a) A sin α

b) A cos α

c) 2A cos α

d) 2A cos α

Answer: b

Explanation: The solid angle is defined by a region by the rays of a sphere, and is measured as A n /r 2 .

3. Likewise the amount of emitted radiation is strongly influenced by the wavelength even if temperature of the body is

a) Constant

b) Increasing

c) Decreasing

d) It is not related with temperature

Answer: a

Explanation: Temperature must remain constant in order to emit radiation.

4. A small body has a total emissive power of 4.5 kW/m 2 . Determine the wavelength of emission maximum

a) 8.46 micron m

b) 7.46 micron m

c) 6.46 micron m

d) 5.46 micron m

Answer: d

Explanation:  max t = 2.8908 * 10 -3 .

5. The sun emits maximum radiation of 0.52 micron meter. Assuming the sun to be a black body, Calculate the emissive ability of the sun’s surface at that temperature

a) 3.47 * 10 7 W/m 2

b) 4.47 * 10 7 W/m 2

c) 5.47 * 10 7 W/m 2

d) 6.47 * 10 7 W/m 2

Answer: c

Explanation: E = σ b t 4 = 5.47 * 10 7 W/m 2 .

6. The law governing the distribution of radiant energy over wavelength for a black body at fixed temperature is referred to as

a) Kirchhoff’s law

b) Planck’s law

c) Wein’s formula

d) Lambert’s law

Answer: b

Explanation: This law gives a relation between energy over wavelength.

7. The Planck’s constant h has the dimensions equal to

a) M L 2 T -1

b) M L T -1

c) M L T -2

d) M L T

Answer: a

Explanation: It has unit equal to J s and its value is 6.626 * 10 -34 .

8. Planck’s law is given by

a)  b = 2 π c 2 h  -5 /[c h/k  T] – 2

b)  b = π c 2 h [exponential [c h/k  T] – 3].

c)  b = 2 π c 2 h  -5 /exponential [c h/k  T] – 1

d)  b = 2 c 2 h  -5 /exponential [c h/k  T] – 6

Answer: c

Explanation: Planck suggested the following law for the spectral distribution of emissive power.

9. A furnace emits radiation at 2000 K. Treating it as a black body radiation, calculate the monochromatic radiant flux density at 1 micron m wavelength

a) 5.81 * 10 7 W/m 2

b) 4.81 * 10 7 W/m 2

c) 3.81 * 10 7 W/m 2

d) 2.81 * 10 7 W/m 2

Answer: d

Explanation:  b = C 1  -5 /exponential [C 2 /  T] – 1.

10. A metal sphere of surface area 0.0225 m 2 is in an evacuated enclosure whose walls are held at a very low temperature. Electric current is passed through resistors embedded in the sphere causing electrical energy to be dissipated at the rate of 75 W. If the sphere surfaces temperature is measured to be 560 K, while in steady state, calculate emissivity of the sphere surface

a) 0.498

b) 0.598

c) 0.698

d) 0.798

Answer: b

Explanation: E = e A σ b T.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Stefan- Boltzman Law”.

1. The Stefan-Boltzmann constant has units of

a) kcal/m 2 hr K 4

b) kcal/m hr K 4

c) kcal/hr K 4

d) kcal/m 2 K 4

Answer: a

Explanation: According to Stefan-Boltzmann law, q = α A T 4 .

2. According to Stefan-Boltzmann law of thermal radiation

a) q = α A T

b) q = α A T 4

c) q = α A T 3

d) q = α A T 5

Answer: b

Explanation: α is Stefan-Boltzmann constant whose value is 5.67 * 10 -8 W/m 2 K 4 .

3. Calculate the radiant flux density from a black surface at 400 degree Celsius?

a) 1631.7 W/m 2

b) 31.7 W/m 2

c) 631.7 W/m 2

d) 11631.7 W/m 2

Answer: d

Explanation: E = σ T 4 = 5.67 * 10 – 8  4 = 11631.7 W/m 2 .

4. If the emitted radiant energy is to be doubled, to what temperature surface of black body needs to be raised? Take radiant flux density as 11631.7 W/m 2.

a) 894.4 K

b) 200.4 K

c) 800.3 K

d) 600.4 K

Answer: d

Explanation: 2 = 5.67 * 10 – 8 T 4 .

5. A furnace having inside temperature of 2250 has a glass circular viewing of 6 cm diameter. If the transmissivity of glass is 0.08, make calculations for the heat loss from the glass window due to radiation

a) 234.54 W

b) 652.32 W

c) 328.53 W

d) 762.32 W

Answer: c

Explanation: Q = σ A T 4  = 328.53 W.

6. The value of radiation coefficient or the Stefan-Boltzmann constant is

a) 5.67 * 10 -8 W/m 2 K 4

b) 5.67 * 10 -7 W/m 2 K 4

c) 5.67 * 10 -6 W/m 2 K 4

d) 5.67 * 10 -5 W/m 2 K 4

Answer: a

Explanation: q = α A T 4 .

7. Measurements were made of the monochromatic absorptivity and monochromatic hemispherical irradiation incident on an opaque surface, and the variation of these parameters with wavelength may be approximated by the result shown below. Determine the total hemispherical absorptivity

heat-transfer-questions-answers-stefan-boltzman-law-q7

a) 0.557

b) 0.667

c) 0.777

d) 0.887

Answer: b

Explanation: Incident flux = 800 – 4800 W/m 2 . Absorptivity = 3200/4800 = 0.667.

8. What is the wavelength band for Ultraviolet rays?

a) 1 * 10 -6 to 3.9 * 10 -1 micron meter

b) 1 * 10 -4 to 3.9 * 10 -1 micron meter

c) 2 * 10 -3 to 3.9 * 10 -1 micron meter

d) 1 * 10 -2 to 3.9 * 10 -1 micron meter

Answer: d

Explanation: This is the maximum and minimum wavelength for Ultraviolet rays.

9. A black body of total area 0.045 m 2 is completely enclosed in a space bounded by 5 cm thick walls. The walls have a surface area 0.5 m 2 and thermal conductivity 1.07 W/ m K. If the inner surface of the enveloping wall is to be maintained at 215 degree Celsius and the outer wall surface at 30 degree Celsius, calculate the temperature of the black body

a) 547.3 K

b) 287.4 K

c) 955.9 K

d) 222.2 K

Answer: c

Explanation: Q r = σ A (T b 4 – T w 4 ), Q c = k A d t/δ = 1979.5 W. So temperature of black body is 955.9 K.

10. What is the wavelength band for solar radiation?

a) 1 * 10 -1 to 3 micron meter

b) 1 * 10 -1 to 2 micron meter

c) 1 * 10 -1 to 1 micron meter

d) 1 * 10 -1 to 10 micron meter

Answer: a

Explanation: This is the maximum and minimum wavelength for solar radiation.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Wein’s Displacement Law”.

1. The relationship  MAX T = constant, between the temperature of a black body and the wavelength at which maximum value of monochromatic emissive power occurs is known as

a) Planck’s law

b) Kirchhoff’s law

c) Lambert’s law

d) Wein’s law

Answer: d

Explanation: This is the Wien’s law. From the spectral distribution of black body emissive power, it is apparent that the wavelength associated with a maximum rate of emission depends upon the absolute temperature of the radiating surface.

2. A body at 500 K cools by radiating heat to ambient atmosphere maintained at 300 K. When the body has cooled to 400 K, the cooling rate as a percentage of original rate is about

a) 31.1

b) 41.5

c) 50.3

d) 80.4

Answer: a

Explanation: Q 2 /Q 1 =  4 –  4 /  4 –  4 = 0.32.

3. Two spheres A and B of same material have radius 1 m and 4 m, and temperatures 4000 K and 2000 K respectively. Then the energy radiated by sphere A is

a) Greater than that of sphere B

b) Less than that of sphere B

c) Equal to that of sphere B

d) Two times that of sphere B

Answer: c

Explanation: E A /E B =  2  4 /  2  4 = 1.

4. A small body has a total emissive power of 4.5 kW/m 2 . Determine its surface temperature of maximum emission

a) 530.77 K

b) 345.65 K

c) 236.54 K

d) 367.8 K

Answer: a

Explanation: E = σ T 4 . So, T = 530.77 K.

5. A small black body has a total emissive power of 4.5 k W/m 2 . In which range of the spectrum does this wavelength fall?

a) Thermal region

b) Cosmic region

c) Visible region

d) Infrared region

Answer: d

Explanation:  T = 2.8908 * 10 -3 . This must be the wavelength of infrared region.

6. The sun emits maximum radiation of 0.52 micron meter. Assuming the sun to be a black body, Calculate the surface temperature of the sun

a) 2345 K

b) 5573 K

c) 9847 K

d) 6492 K

Answer: b

Explanation: T = 2.8908 * 10 -3 /0.52 * 10 -6 = 5573 K.

7. Consider the previous problem, determine the maximum monochromatic emissive power of the sun’s surface

a) 4.908 * 10 13 W/m 2

b) 5.908 * 10 13 W/m 2

c) 6.908 * 10 13 W/m 2

d) 7.908 * 10 13 W/m 2

Answer: c

Explanation:  MAX = 1.285 * 10 -5 T 5 = 6.908 * 10 13 W/m 2 .

8. A furnace emits radiation at 2000 K. Treating it as a black body radiation, calculate the wavelength at which emission is maximum

a) 1.449 * 10 -6 m

b) 2.449 * 10 -6 m

c) 3.449 * 10 -6 m

d) 4.449 * 10 -6 m

Answer: a

Explanation:  T = 2.8908 * 10 -3 . So, wavelength = 1.449 * 10 -6 m.

9. Four identical pieces of copper painted with different colors of paints were heated to the same temperature and then left in the environment to cool. Which of the following paint will give fast cooling?

a) White

b) Rough

c) Black

d) Shining

Answer: c

Explanation: The emissivity of black paint is highest i.e. unity. Consequently, the emitted radiant energy will be maximum when painted black.

10. A surface for which emissivity is constant at all temperatures and throughout the entire range of wavelength is called

a) Opaque

b) Grey

c) Specular

d) Diathermanous

Answer: b

Explanation: Foe grey surface, emissivity is constant. Grey surface radiates much more because of constant wavelength throughout its entire surface.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Kirchoff’s Law”.

1. If radiant energy E B emitted by the black surface strikes the non-black surface. If non-black surface has absorptivity α, it will absorb how much radiations?

a) α E B

b) 2 α E B

c) 3 α E B

d) 4 α E B

Answer: a

Explanation: The remainder  will be reflected back for full absorption at the black surface.

2. If two surfaces are at the same temperature, then the conditions correspond to mobile thermal equilibrium for which the resultant interchange of heat is zero are

a) 3 E – α E B = 0

b) 2 E – α E B = 0

c) E – α E B = 0

d) ½ E – α E B = 0

Answer: c

Explanation: The remainder  will be reflected back for full absorption at the black surface.

3. The absorptivity of black body equals to

a) 2

b) 1

c) 3

d) 4

Answer: b

Explanation: The absorptivity of black body equals to unity.

4. A diathermanous body

a) Shines as a result of incident radiation

b) Gets heated up a result of absorption of incident radiation

c) Allows all the incident radiation to pass through it

d) Partly absorbs and partly reflects the incident radiation

Answer: c

Explanation: It behaves like a body that allows all the incident radiation to pass through it.

5. Choose the false statement

a) Snow is nearly black to thermal radiation

b) Absorption of radiation occurs in a very thin layer of material near the surface

c) Transmissivity varies with the wavelength of incident radiation

d) Most of the engineering materials have rough surfaces, and these rough surfaces give regular  reflections

Answer: d

Explanation: Rough surfaces give diffused reflections. Reflections from highly polished and smooth surfaces have regular characteristics.

6. The emissivity and the absorptivity of a real surface are equal for radiation with identical temperature and wavelength. This law is referred to as

a) Kirchhoff’s law

b) Lambert’s law

c) Planck’s law

d) Wein’s displacement law

Answer: a

Explanation: Emissivity and absorptivity are related by Kirchhoff’s law.

7. With an increase in wavelength, the monochromatic emissive power of a black body

a) Increases

b) Decreases

c) Decreases, reaches a minimum and then increases

d) Increases, reaches a maximum and then decreases

Answer: d

Explanation: It firstly increases to its maximum value and then decreases to zero.

8. The temperature of a solid surface changes from 27 K to 627 K. The emissive power changes would then confirm to the ratio

a) 6:1

b) 9:1

c) 81:1

d) 27:1

Answer: c

Explanation: E 2 /E 1 = (T 2 /T 1 ) 4 = 81.

9. If the temperature of a hot body is increased by 50%, the amount of radiations emitted by it would increase by nearly

a) 200%

b) 500%

c) 50%

d) 100%

Answer: b

Explanation: E 2 /E 1 = (T 2 /T 1 ) 4 = 5.06.

10. Consider two surfaces, one absolutely black and the other non-black. These surfaces are arranged parallel to each other and so close that the radiation of one falls totally on the other. Choose the correct option

heat-transfer-questions-answers-kirchoffs-law-q10

a) 1 denotes the radiant energy E emitted by the non-black surface impinges on the black surface

b) 1 denotes the radiant energy E emitted by the black surface impinges on the non-black surface

c) 2 and 3 denotes the quantity α E b

d) 4 denotes the quantity  E b

Answer: a

Explanation: The radiant energy E b emitted by the black surface strikes the non-black surface. If the non-black surface has absorptivity α, it will absorb α E b radiations.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Gray Body And Selective Emitters”.

1. Consider two bodies, one absolutely back and the other non-black and let these be at same temperature. Which one of the following statement is correct?

a) Black body radiates less intensively than a non-black body

b) Non-black body radiates less intensively than a black body

c) Both will radiates equally

d) None of them will radiates

Answer: b

Explanation: The radiation spectrum for a non-black body may be similar or radically different from that of a black body.

2. When Stefan-Boltzmann law is applied to a black body, it takes the form

a) E = σ T

b) E = σ T 2

c) E = σ T 3

d) E = σ T 4

Answer: d

Explanation: The constant is different for different bodies.

3. When the emissivity of non-black surface is constant at all temperatures and throughout the entire range of wavelength, the surface is called

a) Gray body

b) Transparent body

c) Opaque bodies

d) Perfect black body

Answer: a

Explanation: The radiation spectrum for a grey body, though reduced in vertical scale, is continuous and identical to the corresponding curve for a perfectly black surface.

4. The emissivity of the gray surface may be expressed as

a) σ /2 σ b

b) σ / 3 σ b

c) σ / σ b

d) ½ σ / σ b

Answer: c

Explanation: It is the ratio of radiating coefficient to that of black body radiating coefficient.

5. If a black body at 1000 K and a gray body at 1250 K emit the same amount of radiation, what should be the emissivity of the gray body?

a) 0.3096

b) 0.4096

c) 0.5096

d) 0.6096

Answer: b

Explanation: E = (T b /T G ) 4 = 0.4096.

6. The radiant heat transfer from a plate of 2.5 cm 2 area at 1250 K to a very cold enclosure is 5.0 W. Determine the emissivity of the plate at this temperature

a) 0.444

b) 0.344

c) 0.244

d) 0.144

Answer: d

Explanation: Emissivity = E/σ A T 4 = 0.144.

7. A 100 W light bulb has a tungsten filament  which is required to operate at 2780 K. If the bulb is completely evacuated, calculate the minimum surface area of the tungsten filament

a) 0.98 * 10 -4 m 2

b) 1.98 * 10 -4 m 2

c) 2.98 * 10 -4 m 2

d) 3.98 * 10 -4 m 2

Answer: a

Explanation: E =  σ A T 4 . So, A = 0.98 * 10 -4 m 2 .

8. The monochromatic emissivity € of a diffuse surface at 1600 K varies with wavelength in the following manner

€ = 0.4 for 0 < λ < 2

= 0.8 for 2 < λ < 5

Determine the total emissivity

heat-transfer-questions-answers-gray-body-selective-emitters-q8

a) 0.5558

b) 0.5568

c) 0.5578

d) 0.5588

Answer: c

Explanation: Total emissivity = 0.4  + 0.8  = 0.5578.

9. For a hemisphere solid angle is measured as

a) Radian and its maximum value is π

b) Degree and its maximum value is 180 degree

c) Steradian and its maximum value is 2π

d) Steradian and its maximum value is π

Answer: d

Explanation: It should be measured in Steradian.

10. A gray body  emits the same amount of heat as a black body at 1075 K. Find out the required temperature of the gray body

a) 1146.72 K

b) 1136.72 K

c) 1126.72 K

d) 1116.72 K

Answer: b

Explanation: T b 4 = E T g 4 .

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Intensity Of Radiations”.

1. The solid angle is defined by a region by the rays of a sphere, and is measured as

a) A n /r 2

b) A n /r

c) A n /r 3

d) A n /r 4

Answer: a

Explanation: Solid angle is represented by α. Where, A n is projection of incident surface normal to line of propagation.

2. The plane angle is defined by a region by the rays of a circle, and is measured as

a) 3 L/ r

b) 2 L/ r

c) L/ r

d) 4 L / r

Answer: c

Explanation: It is the ratio of arc of length on the circle to the radius of the circle. Where L is the arc of length and r is the radius of the circle.

3. When the incident surface is a sphere, the projection of surface normal to the line of propagation is the silhouette disk of the sphere which is a circle of the diameter of

a) Parabola

b) Sphere

c) Triangle

d) Hyperbola

Answer: b

Explanation: It must be a circle of a diameter of a sphere.

4. If I n denotes the normal intensity and I α represents the intensity at angle α, then

a) I α = 2 I n cos α

b) I α = 3 I n cos α

c) I α = 4 I n cos α

d) I α = I n cos α

Answer: d

Explanation: The intensity of radiation in a direction from the normal is proportional to cosine of the angle.

5. The intensity of normal radiation I n is how much times the emissive power?

a) 1/π

b) 2/ π

c) 3/ π

d) 4/ π

Answer: a

Explanation: I n = σ T 4 / π and E = σ T 4 .

6. A small surface emits diffusively, and measurements indicate that the total intensity associated with emission in the normal direction I n = 6500 W/square m sr. The emitted radiation is intercepted by three surfaces. Mark calculations for intensity associated with emission

heat-transfer-questions-answers-intensity-radiations-q6

a) 3500 W/m 2 sr

b) 4500 W/m 2 sr

c) 5500 W/m 2 sr

d) 6500 W/m 2 sr

Answer: d

Explanation: For a diffusion emitter, the intensity of the emitted radiation is independent of direction.

7. Consider a deep-space probe constructed as 1 m diameter polished aluminum sphere. Estimate the equilibrium temperature that the probe reaches if the solar energy received is 300 W/m 2 . For solar radiation, absorptivity of aluminum is 0.3 and the average emissivity appropriate for aluminum at low temperature is 0.04

a) 415.67 K

b) 315.67 K

c) 215.67 K

d) 115.67 K

Answer: b

Explanation: Q in = α q A P = 70.7 W. Q out = E σ b A T 4 .

8. The total emissive power of the emitter with area d A and temperature T is given by

a) E = 2 σ T 4 d A

b) E = 3 σ T 4 d A

c) E = σ T 4 d A

d) E = ½ σ T 4 d A

Answer: c

Explanation: E = I n π d A.

9. A black body of 0.2 m 2 area has an effective temperature of 800 K. Calculate the intensity of normal radiations

a) 1234.65 W/m 2 sr

b) 7396.28 W/m 2 sr

c) 3476.74 W/m 2 sr

d) 8739.43 W/m 2 sr

Answer: b

Explanation: I n = α T 4 /π = 7396.28 W/m 2 sr.

10. The energy radiated out decreases with increases in α and becomes zero at an angle of

a) 45

b) 30

c) 0

d) 90

Answer: d

Explanation: I α = I n cos α. So at 90 degree it becomes zero.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Solar Radiations”.

1. The heat energy absorbed by a known area in a fixed time is determined with the help of an instrument called

a) Psychrometer

b) Pyrheliometer

c) Thermometric well

d) Any instrument

Answer: b

Explanation: The effects of absorption by the atmosphere are eliminated by finding the value of the solar constant at various altitudes of the sun on the same day.

2. The observed solar constant S 0 , the true solar constant S and the angular elevation Z of the sun are related by the expression

a) S 0 =S a SEC Z

b) S 0 =2 S a SEC Z

c) S 0 =3 S a SEC Z

d) S 0 =4 S a SEC Z

Answer: a

Explanation: From the straight line graph between log S 0 along y-axis and sec Z along x-axis, the intercept log S on the y-axis is found.

3. The value of solar constant varies between

a) 1123 and 1237 W/m 2

b) 1345 and 1453 W/m 2

c) 1987 and 2346 W/m 2

d) 1335 and 1815 W/m 2

Answer: d

Explanation: When the sun lies at a mean distance from the earth, the heat flux from the sun to the outer edge of the atmosphere has been found to be 1350 W/m 2 .

4. The total amount of heat energy received by the sphere of radius r is

a) 2 π R 2 S

b) 6 π R 2 S

c) 4 π R 2 S

d) π R 2 S

Answer: c

Explanation: Area of sphere is 4 π R 2 .

5. Amount of heat energy radiated by unit surface of the sun in the same time workout as

Take, R = mean distance of earth from the sun and r = radius of sun

a) E = π R 2 S/4 π R 2

b) E = 4 π R 2 S/4 π R 2

c) E = 4 π R 2

d) E = 4 π R 2 S

Answer: b

Explanation: It is the ratio of 4 π R 2 S to 4 π R 2 .

6. Temperature of sun can be workout from

a) Wein’s displacement law

b) Stefan-Boltzmann law

c) Kirchhoff’s law

d) All of the Mentioned

Answer: a

Explanation:  MAX T = 2-89 * 10 -3 m K.

7. The temperature of the photosphere, referred to as the effective temperature of the sun, is usually taken as

a) 3000 K

b) 4000 K

c) 5000 K

d) 6000 K

Answer: d

Explanation: The sun consists of a central hot portion surrounded by the photosphere.

8. The approximate distribution of the flow of sun’s energy to the earth’s surface is

a) 53% is reflected back to space

b) 53% is transmitted to the earth

c) 9% is scattered

d) 22% is absorbed in the atmosphere

Answer: c

Explanation: 33% is reflected back to space, 15% is absorbed in the atmosphere and 43% is transmitted to the earth.

9. The solar radiation that is felt at the earth’s surface includes

 Direct radiation that has passed through the atmosphere

 Diffuse radiation from the sky

 Absorbed radiation from the water

Identify the correct statement

a) i only

b) i and ii

c) i, ii and iii

d) ii and iii

Answer: b

Explanation: Option  must be reflected radiation from water.

