Computational Fluid Dynamic pune university MCQs
Computational Fluid Dynamic pune university MCQs
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Philosophy”.
1. Which among the following is a reason why we do not completely rely upon ground tests for analysing fluid dynamics?
a) Three-dimensional flows cannot be analysed
b) Facilities do not exist in all fight regimes
c) The output generated is not as accurate as theoretical analysis
d) Long run-time
Answer: b
Explanation: Ground test facilities can be used to model three-dimensional flows also and even they produce accurate results in a less run-time. But they cannot be used to test all flight regimes as they want artificial set-up for every single property of the flow.
2. Which one do you think is not possible with wind tunnels for testing trans-atmospheric vehicles?
a) Continuously changing Mach number
b) Transonic flows
c) Simultaneously modelling high Mach numbers and high temperatures
d) Hypersonic flows
Answer: c
Explanation: Continuous change in Mach numbers can be done in wind tunnels. Transonic and hypersonic wind tunnels also exist. If we try to model high speeds at high temperatures, the wind tends to reduce the temperature as we have wind flowing over the stationary model in a wind tunnel.
3. CFD is the third approach for fluid flow analysis. What are the other two approaches?
a) Theoretical and experimental
b) Physical and Mathematical
c) Numerical and experimental
d) Experimental and physical
Answer: a
Explanation: Pure theoretical and pure experimental approaches were the two approaches prior to the advent of CFD. To overcome the disadvantages in both of these approaches, Computational Fluid Dynamics was invented.
4. When were the foundations of experimental fluid dynamics laid?
a) 19 th century
b) 18 th century
c) 16 th century
d) 17 th century
Answer: d
Explanation: Experimental fluid dynamics was started in France and England in the 17 th century when the relation between force and velocity is found from experiments.
5. The eighteenth and nineteenth centuries witnessed the development of theoretical fluid dynamics in ____ countries.
a) Asian
b) American
c) European
d) African
Answer: c
Explanation: Theoretical fluid dynamics was developed in European countries in the 18 th and 19 th centuries first theoretical derivation of drag equation is found.
D ∝ ρ SV 2
6. This invention of the 20 th century and accurate numerical methods have revolutionized the way we analyse Fluid Dynamics.
a) High-speed digital computers
b) Personal computers
c) Submarines
d) Rocketry
Answer: a
Explanation: Invention of high-speed digital computers allowed modelling and simulating fluid flows with high accuracy as the level of computing involved in the numerical methods is very high. Without this, it would have been very difficult to solve the numerical algorithms.
7. Which of the following is not true about CFD?
a) There will be a need for theory and experiments
b) CFD is an equal partner of theoretical and experimental analyses
c) CFD will complement theoretical and experimental Fluid Dynamics
d) CFD will replace the approaches of pure theory and pure experiments
Answer: d
Explanation: The future of fluid dynamics will rest upon a proper balance of pure experiment, pure theory and computational fluid dynamics, each complementing one another in their limitations.
8. The design of this experimental NASA aircraft was aided by CFD in early days.
a) Northrop
b) HiMAT
c) Douglas
d) Rockwell
Answer: b
Explanation: HiMAT is a NASA experimental aircraft designed to test concepts of high manoeuvrability. Wind tunnel tests showed that there will unacceptable drag. Wings of this aircraft is redesigned using CFD to overcome this problem.
9. CFD analyses Fluid Dynamics using this method.
a) Analytical
b) Physical
c) Numerical
d) Experimental
Answer: c
Explanation: As the experimental analysis of fluid flow problems are very expensive, CFD uses theoretical method to analyse them. Among the two theoretical methods stated above , the analytical method uses approximations which makes the theory unreliable. So, CFD uses the numerical method.
10. CFD provides results of ____________
a) Continuous time varying results at discrete locations
b) Discrete points of space and time
c) Continuous spatial results at discrete time points
d) Continuous in time and space
Answer: b
Explanation: CFD discretizes the equations and also the domain and solves the discretized equations for only the points in the discretized domain using numerical methods.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Research and Design Tool”.