10. When the sun lies at a mean distance from the earth, the heat flux from the sun to the outer edge of the atmosphere has been found to be

a) 2350 W/m 2

b) 3350 W/m 2

c) 4350 W/m 2

d) 1350 W/m 2

Answer: d

Explanation: 47% of this would reach the earth’s surface.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Solar Absorptivity”.

1. Which one is having the highest value of solar absorptivity?

a) White marble

b) Concrete

c) Asphalt

d) Snow

Answer: c

Explanation: Solar absorptivity for asphalt is 0.90 while that of white marble, concrete and snow are 0.46, 0.60 and 0.28 respectively.

2. What is the value of solar absorptivity for aluminum foil?

a) 0.15

b) 0.16

c) 0.17

d) 0.18

Answer: a

Explanation: Solar absorptivity doesn’t require a medium and for aluminum, it reaches maximum efficiency in a vacuum.

3. What is the value of solar absorptivity for polished copper?

a) 0.17

b) 0.18

c) 0.19

d) 0.20

Answer: b

Explanation: Solar absorptivity for polished copper depends on the frequency and wavelength of the radiation.

4. Which one is having the lowest value of solar absorptivity?

a) Snow

b) Red brick

c) Aluminum foil

d) Polished aluminum

Answer: d

Explanation: Solar absorptivity for polished aluminum is 0.09 while that of red brick, aluminum foil and snow are 0.63, 0.15 and 0.28 respectively.

5. What is the value of solar absorptivity for polished stainless steel?

a) 0.37

b) 0.36

c) 0.35

d) 0.34

Answer: a

Explanation: There is vibration within the molecules and it can potentially initiate electromagnetic waves.

6. Which one is having the lowest value of solar absorptivity?

a) Black paint

b) Dull stainless steel

c) Concrete

d) White marble

Answer: d

Explanation: Solar absorptivity for white marble is 0.46 while that of black paint, dull stainless steel and concrete are 0.97, 0.50 and 0.60 respectively.

7. What is the value of solar absorptivity for white marble?

a) 0.45

b) 0.46

c) 0.47

d) 0.48

Answer: b

Explanation: It generates and receives electromagnetic waves at the expense of its stored energy.

8. Which one is having the highest value of solar absorptivity?

a) Red brick

b) Asphalt

c) Black paint

d) White paint

Answer: c

Explanation: Emissivity of black paint is 0.97 while that of red brick, asphalt and white paint are 0.63, 0.90 and 0.14 respectively.

9. What is the value of solar absorptivity for asphalt?

a) 0.90

b) 0.80

c) 0.70

d) 0.60

Answer: a

Explanation: Energy emitted by asphalt is not continuous but is in the form of quanta.

10. What is the value of solar absorptivity for snow?

a) 0.25

b) 0.26

c) 0.27

d) 0.28

Answer: d

Explanation: Solar absorptivity for snow varies on the nature of its surface and its temperature.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Heat Exchange Between Black Bodies”.

1. Engineering problems of practical interest are involved with heat exchange between two or more surfaces, and this exchange is strongly dependent upon

 Radiative properties

 Temperature levels

 Surface geometrics

Identify the correct statements

a) i and ii

b) i and iii

c) ii and iii

d) i, ii and iii

Answer: d

Explanation: For black surface, it is necessary to determine what portion of radiation emitted by one will be intercepted by the other.

2. The fraction of the radiative energy that is diffused from one surface element and strikes the other surface directly with no intervening reflections is called

 Radiation shape factor

 Geometrical factor

 Configuration factor

Choose the correct answer

a) i only

b) ii only

c) i, ii and iii

d) iii only

Answer: c

Explanation: Radiation shape factor, geometrical factor and configuration factor are all same.

3. The interchange factor is also known as

a) Equivalent emissivity

b) Irradiation

c) Radiosity

d) Shape factor

Answer: a

Explanation: The interchange factor is also known as equivalent emissivity.

4. For the same type of shapes, the value of the radiation shape factor will be higher when

a) Surfaces are closer

b) Surfaces are larger and held closer

c) Surfaces are moved further apart

d) Surfaces are smaller and held closer

Answer: b

Explanation: Obviously the value of radiation shape factor will be higher when surfaces are larger and held closer.

5. A thin shield of emissivity E 3 on both sides is placed between two infinite parallel plates of emissivities E 1 and E 2 and temperatures T 1 and T 2 . If E 1 = E 2 = E 3 , then the fraction radiant energy transfer without shield takes the value

a) 0.25

b) 0.50

c) 0.75

d) 1.25

Answer: b

Explanation: The ratio of radiant energy transfer without and with shield is given by

(1/E 1 + 1/E 2 – 1)/ [(1/E 1 + 1/E 3 – 1) + (1/E 3 + 1/E 2 – 1)].

6. The grey body shape factor for radiant heat exchange between a small body  in a large enclosure  is

a) 0.1

b) 0.2

c) 0.4

d) 0.5

Answer: c

Explanation:  12 = 1/ (1 – E 1 + 1 + 0).

7. Two long parallel surfaces, each of emissivity 0.7 are at different temperatures and accordingly have radiation exchange between them. It is desired to reduce 75% of this radiant heat transfer by inserting thin parallel shields of equal emissivity 0.7 on both sides. What should be the number of shields?

a) 2

b) 4

c) 1

d) 3

Answer: d

Explanation: Without shields/with shield = 1/N + 1.

8. An enclosure consists of four surfaces 1, 2, 3 and 4. The view factors for radiation heat transfers are

F 11 = 0.1

F 12 = 0.4

F 13 = 0.25

The surface areas A 1 and A 2 are 4 m 2 and 2 m 2 . The view factor F 41 is

a) 0.50

b) 0.75

c) 0.1

d) 0.25

Answer: a

Explanation: F 11 + F 12 + F 13 + F 14 = 1.

9. The value of shape factor depends on how many factors?

a) 4

b) 1

c) 2

d) 3

Answer: c

Explanation: Geometry and orientation.

10. Find the shape factor F 12 for the arrangement shown in the figure. The areas A 1 and A 2 are perpendicular but do not share the common edge

heat-transfer-questions-answers-heat-exchange-black-bodies-q10

a) 0.03

b) 0.04

c) 0.05

d) 0.06

Answer: b

Explanation: A 5 = A 1 + A 3 and A 6 = A 2 + A 4 . The sequence of the solution is, A 5 F 56 = A 1 F 16 + A 3 F 36 .

This set of Heat Transfer Question Bank focuses on “Heat Exchange Between Non Black Bodies”.

1. Consider radiant heat exchange between two non-black parallel surfaces. The surface 1 emits radiant energy E 1 which strikes the surface 2. Identify the correct option

heat-transfer-question-bank-q1

a) The value of B is α E 1

b) The value of C is (1 – α 1 ) E 1

c) The value of D is (1 – α 1 ) (1 – α 2 ) 2 E 1

d) The value of E is (1 – α 1 ) 2 (1 – α 2 ) E 1

Answer: c

Explanation: The surface 1 emits radiant energy E 1 which strikes the surface 2. From it a part α 2 E 1 is absorbed by the surface 2 and the remainder (1 – α 2 ) E 1 is reflected back to surface 1 and so on.

2. A large plane, perfectly insulated on one face and maintained at a fixed temperature T 1 on the bare face, has an emissivity of 0.84 and loses 250 W/m 2 when exposed to surroundings at nearly 0 K. The radiant heat loss from another plane of the same size is 125 W/m 2 when bare face having emissivity 0.42 and is maintained at temperature T 2 is exposed to the same surroundings. Subsequently these two planes are brought together so that the parallel bare faces lie only 1 cm apart and the heat supply to each is so regulated that their respective temperatures T 1 and T 2 remains unchanged. Determine he net heat flux between the planes

a) 0

b) 1

c) 2

d) 3

Answer: a

Explanation: Q 12 = F 12 A 1 σ b (T 1 4 – T 2 4 ). Since T 1 = T 2 , we get Q 12 /A 1 = 0.

3. Interchange factor for body 1 completely enclosed by body 2  is given by

a) 2/ [1/E 1 + A 1 /A 2 (1/E 2 – 1)].

b) 1/ [1/E 1 + A 1 /A 2 (1/E 2 – 1)].

c) 4/ [1/E 1 + A 1 /A 2 (1/E 2 – 1)].

d) 3/ [1/E 1 + A 1 /A 2 (1/E 2 – 1)].

Answer: b

Explanation: This is the interchange factor for the radiation from surface 1 to surface 2.

4. A thermos flask has a double walled bottle and the space between the walls is evacuated so as to reduce the heat flow. The bottle surfaces are silver plated and the emissivity of each surface is 0.025. If the contents of the bottle are at 375 K, find the rate of heat loss from the thermos bottle to the ambient air at 300 K

a) 5.38 W

b) 6.38 W

c) 7.38 W

d) 8.38 W

Answer: d

Explanation: Q 12 = F 12 A 1 σ b (T 1 4 – T 2 4 ). F 12 = 1/ (1/E 1 + 1/E 2 – 1) = 0.01266.

5. A 250 mm by 250 mm ingot casting, 1.5 m high and at 1225 K temperature, is stripped from its mold. The casting is made to stand on end on the floor of a large foundry whose wall, floor and roof can be assumed to be at 300 K temperature. Make calculation for the rate of radiant heat interchange between the casting and the room. The casting material has an emissivity of 0.85

a) 161120 W

b) 171120 W

c) 181120 W

d) 191120 W

Answer: b

Explanation: Q 12 = F 12 A 1 σ b (T 1 4 – T 2 4 ). F 12 = 0.85 and A 1 =  2 + 4  = 1.5625 m 2 .

6. Interchange factor for infinitely long concentric cylinders is given by

a) 1/ [A 1 /A 2 (1/E 2 – 1)].

b) [1/E 1 + A 1 /A 2 (1/E 2 – 1)].

c) 2/ [1/E 1 + A 1 /A 2 (1/E 2 – 1)].

d) 1/ [1/E 1 + A 1 /A 2 (1/E 2 – 1)].

Answer: d

Explanation: This is the interchange factor for the radiation from surface 1 to surface 2.

7. What is the geometric factor for infinitely long concentric cylinders?

a) 1

b) 0.5

c) 0.33

d) 0.75

Answer: a

Explanation: The inner cylinder is completely enclosed by the outer cylinder and as such the entire heat radiations emitted by the emitted by the inner cylinder are intercepted by the outer cylinder.

8. What is the geometric factor for concentric spheres?

a) 0.85

b) 0.33

c) 1

d) 0.95

Answer: c

Explanation: The inner sphere is completely enclosed by the outer sphere and as such the entire heat radiations emitted by the emitted by the inner sphere are intercepted by the outer cylinder.

9. The net heat interchange between non-black bodies at temperature T 1 and T 2 is given by

a) f 12 F 12 σ (T 1 4 – T 2 4 )

b) f 12 F 12 A 1 σ (T 1 4 – T 2 4 )

c) f 12 A 1 σ (T 1 4 – T 2 4 )

d) F 12 A 1 σ (T 1 4 – T 2 4 )

Answer: b

Explanation: The factor f 12 is called the interchanging factor from surface 1 to surface 2.

10. A thermos flask has a double walled bottle and the space between the walls is evacuated so as to reduce the heat flow. The bottle surfaces are silver plated and the emissivity of each surface is 0.025. If the contents of the bottle are at 375 K and temperature of ambient air is 300 K. What thickness of cork  would be required if the same insulating effect is to be achieved by the use of cork?

a) 26.8 cm

b) 25.8 cm

c) 24.8 cm

d) 23.8 cm

Answer: a

Explanation: Q = k A (t 1 – t 2 )/δ. So, δ = 0.268 m = 26.8 cm.

This set of Heat Transfer Questions for entrance exams focuses on “Electrical Network Approach For Radiation Heat Exchange”.

1. The total radiant energy leaving a surface per unit time per unit surface area is represented by

a) Radiation

b) Radiosity

c) Irradiation

d) Interchange factor

Answer: b

Explanation: It comprises the original emittance from the surface plus the reflected portion of any radiation incident upon it.

2. Determine the radiant heat flux between two closely spaced, black parallel plates radiating only to each other if their temperatures are 850 K and 425 K. The plates have an area of 4 m 2

a) .040

b) .030

c) .020

d) .010

Answer: d

Explanation: Q 12 = F 12 A 1 σ b (T 1 4 – T 2 4 ) = .010.

3. What is the value of grey body factor for concentric cylinders?

a) 3/ [1 – e 1 /e 1 + 1 + 1 – e 2 /e 2 (A 1 /A 2 )].

b) 4/ [1 – e 1 /e 1 + 1 + 1 – e 2 /e 2 (A 1 /A 2 )].

c) 1/ [1 – e 1 /e 1 + 1 + 1 – e 2 /e 2 (A 1 /A 2 )].

d) 2/ [1 – e 1 /e 1 + 1 + 1 – e 2 /e 2 (A 1 /A 2 )].

Answer: c

Explanation: Here, F 12 = 1.

4. The net heat exchange between the two grey surfaces may be written as

a) (Q 12 ) NET = E b 1 – E b 2 / (1 – e 1 /A 1 e 1 + 1/A 1 F 12 + 1 – e 2 /A 2 e 2 )

b) (Q 12 ) NET = 2 E b 1 – E b 2 / (1 – e 1 /A 1 e 1 + 1/A 1 F 12 + 1 – e 2 /A 2 e 2 )

c) (Q 12 ) NET = E b 1 – 2 E b 2 / (1 – e 1 /A 1 e 1 + 1/A 1 F 12 + 1 – e 2 /A 2 e 2 )

d) (Q 12 ) NET = 2 E b 1 – 3 E b 2 / (1 – e 1 /A 1 e 1 + 1/A 1 F 12 + 1 – e 2 /A 2 e 2 )

Answer: a

Explanation: This equation gives the electrical network corresponding to surface resistances of two radiating bodies.

5. The net rate at which the radiation leaves the surface is given by

a) e (E b – J)/1 – 4 e

b) e (E b – J)/1 – 3 e

c) e (E b – J)/1 – 2 e

d) e (E b – J)/1 – e

Answer: d

Explanation: The net rate at which the radiation leaves the surface is given by the difference between its radiosity and the incoming irradiation.

6. A ring  of 8 cm inner and 16 cm outer diameter is placed in a horizontal plane. A small element  of 1 cm 2 is placed concentrically 8 cm vertically below the center of the ring. The temperature of the ring is 800 K and that of small area is 400 K. Find the radiant heat gain by the small ring

a) – 10.59 J/hour

b) – 11.59 J/hour

c) – 12.59 J/hour

d) – 13.59 J/hour

Answer: b

Explanation: Q 12 = (F g ) 12 A 1 σ b (T 1 4 – T 2 4 ) = A 1 σ b (T 1 4 – T 2 4 )/ (I/E 1 – 1) + 1/F 12 + (I/E 2 – 1) A 2 /A 1 .

7. Two opposed, parallel, infinite planes are maintained at 420 K and 480 K. Calculate the net heat flux between these planes if one has an emissivity of 0.8 and other an emissivity of 0.7

a) 534.86 W/m 2

b) 634.86 W/m 2

c) 734.86 W/m 2

d) 834.86 W/m 2

Answer: c

Explanation: Q 12 = (F g ) 12 A 1 σ b (T 1 4 – T 2 4 ) and (F g ) 12 = 1/ (I/E 1 – 1) + 1/F 12 + (I/E 2 – 1) A 2 /A 1 .

8. Consider the above problem, if temperature difference is doubled by raising the temperature 480 K to 540 K, then how this heat flux will be affected?

a) 1803.55 W/m 2

b) 1703.55 W/m 2

c) 1603.55 W/m 2

d) 1503.55 W/m 2

Answer: a

Explanation: Q 2 = 0.59 (5.67 * 10 -8 ) (540 4 – 420 4 ).

9. The total radiant energy incident upon a surface per unit time per unit area is known as

a) Shape factor

b) Radiosity

c) Radiation

d) Irradiation

Answer: d

Explanation: Some of it may be reflected to become a part of the radiosity of the surface.

10. Which one of the following is true for the opaque non-black surface?

a) J = E +2 p G

b) J = E + p G

c) J = 2 E + p G

d) J = ½ E + p G

Answer: b

Explanation: For an opaque non-black surface of constant radiation characteristics, the total radiant energy leaving the surface is the sum of its original emittance and the energy reflected by it out of the irradiation impinging on it.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Radiations Shields”.

1. A radiation shield should

a) Have high transmissivity

b) Absorb all the radiations

c) Have high reflexive power

d) Partly absorb and partly transmit the incident radiation

Answer: c

Explanation: Reflexive power is much high for radiation shield.

2. Radiation shield is used between the emitting surfaces such that

a) To reduce overall heat transfer

b) To increase overall heat transfer

c) To increase density

d) To reduce density

Answer: a

Explanation: Many situations are encountered where it is desired to reduce the overall heat transfer between two radiating surfaces.

3. Which of the following can be used as a radiating shield?

a) Carbon

b) Thin sheets of aluminum

c) Iron

d) Gold

Answer: b

Explanation: The shields are thin opaque partitions arranged in the direction perpendicular to the propagation of radiant heat.

4. Two large parallel planes with emissivity 0.4 are maintained at different temperatures and exchange heat only by radiation. What percentage change in net radiative heat transfer would occur if two equally large radiation shields with surface emissivity 0.04 are introduced in parallel to the plates?

heat-transfer-questions-answers-radiations-shields-q4

a) 65.1%

b) 75.1%

c) 85.1%

d) 95.1%

Answer: d

Explanation: When shields are not used, Q 12 = (F g ) 12 A 1 σ b (T 1 4 – T 2 4 ) = 0.2 A 1 σ b (T 1 4 – T 2 4 ) and when shields are used Q 12 = 0.0098 A 1 σ b (T 1 4 – T 2 4 ).

5. Determine the net radiant heat exchange per m 2 area for two infinite parallel plates held at temperature of 800 K and 500 K. Take emissivity as 0.6 for the hot plate and 0.4 for the cold plate

a) 6200 W/m 2

b) 7200 W/m 2

c) 8200 W/m 2

d) 9200 W/m 2

Answer: a

Explanation: Q 12 = (F g ) 12 A 1 σ b (T 1 4 – T 2 4 ) and (F g ) 12 = 0.135. Therefore, Q 12 = 6200 W/m 2 .

6. Consider the above problem, what should be the emissivity of a polished aluminum shield placed between them if heat flow is to be reduced to 40 percent of its original value?

a) 0.337

b) 0.347

c) 0.357

d) 0.367

Answer: b

Explanation: (F g ) 12 = 1/E 1 +1/E 2 +2/E 3 – 2 = 7.936.

7. Consider radiative heat transfer between two large parallel planes of surface emissivities 0.8. How many thin radiation shields of emissivity 0.05 be placed between the surfaces is to reduce the radiation heat transfer by a factor of 75?

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: (Q 12 ) ONE SHIELD = A σ b (T 1 4 – T 2 4 )/ 1/E 1 +1/E 2 +2/E 3 – 2 and 75 = (Q 12 ) NO SHIELD / (Q 12 ) N SHIELD .

8. Two parallel square plates, each 4 m 2 area, are large compared to a gap of 5 mm separating them. One plate has a temperature of 800 K and surface emissivity of 0.6, while the other has a temperature of 300 K and surface emissivity of 0.9. Find the net energy exchange by radiations between the plates

a) 61.176 k W

b) 51.176 k W

c) 41.176 k W

d) 31.176 k W

Answer: b

Explanation: Q 12 = (F g ) 12 A 1 σ b (T 1 4 – T 2 4 ).

9. The furnace of a boiler is laid from fire clay brick with outside lagging from the plate steel, the distance between the two is quite small compared with the size of the furnace. The brick setting is at an average temperature of 365 K whilst the steel lagging is at 290 K. Calculate the radiant heat flux. Assume the following emissivity values

For brick = 0.85

For steel = 0.65

a) 352.9 W/m 2

b) 452.9 W/m 2

c) 552.9 W/m 2

d) 652.9 W/m 2

Answer: a

Explanation: Q 12 = (F g ) 12 A 1 σ b (T 1 4 – T 2 4 ).

10. Consider the above problem, find the reduction in heat loss if a steel screen having an emissivity value of 0.6 on both sides is placed between the brick and steel setting

a) 5.56

b) 4.46

c) 3.36

d) 2.36

Answer: d

Explanation: (F g ) 12 = 0.247 and Q = 149.51 W/m 2 .

This set of Heat Transfer Questions for campus interviews focuses on “Adiabatic And Reradiating Surfaces”.

1. Two black discs each of diameter 50 cm are placed parallel to each other concentrically at a distance of one meter. The discs are maintained at 1000 K and 500 K. Calculate the heat flow between the discs when no other surface is present

a) 317.27 W

b) 417.27 W

c) 517.27 W

d) 617.27 W

Answer: b

Explanation: Q = F 12 A 1 σ B (T 1 4 – T 2 4 ).

2. Two black discs each of diameter 50 cm are placed parallel to each other concentrically at a distance of one meter. The discs are maintained at 1000 K and 500 K. Calculate the heat flow between the discs when the disks are connected by a cylindrical black no-flux surface

a) 2417.68 W

b) 3417.68 W

c) 4417.68 W

d) 5417.68 W

Answer: d

Explanation: Q = F 12 A 1 σ B (T 1 4 – T 2 4 ).

3. Heat exchange between two black surfaces enclosed by an insulated surface is given by

a) Q 12 = A 1 σ b (T 1 4 – T 2 4 ) [A 2 – A 1 F 12 2 /A 1 + A 2 – 2 A 1 F 12 ].

b) Q 12 = 2 A 1 σ b (T 1 4 – T 2 4 ) [A 2 – A 1 F 12 2 /A 1 + A 2 – 2 A 1 F 12 ].

c) Q 12 = 3 A 1 σ b (T 1 4 – T 2 4 ) [A 2 – A 1 F 12 2 /A 1 + A 2 – 2 A 1 F 12 ].

d) Q 12 = 4 A 1 σ b (T 1 4 – T 2 4 ) [A 2 – A 1 F 12 2 /A 1 + A 2 – 2 A 1 F 12 ].

Answer: a

Explanation: This is the net heat exchange between two black surfaces enclosed by an insulated surface.

4. Heat exchange between two gray surfaces enclosed by an adiabatic surface is given by

a) Q 12 = A (T 1 4 – T 2 4 ) / [1/E 1 + 1/E 2 – 2 + 2/1 + F 12 ].

b) Q 12 = A σ b (T 1 4 – T 2 4 ) / [1/E 1 + 1/E 2 + 2/1 + F 12 ].

c) Q 12 = A σ b (T 1 4 – T 2 4 ) / [1/E 1 + 1/E 2 – 2 + 2/1 + F 12 ].

d) Q 12 = A σ b (T 1 4 – T 2 4 ) / [1/E 1 + 1/E 2 – 2].

Answer: c

Explanation: This is the net heat exchange between two gray surfaces enclosed by an adiabatic surface.

5. A blind cylindrical hole of 2 cm diameter and 3 cm length is drilled into a metal slab having emissivity 0.7. If the metal slab is maintained at 650 K, make calculations for the radiation heat escape from the hole

heat-transfer-questions-campus-interviews-q5

a) 7 W

b) 3 W

c) 1 W

d) 9 W

Answer: b

Explanation: Q = E 1 A 1 σ b T 1 4 [1 – F 11 /1 – (1 – E 1 ) F 11 ].

6. A cavity in the shape of a frustum of a cone has diameter 30 cm and 60 cm and the height is 80 cm. If the cavity is maintained at temperature of 800 K, determine the heat loss from the cavity when the smaller diameter is at the bottom

a) 6577 W

b) 2367 W

c) 8794 W

d) 3675 W

Answer: a

Explanation: Q = E 1 A 1 σ b T 1 4 [1 – F 11 /1 – (1 – E 1 ) F 11 ].

7. Consider the above problem, find how this heat loss would be affected if the cavity is positioned with bigger diameter at the base

a) 75.06 % 

b) 55.06 % 

c) 65.06 % 

d) 75.06 % 

Answer: d

Explanation: Percentage change in heat flow = 6577 – 1640/6577 = 0.7506.

8. A conical cavity of base diameter 15 cm and height 20 cm has inside surface temperature 650 K. If emissivity of each surface is 0.85, determine the net radiative heat transfer from the cavity

a) 168.3 W

b) 158.3 W

c) 148.3 W

d) 138.3 W

Answer: a

Explanation: Q = E 1 A 1 σ b T 1 4 [1 – F 11 /1 – (1 – E 1 ) F 11 ]. Here, F 11 = 0.649 and A 1 = 0.0503 m 2 .

9. A cylindrical cavity of base diameter 15 cm and height 20 cm has inside surface temperature 650 K. If emissivity of each surface is 0.85, determine the net radiative heat transfer from the cavity

a) 194 W

b) 184 W

c) 174 W

d) 164 W

Answer: c

Explanation: Q = E 1 A 1 σ b T 1 4 [1 – F 11 /1 – (1 – E 1 ) F 11 ]. Here, F 11 = 0.842 and A 1 = 0.11186 m 2 .

10. What is the unit of coefficient of radiant heat transfer?

a) W/K

b) W/m 2 K

c) W/m 2

d) W/m K

Answer: b

Explanation: Its value can be calculated from the heat flux equation for any configuration.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Gaseous Radiations”.

1. Which of the following have a continuous spectrum?

a) Solids

b) Liquids

c) Gases

d) Solids and liquids

Answer: d

Explanation: They emit and absorb the radiant energy of all wavelength.

2. Which of the following is/are known as selective absorbers?

a) Solids

b) Liquids

c) Gases and vapors

d) Vapors

Answer: c

Explanation: They emit and absorb radiant energy in definite parts of the spectrum called bands.

3. Which layers participated in the process of thermal radiation through solids and liquids?

a) 1 micron to 1 mm thick

b) 2 micron to 2 mm thick

c) 1 micron to 0.5 mm thick

d) 2 micron to 4.57 mm thick

Answer: a

Explanation: Gases possess a much small radiating power and all their volumes participate in the radiation.

4. Some of the gases have low emissive power and absorptivity. They are considered practically diathermanous. Which of the following option is true?

a) Ammonia

b) Helium

c) Carbon-dioxide

d) Methane

Answer: b

Explanation: Helium is monoatomic gas and is extremely inert to thermal radiation.

5. Consider an enclosure formed by three surfaces having the following values of shape factors, emissivities and temperatures

heat-transfer-questions-answers-gaseous-radiations-q5

Surface 1 i.e. curved cylindrical has an emissivity 0.75 and temperature 800 K

Surface 2 i.e. closing disc has an emissivity 0.8 and temperature 700 K

Surface 3 i.e. closing disc has an emissivity 0.8 and temperature 700 K

The closing flat discs are 25 mm in diameter and they have interspacing distance equal to 100 mm. If the shape factor between these two identical discs is 0.05, calculate the net rate of radiant heat flow from the curve surface to each of the closing end surface

a) 1.561 W

b) 2.561 W

c) 3.561 W

d) 4.561 W

Answer: c

Explanation: F 21 + F 23 + 1 and A 1 = 0.00785 m 2 , A 2 = 0.000491 m 2 . Then from reciprocity theorem, F 12 = 0.0594. Q 12 = F 12 A 1 σ b (T 1 4 – T 2 4 ) = 3.561 W.

6. Consider a beam of monochromatic radiation at wavelength λ that enters a layer of absorbing gas. As the beam passes through the gas layer, its intensity

a) Decreases

b) Increases

c) Become twice

d) Remains same

Answer: a

Explanation: The decrease is given by d I λ X = – K λ I λ X d x.

7. When a gas or vapor is in the process of oxidation and combustion, it is called as

a) Oxidation

b) Flame

c) Reduction

d) Combustion

Answer: b

Explanation: It is known as a flame.

8. The monochromatic absorption coefficient depends upon

 Temperature

 Time

 Pressure

 Wavelength

Identify the correct option

a) i, ii and iv

b) ii, iii and iv

c) i and ii

d) i, iii and iv

Answer: d

Explanation: It depends upon the state of gas and the wavelength.

9. Which one of the following is not extremely inert to thermal radiation?

a) Orgon

b) Oxygen

c) Methane

d) Nitrogen

Answer: c

Explanation: Methane is a polyatomic gas.

10. Consider a beam of monochromatic radiation at wavelength λ that enters a layer of absorbing gas. As the beam passes through the gas layer, its intensity is given by

a) d I λ X = – K λ I λ X d x

b) d I λ X = – 2 K λ I λ X d x

c) d I λ X = – 3 K λ I λ X d x

d) d I λ X = – 4 K λ I λ X d x

Answer: a

Explanation: Its intensity gets reduced.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Fouling Factor”.

1. Which one is having the highest value of fouling factor?

a) Sea water

b) Distilled water

c) Liquid gasoline

d) Clean river

Answer: c

Explanation: Fouling factor for liquid gasoline is 0.0008 while that of clean river, sea water and distilled water are 0.0002, 0.0001 and 0.0001 respectively. Fouling factor = 1/h S .

2. What is the value of fouling factor for sea water?

a) 0.0001-0.0002 m 2 K/W

b) 0.0002-0.0003 m 2 K/W

c) 0.0003-0.0004 m 2 K/W

d) 0.0004-0.0005 m 2 K/W

Answer: a

Explanation: There are strong hydrogen bonding between its molecules.

3. What is the value of fouling factor for clean river?

a) 0.0014-0.0018 m 2 K/W

b) 0.0010-0.0014 m 2 K/W

c) 0.0006-0.0010 m 2 K/W

d) 0.0002-0.0006 m 2 K/W

Answer: d

Explanation: It has low electrical conductivity but this increases with the dissolution of small amount of ionic materials.

4. Which one is having the lowest value of fouling factor?

a) Lake river

b) Distilled water

c) Transformers

d) Lubricating oil

Answer: b

Explanation: Fouling factor for distilled water is 0.0001 while that of Lake River, transformer and lubricating oil are 0.0002, 0.0002 and 0.0002 respectively.

5. What is the value of fouling factor for treated boiler feed water?

a) 0.0001-0.0002 m 2 K/W

b) 0.0002-0.0003 m 2 K/W

c) 0.0003-0.0004 m 2 K/W

d) 0.0004-0.0005 m 2 K/W

Answer: a

Explanation: It is a good polar solvent and is referred to as the universal solvent. Fouling factor = 1/h S .

6. Which one is having the highest value of fouling factor?

a) Industrial liquids

b) Fuel oil

c) Turbine exhaust

d) Engine exhaust

Answer: d

Explanation: Fouling factor for engine exhaust is 0.002 while that of industrial liquids, fuel oil and turbine exhaust are 0.0002, 0.0010 and 0.0002 respectively.

7. What is the value of fouling factor for fuel oil?

a) 0.0009 m 2 K/W

b) 0.0010 m 2 K/W

c) 0.0011 m 2 K/W

d) 0.0012 m 2 K/W

Answer: b

Explanation: It is a neutral, non-polar chemical substance that is viscous liquid at ambient temperature.

8. Which one is having the lowest value of fouling factor?

a) Fuel gases

b) Oil bearing system

c) Non-oil bearing system

d) Engine exhaust

Answer: c

Explanation: Fouling factor for non-oil bearing system is 0.0001 while that of fuel gases, oil bearing system and engine exhaust are 0.0010, 0.0002 and 0.002 respectively.

9. What is the value of fouling factor for oil bearing steam?

a) 0.0002 m 2 K/W

b) 0.0003 m 2 K/W

c) 0.0004 m 2 K/W

d) 0.0005 m 2 K/W

Answer: a

Explanation: Fouling factor = 1/h S .

10. Which one is having the highest value of fouling factor?

a) Clean water

b) Sea water

c) Liquid gasoline

d) Distilled water

Answer: c

Explanation: Fouling factor for liquid gasoline is 0.0008 while that of clean water, sea water and distilled water are 0.0002, 0.0001 and 0.0001 respectively.

This set of Advanced Heat Transfer Questions & Answers focuses on “Symbols Of Geomertical And Thermo Physical Properties”.

1. What is the symbol used for mass?

a) M

b) L

c) A

d) V

Answer: a

Explanation: Unit of mass is kg.

2. What is the symbol used for length?

a) W

b) M

c) L

d) K

Answer: c

Explanation: Unit of length is meter.

3. What is the symbol used for time?

a) M

b) H

c) K

d) T

Answer: d

Explanation: Unit of time is second.

4. What is the symbol used for gravity?

a) g

b) h

c) k

d) v

Answer: a

Explanation: Unit of gravity is m/s 2 .

5. What is the symbol used for dynamic viscosity?

a) σ

b) µ

c) λ

d) δ

Answer: b

Explanation: Unit of dynamic viscosity is kg/m s.

6. What is the symbol used for thermal conductivity?

a) v

b) g

c) h

d) k

Answer: d

Explanation: Unit of thermal conductivity is W/m Degree.

7. What is the symbol used for specific heat?

a) C S

b) C R

c) C P

d) C M

Answer: c

Explanation: Unit of specific heat is k J/kg degree.

8. What is the symbol used for the convective film coefficient?

a) v

b) h

c) σ

d) µ

Answer: b

Explanation: Unit of convective film coefficient is W/m 2 degree.

9. What is the symbol used for kinematic viscosity?

a) v

b) λ

c) µ

d) σ

Answer: a

Explanation: Unit of kinematic viscosity is m 2 /s.

10. What is the symbol used for the coefficient of volumetric expansion?

a) λ

b) µ

c) δ

d) β

Answer: d

Explanation: Unit of a coefficient of volumetric expansion is per degree.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on ” Types Of Flow”.

1. How many types of fluid flow are characterized in the realms of fluid mechanics?

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: There are two types of flow i.e. laminar and turbulent flow.

2. In which fluid flow, the motion of fluid particles is irregular?

a) Turbulent

b) Laminar

c) One dimensional

d) Two dimensional

Answer: a

Explanation: It proceeds along erratic and unpredictable paths.

3. Following are the characteristics of turbulent flow

 Eddying

 Sinuous

 Rectilinear

Identify the correct option

a) ii and iii

b) i and iii

c) i, ii and iii

d) i and ii

Answer: d

Explanation: Obviously a turbulent flow is eddying and sinuous rather than rectilinear in character.

4. The nature of the fluid flow is governed by the following parameters

 Mean flow velocity

 Density of fluid

 Dynamic viscosity of the fluid

Identify the correct statements

a) i and iii

b) i only

c) i, ii and iii

d) ii and iii

Answer: c

Explanation: Osborne Reynolds, an English scientist confirmed the existence of these two regimes experimentally.

5. The value of convective coefficient of air in case of free convection is

a) 3-7 W/m 2 K

b) 3-4 W/m 2 K

c) 8-9 W/m 2 K

d) 9-9.5 W/m 2 K

Answer: a

Explanation: In case of air, the value of convective coefficient is small i.e. 3 W/m 2 K to 7 W/m 2 K. This is due to presence of some moisture in air.

6. The fluid particles move in flat or curved un-mixing layers or streams and follow a smooth continuous path. This type of flow is known as

a) Steady flow

b) Stream flow

c) Turbulent flow

d) Laminar flow

Answer: d

Explanation: The paths of fluid movement are well defined so it is laminar flow.

7. The characteristic dimension d in the relation R E = V d p/δ is the equivalent diameter and is defined as how many times the cross-sectional flow area divided by the wetted perimeter

a) 7

b) 4

c) 1

d) 6

Answer: b

Explanation: It is defined as four times the equivalent diameter and is defined as how many times the cross-sectional flow area divided by wetted perimeter.

8. For a duct of rectangular cross-section with length l and breadth b, the value of d e is

a) l b / l + b

b) 2 l b

c) 2 l b / l + b

d) 4 l b / l + b

Answer: c

Explanation: d e = 4  /2 l + 2 b.

9. In many flow situations, the duct can be

 Circular

 Rectangle

 Trapezoidal

 Annulus

Identify the correct option

a) i and ii

b) i, ii, iii and iv

c) i, ii and iii

d) iii and iv

Answer: b

Explanation: In many flow situations, the duct is not circular but is a rectangle, trapezoidal or even annulus formed by a tube within another tube.

10. If an annulus has an inner diameter of d 1 and an outer diameter of d 2 then the equivalent diameter is

a) 2 d 2 – d 1

b) d 2 – 2 d 1

c) d 1 – d 2

d) d 2 – d 1

Answer: d

Explanation: 4  (d 2 2 – d 1 2 )/π (d 1 + d 2 ).

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Reynolds Number”.

1. The ratio of inertia force to viscous force is known as

a) Grashof number

b) Reynolds number

c) Fourier number

d) Nusselt number

Answer: b

Explanation: Reynolds number is indicative of the relative importance of inertial and viscous effects in a fluid motion.

2. Velocity within the given fields would be similar in magnitude, direction and turbulence pattern when

a) Nusselt number are different

b) Nusselt number are same

c) Reynolds number are different

d) Reynolds number are same

Answer: d

Explanation: Reynolds number constitutes an important criterion of kinematic and dynamic similarity in forced convection heat transfer.

3. Reynolds number is given by the quantity

a) p V l/δ

b) 2 p V l/δ

c) 3 p V l/δ

d) 4 p V l/δ

Answer: a

Explanation: p V 2 l 2 / δ V l.

Where, p is density

δ is viscosity

V is volume

l is length.

4. Air enters a rectangular duct measuring 30 cm by 40 cm with a velocity of 8.5 m/s and a temperature of 40 degree Celsius. The flowing air has a thermal conductivity 0.028 W/m K, kinematic viscosity 16.95 * 10 -6 m 2 /s and from empirical correlations the Nusselt number has been approximated to be 425. Find out the flow Reynolds number

a) 1.1719 * 10 6

b) 2.1719 * 10 6

c) 0.1719 * 10 6

d) 4.1719 * 10 6

Answer: c

Explanation: R e = V d e /v.

5. Consider the above problem, find the convective heat flow coefficient

a) 24.71 W/m 2 K

b) 34.71 W/m 2 K

c) 44.71 W/m 2 K

d) 54.71 W/m 2 K

Answer: b

Explanation: h = N U k/d e .

6. For laminar flow, Reynolds number should be

a) Less than 2300

b) Equal to 2300

c) Greater than 2300

d) Less than 4300

Answer: a

Explanation: In laminar flow, the fluid particles move in flat or curved un-mixing layers or streams and follow a smooth continuous path.

7. For turbulent flow, Reynolds number must be

a) Less than 5000

b) Equal to 6000

c) Less than 6000

d) Greater than 6000

Answer: d

Explanation: In turbulent flow, the motion of fluid particles is irregular, and it proceeds along erratic and unpredictable paths.

8. What is the value of convective coefficient of air and superheated steam in case of forced convection?

a) 30-900 W/m 2 K

b) 30-700 W/m 2 K

c) 30-300 W/m 2 K

d) 30-400 W/m 2 K

Answer: c

Explanation: In forced convection, flow of fluid is caused by a pump, a fan or by the atmospheric winds. These mechanical devices provide a definite circuit for the circulating currents.

9. Heat is being transferred by convection from water at 48 degree Celsius to glass plate whose surface is exposed to water at 40 degree Celsius. The thermal conductivity of water is 0.6 W/m K and the thermal conductivity of glass is 1.2 W/m K. The spectral gradient of temperature in the water at the water glass interface is 10 -4 K/m. The heat transfer coefficient in W/m 2 K is

heat-transfer-questions-answers-reynolds-number-q9

a) 0.0

b) 750

c) 6.0

d) 4.8

Answer: b

Explanation: q = h d t = 6000. Therefore, h = 6000/48 – 40.

10. For transient flow, the value of Reynolds number may vary between

a) 1450-9870

b) 1200-4500

c) 2300-6000

d) 6000-9000

Answer: c

Explanation: The term transient designates a phenomenon which is time dependent.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Types of Convection”.

1. Conduction plus fluid flow in motion is known as

a) Radiation

b) Conduction

c) Convection

d) Heat exchanger

Answer: c

Explanation: It is convection i.e. conduction plus some velocity. It is the process of energy transport affected by the mixing of a fluid medium.

2. How many types of convection are there?

a) 4

b) 3

c) 2

d) 1

Answer: b

Explanation: It is of three types i.e. forced convection, natural convection and mixed convection.

3. Which of the following heat flow situations pertains to free or natural convection?

a) Air conditioning installations and nuclear reactors

b) Flow of water inside the condenser tubes

c) Cooling of internal combustion engine

d) Cooling of billets in atmosphere

Answer: d

Explanation: Cooling of billets in atmosphere is both free and natural convection.

4. Mark the system where heat transfer is given by forced convection

a) Chilling effect of cold wind on warm body

b) Fluid passing through the tubes of a condenser and other heat exchange equipment

c) Heat flow from a hot pavement to surrounding atmosphere

d) Heat exchange on the outside of cold and warm pipes

Answer: b

Explanation: If the fluid motion involve in the process is induced by some external means then it is called forced convection.

5. Forced convection in a liquid bath is caused by

a) Intense stirring by an external agency

b) Molecular energy interactions

c) Density difference brought about by temperature gradients

d) Flow of electrons in a random fashion

Answer: a

Explanation: If the fluid motion involve in the process is induced by some external means then it is called forced convection.

6. A finned tube hot water radiator with a fan blowing air over it is kept in rooms during winter. The major portion of the heat transfer from the radiation is due to

a) Combined conduction and radiation

b) Radiation to the surroundings

c) Better conduction

d) Convection to the air

Answer: d

Explanation: Convection is a process in which thermal energy is transferred between solid and fluid flowing through it.

7. A body cooling from 80 degree Celsius to 70 degree Celsius takes 10 minutes when left exposed to environmental conditions. If the body is to cool further from 70 degree Celsius to 60 degree Celsius under the same external conditions, it will take

a) Same time of 10 minutes

b) More than 10 minutes

c) Less than 10 minutes

d) Time will depend upon the environmental conditions

Answer: b

Explanation: Q = h A (t b – t a ). Apparently, the cooling depends upon the same temperature difference.

8. On a summer day, a scooter rider feels more comfortable while on the move than while at a stop light because

a) An object in motion captures less radiation

b) Air has a low specific heat and hence it is cooler

c) More heat is loss by convection and radiation while in motion

d) Air is transparent to radiation and hence it is cooler than the body

Answer: d

Explanation: The situation corresponds to forced convection when the scooter is in motion and the convective heat transfer coefficient for forced convection is greater than that for free convection.

9. What is the value of convective coefficient of oil in case of forced convection?

a) 1460-3000 W/m 2 K

b) 460-3000 W/m 2 K

c) 60-3000 W/m 2 K

d) 160-3000 W/m 2 K

Answer: c

Explanation: In forced convection, the flow of fluid is caused by a pump, fan or by atmospheric winds. Convection mechanism involving phase changes leads to the important fields of boiling and condensation.

10. A sphere, a cube and a thin circular plate, all made of the same material and having the same mass are initially heated to the temperature of 250 degree Celsius. When left in air at room temperature, what will be their response to cooling?

a) Cube will cool faster than sphere but slower than the circular plate

b) They will cool at the same rate

c) Sphere will cool faster

d) Circular plate will cool at the slower rate

Answer: a

Explanation: Q = h A . Here cooling depends upon the surface area of the body.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Nusselt Number”.

1. Which quantity signifies the ratio of temperature gradient at the surface to a reference temperature gradient?

a) Reynolds number

b) Nusselt number

c) Fourier number

d) Stanton number

Answer: b

Explanation: It is given by h l/k.

2. The determination of a value of Nusselt number or the convective film coefficient forms a basis for the computation of heat transfer by convection. Towards that end, following approaches have been suggested

 Non-dimensional analysis and experimental correlations

 Hydrodynamic concept of velocity boundary layer

 Reynolds similarity between the mechanism of fluid friction in the boundary layer and the transfer of heat by convection

Identify the correct one

a) i, ii and iii

b) i and ii

c) ii and iii

d) i and iii

Answer: c

Explanation: It should be dimensional analysis and experimental correlations.

3. Nusselt number is given by

a) h l/k

b) 2 h l/k

c) 3 h l/k

d) 4 h l/k

Answer: a

Explanation: The length parameter l specifies the geometry of solid body.

4. The temperature profile at a particular location in a thermal boundary layer is prescribed by n expression of the form

t  = A- B y + C y 2

Where, A, B and Care constants. What is the value of heat transfer coefficient?

a) B k/ (t s – t infinity )

b) 2 B k/ (t s – t infinity )

c) 3 B k/ (t s – t infinity )

d) 4 B k/ (t s – t infinity )

Answer: a

Explanation: h = – k/ (t s – t infinity ) [d t/d y] y = 0 and  y = 0 = – B.

5. The temperature profile at a particular location on a surface is prescribed by the identity

(t s – t) / (t s – t infinity ) = sin 

If thermal conductivity of air is stated to be 0.03 W/m K, determine the value of convective heat transfer coefficient

a) 6.48 W/m 2 K

b) 6.38 W/m 2 K

c) 6.28 W/m 2 K

d) 6.18 W/m 2 K

Answer: c

Explanation: h = – k/ (t s – t infinity ) [d t/d y] y = 0 . Therefore h = – k/ (t s – t infinity ) [π (t s – t infinity )/0.015].