1. Computational fluid dynamic results are _________ wind tunnel results.
a) Better than
b) Analogous to
c) More reliable than
d) Energy consuming when compared to
Answer: b
Explanation: CFD results completely depend on the problem and solution models we opt to simulate the flow. So, if the chosen models are correct, the CFD results will be analogous to experimental results. In fact, by comparing the results with wind tunnel results, CFD models are changed.
2. Which of these characteristics does not apply for a CFD tool?
a) Unwieldy
b) Easy to carry around
c) Can be remotely accessed
d) Transportable
Answer: a
Explanation: A CFD tool just means a computer program that makes it very easy to carry or transfer. Moreover, the source program can even be remotely accessed. This makes it a readily transportable tool. A wind tunnel is a tool which is unwieldy.
3. CFD can be used to ___________ the experimental results.
a) Improve
b) Replace
c) Interpret
d) Convert
Answer: c
Explanation: CFD does not just provide a quantitative comparison with the wind tunnel results. It provides a means to interpret the experimental conditions. For example, to determine whether the flow is laminar or turbulent in a quantitative experiment carried out in a wind tunnel.
4. CFD carries out ___________ experiments.
a) Observational
b) Analytical
c) Field
d) Numerical
Answer: d
Explanation: CFD uses numerical experimentation to obtain the flow properties at discrete points. Computer programs are used to carry out these experiments.
5. ___________ technique is used in a wind tunnel to find whether the flow is laminar or turbulent.
a) Pressure sensitive paint
b) Force measurement
c) Flow visualization
d) Quantitative
Answer: c
Explanation: A flow can be categorized into laminar or turbulent by seeing it. So, the flow of air should be made visible by some flow visualization techniques. In the case of CFD, it can be easily seen from the resulting plot of flow properties.
6. In the early days, CFD simulations were limited to two-dimensional analyses. Three-dimensional analyses could not be performed because of _________
a) Complex mathematical models were not resolved
b) Governing equations were not developed for three-dimensions
c) Approximations for three-dimensions did not exist
d) The type of computers and algorithms that existed
Answer: d
Explanation: The computers and algorithms that existed that time was not suitable for three-dimensional analyses. Storage and speed capacities of computers were not enough.
7. Which of these problems does not require three-dimensional analysis?
a) Internal flow in SCRAM jet engines
b) Flow over an aircraft wing
c) Flow past gas turbine compressors
d) Flow over airfoils
Answer: d
Explanation: An airfoil is a two-dimensional profile of a wing. Flow over this two-dimensional profile will not need any three-dimensional analysis.
8. The knowledge of pressure distribution is required for ____________ engineers.
a) Aerodynamic
b) Thermal
c) Structural
d) Avionics
Answer: a
Explanation: An aerodynamic engineer needs the pressure distribution, vortex formation and other aerodynamic details to make an aircraft aerodynamically efficient.
9. The knowledge of aerodynamic loads on an aircraft is needed for ___________ engineers.
a) Aerodynamic
b) Thermal
c) Structural
d) Avionics
Answer: c
Explanation: A structural engineer is the one who analyses the loads to be carried and decides the materials for various aircraft parts. So, the loads acting on the parts must be known to him.
10. Aerodynamics engineers obtain the lift and pressure drag by integrating the ___________ distribution over a surface.
a) Velocity
b) Pressure
c) Temperature
d) Viscosity
Answer: b
Explanation: Lift force can be obtained by the difference in pressure between the top and bottom surfaces of a body. Similarly, the pressure drag of a body can be obtained by finding the pressure difference between the front and back halves of a body. Both of these quantities depend upon the pressure distribution.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “CFD Solution Procedure”.
1. Which of these will not come under the three main elements of CFD packages?
a) Pre-processor
b) Post-processor
c) Code creator
d) Solver
Answer: c
Explanation: In order to provide their users with easy access to its solver, CFD packages have sophisticated input and output interfaces. The three main elements of CFD packages are pre-processor, solver and post-processor.
2. The region of interest for analysis in CFD is called as _______________
a) Cell
b) Domain
c) Mesh
d) Grid
Answer: b
Explanation: The region of interest for solving a particular fluid flow problem is called domain. The first step in pre-processing is to define the geometry of this domain.