6. Air at 20 degree Celsius flows over a flat plate maintained at 75 degree Celsius. Measurements shows that temperature at a distance of 0.5 mm from the surface of plate is 50 degree Celsius. Presuming thermal conductivity of air is 0.0266 W/m K, estimate the value of local heat transfer coefficient

a) 23.18 W/m 2 K

b) 24.18 W/m 2 K

c) 25.18 W/m 2 K

d) 26.18 W/m 2 K

Answer: b

Explanation: h = – k/ (t s – t infinity ) [d t/d y] y = 0 and d t/d y = – 50 * 10 3 degree Celsius/m.

7. Air at 20 degree Celsius flows over a flat surface maintained at 80 degree Celsius. Estimate the value of local heat transfer coefficient if the local heat flow at a point was measured as 1250 W/m 2 . Take thermal conductivity of air as 0.028 W/m K

a) 23.83 W/m 2 K

b) 22.83 W/m 2 K

c) 21.83 W/m 2 K

d) 20.83 W/m 2 K

Answer: d

Explanation: Q = h A (t s – t infinity ).

8. Consider the above problem, calculate the temperature gradient at the surface

a) – 44636 degree Celsius/m

b) – 34636 degree Celsius/m

c) – 24636 degree Celsius/m

d) – 14636 degree Celsius/m

Answer: a

Explanation:  y = 0 = – h (t s – t infinity )/k.

9. At the interface of solid body, heat flows by conduction and is given by

a) A (t s – t infinity )

b) h A (t s – t infinity )

c) h (t s – t infinity )

d) h A

Answer: b

Explanation: Q = – k A  y = 0 .

10. For a given value of Nusselt number, the convective surface coefficient h is directly proportional to

a) Length

b) Mass

c) Thermal conductivity

d) Density

Answer: c

Explanation: It is directly proportional to k and inversely proportional to significant length.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Newton- Rikhman Law”.

1. Newton-Rikhman law is given by

a) Q = h A (t s – t f )

b) Q = 2 h A (t s – t f )

c) Q = 3 h A (t s – t f )

d) Q = 4 h A (t s – t f )

Answer: a

Explanation: Regardless of the particular nature, the appropriate rate equation for the convective heat transfer is prescribed by Newton’s law of cooling.

2. The value of film coefficient is dependent upon

 Boundary layer configuration

 Geometry and orientation of the surface

 Surface conditions

a) i and ii

b) ii and iii

c) i and ii

d) i, ii and iii

Answer: d

Explanation: It depends upon surface conditions i.e. roughness and cleanliness, geometry and orientation of the surface i.e. plate, tube and cylinder placed vertically or horizontally.

3. The convection coefficients for boiling and condensation lie in the range

a) 5000-12500 W/m 2 K

b) 2500-100000 W/m 2 K

c) 2500-5000 W/m 2 K

d) 2500-12500 W/m 2 K

Answer: b

Explanation: Convection mechanisms involving phase changes lead to important field of boiling and condensatio.

4. Forced air flows over a convection heat exchanger in a room heater, resulting in a convective heat transfer coefficient 1.136 k W/m 2 K. The surface temperature of heat exchanger may be considered constant at 65 degree Celsius, and the air is at 20 degree Celsius. Determine the heat exchanger surface area required for 8.8 k W of heating

a) 0.272 m 2

b) 0.472 m 2

c) 0.172 m 2

d) 0.672 m 2

Answer: c

Explanation: Q = h A (t s – t f ). So. A = 0.172 m 2 .

5. A region of fluid motion near a plate in which temperature gradient exist is

a) Thermal boundary layer

b) Diathermia boundary layer

c) Turbulent flow

d) Laminar flow

Answer: a

Explanation: The fluid velocity decreases as it approaches the solid surface reaching to zero in the fluid layer immediately next to the surface. The thin layer of stagnated fluid is called thermal boundary layer.

6. Thermo-physical properties of the fluid are represented by

 Density

 Viscosity

 Specific heat

 Thermal conductivity

Identify the correct option

a) i and ii

b) i, ii, iii and iv

c) ii, iii and iv

d) i, ii and iii

Answer: b

Explanation: The value of film coefficient is dependent upon thermos-physical properties of he fluid i.e. density, viscosity, specific heat, coefficient of expansion and thermal conductivity.

7. A motor cycle cylinder consists of ten fins, each 150 mm outside diameter and 75 mm inside diameter. The average fin temperature is 500 degree Celsius and the surrounding air is at 20 degree Celsius temperature. Make calculations for the rate of heat dissipation from the cylinder fins by convection when motor cycle is stationary and convective coefficient is 6 W/m 2 K

a) 432.2 W

b) 532.2 W

c) 632.2 W

d) 763.2 W

Answer: d

Explanation: A = 0.265 m 2 and Q =    = 763.2 W.

8. Consider the above problem, make calculations for the rate of heat dissipation from the cylinder fins by convection when motor cycle is moving at 60 km/hr and convective coefficient is 75 W/m 2 K

a) 9640 W

b) 9540 W

c) 9440 W

d) 9340 W

Answer: b

Explanation: A = 0.265 m 2 and Q =    = 9540 W.

9. The temperature profile at a particular location on a surface is prescribed by the identity

(t s – t) / (t s – t infinity ) =   3 +  

If thermal conductivity of air is stated to be 0.03 W/m K, determine the value of convective heat transfer coefficient

a) 4 W/m 2 K

b) 5 W/m 2 K

c) 6 W/m 2 K

d) 7 W/m 2 K

Answer: c

Explanation: h = – k/ (t s – t infinity ) [d t/d y] y = 0 .

10. Air at 20 degree Celsius flows over a flat surface maintained at 80 degree Celsius. The local heat flow at a point was measured as 1250 W/m 2 .Take thermal conductivity of air as 0.028 W/m K, calculate the temperature at a distance 0.5 mm from the surface

a) 57.682 degree celsius

b) 67.682 degree celsius

c) 77.682 degree celsius

d) 87.682 degree celsius

Answer: a

Explanation: Temperature at 0.5 mm from the surface is 80 +  y = 0  = 57.682 degree celsius.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “System of dimensions”.

1. What is the dimension of heat?

a) M L T -2

b) M L 2 T -2

c) M L 2 T -1

d) M L 2 T

Answer: b

Explanation: Unit of heat is Joule. It is a form of energy that can be transferred from an object of high temperature to the lower one.

2. What is the dimension of dynamic viscosity?

a) M L -1 T -1

b) L -1 T -1

c) M L -2 T -1

d) M L -1 T -2

Answer: a

Explanation: Unit of dynamic viscosity is kg/meter second. It is defined as a quantity measuring the force needed to overcome internal friction of a moving liquid.

3. What is the dimension of kinematic viscosity?

a) M L 2 T -2

b) L 2 T -2

c) L 2 T -1

d) M L 2 T -1

Answer: c

Explanation: Unit of kinematic viscosity is m 2 /s. It is defined as dynamic viscosity per unit density.

4. What is the dimension of energy?

a) M L 2 T -2

b) M L 2 T -1

c) M L 1 T -2

d) M L T -2

Answer: a

Explanation: Unit of energy is m N. It is a form of energy that can be transferred from one object to another.

5. What is the dimension of force?

a) M L T -1

b) L T -2

c) M T -2

d) M L T -2

Answer: d

Explanation: Unit of force is N. It is a push or pull that can stop a moving object.

6. What is the dimension of density?

a) M L -3 T

b) M L -3

c) M L -3 T 2

d) M L -2

Answer: b

Explanation: Unit of density is kg/m 3 . It is defined as mass per unit volume.

7. What is the dimension of acceleration?

a) M L T -2

b) L T -4

c) M L T -1

d) L T -2

Answer: d

Explanation: Unit of acceleration is m/s 2 . It is defined as force per unit mass.

8. What is the dimension of specific heat?

a) L T -1 α -1

b) M L T -1 α -1

c) L T -2 α -1

d) M L T -2 α -1

Answer: c

Explanation: Unit of specific heat is k J/kg K. The specific heat is the amount of heat per unit mass required to raise the temperature by one degree celsius.

9. What is the dimension of thermal conductivity?

a) M L T -3 α -1

b) M L T -3

c) M L T -2 α -1

d) M L T -3 α -2

Answer: a

Explanation: Unit of thermal conductivity is W/m K. It is the amount of heat per unit time per unit area that can be conducted through a plate of unit thickness of a given material.

10. What is the dimension of gravity?

a) M L T -2

b) L T -2

c) M L T -1

d) L T -4

Answer: b

Explanation: Unit of gravity is m/s 2 . It is a force that attracts a body towards itself.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Dimensional Homogeneity”.

1. How is dimensional homogeneity related with fundamental units of measurements?

a) Independent

b) Dependent

c) Dependent but can vary

d) Twice

Answer: a

Explanation: This implies that the length dimension can be added to subtract from only a length dimension.

2. What is the time period of oscillation of a simple pendulum of length L and mass m?

a) 2 π  3/2

b) 2 π 

c) 2 π  2

d) 2 π  1/2

Answer: d

Explanation: The time period of a simple pendulum means the time it takes for one complete revolution.

3. The principle of dimensional homogeneity serves the following useful concepts

 It helps to check whether an equation of any physical phenomenon is dimensionally homogenous or not

 It helps to determine the dimensions of a physical quantity

 It helps to convert the units from one system to another

Identify the correct statements

a) i and ii

b) ii and iii

c) i, ii and iii

d) ii only

Answer: c

Explanation: It serves all the aspects.

4. The equation of friction loss in a pipe of length l and diameter d through which fluid flows with velocity v is

a) h i = 4 f V 2 /d g

b) h i = 4 f V 2 /d 2 g

c) h i = 4 f V 2 /d 2 g

d) h i = 4 f V 2 /2 g

Answer: b

Explanation: Here f is any constant with no dimensions whatsoever.

5. Bernoulli’s equation for fluid flow along a stream line is given as

a) p/w + V 2 /2 g + y = 2

b) p/w + V/2 g + y = constant

c) p/w + V 2 /2 g + y = 1

d) p/w + V 2 /2 g + y = constant

Answer: d

Explanation: This is the Bernoulli’s equation and is dimensionally homogenous.

6. The convective film coefficient in k cal/m 2 hr degree can be converted to J/m 2 s degree by multiplying it with a factor

a) 1.1627

b) 1.1527

c) 1.1427

d) 1.1327

Answer: a

Explanation: Both k cal/m 2 hr degree and J/m 2 s are the units of convective film coefficient.

7. The pressure in kg/cm 2 can be converted to N/m 2 by multiplying it with a factor

a) 9.807 * 10 2

b) 9.807 * 10 3

c) 9.807 * 10 4

d) 9.807 * 10 5

Answer: c

Explanation: Both kg/cm 2 and N/m 2 are the units of pressure.

8. How many Newton’s are there in one kg?

a) 9.507

b) 9.607

c) 9.707

d) 9.807

Answer: d

Explanation: 1 kg = 9.807 N. Newton is the unit of force and kg is the unit of mass.

9. How many Joule are there in 1 k cal?

a) 3186

b) 4186

c) 5186

d) 6186

Answer: b

Explanation: 1 k cal = 4186 J. Joule is the unit of work done.

10. How many fundamental quantities are there?

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: These are M, L and T. M is mass, L is length and T is time.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Rayleigh’s Method”.

1. What is the dimension of coefficient of volumetric expansion?

a) α

b) α 1

c) α -2

d) α -1

Answer: d

Explanation: Its unit is per degree. Coefficient of volumetric expansion is defined as the percentage increase in volume.

2. What is the dimension of convective film coefficient?

a) M T -3 α -1

b) M T -3 α -2

c) M T -2 α -1

d) M T -1 α -1

Answer: a

Explanation: Its unit is k cal/m 2 hr degree. It is used in thermodynamics to calculate the heat and it is denoted by h.

3. What is the dimension of velocity?

a) L T -2

b) L T 1

c) L T -1

d) L T

Answer: c

Explanation: Its unit is m/s. Velocity is defined as a speed in a particular direction.

4. What is the dimension of area?

a) M L 2

b) L 2

c) L 1

d) L 3

Answer: b

Explanation: Its unit is m 2 . It is defined as a range of activity or an interest.

5. What is the dimension of volume?

a) L 2

b) M L 3

c) L

d) L 3

Answer: d

Explanation: Its unit is m 3 . It is defined as a region or part of a town, a country or the world.

6. What is the dimension of work?

a) M L 2 T -2

b) M L 2 T -1

c) M L 2 T

d) M L 1 T -2

Answer: a

Explanation: Its unit is m N. Work can be defined as transfer of energy.

7. What is the dimension of temperature?

a) α -2

b) α 2

c) α

d) M α

Answer: c

Explanation: Its unit is Kelvin. It refers to how cold or hot something is.

8. What is the dimension of mass?

a) M

b) M L

c) L

d) M L T

Answer: a

Explanation: Its unit is kg. It is the property of a physical body which determines the strength of its mutual gravitational attraction to the earth.

9. What is the dimension of length?

a) T

b) M

c) L

d) M L

Answer: c

Explanation: Its unit is a meter. in geometrical measurements, it is the most extended dimension of an object.

10. What is the dimension of time?

a) M T

b) T

c) M

d) M L

Answer: b

Explanation: Its unit is second. Time is a measure in which events can be ordered from the past through the present into the future.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Buckingham’s Pi- Method”.

1. If there are n variables in a dimensionally homogeneous equation and if these variables contain m primary dimensions, then the variables can be grouped into how many non-dimensional parameters?

a) m

b) n-m

c) n-2m

d) n

Answer: b

Explanation: The non-dimensional groups are called pi-terms. If there are n variables in a dimensionally homogenous equation and if these variables contain m primary dimensions, then the variables can be grouped into  non dimensional parameters.

2. Which of the following is not known as a core group?

a) Time

b) Geometric property

c) Fluid property

d) Flow characteristics

Answer: a

Explanation: These are also known as repeated variables.

3. What is the dimension of resistance?

a) M L T 2

b) M L T

c) M L T -2

d) M L T -1

Answer: c

Explanation: Unit of resistance is N. Resistance is the stopping effect exerted by one material thing to another.

4. The resistance R experienced by a partially submerged body depends upon the velocity V, length of the body l, viscosity of the fluid µ, density of the fluid p and gravitational acceleration g. Establish a relation between involving non-dimensional groups

a) R = p V l 2 function [p V l/ µ, V/  1/2 ].

b) R = p l 2 function [p V l/ µ, V/  1/2 ].

c) R = V 2 l 2 function [p V l/ µ, V/  1/2 ].

d) R = p V 2 l function [p V l/ µ, V/  1/2 ].

Answer: d

Explanation: This can be solved by Buckingham pi theorem.

5. If there are 6 physical quantities and 3 fundamental units, then the number of pi terms are

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: It should be 6-3 = 3.

6. The study of predicting prototype conditions from model observations is known as

a) Similitude

b) Geometrical similarity

c) Prototype

d) Model

Answer: a

Explanation: It prescribes the relationship between a full scale flow and a flow involving smaller but geometrically similar boundaries.

7. S.I unit of heat is

a) J

b) W

c) J s

d) W/K

Answer: a

Explanation: Its unit is Joule. Heat is a form of energy that can be transferred between two objects.

8. S.I unit of the coefficient of volumetric expansion is

a) Degree

b) Per meter

c) Per second

d) Per degree

Answer: d

Explanation: Its unit is per degree. Coefficient of volumetric expansion is defined as increase in volume.

9. S.I unit of specific heat is

a) K kg/J

b) kg/K

c) k J/kg K

d) J/K

Answer: c

Explanation: Its unit is k J/kg K. It is another physical property of matter. All the matter has a temperature associated with it.

10. S.I unit of dynamic viscosity is

a) kg/m

b) kg/m s

c) m/kg

d) m s /kg

Answer: b

Explanation: Its unit is kg/m s. It is the kinematic viscosity per unit density.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Model Studies And Similitude”.

1. Which is a full size structure employed in the actual engineering design?

a) Proton

b) Prototype

c) Electron

d) Neutron

Answer: b

Explanation: The prototype operates under the actual working conditions.

2. Which term refers to the theory and art of predicting prototype conditions from model observations?

a) Nusselt number

b) Dimensional homogeneity

c) Thermal boundary layer

d) Similitude

Answer: d

Explanation: It prescribes the relationship between a full scale flow and a flow involving smaller but geometrically similar boundaries.

3. The results obtained from experiments on models can be applied to prototype only if a complete similarity exists between the model and prototype and for that the two systems may be

 Geometrically similar

 Kinematically similar

 Dynamically similar

Identify the correct statements

a) i and ii

b) ii and iii

c) i, ii and iii

d) i and iii

Answer: c

Explanation: It should be geometrically, cinematically and dynamically similar.

4. Geometrically similarity prescribes that the ratio of the corresponding linear dimensions of the two systems are

a) Unity

b) Same

c) Never same

d) May be twice

Answer: b

Explanation: It refers to the similarity of shape and form.

5. Thermal similarity refers to the comparison of two systems made on the basis of their

a) Temperature

b) Specific heat

c) Heat flux

d) Length

Answer: d

Explanation: A similarity in thermal quantities is achieved when Prandtl number is same for both the fields.

6. The comparison of two systems made on the basis of their temperature, specific heat and heat flu is known as

a) Dynamic similarity

b) Kinematic similarity

c) Thermal similarity

d) Geometrical similarity

Answer: c

Explanation: It is known as thermal similarity.

7. The similarity of masses and forces of the corresponding particles of flow is known as

a) Kinematic similarity

b) Dynamic similarity

c) Geometrical similarity

d) Thermal similarity

Answer: b

Explanation: Systems are dynamically similar if the corresponding particles experience the similar force.

8. The similarity of motion is known as

a) Thermal similarity

b) Dynamic similarity

c) Geometrical similarity

d) Kinematic similarity

Answer: d

Explanation: Both the systems undergo similar rates of change of motion.

9. The similarity of shape and form is known as

a) Geometrical similarity

b) Thermal similarity

c) Geometrical similarity

d) Kinematic similarity

Answer: a

Explanation: They may differ in size but are identical in shape.

10. A similarity in thermal quantities is achieved when

a) Nusselt number is same for both the fields

b) Nusselt number is different for both the fields

c) Prandtl number is same for both the fields

d) Prandtl number is different for both the fields

Answer: c

Explanation: A similarity in thermal quantities is achieved when Prandtl number is same for both the fields.

This set of Heat Transfer aptitude tests focuses on “Dimensional Analysis – Advantages And Limitations”.

1. Ratio of actual velocity to sonic velocity is known as

a) Mach number

b) Peclet number

c) Reynolds number

d) Grashof number

Answer: a

Explanation: It is the ratio of the speed of a body to the speed of sound in the surrounding medium.

2. The value of Prandtl number for air is about

a) 0.1

b) 0.4

c) 0.7

d) 1.1

Answer: c

Explanation: It is about 0.7. It is indicative of the relative ability of the fluid to diffuse momentum and internal energy by molecular mechanisms.

3. Let us say Mach number , the flow is

a) Sonic

b) Subsonic

c) Supersonic

d) No flow

Answer: c

Explanation: Compressibility effects are important for supersonic aircraft, and shock waves are generated by the surface of the object. It is supersonic.

4. Free convection heat flow does not depend on

a) Density

b) Coefficient of viscosity

c) Gravitational force

d) Velocity

Answer: b

Explanation: It does not depend on coefficient of viscosity.

5. Which dimensionless number has a significant role in forced convection?

a) Prandtl number

b) Peclet number

c) Mach number

d) Reynolds number

Answer: d

Explanation: It is indicative of the relative importance of inertial and viscous effects in a fluid motion. Reynolds number is important in forced convection.

6. The non-dimensional parameter known as Stanton number is used in

a) Forced convection heat transfer

b) Condensation heat transfer

c) Natural convection heat transfer

d) Unsteady state heat transfer

Answer: a

Explanation: It is the ratio of heat transfer coefficient to the flow of heat per unit temperature rise due to velocity of the fluid. It is used only in forced convection heat transfer.

7. The Prandtl number will be lowest for

a) Water

b) Liquid metal

c) Lube oil

d) Aqueous solution

Answer: b

Explanation: It is minimum in case of liquid metal.

8. The dimensionless parameter l 3 p 2 β g d t/µ 2 is referred to as

a) Stanton number

b) Schmidt number

c) Grashof number

d) Peclet number

Answer: c

Explanation: It indicates the relative strength of the buoyant to viscous forces. It is the Schmidt number.

9. Figure depicts the variation of which two numbers?

heat-transfer-aptitude-test-q9

a) Nusselt number and Reynolds number

b) Stanton number and Reynolds number

c) Peclet number and Grashof number

d) Nusselt number and Stanton number

Answer: a

Explanation: It depicts variation of Nusselt number and Reynolds number for flow of air over a pipe of 25 mm outside diameter. This correlation curve permits the evaluation of Nusselt number for air flow over any size of pipe as long as Reynolds number of that arrangement lies within the range.

10. Heat is loss from a 100 mm diameter steam pipe placed horizontally in ambient air at 30 degree Celsius. If the Nusselt number is 25 W/m 2 K and thermal conductivity of the air is 0.03 W/m K, then heat transfer coefficient will be

a) 7.5 W/m 2 K

b) 16.5 W/m 2 K

c) 25 W/m 2 K

d) 30 W/m 2 K

Answer: a

Explanation: N = h d/k. So, h = N k/d = 7.5 W/m 2 K.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Significance Of Dimensionless Groups”.

1. Ratio of inertia force to viscous force is known as

a) Grashof number

b) Reynolds number

c) Peclet number

d) Stanton number

Answer: b

Explanation: It is given by p l v/µ. It is the indicative of the relative importance of inertial and viscous effects in a fluid motion.

2. The ratio of heat flow rate by convection to flow rate by conduction is known as

a) Stanton number

b) Graetz number

c) Fourier number

d) Peclet number

Answer: d

Explanation: It is given by I V/α. It is a function of Reynolds number and Prandtl number.

3. The ratio of heat capacity of fluid flowing through the pipe per unit of length to the conductivity of pipe material is known as

a) Graetz number

b) Reynolds number

c) Peclet number

d) Fourier number

Answer: a

Explanation: It is given by Pe . It represents the ratio of the heat capacity of fluid flowing through the pipe per unit of length to the conductivity of pipe material.

4. Identify the correct statement

a) Peclet number =  

b) Peclet number =  

c) Peclet number =  

d) Peclet number =  

Answer: c

Explanation: Peclet number is a function of Reynolds number and Prandtl number. Peclet number = p c l V/ k = p l V/µ µ =  .

5. What is the value of Prandtl number for highly viscous oils?

a) 100-1000

b) 0-100

c) 10-100

d) 100-10000

Answer: d

Explanation: It indicates the rapid diffusion of momentum by viscous action compared to the diffusion of energy.

6. What is the value of Prandtl number for liquid metals?

a) 0.003-0.01

b) 0.01-0.1

c) 0.1-0.5

d) 0.5-0.95

Answer: a

Explanation: It indicates more rapid diffusion of energy compared to the momentum diffusion rate.

7. The product of buoyant force and inertia force to the square of the viscous force is known as

a) Stanton number

b) Grashof number

c) Fourier number

d) Peclet number

Answer: b

Explanation: It indicates the relative strength of the buoyant to the viscous forces. It represents the ratio of the product of buoyant and inertia forces to the square of the viscous forces.

8. The ratio of heat transfer coefficient to the flow of heat per unit temperature rise due to the velocity of the fluid is known as

a) Fourier number

b) Grashof number

c) Peclet number

d) Stanton number

Answer: d

Explanation: Stanton number can be used in correlating forced convection data. This becomes obvious when we observe the velocity V contained in the expression of Stanton number. It is the ratio of heat transfer coefficient to the flow of heat per unit temperature rise due to the velocity of the fluid.

9. Which number indicates the relative ability of the fluid to diffuse momentum and internal energy by molecular mechanisms?

a) Nusselt number

b) Prandtl number

c) Peclet number

d) Stanton number

Answer: b

Explanation: It is the ratio of kinematic viscosity to thermal diffusivity of the fluid.

10. Which number establishes the relation between convective film coefficient, thermal conductivity of the fluid and a significant length parameter?

a) Nusselt number

b) Stanton number

c) Peclet number

d) Fourier number

Answer: a

Explanation: The Nusselt number may be interpreted as the ratio of temperature gradient to an overall reference temperature gradient.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Correlations For Forced Convection”.

1. The convective heat transfer coefficient in laminar flow over a flat plate

a) Increases with distance

b) Increases if a higher viscosity fluid is used

c) Increases if a denser fluid is used

d) Decreases with increase in free stream velocity

Answer: c

Explanation: It mostly increases if a denser fluid is used.

2. For laminar flow over a flat plate, the average value of a Nusselt number is prescribed by the relation

Nu = 0.664  0.5  0.33

Which of the following is then a false statement?

a) Density has to be increased four times

b) Plate length has to be decreased four times

c) Specific heat has to be increased four times

d) Dynamic viscosity has to be decreased sixteen times

Answer: d

Explanation: The dynamic viscosity has an inverse relation to 1/6 power. To double the convective heat transfer coefficient, the dynamic viscosity has to be decreased 64 times.

3. For turbulent flow over a flat plate, the average value of Nusselt number is prescribed by the relation

Nu = 0.664  0.5  0.33

Which of the following is then a false statement?

The average heat transfer coefficient increases as

a) 1/5 power of plate length

b) 2/3 power of thermal conductivity

c) 1/3 power of specific heat

d) 4/5 power of a free stream velocity

Answer: a

Explanation: The average heat transfer coefficient reduces with length as 1/5th power of the length.

4. A nuclear reactor with its core constructed of parallel vertical plates 2.25 m high and 1.5 m wide has been designed on free convection heating of liquid bismuth. Metallurgical considerations limit the maximum surface temperature of the plate to 975 degree Celsius and the lowest allowable temperature of bismuth is 325 degree Celsius. Estimate the maximum possible heat dissipation from both sides of each plate. The appropriate correlation for the convection coefficient is

Nu = 0.13  1/3

a) 143 MW

b) 153 MW

c) 163 MW

d) 173 MV

Answer: b

Explanation: Q = 2 h A d t = 153 MW.

5. Consider the above problem, find the value of Grashoff number

a) 101.3 * 10 12

b) 102.3 * 10 12

c) 103.3 * 10 12

d) 104.3 * 10 12

Answer: d

Explanation: Grashof number = l 3 p 2 β g d t/µ 2 .

6. A thin walled duct of 0.5 m diameter has been laid in an atmosphere of quiescent air at 15 degree Celsius and conveys a particular gas at 205 degree Celsius. Base your calculations on one meter length of the duct, estimate the convective coefficient of heat transfer

a) 5.086 W/m 2 K

b) 6.086 W/m 2 K

c) 7.086 W/m 2 K

d) 8.086 W/m 2 K

Answer: a

Explanation: h = 1.37  0.25 = 5.086 W/m 2 K.

7. Free correction modulus is given by

a) p 2 β g c P /µ

b) p 2 β g c P /k

c) p 2 β g c P /µ k

d) p 2 β g c P

Answer: c

Explanation: It contains only fluid properties and is called the free convection modulus.

8. The free convection coefficient is given by

h = C 1 d t m /l 1 – 3m

The value of exponent for laminar flow is

a) 0.5

b) 0.6

c) 0.7

d) 0.8

Answer: a

Explanation: For laminar flow h = C 1  0.25 .

9. For inclined plates we multiply Grashoff number with

a) Cos 2 α

b) Sin 2 α

c) Sin α

d) Cos α

Answer: d

Explanation: It should be multiplied with cos α, as α is angle with the horizontal.

10. The free convection coefficient is given by

h = C 1 d t m /l 1 – 3m

The value of exponent for turbulent flow is

a) 0.43

b) 0.33

c) 0.23

d) 0.13

Answer: b

Explanation: For turbulent flow h = C .

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Laminar Flow”.

1. The Nusselt number is related to Reynolds number in laminar and turbulent flows respectively as

a) R e -1/2 and R e 0.8

b) R e 1/2 and R e 0.8

c) R e -1/2 and R e -0.8

d) R e 1/2 and R e -0.8

Answer: b

Explanation: Nusselt number = h l/k and Reynolds number = p V l/µ.

2. A hot plate of 15 cm 2 area maintained at 200 degree celsius is exposed to still air at 30 degree Celsius temperature. When the smaller side of the plate is held vertical, convective heat transfer rate is 15 per cent higher than bigger side of the plate is held vertical. Find size of the plate. The appropriate correlation for the convection coefficient is

heat-transfer-questions-answers-laminar-flow-q2

Nu = 0.60  0.25

a) l = 5.123 cm and b = 2.928 cm

b) l = 6.123 cm and b = 3.928 cm

c) l = 7.123 cm and b = 4.928 cm

d) l = 8.123 cm and b = 5.928 cm

Answer: a

Explanation: t f = 200 + 30/2 = 115 degree celsius. Pr = µ c/k, Gr = b 3 p 2 β g d t/µ 2 and Nu = h b/k.

3. Which of the following is true for laminar flow?

a) 10 4 < G r P r < 10 7

b) 10 4 < G r P r < 10 8

c) 10 4 < G r P r < 10 9

d) 10 4 < G r P r < 10 10

Answer: c

Explanation: The product G r P r is often referred to as Rayleigh number, and its value sets the criterion of laminar character of flow.

4. A horizontal heated plate at 200 degree Celsius and facing upwards has been placed in still air at 20 degree Celsius. If the plate measures 1.25 m by 1 m, make calculations for the heat loss by natural convection. The convective film coefficient for free convection is given by the following empirical relation

h = 0.32  0.25 W/m 2 K

Where α is the mean film temperature in degrees kelvin

a) 6006 W

b) 5006 W

c) 4006 W

d) 3006 W

Answer: a

Explanation: α = 273 + 200 + 20/2 = 383 degree Celsius. Therefore, rate of heat loss = h A d t = 3006 W.

5. For laminar flow, Reynolds number must not be less than

a) 1000

b) 2000

c) 20000

d) 40000

Answer: d

Explanation: It should not be less than 40000.

6. For laminar flow, Prandtl number must be more than

a) 0.05

b) 0.2

c) 0.6

d) 0.3

Answer: c

Explanation: It must be more than 0.6. It is indicative of the relative ability of the fluid to diffuse momentum and internal energy by molecular mechanisms.

7. Air at atmospheric pressure and 20 degree Celsius flows with 6 m/s velocity through main trunck duct of air conditioning system. The duct is rectangular in cross-section and measures 40 cm by 80 cm. Determine heat loss per meter length of duct corresponding to unit temperature difference. The relevant thermos-physical properties of air are

v = 15 * 10 -6 m 2 /s

α = 7.7 * 10 -2 m 2 /hr

k = 0.026 W/m degree

a) 32.768 W

b) 42.768 W

c) 52.768 W

d) 62.768 W

Answer: b

Explanation: Q = h A d t, Area = 2   = 2.4 m 2 , h = 365.34 k/l.

8. The local film coefficient for laminar flow past a flat plate may be obtained from the correlation

a) Nu = 0.332  0.5  0.33

b) Nu = 0.332  0.5  0.43

c) Nu = 0.332  0.5  0.53

d) Nu = 0.332  0.5  0.63

Answer: a

Explanation: Fluid properties are evaluated at mean film temperature.

9. For a plate of length l, an average value of Nusselt number is given by

a) Nu = 0.664  0.5  0.54

b) Nu = 0.664  0.5  0.74

c) Nu = 0.664  0.5  0.27

d) Nu = 0.664  0.5  0.33

Answer: d

Explanation: For a plate of length l, an average value of Nusselt number or convection coefficient may be obtained by integration.

10. The correlation for liquid metal is given by

a) Nu = 0.465  

b) Nu = 0.565  

c) Nu = 0.665  

d) Nu = 0.765  

Answer: b

Explanation: It is valid for Pr less than or equal to 0.05.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Turbulent Flow”.

1. Which of the following is true for turbulent flow?

a) G r P r > 10 8

b) G r P r > 10 9

c) G r P r > 10 3

d) G r P r > 10 15

Answer: b

Explanation: The product G r P r is often referred to as Rayleigh number, and its value sets the criterion of turbulent character of flow.

2. Mc Adam relation is given by

a) Nu = 0.023  0.8  0.4

b) Nu = 0.023  0.8  0.3

c) Nu = 0.023  0.8  0.2

d) Nu = 0.023  0.8  0.1

Answer: a

Explanation: Here, n = 0.4 if the fluid is being heated.

3. Investigate the effect of following condition on the average value of heat transfer coefficient in flow through a tube

Two fold increases in flow velocity by varying mass flow rate

It may be presumed that there is no change in the temperature of the liquid and the tube wall, and that the flow through the tube is turbulent in character

a) 64.1% decrease

b) 64.1% increase

c) 74.1%decrease

d) 74.1% increases

Answer: d

Explanation: h 2 /h 1 = (V 2 /V 1 ) 0.8 = .0741 = 74.1%

4. Consider the above problem for two fold increase in the diameter of the tube, the flow velocity is maintained constant by change in the rate of liquid flow

a) 15%

b) 14%

c) 13%

d) 12%

Answer: c

Explanation: h 2 /h 1 = (d 1 /d 2 ) 0.2 .

5. Calculate the rate of heat loss from a human body which may be considered as a vertical cylinder 30 cm in diameter, and 175 cm high while standing in a 30 km/hr wind at 15 degree Celsius. Human has a surface temperature of 35 degree Celsius

a) 588.86 W

b) 688.86 W

c) 788.86 W

d) 888.86 W

Answer: b

Explanation: Q = h A d t, h = Nu k/d.

6. A heat treat steel plate measures 3 m by 1 m and is initially at 30 degree Celsius. It is cooled by blowing air parallel to 1 m edge at 9 km/hr. If the air is at 10 degree Celsius. Estimate the convection heat transfer from both sides of the plate

a) 477.7 W

b) 547.7 W

c) 647.7 W

d) 747.6 W

Answer: d

Explanation: Q = h A d t, h = Nu k/d.

7. Consider the above problem, find the value of Reynolds number

a) 166003

b) 177003

c) 188003

d) 199003

Answer: a

Explanation: Re = p V l/µ = 16603

8. The oil pan of 1.6 engine approximates a flat plate 0.3 m wide by 0.45 m long and protrudes below the framework of the automobile. The engine oil is at 95 degree Celsius and the ambient air temperature is 35 degree Celsius. If the automobile runs at 36 km/hr, make calculations for the rate of heat transfer from the oil-pan surface. Assume negligible resistance to conduction through the oil pan

a) 190.37 W

b) 180.37 W

c) 170.37 W

d) 160.37 W

Answer: c

Explanation: Q = h A d t, where h =  /l and Nu = 0.664  0.5  .33 .

9. Air flows through a 10 cm internal diameter tube at the rate of 75 kg/hr. Measurements indicate that at a particular point in the tube, the pressure and temperature of air are 1.5 bar and 325 K whilst the tube wall temperature is 375 K. Find heat transfer rate from one meter length in the region of this point

The general non-dimensional correlation for turbulent flow in the tube is

Nu = 0.023  0.8  0.4

Where the fluid properties are evaluated at the bulk temperature

a) 147.35 W

b) 157.35 W

c) 167.35 W

d) 177.35 W

Answer: d

Explanation: Q = h A d t, where h =  /l = 177.35 W.

10. Consider the above problem, find the value of Nusselt number

a) 40.46

b) 50.56

c) 60.66

d) 70.76

Answer: a

Explanation: Nu = 0.023  0.8  0.4 = 40.46.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Boiling”.

1. Boiling refers to a change from the

a) Solid to a liquid phase

b) Vapor to a liquid phase

c) Liquid to a solid phase

d) Liquid to a vapor phase

Answer: d

Explanation: Boiling is a convective heat transfer process that is associated with a change in the phase of a fluid.

2. The boiling process has wide-spread applications in

 Production of steam in nuclear and steam power plants for generation and for industrial processes and space heating

 Absorption of heat in refrigeration and air-conditioning systems

 Concentration, dehydration and drying of foods and materials

Identify the correct statements

a) i and ii

b) ii and iii

c) i, ii and iii

d) i and iii

Answer: c

Explanation: Boiling constitutes the convective heat transfer process that involves a phase change from liquid to vapor state.

3. Maximum heat transfer rate in a modern boiler is about

a) 2 * 10 5 W/m 2

b) 3 * 10 5 W/m 2

c) 4 * 10 5 W/m 2

d) 5 * 10 5 W/m 2

Answer: a

Explanation: Greater importance has recently been given to the boiling heat transfer.

4. Which type of boiling occurs in steam boilers employing natural convection?

a) Forced convection

b) Pool

c) Local

d) Saturated

Answer: b

Explanation: The liquid above the hot surface is essentially stagnant and the only motion near the surface is because of free convection.

5. In which type of boiling the fluid motion is induced by external means?

a) Pool

b) Local

c) Forced convection

d) Subcooled

Answer: c

Explanation: The liquid is pumped and forced across the surface in a controlled manner.

6. The temperature of the liquid is below the saturation temperature and boiling takes place only in vicinity of the heated surface. This type of boiling is known as

a) Subcooled

b) Forces

c) Saturated

d) Pool

Answer: a

Explanation: The vapor bubbles travel a short path and then vanish, apparently they condense in the bulk of the liquid which is at a temperature less than the boiling point.

7. In which type of boiling the temperature of the liquid exceeds the saturation temperature?

a) Forced

b) Saturated

c) Pool

d) Saturated

Answer: d

Explanation: The vapor bubbles generated at the solid surface are transported through the liquid by buoyancy effects and eventually escape from the surface.

8. The phenomenon of stable film boiling is referred to as

a) Nucleate effect

b) Boiling regimes

c) Leiden frost effect

d) Von karma effect

Answer: c

Explanation: This is the region of stable film boiling.

9. For water evaporating at atmospheric pressure, the burnout occurs at temperature excess slightly above

a) 25 K

b) 55 K

c) 75 K

d) 105 K

Answer: b

Explanation: The burnout point on the boiling curve represents the point of maximum heat flux at which transition occurs from nucleate to film boiling.

10. The boiling phenomenon is known to occur in how many forms?

a) 1

b) 2

c) 3

d) 4

Answer: d

Explanation: Pool, forced convection, subcooled and saturated boiling.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Bubble Growth”.

1. The bubble diameter at the time of detachment from the surface can be worked out from the relation proposed by

a) Stanton

b) Fritz

c) Fourier

d) Nusselt

Answer: b

Explanation: It is given by Fritz, an American scientist.

2. The nucleate pool boiling is influenced by the following factors

 Pressure

 Liquid properties

 Material

Identify the correct statements

a) i, ii and iii

b) ii only

c) i and ii

d) ii and iii

Answer: a

Explanation: It depends on pressure, material and liquid properties.

3. Consider the following phenomena

 Boiling

 Free convection in air

 Forced convection in air

 Conduction in air

Identify the correct sequence 

a) iii – iv – i – ii

b) iv – i – iii – ii

c) iv – iii – ii – i

d) iv – ii – iii – i

Answer: d

Explanation: Heat transfer coefficient is maximum for conduction in air and is least for boiling.

4. The bubble diameter at the time of detachment from the surface can be worked out from the relation proposed by Fritz and is given by

a) C d β [2 σ/g (p t – p v )].

b) C d β [2 σ/g (p t – p v )] 3/2

c) C d β [2 σ/g (p t – p v )] 1/2

d) C d β [2 σ/g (p t – p v )] 5/2

Answer: c

Explanation: The constant C d has the value 0.0148 for water bubbles.

5. In spite of large heat transfer coefficient in boiling liquids, fins are used advantageously when the entire surface is exposed to

a) Film boiling

b) Transition boiling

c) Nucleate boiling

d) All modes of boiling

Answer: d

Explanation: Here all modes of boiling i.e. film, transition and nucleate are of great importance.

6. With increase in excess temperature, the heat flux in boiling

a) Increases continuously

b) Decreases and then increases

c) Decreases, then increases and again decreases

d) Increases, then decreases and again increases

Answer: d

Explanation: It first increases to 50% then decreases not to zero and again increases to its maximum value.

7. The excess temperature range 50 degree Celsius < d t < 200 degree Celsius is indicative of the region of

a) Interface evaporation

b) Nuclear boiling

c) Partial film boiling

d) Stable film boiling

Answer: c

Explanation: The physical phenomenon of pool boiling is generally divided into four different regions based on the excess temperature.

8. Heat flux increases with temperature excess beyond the Leiden-frost point due to

a) Radiation effect becomes predominant

b) Occurrence of subcooled boiling

c) Vapor space become large

d) Promotion of nucleate boiling

Answer: a

Explanation: Heat flux increases with temperature excess beyond the Leiden-frost point due to radiation effect becomes predominant.

9. Leiden-frost expansion is related to

a) Condensation of vapor on a cold surface

b) Exchange of heat between two solids

c) Evaporation of a solution

d) Boiling of liquid on a hot surface

Answer: d

Explanation: It is related to boiling of liquid on a hot surface.

10. Consider the following statements regarding nucleate boiling

 The temperature of the surface is greater than the saturation temperature of the liquid

 Bubbles are created by the expansion of entrapped gas oil vapor at small cavities in the surface

 The temperature is greater than that in film boiling

 The heat transfer from the surface to the liquid is greater than that in the film boiling

Which of these statements are correct?

a) i and iii

b) i, ii and iv

c) ii, iii and iv

d) i, ii and iii

Answer: b

Explanation: The temperature must be less than that in film boiling.

This set of Heat Transfer Problems focuses on “Nucleate Boiling”.

1. Reynolds number is replaced by a modulus significant of the agitation of the fluid particles in nucleate boiling. Such a dimensionless modulus is defined by the relation

a) Re b = 2 D b G b /δ f

b) Re b = D b G b /δ f

c) Re b = ½ D b G b /δ f

d) Re b = 3 D b G b /δ f

Answer: b

Explanation: This nucleate boiling regime is of great importance because of the very high heat fluxes possible with moderate temperature differences. Where, D b is the average bubble diameter, G b is the mass velocity of the bubble per unit area and δ f is the fluid viscosity.

2. The bubble diameter has been expressed by Fritz as

a) D b = C d β [2 σ/g (p f – p g )] 1/2

b) D b = C d β [2 σ/g (p f – p g )] 3/2

c) D b = C d β [2 σ/g (p f – p g )] 5/2

d) D b = C d β [2 σ/g (p f – p g )] 7/2

Answer: a

Explanation: C d is a constant which has been evaluated as 0.0148 for hydrogen and water bubbles. Where, C d is a constant, σ is surface tension of the liquid and β is the bubble contact angle measured through liquid in degrees.

3. What is the value of surface fluid constant for water-copper combination?

a) 0.010

b) 0.011

c) 0.012

d) 0.013

Answer: d

Explanation: The surface fluid regime is of great importance because of the very high heat fluxex possible with moderate temperature differences.

4. What is the value of surface fluid constant for water-brass combination?

a) 0.004

b) 0.005

c) 0.006

d) 0.007

Answer: c

Explanation: For H 2 O and Brass combination, this value must lie between 0.0056 to 0.00062.

5. Spherical bubbles of 3 mm diameter are observed in the bulk fluid boiling of water at standard atmospheric pressure. Assuming pure water vapor in the bubble and vapor pressure equal to 101.325 k N/m 2 , calculate the temperature of the vapor

a) 100.217 degree Celsius

b) 200.217 degree Celsius

c) 300.217 degree Celsius

d) 400.217 degree Celsius

Answer: a

Explanation: T v – T sat = (2 σ/r – p g ) R v T V 2 /p v h f g.

6. An electric wire of 1.25 mm diameter and 250 mm long is laid horizontally and submerged in water at 7 bar. The wire has an applied voltage of 2.2 V and carries a current of 130 amperes. If the surface of the wire is maintained at 200 degree Celsius, make calculations for the heat flux

a) 0.0915 * 10 6 W/m 2

b) 0.1915 * 10 6 W/m 2

c) 0.2915 * 10 6 W/m 2

d) 0.3915 * 10 6 W/m 2

Answer: c

Explanation: Q= V I = 286 W and A = 9.81 * 10 -4 m 2 . Therefore heat flux = Q/A.

7. Consider the above problem, find the boiling heat transfer coefficient

a) 5330 W/m 2 K

b) 6330 W/m 2 K

c) 7330 W/m 2 K

d) 8330 W/m 2 K

Answer: d

Explanation: Q = h A d t. So, h = 8330 W/m 2 K.

8. Which of the following parameters affect burnout heat flux in the nucleate boiling region

 Heat of evaporation

 Temperature difference

 Density of vapor

 Density of liquid

 Surface tension at the vapor-liquid interface

Mark the correct answer from the codes indicated below

a) i, ii, iii and v

b) i, iii, iv and v

c) i, ii, iii and iv

d) i, iii and v

Answer: b

Explanation: Acc to Zuber relation, burn out  = 0.18 p g h f g [p (p f – p g )/p g 2 ] 0.25 [p f /p g + p f ] 0.5 .