3. Over 50% of the time spent in the industry on a CFD project is devoted to the definition of the domain geometry and grid generation. Which one will be the reason for this?
a) More grids will give better results
b) Calculation time is directly proportional to the number of cells
c) To generate non-uniform grids
d) To generate an optimal grid which is a compromise between desired accuracy and solution cost
Answer: d
Explanation: Accuracy of a CFD solution directly depends on the number of grids. On the other hand, if there are more grids, the cost of computation will increase. To overcome this, a grid which is a compromise between both of these should be generated. So, a lot of time is spent on grid generation.
4. Which of these could be an optimal mesh?
a) Non-uniform
b) Uniform
c) Grids with increasing lengths
d) Grids with decreasing lengths
Answer: a
Explanation: A uniform grid has all the cells with the same dimensions. The flow properties may not vary uniformly in the domain. Therefore, a non-uniform grid with more cells in the areas with large variations is an optimal way of grid generation.
5. The solution of a flow problem is defined at discrete points in the domain is called as _________
a) Elements
b) Cells
c) Grids
d) Nodes
Answer: d
Explanation: Nodes are the intersection of cells in a domain. These are the points where the flow properties are defined after solving the problem. The flow is not analysed in continuous points of the domain.
6. CFD packages solve the algebraic equations of flow using ____________ method.
a) Direct
b) Iterative
c) Analytical
d) Trial and error
Answer: b
Explanation: The physical phenomena makes the algebraic equations complex and non-linear. Hence, an iterative method is used in CFD packages to solve these equations.
7. Validation of a CFD code requires information about ____________
a) Boundary conditions
b) Domain
c) Grids
d) Cells
Answer: a
Explanation: Validation is the process of checking the accuracy of a CFD analysis. This needs highly detailed information about the boundary conditions.
8. Which of these will fall into the post-processing category?
a) Definition of boundary conditions
b) Grid generation
c) Flow visualization
d) Discretization
Answer: c
Explanation: Post-processing is the final step in CFD after pre-processing and solving. This involves various methods of visualizing the flow as well as getting quantitative information.
9. The step – specification of boundary conditions – in CFD comes under ________________
a) Post-processing
b) Solving
c) Discretizing
d) Pre-processing
Answer: d
Explanation: Boundary conditions at the cells which are in touch with the domain boundaries must be specified before starting to solve a problem in CFD. So, it comes under pre-processing.
10. Which is the input part of a CFD problem?
a) Post-processing
b) Flow visualization
c) Pre-processing
d) Solving
Answer: c
Explanation: The pre-processing part of a CFD problem is the first step before starting to process the problem. This is where the inputs are specified.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Applications”.
1. For which of the following purposes can an automobile company not use the CFD tool?
a) Study heat transfer between its parts
b) Increase aerodynamic performance
c) Increasing load capacity
d) Increasing fuel economy
Answer: c
Explanation: For increasing the load capacity, the strength of the vehicle should be more. Therefore, a structural analysis would be ideal for that purpose. There will not be a need for fluid flow analysis.
2. The internal flow analysis of an automobile running based on Otto-cycle will need a ___________ analysis.
a) Transient
b) Steady
c) Finite difference
d) Finite element
Answer: a
Explanation: The piston inside an internal combustion Otto-engine moves up and down continuously which makes the flow unsteady. A flow of fluid can be called steady if its properties do not vary with time.
3. Which of these forces will have to be analysed using CFD to improve the aerodynamic performance of a vehicle?
a) Lift
b) Drag
c) Thrust
d) Weight
Answer: b
Explanation: The drag force is the one which pushes a body backward during its motion. Therefore, to increase the efficiency of a vehicle, there should be less drag.
4. Which of these will not be applicable for CFD in naval applications?
a) Propeller Design
b) Wind loads
c) Lift analysis
d) Stability in manoeuvring
Answer: c
Explanation: Hydrodynamic analysis on the propeller of a naval vehicle is done. As a vehicle will be moving through the air, wind loads should also be analysed. The stability of a vehicle when it changes its position is also analysed. But, the lift force need not be analysed.