9. All the following statements are correct, except

a) Nucleate boiling gets promoted on a smooth surface

b) In subcooled heating, the temperature of the heating surface is more than the boiling point of the liquid

c) Film boiling region is usually avoided in commercial equipment

d) There occurs transition from nucleate to film boiling burn-out point on the boiling curve

Answer: a

Explanation: A rough surface gives a better heat transmission than when the surface is either smooth or has been coated to weak its tendency to get wetted.

10. Milk spills over when it is boiled in an open vessel. The boiling of milk at this instant is referred to as

a) Interface evaporation

b) Sub-cooled boiling

c) Film boiling

d) Saturated nucleate boiling

Answer: b

Explanation: This is an application of sub-cooled boiling.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Free Convection Boiling”.

1. When evaporation takes place at the liquid-vapor interface, the heat transfer is solely due to free convection and the film coefficient follows the relation

a) Nu = f 1  f 2 

b) Nu = 2 f 1  f 2 

c) Nu = 3 f 1  f 2 

d) Nu = 4 f 1  f 2 

Answer: a

Explanation: The functions f 1 and f 2 depend upon the geometry of the heating surface.

2. Fritz criterion is given by

a) h = 1.973  0.45

b) h = 1.973  0.55

c) h = 1.973  0.65

d) h = 1.973  0.75

Answer: d

Explanation: Fritz formulated the following formula for water boiling at atmospheric pressure in free convection in a vertical tube headed from outside.

3. A 0.10 cm diameter and 15 cm long wire has been laid horizontally and submerged in water at atmospheric pressure. The wire has a steady state voltage drop of 14.5 V and a current of 42.5 A. Determine the heat flux of the wire.

The following equation applies for water boiling on a horizontal submerged surface

H = 1.54  0.75 = 5.58  3 W/m 2 K where Q/A is the heat flux rate in W/m 2 and d t is the temperature difference between surface and saturation

a) 1.308 * 10 8 W/m 2

b) 1.308 * 10 7 W/m 2

c) 1.308 * 10 6 W/m 2

d) 1.308 * 10 5 W/m 2

Answer: c

Explanation: Q = E I = 616.25 W and A = 4.71 * 10 -4 m 2 .

4. Consider the above problem, find the excess temperature of the wire

a) 18.01 degree Celsius

b) 19.01 degree Celsius

c) 20.01 degree Celsius

d) 21.01 degree Celsius

Answer: b

Explanation: 1.54 (1.308 * 10 6 ) 2 = 5.58  3 .

5. Natural convection heat transfer coefficients over surface of a vertical pipe and a vertical flat plate for same height. What is/are the possible reasons for this?

 Same height

 Both vertical

 Same fluid

 Same fluid flow pattern

Select the correct answer

a) iv

b) i and ii

c) i

d) iii and iv

Answer: d

Explanation: The fluids must be same so their flow pattern.

6. The heat flux in nucleate boiling varies in accordance with

a) h f g

b) (h f g ) 0.5

c) 1/(h f g ) 2

d) (h f g ) 3

Answer: c

Explanation: Q/A = δ f h f g [(p f – p g ) g/σ] 0.5 [C f d t/h f g p C s f ] 3 .

7. In nucleate pool boiling, the heat flux depends on

a) Liquid properties, material and condition of the surface

b) Material of the surface only

c) Material and roughness of the surface

d) Liquid properties and material of the surface

Answer: a

Explanation: The heat flux must depends on liquid properties material and condition of the surface.

8. Identify the wrong statement with respect to boiling heat transfer?

a) The steam boilers employing natural convection have steam raised through pool boiling

b) Boiling occurs when a heated surface is exposed to a liquid and maintained at a temperature lower than the saturation temperature of the liquid

c) Leiden-frost effect refers to the phenomenon of stable film boiling

d) The nucleation boiling is characterized by the formation of bubbles at the nucleation sites and the resulting liquid agitation

Answer: b

Explanation: For boiling to occur, the heated surface must be exposed to a liquid and maintained at a temperature higher than the saturation temperature of the liquid.

9. Estimate the peak heat flux for water boiling at normal atmospheric pressure. The relevant thermo-physical properties are

p f  = 958.45 kg/m 3

p g  = 0.61 kg/m 3

h f g = 2.25 * 10 6 J/kg

σ = 0.0585 N/m

a) 1.53 * 10 8 W/m 2

b) 1.53 * 10 7 W/m 2

c) 1.53 * 10 6 W/m 2

d) 1.53 * 10 5 W/m 2

Answer: c

Explanation:  = 0.18 p g h f g [σ (p f – p g )/p g 2 ] 0.25 .

10. A 1.0 mm diameter and 300 mm long nickel wire is submerged horizontal in water at atmospheric pressure. At burnout, the wire has a current of 195 A. Calculate the voltage at burnout. The relevant thermos-physical properties are

p f  = 959.52 kg/m 3

p g  = 0.597 kg/m 3

h f g = 2257000 J/kg

σ = 0.0533 N/m

a) 6.15 V

b) 7.15 V

c) 8.15 V

d) 9.15 V

Answer: b

Explanation:  MAX = 1480000 W/m 2 . Let E b be the voltage at burnout. Then electric energy input to wire is E b I = 195 E b W.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Condensation”.

1. Condensation refers to a change from the

a) Solid to a liquid phase

b) Vapor to a liquid phase

c) Liquid to a solid phase

d) Liquid to a vapor phase

Answer: b

Explanation: Condensation is a convective heat transfer process that is associated with a change in the phase of a fluid.

2. Condensation process is very common in

 Boilers

 Condensers

 Evaporators

Identify the correct statements

a) i and ii

b) ii and iii

c) i, ii and iii

d) i and iii

Answer: c

Explanation: This process is very common in power plants and refrigeration systems.

3. The convective coefficients for condensation usually lie in the range

a) 30-300 W/m 2 K

b) 60-3000 W/m 2 K

c) 300-10000 W/m 2 K

d) 2500-10000 W/m 2 K

Answer: d

Explanation: The convective coefficient for condensation should be high because condensation refers to a change from the vapor to a liquid phase.

4. Drop wise condensation usually occurs on

a) Oily surface

b) Glazed surface

c) Smooth surface

d) Coated surface

Answer: a

Explanation: It generally occurs on oily surface. It is the convective heat transfer process that is associated with a change in the phase of a fluid.

5. Consider the following statements

 If a condensing liquid does not wet a surface, then drop wise condensation will not take place on it

 Drop wise condensation gives a higher transfer rate than film wise condensation

 Reynolds number of condensing liquid is based on its mass flow rate

 Suitable coating or vapor additive is used to promote film wise condensation

Identify the correct statement

a) i and ii

b) ii, iii and iv

c) iv only

d) i, ii and iii

Answer: d

Explanation: All are correct except the last one because suitable coating or vapor additive is not used to promote film wise condensation.

6. Depending upon the behavior of condensate up on the cooled surface, the condensation process are classified into how many distinct modes?

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: Film condensation and drop wise condensation.

7. A plate condenser was designed to be kept vertical. How would the condensation coefficient be effected if due to site constraints, it has to be kept at 60 degree to the horizontal?

a) 1.53% reduction in condensation coefficient

b) 2.53% reduction in condensation coefficient

c) 3.53% reduction in condensation coefficient

d) 4.53% reduction in condensation coefficient

Answer: c

Explanation: h VER = 0.943 [k 3 p 2 g h f g /δ l (t sat – t s )] 0.25 , h INC = 0.943 [k 3 p 2 g sin α h f g /δ l (t sat – t s )] 0.25 .

8. Saturated steam is allowed to condense over a vertical flat surface and the condensate film flows down the surface. The local coefficient of heat transfer for condensation

a) Remains constant at all heights of the surface

b) Decreases with increasing distance from the top of the surface

c) Increases with increasing thickness of film

d) Increases with increasing temperature differential between the surface and vapour

Answer: b

Explanation: It decreases with increasing thickness of condensate film.

9. In condensation over a vertical surface, the value of convection coefficient varies as

a) k 0.25

b) k 0.33

c) k 0.75

d) k -0.5

Answer: a

Explanation: h = 0.943 [k 3 p 2 g h f g /δ l (t sat – t s )] 0.25 .

10. For film wise condensation on a vertical plane, the film thickness δ and heat transfer coefficient h vary with distance x from the leading edge as

a) δ decreases, h increases

b) Both δ and h increases

c) δ increases, h decreases

d) Both δ and h decreases

Answer: c

Explanation: Thickness increases and heat transfer coefficient decreases.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Laminar Film Condensation”.

1. For laminar film condensation on a vertical plate, the velocity distribution at a distance δ from the top edge is given by

a) p g /σ

b) p g (δ y – y 2 )/σ

c) p g (δ y – y 2 /2)/σ

d) p g (δ – y 2 /2)/σ

Answer: c

Explanation: An equation for the velocity distribution as a function of some distance from the wall surface can be set up by considering the equilibrium between the gravity and viscous forces on an elementary volume of the liquid film.

2. For laminar film condensation on a vertical plate, the film thickness is given by

a) [4 k δ (t sat – t s ) x/p 2 g h f g ] 0.25

b) [4 k δ (t sat – t s ) x/p 2 g h f g ] 0.5

c) [4 k δ (t sat – t s ) x/p 2 g h f g ]

d) [4 k δ (t sat – t s ) x/p 2 g h f g ] 1.5

Answer: a

Explanation: The film thickness increases as the fourth root of the distance down the surface.

3. For laminar film condensation on a vertical plate, the mass flow rate of the condensate per unit depth of the film at any position x is given by

a) p g δ 3 / 3

b) p g δ 3 / 3 σ

c) p g δ 2 / 3 σ

d) p g δ 2 / 3

Answer: b

Explanation: Mass flow rate = mean flow velocity * flow area * density.

4. For laminar film condensation on a vertical plate, the gravitational acceleration g is replaced by

a) 4 g sin α

b) 3 g sin α

c) 2 g sin α

d) g sin α

Answer: d

Explanation: It is replaced by sin α. Where, α is the inclination angle with the horizontal.

5. For laminar film condensation on a vertical plate, the local heat transfer coefficient at the lower edge of the plate is given by

a) [k 3 p 2 g h f g /4 δ l (t sat – t s )] 0.25

b) [k 3 p 2 g h f g /4 δ l (t sat – t s )] 0.5

c) [k 3 p 2 g h f g /4 δ l (t sat – t s )] 0.1

d) [k 3 p 2 g h f g /4 δ l (t sat – t s )] 0.125

Answer: a

Explanation: The rate of condensation heat transfer is higher at the upper end of the plate than at the lower end.

6. Mark the wrong statement with respect to laminar flow condensation on a vertical plate

a) The rate of condensation heat transfer is maximum at the upper edge of the plate and progressively decreases as the lower edge is approached

b) The average heat transfer coefficient is two third of the local heat transfer coefficient at the lower edge of the plate

c) At a definite point on the heat transfer, the film coefficient is directly proportional to thermal conductivity and inversely proportional to thickness of film at the point

d) The film thickness increases as the fourth root of the distance down the upper edge

Answer: b

Explanation: The average heat transfer coefficient is 4/3 of the local heat transfer coefficient at the lower edge of the plate.

7. A plate condenser of dimensions l * b has been designed to be kept with side l in the vertical position. However due to oversight during erection and installation, it was fixed with side b vertical. How would this affect the heat transfer? Assume laminar conditions and same thermos-physical properties and take b = l/2

a) The condenser should be installed with shorter side horizontal

b) The condenser should be installed with longer side horizontal

c) The condenser should be installed with longer side vertical

d) The condenser should be installed with shorter side vertical

Answer: d

Explanation: h 1 = 0.943 [k 3 p 2 g h f g /δ l (t sat – t s )] 0.25 , h 2 = 0.943 [k 3 p 2 g h f g /δ b (t sat – t s )] 0.25 . So, h 1 /h 2 = 0.8409.

8. Determine the length of a 25 cm outer diameter tube if the condensate formed on the surface of the tube is to be same whether it is kept vertical or horizontal

a) 61.5 cm

b) 71.5 cm

c) 81.5 cm

d) 91.5 cm

Answer: b

Explanation: h v = 0.943 [k 3 p 2 g h f g /δ l (t sat – t s )] 0.25 , h H = 0.943 [k 3 p 2 g h f g /δ d (t sat – t s )] 0.25 , l/d = 2.86.

9. The critical Reynolds number for transition from laminar to turbulent film condensation is

a) 2000

b) 1900

c) 1800

d) 1700

Answer: c

Explanation: This should be 1800 for perfect transition from laminar to turbulent film condensation.

10. Which of the following is a wrong statement in the context of a convective heat transfer coefficient in laminar film condensation?

The heat transfer coefficient varies as

a) – ¼ power of acceleration due to gravity

b) ½ power of density of liquid

c) ¼ power of enthalpy of evaporation

d) – ½ power of dynamic viscosity

Answer: a

Explanation: h = 0.943 [k 3 p 2 g h f g /δ l (t sat – t s )] 0.25 . Further the heat transfer coefficient varies as ¼ power of acceleration due to gravity.

This set of tricky Heat Transfer Questions & Answers focuses on “Turbulent Film Condensation”.

1. Which parameter is responsible for the commencement of the turbulent flow?

a) Fourier number

b) Reynolds number

c) Stanton number

d) Nusselt number

Answer: b

Explanation: The character of condensate film can range from laminar to highly turbulent.

2. The transition from laminar to turbulent flow occurs at a critical Reynolds number of

a) 1800

b) 2200

c) 2600

d) 3000

Answer: a

Explanation: For turbulent film condensation on vertical surfaces, Kirk bride has suggested the correlation for the average heat transfer coefficient which is valid for Reynolds number greater than 1800.

3. A condenser is to be designed to condense 225.0 kg of steam per hour at a pressure of 0.15 bar. A square array of 400 tubes, each of 6 mm in diameter, is available for the task. If the tube surface temperature is to be maintained at 26 degree Celsius, make calculations for the length of the tube

a) 4.353 m

b) 3.353 m

c) 2.353 m

d) 1.353 m

Answer: d

Explanation: t sat = 54 degree Celsius and h f g = 2373000 J/kg, t f = 40 degree Celsius and h v = 0.725 [k 3 g p 2 h f g /l δ (t sat – t s )] 0.25 = 5202.34 W/m 2 K. Total S.A of 400 tubes = 7.536l and heat flow rate = m h f g .

4. The outer surface of a vertical tube is 1.25 m long and outer diameter is 50 mm is exposed to saturated steam at atmospheric pressure. If the tube surface is maintained at 80 degree Celsius by the flow of cooling water through it, determine the rate of heat transfer to the coolant

a) 47648 W

b) 12345 W

c) 19879 W

d) 97123 W

Answer: c

Explanation: h v = 0.943 [k 3 p 2 g h f g /δ l (t sat – t s )] 0.25 = 5046.8 W/m 2 k and heat flow rate = 19879 W.

5. Consider the above problem, find the rate at which steam is condensed at the tube surface

a) 7.7 * 10 -3

b) 8.7 * 10 -3

c) 9.7 * 10 -3

d) 10.7 * 10 -3

Answer: b

Explanation: Steam condensation rate = 198789/2285 * 10 3 = 8.7 * 10 -3 .

6. What is the value of characteristics length in turbulent film condensation?

a) δ

b) 3 δ

c) 2 δ

d) 4 δ

Answer: d

Explanation: d = 4 A/P = 4 b δ/b = 4 δ. Where δ is length of the plate.

7. What is the value of Reynolds number in terms of mass flow rate?

a) 4 m/δ b

b) 2 m/δ b

c) 3 m/δ b

d) m /δ b

Answer: a

Explanation: R = V 4 δ p/δ = 4 m/δ b. Where δ is the length of the plate and b id width of the plate.

8. Kirk bride criterion is given by

a) h = 0.0057  0.4 [k 3 p 2 g/δ 2 ] 1/3

b) h = 0.0067  0.4 [k 3 p 2 g/δ 2 ] 1/3

c) h = 0.0077  0.4 [k 3 p 2 g/δ 2 ] 1/3

d) h = 0.0087  0.4 [k 3 p 2 g/δ 2 ] 1/3

Answer: c

Explanation: This is valid for Re greater than 1800.

9. The relation between relative effectiveness of horizontal and vertical tubes as condensing surfaces is given by

a) h h /h v = 0.768 

b) h h /h v = 0.768  0.25

c) h h /h v = 0.768  0.5

d) h h /h v = 0.768  1.25

Answer: b

Explanation: h h /h v = 0.725/0.943 [ 0.25 ].

10. For n-tubes in a vertical column of the tube bank pattern, identify the correct statement

a) D e = n D

b) D e = n 2 D

c) D e =  D

d) D e = D

Answer: a

Explanation: It should be n times the diameter of a single tube in the bank.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Thermal Boundary Layer”.

1. Which field is set up when a fluid flows past a heated or cold surface?

a) Energy

b) Temperature

c) Mass

d) Time

Answer: b

Explanation: The temperature field encompasses a very small region of fluid.

2. The zone or thin layer wherein the temperature field exists is called the

a) Single boundary layer

b) Multi boundary layer

c) Hydrodynamic boundary layer

d) Thermal boundary layer

Answer: d

Explanation: It is known as the thermal boundary layer. The temperature gradient results due to heat exchange between the plate and the fluid.

3. The thickness of thermal boundary layer is arbitrarily defined as the distance from the plate surface at which

a) t S – t/t S – t INFINITY = 0.34

b) t S – t/t S – t INFINITY = 0.10

c) t S – t/t S – t INFINITY = 0.99

d) t S – t/t S – t INFINITY = 0.87

Answer: c

Explanation: It should be 0.99. The convection of energy reduces the outward conduction in the fluid and consequently the temperature gradient decreases away from the surface.

4. The convection of energy reduces the outward conduction in the fluid so temperature gradient

a) Decreases

b) Increases

c) Constant

d) Becomes twice

Answer: a

Explanation: The temperature gradient is infinite at the leading edge and approaches zero as the layer develops downstream.

5. The velocity profile of the hydrodynamic boundary layer is dependent upon

a) Time

b) Viscosity

c) Temperature

d) Mass

Answer: b

Explanation: It depends upon viscosity as the fluid velocity varies from zero at the solid surface to the velocity of free stream flow at a certain distance away from the solid surface.

6. Consider an elemental control volume for mass balance continuity equation. Which one of the following is correct?

heat-transfer-questions-answers-thermal-boundary-equation-q6

a) Value of 1 is 2 v + d u/d x 

b) Value of 2 is u + d u/d y 

c) Value of 1 is v + d u/d x 

d) Value of 2 is 2 u + d v/d x 

Answer: c

Explanation: The flow velocity changes in the direction of x axis and the rate of change is d u/d x.

7. Which is true for two dimensional boundary layer?

a) d u/d x – d v/d y = 1

b) d u/d x – d v/d y = 0

c) d u/d x + d v/d y = 1

d) d u/d x + d v/d y = 0

Answer: d

Explanation: This a two dimensional general equation. This equation also represents the continuity equation.

8. The differential energy equation for flow past a flat plate is given by

a) u d t/d x + v d t/d y = µ/p c  2 + k d 2 t/p c d y 2

b) u d t/d x + v d t/d y = µ/p c  + k d 2 t/p c d y 2

c) u d t/d x + v d t/d y = k d 2 t/p c d y 2 + µ/p c  -2

d) u d t/d x + v d t/d y = µ/p c  -1 + k d 2 t/p c d y 2

Answer: a

Explanation: Here heat generation due to viscous effects is not neglected.

9. The assumptions for thermal boundary layer are

 Steady compressible flow

 Negligible body forces, viscous heating and conduction in the flow direction

 Constant fluid properties evaluated at the film temperature

Identify the correct option

a) i and iii

b) i, ii and iii

c) ii and iii

d) i and ii

Answer: c

Explanation: It is steady incompressible flow. The temperature of the fluid changes from a minimum at the plate surface to the temperature of the mainstream at a certain distance from the surface.

10. The relationship between the thermal and hydrodynamic boundary layer thickness is governed by the

a) Peclet number

b) Prandtl number

c) Stanton number

d) Fourier number

Answer: b

Explanation: It is governed by the Prandtl number, which is indicative of the relative ability of the fluid to diffuse momentum and internal energy by molecular mechanisms.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Pohlhausen Equation”.

1. A small thermo-couple is positioned in a thermal boundary layer near a flat plate past which water flows at 30 degree Celsius and 0.15 m/s. The plate is heated to a surface temperature of 50 degree Celsius and at the location of the probe, the thickness is 15 mm. The probe is well-represented by

t – t S /t INFINITY – t S = 1.5  – 0.5  3

Determine the heat transfer coefficient

a) 33.3 W/m 2 K

b) 43.3 W/m 2 K

c) 53.3 W/m 2 K

d) 63.3 W/m 2 K

Answer: d

Explanation: h = Q/A (t INFINITY – t S ) = 63.3 W/m 2 K.

2. Air at 25 degree Celsius approaches a 0.9 m long and 0.6 m wide flat plate with a velocity 4.5 m/s. Let the plate is heated to a surface temperature of 135 degree Celsius. Find local heat transfer coefficient from the leading edge at a distance of 0.5 m

a) 5.83 W/m 2 K

b) 6. 83 W/m 2 K

c) 7. 83 W/m 2 K

d) 8. 83 W/m 2 K

Answer: a

Explanation: h = Nu k/x = 5. 83 W/m 2 K.

3. Consider the above problem, find the total rate of heat transfer from the plate to the air

a) 316.78 W

b) 416.78 W

c) 516.78 W

d) 616.78 W

Answer: c

Explanation: Q = h A d t = 516.78 W.

4. A small thermo-couple is positioned in a thermal boundary layer near a flat plate past which water flows at 30 degree Celsius and 0.15 m/s. The plate is heated to a surface temperature of 50 degree Celsius and at the location of the probe, the thickness of thermal boundary layer is 15 mm. If the temperature profile as measured by the probe is well-represented by

t – t S /t INFINITY – t S = 1.5 (y/δ t ) – 0.5 (y/δ t ) 3

Determine the heat flux from plate to water

a) 266 W/m 2

b) 1266 W/m 2

c) 2266 W/m 2

d) 3266 W/m 2

Answer: b

Explanation: Q/A = – k (t INFINITY – t S ) d/d y [t – t S /t INFINITY – t S ] Y = 0 . So, heat flux = 1266 W/m 2 .

5. Atmospheric air at 30 degree Celsius temperature and free stream velocity of 2.5 m/s flows along the length of a flat plate maintained at a uniform surface temperature of 90 degree Celsius. Let length = 100 cm, width = 50 cm and thickness = 2.5 cm. Thermal conductivity of the plate material is 25 W/m K, find heat lost by the plate

a) 155.88 W

b) 165.88 W

c) 175.88 W

d) 185.88 W

Answer: d

Explanation: Q = h A d t where, Nu = h l/k. So, Q = 185.88 W.

6. Consider the above problem, find the temperature of bottom surface of the plate for steady state condition

a) 90.372 degree Celsius

b) 80.372 degree Celsius

c) 70.372 degree Celsius

d) 60.372 degree Celsius

Answer: a

Explanation: Q = – k A (t S – t B )/δ.

7. Ambient air at 20 degree Celsius flows past a flat plate with a sharp leading edge at 3 m/s. The plate is heated uniformly throughout its entire length and is maintained at a surface temperature of 40 degree Celsius. Calculate the distance from the leading edge at which the flow in the boundary layer changes from laminar to turbulent conditions. Assume that transition occurs at a critical Reynolds number of 500000

a) 4.67 m

b) 3.67 m

c) 2.67 m

d) 1.67 m

Answer: c

Explanation: Re = x U INFINITY /v.

8. Ambient air at 20 degree Celsius flows past a flat plate with a sharp leading edge at 3 m/s. The plate is heated uniformly throughout its entire length and is maintained at a surface temperature of 40 degree Celsius. Calculate the thickness of the hydrodynamic boundary layer. Assume that transition occurs at a critical Reynolds number of 500000

a) 16.5 mm

b) 17.5 mm

c) 18.5 mm

d) 19.5 mm

Answer: b

Explanation: Thickness = 4.64/  ½ .

9. Ambient air at 20 degree Celsius flows past a flat plate with a sharp leading edge at 3 m/s. The plate is heated uniformly throughout its entire length and is maintained at a surface temperature of 40 degree Celsius. Calculate the thickness of the thermal boundary layer. Assume that transition occurs at a critical Reynolds number of 500000

a) 19.23 mm

b) 18.23 mm

c) 17.23 mm

d) 16.23 mm

Answer: a

Explanation: Thickness = 0.976 /  1/3 .

10. Ambient air at 20 degree Celsius flows past a flat plate with a sharp leading edge at 3 m/s. The plate is heated uniformly throughout its entire length and is maintained at a surface temperature of 40 degree Celsius. Calculate the local convective heat transfer coefficient. Assume that transition occurs at a critical Reynolds number of 500000

a) 4.519 k J/m 2 hr degree

b) 5.519 k J/m 2 hr degree

c) 6.519 k J/m 2 hr degree

d) 7.519 k J/m 2 hr degree

Answer: d

Explanation: h = Nu k/x.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Reynolds Analogy”.

1. Temperature and velocity profiles are identical when the dimensionless Prandtl number is

a) 1

b) 2

c) 3

d) 4

Answer: a

Explanation: They are identical when Prandtl number is unity.

2. Reynolds analogy is given by

a) Nu x / (Re x ) (Pr x ) = 5 St X = – 2 C F x

b) Nu x / 2 (Re x ) (Pr x ) = 4 St X = – C F x /3

c) Nu x / (Re x ) (Pr x ) = St X = – ½ C F x

d) Nu x / (Re x ) (Pr x ) = 2 St X = – C F x /4

Answer: c

Explanation: It is an excellent example of the similar nature of energy and momentum transfer.

3. The average drag coefficient for turbulent boundary layer flow past a thin plate is given by

C f = 0.455/ (log 10 R el ) 2.58

Where R el is the Reynolds number based on plate length. A plate 50 cm wide and 5 m long is kept parallel to the flow of water with free stream velocity 3 m/s. Calculate the drag force on both sides of the plate. For water, kinematic viscosity = 0.01 stokes

a) 53.38 N

b) 63.38 N

c) 73.38 N

d) 83.38 N

Answer: b

Explanation: Drag force = 2 C f (p U INFINITY /2)   = 25.42 N per unit width.

4. Consider the above problem, estimate the value of Reynolds number

a) 0.12

b) 0.13

c) 0.14

d) 0.15

Answer: d

Explanation: Re = l U INFINITY /v = 0.15.

5. During test-run, air flows at 215 m/s velocity and 25 degree Celsius temperature past a smooth thin model airfoil which can be idealized as a flat plate. If the chord length of the airfoil is 15 cm, find drag per unit width. The relevant physical properties of air are

p = 1.82 kg/m 3

v = 15.53 * 10 -6 m 2 /s

a) 25.42 N per unit width

b) 35.42 N per unit width

c) 45.42 N per unit width

d) 55.42 N per unit width

Answer: a

Explanation: Drag force = 2 C f (p U INFINITY /2)   = 25.42 N per unit width.

6. A flat plate was positioned at zero incidence in a uniform flow stream of air. Assuming boundary layer to be turbulent over the entire plate, workout the ratio of skin-friction forces on the front and rear half part of the plate

a) 1.557

b) 1.447

c) 1.347

d) 1.247

Answer: c

Explanation: F 1 /F 2 = 0.574/1 – 0.574 = 1.347.

7. For a particular engine, the underside of the crankcase can be idealized as a flat plate measuring 80 cm by 20 cm. The engine runs at 80 km/hr and the crankcase is cooled by the air flowing past it at the same speed. Find loss of the heat from the crank case surface (t S = 25 degree Celsius). Assume the boundary layer to be turbulent

a) 465.04 W

b) 565.04 W

c) 665.04 W

d) 765.04 W

Answer: b

Explanation: Heat loss by crankcase = h A d t = 565.4 W.

8. With respect to above problem, find the value of Nusselt number

a) 2000.89

b) 3000.89

c) 4000.89

d) 5000.89

Answer: a

Explanation: Nusselt number = 0.036  0.8  0.33 = 2000.89.

9. A flat plate 1 m by 1 m is placed in a wind tunnel. The velocity and temperature of free stream air are 80 m/s and 10 degree Celsius. The flow over the whole length of the plate is made turbulent by turbulizing grid placed upstream of the plate. Find the thickness of the hydrodynamic boundary layer at trailing edge of the plate

a) 19.55 mm

b) 18.55 mm

c) 17.55 mm

d) 16.55 mm

Answer: d

Explanation: Thickness =  /  0.2 = 0.01655 m.

10. A flat plate 1 m by 1 m is placed in a wind tunnel. The velocity and temperature of free stream air are 80 m/s and 10 degree Celsius. The flow over the whole length of the plate is made turbulent by turbulizing grid placed upstream of the plate. Find the heat flow from the surface of the plate

a) 9424.5 W

b) 8424.5 W

c) 7424.5 W

d) 6424.5 W

Answer: c

Explanation: Heat flow from the plate = h A d t = 7424.5 W.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Hydrodynamic Boundary Layer”.

1. The concept of the hydrodynamic boundary layer was first suggested by

a) Isaac Newton

b) Ludwig Prandtl

c) Rodridge

d) Fourier

Answer: b

Explanation: It was suggested by Ludwig Prandtl in 1904. This is the thin layer where velocity changes continuously.

2. The free stream undisturbed flow has a uniform velocity U INFINITY in the

a) X-direction

b) Y-direction

c) Z-direction

d) Any direction

Answer: a

Explanation: It is in the x-direction only. Particles of fluid adhere to the plate surface as they approach it and the fluid is slowed down considerably.

3. The thin layer where velocity changes continuously is called

a) Differential layer

b) Thermal boundary layer

c) Hydrodynamic boundary layer

d) Velocity distribution layer

Answer: c

Explanation: In this layer the flow is rotational and shear stresses are present.

4. The conditions for flow beyond the boundary layer and its outer edge are

a) d u/d y = 0 and u = U 0

b) d u/d y = Infinity and u = U INFINITY

c) d u/d y = 1 and u = U 0

d) d u/d y = 0 and u = U INFINITY

Answer: d

Explanation: All the variation in fluid velocity is concentrated in a comparatively thin layer in immediate vicinity of the plate surface.

5. The pattern of flow in the boundary layer is judged by the

a) Reynolds number

b) Fourier number

c) Peclet number

d) Grashof number

Answer: a

Explanation: Re = U INFINITY x/v.

6. Consider the diagram given below and identify the correct option

heat-transfer-questions-answers-hydrodynamic-boundary-layer-q6

a) The velocity gradient is zero everywhere

b) The velocity profile changes at every instant of time

c) Boundary layers from the pipe walls meet the pipe anywhere

d) Thickness of the boundary layer is limited to the pipe radius

Answer: d

Explanation: Thickness of the boundary layer is limited to the pipe radius because of the flow being within a confined passage.

7. The transition from laminar to turbulent pattern of flow occurs at values of Reynolds number between

a) 1000-2000

b) 300000-500000

c) 500000-700000

d) 35750-45678

Answer: b

Explanation: It occurs at a higher value. The pattern of flow in the boundary layer is judged by the Reynolds number, Re = U INFINITYY x/v.

8. The entrance length required for the flow to become fully-developed turbulent flow is dependent on

 Surface finish

 Downstream conditions

 Fluid properties

Identify the correct answer

a) ii and iii

b) i and iii

c) i, ii and iii

d) i and ii

Answer: c

Explanation: It depends on surface finishing, various upstream and downstream conditions and various fluid properties.

9. What is the value of the thickness of the boundary layer at leading edge of the plate?

a) 0.33

b) 1

c) 0.5

d) 0

Answer: d

Explanation: The thickness of the boundary layer is variable along the flow direction.

10. The boundary layer thickness is taken to be at a distance from the plate surface to a point at which the velocity is given by

a) u = 0.99 U INFINITY

b) u = 0.75 U INFINITY

c) u = 0.50 U INFINITY

d) u = 0.33 U INFINITY

Answer: a

Explanation: Velocity is one per cent of the asymptotic limit.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Von- Karmal Equation”.

1. Glycerin at 10 degree Celsius flows past a flat plate at 20 m/s. Workout the velocity components at a point P in the fluid flow where

x = 2 m from the leading edge of the plate

y = 5 cm from the plate surface

For glycerin at 10 degree Celsius, kinematic viscosity = 2.79 * 10 -3 m 2 /s

a) 15.92 m/s and 0.0952 m/s

b) 16.92 m/s and 0.0952 m/s

c) 17.92 m/s and 0.0752 m/s

d) 18.92 m/s and 0.0752 m/s

Answer: b

Explanation: u/U INFINITY = 0L.846 and v/U INFINITY  ½ = 0.57.

2. A plate 0.3 m long is placed at zero angle of incidence in a stream of 15 degree Celsius water moving at 1 m/s. Find out the stream wise velocity component at the mid-point of the boundary layer. For water at 15 degree Celsius

p = 998.9 kg /m 3

µ = 415.85 * 10 -2 kg/hr m

a) 0.736 m/s

b) 0.636 m/s

c) 0.536 m/s

d) 0.436 m/s

Answer: a

Explanation: n = y (U INFINITY /v x) ½ = 2.5.

3. Air at 25 degree Celsius flows over a flat surface with a sharp leading edge at 1.5 m/s. Find the boundary layer thickness at 0.5 from the leading edge. For air at 25 degree Celsius, kinematic viscosity = 15.53* 10n -6 m 2 /s

a) 4.1376 cm

b) 3.1376 cm

c) 2.1376 cm

d) 1.1376 cm

Answer: d

Explanation: δ = 5 x/  ½ = 1.1376 m.

4. Local skin friction coefficient is given by

a) 0.646/  1/2

b) 1.646/  1/2

c) 2.646/  1/2

d) 3.646/  1/2

Answer: a

Explanation: It should be 0.646/  ½ .

5. A plate 0.3 m long is placed at zero angle of incidence in a stream of 15 degree Celsius water moving at 1 m/s. Find out the maximum boundary layer thickness. For water at 15 degree Celsius. For water at 15 degree Celsius

p = 998.9 kg /m 3

µ = 415.85 * 10 -2 kg/hr m

a) 4.945 m

b) 3.945 m

c) 2.945 m

d) 1.945 m

Answer: c

Explanation: Re = l p U INFINITY /µ.

6. Shear stress at the middle of the plate is given by

a) T W = 0.964 p U INFINITY 2 /2  1/2

b) T W = 0.864 p U INFINITY 2 /2  1/2

c) T W = 0.764 p U INFINITY 2 /2  1/2

d) T W = 0.664 p U INFINITY 2 /2  1/2 .

Answer: d

Explanation: T W = 3 µ U INFINITY /2 δ = 0.664 p U INFINITY 2 /2  ½ .

7. Boundary layer thickness is given by

a) δ = 5.64 x/  ½

b) δ = 5.64 x/  ½

c) δ = 6.64 x/  ½

d) δ = 7.74 x/  ½

Answer: a

Explanation: δ/x =  ½ (µ/x p U INFINITY ) ½ .

8. Air at 25 degree Celsius flows over a flat surface with a sharp leading edge at 1.5 m/s. Find the value of Reynolds number. For air at 25 degree Celsius, kinematic viscosity = 15.53* 10n -6 m 2 /s

a) 38694

b) 12846

c) 48294

d) 76386

Answer: c

Explanation: Re = x U INFINITY /v = 48294.

9. A plate 0.3 m long is placed at zero angle of incidence in a stream of 15 degree Celsius water moving at 1 m/s. Find out the maximum value of the normal component of velocity at thr trailing edge of the plate. For water at 15 degree Celsius

p = 998.9 kg /m 3

µ = 415.85 * 10 -2 kg/hr m

a) 1.6885 * 10 -2 m/s

b) 1.6885 * 10 -3 m/s

c) 1.6885 * 10 -4 m/s

d) 1.6885 * 10 -5 m/s

Answer: b

Explanation: v/U INFINITY  ½ =0.860.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Classification of Heat Exchanger”.

1. Some examples of heat exchanger are

 Condensers and evaporators in refrigeration units

 Evaporator of an ice plant and milk chiller of a pasteurizing plant

 Automobile radiators and oil coolers of heat engines

Identify the correct answer

a) i only

b) ii and iii

c) i, ii and iii

d) i and ii

Answer: c

Explanation: All are the examples of heat exchanger.

2. Heat exchangers are classified into how many categories?

a) 1

b) 2

c) 3

d) 4

Answer: d

Explanation: Nature of heat exchange process, relative direction of motion of fluid, mechanical design of heat exchange surface and physical state of heat exchanging fluids.

3. Based upon the nature of heat exchange process, the heat exchangers are classified into how many categories

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: Direct contact, regenerators and recuperators.

4. The energy transfer between the hot fluid and cold fluids is brought about by their complete physical mixing in

a) Direct contact heat exchanger

b) Regenerators

c) Recuperators

d) Boilers

Answer: a

Explanation: In this type of heat exchanger, there is a simultaneous transfer of heat and mass.

5. Which type of flow arrangement is this?

heat-transfer-questions-answers-classification-heat-exchanger-q5

a) Counter flow

b) Parallel flow

c) Regenerator

d) Shell and tube

Answer: b

Explanation: In this type of arrangement, the fluids enter the unit from the same side, flow in the same direction and leave from the same side.

6. Which of the following is not an example of recuperators type heat exchanger?

a) Automobile radiators

b) Condensers

c) Chemical factories

d) Oil heaters for an aero plane

Answer: c

Explanation: Recuperators are not used in chemical factories.

7. In how many categories heat exchangers are classified on the basis of direction of flow of fluids?

a) 4 categories

b) 3 categories

c) 2 categories

d) 1 categories

Answer: b

Explanation: Parallel, counter and cross flow.

8. In how many categories heat exchangers are classified on the basis of mechanical design of heat exchanger surface?

a) 2

b) 4

c) 1

d) 3

Answer: d

Explanation: Concentric tubes, shell and tube and multiple shell.

9. In how many categories heat exchangers are classified on the basis of physical state of heat exchanging fluids?

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: Condenser and evaporator.

10. Many types of heat exchangers have been developed to meet the widely varying applications. Based upon their

 Operating principle

 Arrangement of flow path

 Design

Identify the correct statements

a) i, ii and iii

b) i and ii

c) ii and iii

d) i and iii

Answer: a

Explanation: Heat exchanger is a process equipment designed for the effective transfer of heat energy between two fluids.

This set of Basic Heat Transfer Questions & Answers focuses on “Mean Value of Capacity Ratio”.

1. How many fluids remain at a constant temperature during the process of boiling and condensation?

a) 1

b) 2

c) 3

d) 4

Answer: a

Explanation: During the process of boiling and condensation, only a phase change takes place and one fluid remain at a constant temperature.

2. Identify the correct expression for effectiveness

a) 4 – exponential 

b) 3 – exponential 

c) 2 – exponential 

d) 1 – exponential 

Answer: d

Explanation: Here, C MAX = infinity and C MIN /C MAX = 0.

3. In a gas turbine recuperator, the capacity ratio becomes

a) 2

b) 1

c) 4

d) 3

Answer: b

Explanation: In a gas turbine recuperator, the exhaust gases after expansion in the turbine are used to heat the compressed air.

4. In a gas turbine recuperator, the expression for effectiveness for the parallel flow configuration reduces to

a) 1 – exponential /3

b) 1 – exponential /4

c) 1 – exponential /2

d) 1 – exponential /6

Answer: c

Explanation: In a gas turbine recuperator, the exhaust gases after expansion in the turbine are used to heat the compressed air.

5. What is the maximum efficiency for parallel flow heat exchanger?

a) 5%

b) 10%

c) 20%

d) 50%

Answer: a

Explanation: No matter how large the exchanger be or how high be the flow of overflow at transfer coefficient, the maximum efficiency for parallel flow heat exchanger is 5%.

6. A two pass surface condenser is required to handle the exhaust from a turbine developing 15 MW with specific steam consumption of 5 kg/k W h. The quality of exhaust steam is 0.9, the condenser vacuum is 66 cm of mercury while the bar meter reads 76 cm of mercury. The condenser tubes are 28 mm inside diameter, 4 mm thick and water flows through tubes with a speed of 3 m/s and inlet temperature 20 degree Celsius. All the steam is condensed, the condensate is saturated water and temperature of cooling water at exit is 5 degree Celsius less than the condensate temperature. Assuming that overall coefficient of heat transfer is 4 k W/m 2 degree, determine the mass of cooling water circulated

basic-heat-transfer-questions-answers-q6

a) 746.13 kg/s

b) 646.13 kg/s

c) 546.13 kg/s

d) 446.13 kg/s

Answer: d

Explanation: P S = 76 – 66/76  = 0.133 bar. From energy balance, heat loss by steam = heat gained by cooling water.

7. The curve of effectiveness versus NTU parameters indicates the relationship between

 Effectiveness

 NTU MAX

 C MIN /C MAX

Identify the correct statements

a) i and ii

b) i, ii and iii

c) ii only

d) iii only

Answer: b

Explanation: When any two of three parameters are known, the third can be find out.

8. Hot water having specific heat 4200 J/kg K flows through a heat exchanger at the rate of 4 kg/min with an inlet temperature of 100 degree Celsius. A cold fluid having a specific heat 2400 J/kg K flows in at a rate of 8 kg/min and with inlet temperature 20 degree Celsius. Make calculations for maximum possible effectiveness if the fluid flow conforms to parallel flow arrangement

a) 0.533

b) 0.633

c) 0.733

d) 1

Answer: a

Explanation: Maximum possible effectiveness = 1 – exponential /1 + C = 0.533.

9. Consider the above problem, make calculations for maximum possible effectiveness if the fluid flow conforms to counter flow arrangement

a) 1.23

b) 0.5

c) 1

d) 0.465

Answer: c

Explanation: Maximum possible effectiveness = 1 – exponential / 1 – C exponential  = 1.

10. A counter flow heat exchanger is used to col 2000 kg/hr of oil (c p = 2.5 k J/kg K) from 105 degree Celsius to 30 degree Celsius by the use of water entering at 15 degree Celsius. If the overall heat transfer coefficient is expected to be 1.5 k W/m 2 K, find out the water flow rate. Presume that the exit temperature of the water is not to exceed 80 degree Celsius

a) 1680.2 kg/hr

b) 1580.2 kg/hr

c) 1480.2 kg/hr

d) 1380.2 kg/hr

Answer: d

Explanation: Mass flow rate of coolant = 2000  /4.18  = 1380.2 kg/hr.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Mean Temperature Difference”.

1. Assumptions made for calculation of logarithmic mean temperature difference are

 Constant overall heat transfer coefficient

 The kinetic and potential energy changes are negligible

 There is no conduction of heat along the tubes of heat exchanger

Identify the correct statements

a) i, ii and iii

b) i and iii

c) i and ii

d) ii and iii

Answer: a

Explanation: These assumptions are made for simplicity. During heat exchange between two fluids, the temperature of the fluids change in the direction of flow and consequently there occurs a change in the thermal head causing the flow of heat.

2. A cold fluid  at 10 kg/min is to be heated from 25 degree Celsius to 55 degree Celsius in a heat exchanger. The task is accomplished by extracting heat from hot water  available at mass flow rate 5 kg/min and inlet temperature 85 degree Celsius. Identify the type of arrangement of the heat exchanger

a) Concentric tubes

b) Parallel flow

c) Counter flow

d) Shell and tubes

Answer: c

Explanation: m h c h (t h 1 – t h 2 ) = m c c c (t c 2 – t c 1 ).

3. In a food processing plant, a brine solution is heated from – 12 degree Celsius to – 65 degree Celsius in a double pipe parallel flow heat exchanger by water entering at 35 degree Celsius and leaving at 20.5 degree Celsius. Let the rate of flow is 9 kg/min. Estimate the area of heat exchanger for an overall heat transfer coefficient of 860 W/m 2 K. For water c P = 4.186 * 10 3 J/kg K

a) 1. 293 m 2

b) 0.293 m 2

c) 7. 293 m 2

d) 8. 293 m 2

Answer: b

Explanation: Q = m c P d t = 9104.5 J/s. A = Q/ U α m .

4. Exhaust gases (c P = 1.12 k J/kg K) flowing through a tubular heat exchanger at the rate of 1200 kg/hr are cooled from 400 degree Celsius to 120 degree Celsius. This cooling is affected by water (c P = 4.18 k J/kg K) that enters the system at 10 degree Celsius at the rate of 1500 kg/hr. If the overall heat transfer coefficient is 500 k J/m 2 hr degree, what heat exchanger area is required to handle the load for parallel flow arrangement?

a) 7.547 m 2

b) 6.547 m 2

c) 5.547 m 2

d) 4.547 m 2

Answer: d

Explanation: m h c h (t h 1 – t h 2 ) = m c c c (t c 2 – t c 1 ).

5. A steam condenser is transferring 250 k W of thermal energy at a condensing temperature of 65 degree Celsius. The cooling water enters the condenser at 20 degree Celsius with a flow rate of 7500 kg/hr. Calculate the log mean temperature difference

a) 28.25 degree Celsius

b) 29.25 degree Celsius

c) 30.25 degree Celsius

d) 31.25 degree Celsius

Answer: a

Explanation: Q = m c c c (t c 2 – t c 1 ) and log mean temperature difference = α 1 – α 2 / log (α 1 /α 2 ).

6. Consider the above problem, find what error would be introduced if the arithmetic mean temperature difference is used rather than the log-mean temperature difference? Take overall heat transfer coefficient for the condenser surface as 1250 W/m 2 K

a) 7.61%

b) 7.71%

c) 7.81%

d) 7.91%

Answer: d

Explanation: α = α 1 + α 2 /2. Error = 7.08 – 6.52/7.08 = 7.91%.

7. For what value of end temperature difference ratio, is the arithmetic mean temperature difference 5% higher than the log-mean temperature difference?

a) 2.4

b) 2.3

c) 2.2

d) 2.1

Answer: c

Explanation: α 1 / α 2 = 2.2.

8. A company is heating a gas by passing it through a pipe with steam condensing on the outside. What percentage change in length would be needed if it is proposed to triple the heating capacity?

a) 200%

b) 400%

c) 600%

d) 800%

Answer: a

Explanation: Present capacity, Q 1 = U 1 A 1 α 1 and new capacity, Q 2 = U 2 A 2 α 2 . According to the given condition, U 2 A 2 α 2 = 3 U 1 A 1 α 1 .

9. A steam condenser is transferring 250 k W of thermal energy at a condensing temperature of 65 degree Celsius. The cooling water enters the condenser at 20 degree Celsius with a flow rate of 7500 kg/hr. If overall heat transfer coefficient for the condenser surface is 1250 W/m 2 K, what surface area is required to handle this load?

a) 4.08 m 2

b) 5.08 m 2

c) 6.08 m 2

d) 7.08 m 2

Answer: d

Explanation: Q = U A α m . So, A = 7.08 m 2 .

10. Exhaust gases (c P = 1.12 k J/kg K) flowing through a tubular heat exchanger at the rate of 1200 kg/hr are cooled from 400 degree Celsius to 120 degree Celsius. This cooling is affected by water (c P = 4.18 k J/kg K) that enters the system at 10 degree Celsius at the rate of 1500 kg/hr. If the overall heat transfer coefficient is 500 k J/m 2 hr degree, what heat exchanger area is required to handle the load for counter flow arrangement?

a) 2.758 m 2

b) 3.758 m 2

c) 4.758 m 2

d) 5.758 m 2

Answer: b

Explanation: m h c h (t h 1 – t h 2 ) = m c c c (t c 2 – t c 1 ).

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Specific Heat”.

1. What is the value of specific heat for brick in J/kg K?

a) 835

b) 735

c) 635

d) 535

Answer: a

Explanation: It is a block of ceramic materials. It is built to withstand high temperature, but a low thermal conductivity.

2. Which one is having the highest value of specific heat?

a) Asphalt

b) Bakelite

c) Chrome brick

d) Fire clay

Answer: b

Explanation: Specific heat of bakelite is 1465 J/kg K while that of asphalt, chrome brick and fire clay are 920 J/kg K, 835 J/kg K and 960 J/kg K respectively.

3. What is the value of specific heat for asphalt in J/kg K?

a) 920

b) 820

c) 720

d) 620

Answer: a

Explanation: Asphalt is a composite material used to surface roads and airports. It is produced by heating the asphalt binder to decrease its viscosity.

4. What is the value of specific heat for Bakelite in J/kg K?

a) 1665

b) 1565

c) 1465

d) 1365

Answer: c

Explanation: It is also known as polyoxybenzylmethylenglycolanhysride. It is a thermosetting phenol formaldehyde resin. Bakelite was used for its electrical non conductivity.

5. Which one is having the lowest value of specific heat?

a) Coal

b) Clay

c) Banana

d) Concrete

Answer: d

Explanation: Specific heat of concrete is 880 J/kg K while that of coal, banana and clay are1260 J/kg K, 3350 J/kg K and 890 J/kg K respectively.

6. What is the value of specific heat for coal in J/kg K?

a) 1260

b) 1360

c) 1460

d) 1560

Answer: a

Explanation: The carbon content of coal is around 60% – 80%. The rest is composed of water, air, hydrogen and Sulphur.

7. What is the value of specific heat for cotton in J/kg K?

a) 1600

b) 1500

c) 1400

d) 1300

Answer: d

Explanation: It is also known as polyethylene. These are highly strainly resistant and is used as cushioning and insulating materials in pillows.

8. Which one is having the highest value of specific heat?

a) Sand

b) Berea

c) Magnesite

d) Limestone

Answer: c

Explanation: Specific heat of magnesite is 1130 J/kg K while that of sand, barea and limestone are 800 J/kg K, 745 J/kg K and 810 J/kg K respectively.

9. What is the value of specific heat for sand in J/kg K?

a) 800

b) 700

c) 600

d) 500

Answer: a

Explanation: There is a weak bonding between the molecules of sand particles due to which it is exposed to high temperature when left free in the atmosphere.

10. Which one is having the lowest value of specific heat?

a) Ice

b) Pyrex

c) Paraffin

d) Rubber

Answer: b

Explanation: Specific heat of pyrex is 835 J/kg K while that of ice, paraffin and rubber are 2040 J/kg K, 2890 J/kg K and 2010 J/kg K respectively.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Heat Transfer Coefficient”.

1. Which one is having highest value of overall heat transfer coefficient?

a) Steam condensers

b) Feed water heaters

c) Alcohol condensers

d) Steam

Answer: b

Explanation: Overall heat transfer coefficient for feed water heaters is 8500 W/m 2 K while that of steam, alcohol condensers and ammonia condensers are 5000 W/m 2 K, 630 W/m 2 K and 1400 W/m 2 K.

2. What is the value of overall heat transfer coefficient for air to heavy tars and liquid?

a) As low as 45 W/m 2 K

b) As low as 40 W/m 2 K

c) As low as 35 W/m 2 K

d) As low as 30 W/m 2 K

Answer: a

Explanation: It is an incompressible fluid which has constant volume independent of pressure.

3. What is the value of overall heat transfer coefficient for air to low viscosity liquid?

a) As high as 900 W/m 2 K

b) As high as 800 W/m 2 K

c) As high as 700 W/m 2 K

d) As high as 600 W/m 2 K

Answer: d

Explanation: It is made up of tiny vibrating particles of matter which are held together by intermolecular bonding.

4. Which one is having lowest value of overall heat transfer coefficient?

a) Steam

b) Air condensers

c) Air to heavy tars

d) Ammonia condensers

Answer: c

Explanation: Overall heat transfer coefficient for air to heavy tars is 45 W/m 2 K while that of steam, air condensers and ammonia condensers are 340 W/m 2 K, 780 W/m 2 K and 1400 W/m 2 K.

5. What is the value of overall heat transfer coefficient for air condensers?

a) 350-780 W/m 2 K

b) 250 -900 W/m 2 K

c) 200-350 W/m 2 K

d) 200-1950 W/m 2 K

Answer: a

Explanation: It consists of an air coil which is used to remove heat from fluids.

6. Which one is having highest value of overall heat transfer coefficient?

a) Steam

b) Alcohol condensers

c) Air condensers

d) Air to various gases

Answer: c

Explanation: Overall heat transfer coefficient for air condensers is 780 W/m 2 K while that of steam, alcohol condensers and air to various gases are 340 W/m 2 K, 700 W/m 2 K and 550 W/m 2 K.

7. What is the value of overall heat transfer coefficient ammonia condensers?

a) 800-1400 W/m 2 K

b) 200-750 W/m 2 K

c) 250-2500 W/m 2 K

d) 1500-1750 W/m 2 K

Answer: a

Explanation: It is an electrical device that stores a vast amount of energy.

8. Which one is having the lowest value of overall heat transfer coefficient?

a) Air condensers

b) Air to low viscosity liquids

c) Steam condensers

d) Feed water heaters

Answer: b

Explanation: Overall heat transfer coefficient for air condensers is 780 W/m 2 K while that of steam condensers, air to low viscosity liquids and feed water heaters are 5000 W/m 2 K, 600 W/m 2 K and 8500 W/m 2 K.

9. What is the value of overall heat transfer coefficient for steam condensers?

a) 200-9000 W/m 2 K

b) 3000-5500 W/m 2 K

c) 2000-9500 W/m 2 K

d) 1500-5000 W/m 2 K

Answer: d

Explanation: It is a rotating machine similar to a motor used to control power flow in electric power transmission.

10. Which one is having highest value of overall heat transfer coefficient?

a) Feed water heaters

b) Steam condensers

c) Alcohol condensers

d) Ammonia condensers

Answer: a

Explanation: Overall heat transfer coefficient for feed water heaters is 8500 W/m 2 K while that of steam condensers, alcohol condensers and ammonia condensers are 5000 W/m 2 K, 700 W/m 2 K and 1400 W/m 2 K.

This set of tough Heat Transfer Questions & Answers focuses on “Heat Flux Through A Cylindrical Wall And Plate”.

1. When two fluids of the heat exchanger are separated by a plane wall, the thermal resistance comprises

 Convection resistance due to the fluid film at the inside surface

 Conduction resistance

 Convection resistance due to the fluid film at the outside surface

Identify the correct option

a) i and ii

b) i, ii and iii

c) ii and iii

d) i and iii

Answer: b

Explanation: It consists of all the above i.e. convection resistance and conduction resistance.

2. Figure represents the block diagram of a heat exchanger. There were some aspects in the design and performance analysis of a heat exchanger. Identify the correct one

tough-heat-transfer-questions-answers-q2

a) The coolant picks up heat = m h c c (t c1 – t c2 )

b) The hot fluid gives up heat = m h c h (t h1 – t c2 )

c) The coolant picks up heat = m c c h (t h1 – t c2 )

d) The hot fluid gives up heat = m h c h (t h1 – t h2 )

Answer: d

Explanation: The structure of the heat exchanger transfers heat from the hot fluid to the coolant.

3. The heat loss from unpainted aluminum side of a house has been calculated on the presumption that overall coefficient of heat transfer is 5 W/m 2 K. Later, it was discovered that the air pollution levels are such that fouling factor on this side is of the order of 0.0005 m 2 K/W. Find overall heat transfer coefficient of dirty side

a) 2.9875 W/m 2 K

b) 3.9875 W/m 2 K

c) 4.9875 W/m 2 K

d) 5.9875 W/m 2 K

Answer: c

Explanation: R = 1/U DIRTY – 1/U CLEAN .

4. What is the value of fouling factor for engine exhaust?

a) 0.002 m 2 K/W

b) 0.003 m 2 K/W

c) 0.004 m 2 K/W

d) 0.005 m 2 K/W

Answer: a

Explanation: Its unit is m 2 hr K/kcal and it represents the reciprocal of the scale coefficient i.e. heat transfer.

5. What is the value of fouling factor for industrial liquids?

a) 0.0004 m 2 K/W

b) 0.0003 m 2 K/W

c) 0.0002 m 2 K/W

d) 0.0001 m 2 K/W

Answer: c

Explanation: Its unit is m 2 hr K/kcal and it represents the reciprocal of the scale coefficient i.e. heat transfer.

6. After being in service for a period of six months, a heat exchanger transforms 10% less heat than it does what new. Determine the effective fouling factor in terms of its clean overall heat transfer coefficient. It may be presumed that the heat exchanger operates between the same temperature differentials and that there is no change in the effective surface area due to scale build up

a) 0.13/ U CLEAN

b) 0.11/ U CLEAN

c) 0.09/ U CLEAN

d) 0.07/ U CLEAN

Answer: b

Explanation: Q CLEAN /Q DIRTY = U CLEAN /U DIRTY .

7. In a counter flow heat exchanger, water flowing through a tube of 10 cm inner diameter is heated by steam condensing on the outside of the tube. The convective film coefficient on the water and steam side are estimated to be 12000 and 20000 k J/m 2 hr degree. Neglecting tube thickness and its resistance to heat flow, workout the overall heat transfer coefficient for the heat exchanger

a) 4500 k J/m 2 hr degree

b) 5500 k J/m 2 hr degree

c) 6500 k J/m 2 hr degree

d) 7500 k J/m 2 hr degree

Answer: d

Explanation: U = 1/h i + h o .

8. A heat exchanger to preheat oil for a furnace was designed without considering the possibility of scale formation, and the overall heat transfer coefficient based on the fuel oil side was 3200 k J/m 2 hr degree. What would be the corrected coefficient of heat transfer if a fouling factor of 0.00025 m 2 hr degree/k J for the fuel oil is taken into account?

a) 1777.78 k J/m 2 hr degree

b) 1666.78 k J/m 2 hr degree

c) 1555.78 k J/m 2 hr degree

d) 1444.78 k J/m 2 hr degree

Answer: a

Explanation: U = 1/ (R + 1/h S ).

9. A copper pipe  having inner diameter of 1.75 cm and 2.0 cm outside diameter conveys water and the oil flows through the annular passage between this pipe and a steel pipe. On the water side, the film coefficient is 4600 W/m 2 K and the fouling factor is 0.00034 m 2 K/W. The corresponding values for the oil side are 1200 W/m 2 K and 0.00086 m 2 K/W. Estimate the overall heat transfer coefficient between the oil and water

a) 235.16 W/m 2 K

b) 335.16 W/m 2 K

c) 435.16 W/m 2 K

d) 535.16 W/m 2 K

Answer: c

Explanation: A fouling factor represents the reciprocal of the scale coefficient i.e. heat transfer.

10. Unit of fouling factor is

a) m 2 K/kcal

b) m 2 hr K/kcal

c) m 2 hr/kcal

d) m 2 hr K

Answer: b

Explanation: The fouling factors are used in the design of heat exchangers. It is the overall heat transfer coefficient.

This set of Heat Transfer Multiple Choice Questions & Answers  focuses on “Heat Exchanger Effectiveness”.

1. Capacity ratio is defined as the product of

a) Mass and temperature

b) Mass and specific heat

c) Specific heat and temperature

d) Time and temperature

Answer: b

Explanation: The product mass and specific heat of a fluid flowing in a heat exchanger is known as capacity ratio.

2. A single pass shell and tube heat exchanger, consisting of a bundle of 100 tubes  is used for heating 28 kg/s of water from 25 degree Celsius to 75 degree Celsius with the help of a steam condensing at atmospheric pressure on the shell side with condensing heat transfer coefficient 5000 W/m 2 degree. Make calculation for overall heat transfer coefficient based on the inner area. Take fouling factor on the water side to be 0.002 m 2 degree/W per tube and neglect effect of fouling factor on the shell side and thermal resistance of the tube wall

heat-transfer-questions-answers-heat-exchanger-effectiveness-q2

a) 647.46 W/m 2 degree

b) 747.46 W/m 2 degree

c) 847.46 W/m 2 degree

d) 947.46 W/m 2 degree

Answer: c

Explanation: Q = m c c c (t c2 – t c1 ). Re = 7394, Pr = 3.53 and Nu = 47.41. I/U = I/h I + R + r i /(r 0 ) (h 0 ).

3. Which of the following is not associated with a heat exchanger?

a) Fouling

b) NTU

c) Capacity ratio

d) Mc Adam’s correction factor

Answer: d

Explanation: The correction factor i.e. Mc Adam’s is associated with laminar film condensation on a vertical plate.

4. The engine oil at 150 degree Celsius is cooled to 80 degree Celsius in a parallel flow heat exchanger by water entering at 25 degree Celsius and leaving at 60 degree Celsius. Estimate the exchanger effectiveness

a) 0.56

b) 0.66

c) 0.76

d) 0.86

Answer: a

Explanation: Effectiveness = (t h 1 – t h 2 ) C h /C MIN (t h 1 – t c 2 ).

5. Consider the above problem, if the fluid flow rates and the inlet conditions remain unchanged, workout the lowest temperature to which the oil may be cooled by increasing length of the exchanger

a) 46.62 degree Celsius

b) 56.62 degree Celsius

c) 66.62 degree Celsius

d) 76.62 degree Celsius

Answer: c

Explanation: Effectiveness = 1 – [exponential [- NTU ]/1 + C].

6. In a surface condenser, the water flowing through a series of tubes at the rate of 200 kg/hr is heated from 15 degree Celsius to 75 degree Celsius. The steam condenses on the outside surface of tubes at atmospheric pressure and the overall heat transfer coefficient is estimated at 860 k J/m 2 hr degree. Find the effectiveness of the heat exchanger. At the condensing pressure, stream has a saturation temperature 0f 100 degree Celsius and the latent heat of vaporization is 2160 k J/kg. Further, the steam is initially just saturated and the condensate leaves the exchanger without sub-cooling i.e. only latent heat of condensing steam is transferred to the water. Take specific heat of water as 4 k J/kg K

a) 0.224

b) 0.706

c) 2.224

d) 3.224

Answer: b

Explanation: Effectiveness = 1 – exponential  and Effectiveness = C h (t h 1 – t h 2 )/C MIN (t h 1 – t c 2 ).

7. Consider the above problem, find the tube length. Let the diameter of tube is 25 mm

a) 14.5 m

b) 15.5 m

c) 16.5 m

d) 17.5 m

Answer: a

Explanation: NTU = U /C.

8. For evaporators and condensers, for the given conditions, the logarithmic mean temperature difference for parallel flow is

a) Does not depend on counter flow

b) Smaller than counter flow

c) Greater than counter flow

d) Equal to counter flow

Answer: d

Explanation: The temperature of one of the fluid remains constant during the flow passage.

9. Water  enters a cross flow exchanger  at 15 degree Celsius and flows at the rate of 7.5 kg/s. It cools air (C P = 1 k J/kg K) flowing at the rate of 10 kg/s from an inlet temperature of 120 degree Celsius. For an overall heat transfer coefficient of 780 k J/m 2 hr degree and the surface area is 240 m 2 , determine the NTU

a) 4.2

b) 5.2

c) 6.2

d) 7.2

Answer: b

Explanation: NTU = U A/C MIN = 5.2.

10. Consider the above problem, find the capacity ratio of the heat exchanger

a) 0.555

b) 0.444

c) 0.333

d) 0.222

Answer: c

Explanation: Capacity ratio = 10/30 = 0.333.

This set of Heat Transfer Puzzles focuses on “Number Of Transfer Units”.

1. Pick the odd one out

a) Open feed water heaters

b) Jet condensers

c) De-super heaters

d) Surface condensers

Answer: d

Explanation: In surface condenser, the heat transfer occurs between the fluid streams without mixing or physical contact with each other.

2. The lubricating oil for a large industrial gas turbine engine is cooled in a counter flow, concentric tube heat exchanger. The cooling water flows through the inner tube  with inlet temperature 25 degree celsius and mass flow rate 0.2 kg/s. The oil flows through the annulus  with mass flow rate 0.125 kg/s and its temperature at entry and exit are 90 degree Celsius and 60 degree Celsius. Find outlet temperature of cooling water

heat-transfer-questions-puzzles-q2

a) 14.58 degree celsius

b) 24.58 degree celsius

c) 34.58 degree celsius

d) 44.58 degree celsius

Answer: c

Explanation: m h c c (t h1 – t h2 ) = m c c c (t c2 – t c1 ).

3. A cross flow type air heater has an area of 50 cm 2 . The overall heat transfer coefficient is 100 W/m 2 K and heat capacity of both hot and cold stream is 1000 W/m K. The value of NTU is

a) 1000

b) 500

c) 5

d) 0.2

Answer: b

Explanation: NTU = A U/C MIN = 5.

4. In a balanced counter flow heat exchanger with m h c h = m c c c , the NTU is equal to unity. What is the effectiveness of heat exchanger?

a) 0.5

b) 1.5

c) 0.33

d) 0.2

Answer: a

Explanation: Effectiveness = NTU/NTU – 1 = 0.5.

5. After expansion from a gas turbine, the hot exhaust gases are used to heat the compressed air from a compressor with the help of a cross flow heat exchanger of 0.8 effectiveness. What is the number of transfer units of the heat exchanger?

a) 2

b) 4

c) 8

d) 16

Answer: b

Explanation: Effectiveness = NTU/NTU – 1 = 0.5.

6. E – NTU method is particularly useful in thermal design of heat exchangers when

a) Inlet temperature of the fluid streams are not known as a priori

b) The outlet temperatures of the hot fluid streams is known but that of cold fluid stream is not known as a priori

c) The outlet temperatures of the fluid streams is known as a priori

d) The outlet temperatures of the fluid streams are not known as a priori

Answer: d

Explanation: It is useful when outlet temperatures of the fluid streams are not known as a priori.

7. The engine oil at 150 degree Celsius is cooled to 80 degree Celsius in a parallel flow heat exchanger by water entering at 25 degree Celsius and leaving at 60 degree Celsius. Estimate the number of transfer units

a) 1.221

b) 2.221

c) 3.221

d) 4.221

Answer: a

Explanation: Effectiveness = 1 – exponential [- NTU ]/1 + C.

8. In a surface condenser, the water flowing through a series of tubes at the rate of 200 kg/hr is heated from 15 degree Celsius to 75 degree Celsius. The steam condenses on the outside surface of tubes at atmospheric pressure and the overall heat transfer coefficient is estimated at 860 k J/m 2 hr degree. Find the number of transfer units of the heat exchanger. At the condensing pressure, stream has saturation temperature 0f 100 degree Celsius and the latent heat of vaporization is 2160 k J/kg. Further, the steam is initially just saturated and the condensate leaves the exchanger without sub-cooling i.e. only latent heat of condensing steam is transferred to water. Take specific heat of water as 4 k J/kg K

a) 3.224

b) 2.224

c) 1.224

d) 0.224

Answer: c

Explanation: Effectiveness = 1 – exponential [- NTU].

9. Consider the above problem, find the steam condensation rate

a) 52.22 kg/hr

b) 42.22 kg/hr

c) 32.22 kg/hr

d) 22.22 kg/hr

Answer: d

Explanation: m h f g = m c c c (t c 2 – t c 1 ).

10. NTU is a number of transfer units, dimensionless parameter defined as

a) U A/C MIN

b) 2 U A/C MIN

c) 3 U A/C MIN

d) 4 U A/C MIN

Answer: a

Explanation: The number of transfer units is a measure of the size of the heat exchanger, it provides some indication of the size of the heat exchanger.