5. CFD can be used to understand the flow behaviour of liquid metal during mould filling. This can be used to ____________
a) Change the mould according to fluid flow
b) Choose the best metal
c) Improve casting techniques
d) Change temperature
Answer: c
Explanation: The mould cannot be changed according to fluid flow. It should have the shape of the product needed. Choosing the metal depends upon the application and structural concern. The temperature change may affect the quality of the product. By knowing the flow pattern, casting techniques can be improved.
6. Which of these models would be the best for flow over a submarine?
a) 3-D Navier-Stokes equation for compressible flow without a turbulence model
b) 3-D Navier-Stokes equation for incompressible flow without a turbulence model
c) 3-D Navier-Stokes equation for compressible flow with a turbulence model
d) 3-D Navier-Stokes equation for incompressible flow with a turbulence model
Answer: d
Explanation: Incompressible flow is chosen as the flow of water will mostly be incompressible unless the flow velocity is very high. Turbulence model is chosen as the flow properties will get abrupt change due to high Reynolds number.
7. CFD applications provide information for the design of furnaces with ____ thermal efficiency and ____ emissions of pollutants.
a) Increased, reduced
b) Reduced, increased
c) Reduced, reduced
d) Increased, increased
Answer: a
Explanation: Thermal efficiency should be increased to get the best out of the input energy. Considering the environmental effects, the emission of pollutants should be less.
8. Which is not an internal analysis?
a) Combustion
b) Turbulence
c) Flow over compressor
d) Exhaust pipes
Answer: c
Explanation: Flow over the compressors actually takes place inside a gas turbine engine. But, analysing the flow over the compressor blades is an external flow analysis.
9. What would be the major difference between aerodynamic and hydrodynamic analyses?
a) Temperature
b) Reynolds Number
c) Velocity
d) Domain
Answer: b
Explanation: Reynolds number of the flow would be the major change as the density and viscosity of water will be higher than that of air. Therefore, while modelling a hydrodynamic flow, care should be taken.
10. This created a problem in modelling supersonic blunt nose.
a) Change of flow equations from elliptic to hyperbolic
b) High speed with high temperature
c) Supersonic Mach number
d) High temperatures
Answer: a
Explanation: Straight to the nose of the vehicle, the shock is normal. This results in a subsonic region and elliptic flow equations. But, downstream the shock is oblique which creates a supersonic region resulting in the elliptic equation. This was the problem in modelling supersonic blunt nose. In later years this was overcome by a better model.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “CFD Advantages and Disadvantages”.
1. Computational investigation is _____________ experimental investigation.
a) Faster than
b) At the same speed of
c) Slower than
d) Cannot be compared
Answer: a
Explanation: Computational investigation can be carried out at a higher speed when compared to experiments especially when the models are simple and suitable for computation.
2. What is the disadvantage in predicting a complex problem with a very limited objective in CFD?
a) Time-consuming
b) Impossible to solve
c) Slower
d) Expensive
Answer: d
Explanation: For difficult problems with complex geometry, strong non-linearity, etc., if the objective is very limited, then use of CFD will become costlier than experimenting. For such cases, experimental results are suitable.
3. What is the advantage of numerical methods over analytical method?
a) Speed
b) Cost
c) Flexibility
d) Time
Answer: c
Explanation: Analytical methods of solving a problem is not applicable for many of the cases. Only a tiny part of the range of practical problems can be solved using analytical methods. Numerical methods can be used to solve a very wide range of practical problems.
4. Why ideal conditions should be simulated in CFD?
a) Experimental analysis is impossible for ideal conditions
b) Experimental set-up will have imperfections
c) Experimental set-up will lead to wrong solutions
d) Experimental analysis can be done for ideal cases
Answer: b
Explanation: Some research purposes will need non-practical processes to be analysed. Setting up experimental analysis for such cases involves a lot of steps and that may not yield proper results as the set-up cannot be perfect.
5. CFD can give ____________ results than experiments.
a) Detailed
b) Accurate
c) Reliable
d) Approximate
Answer: a
Explanation: Flow properties at only particular points can be obtained in experiments. CFD can provide information about multiple flow properties at many points in a single analysis.
6. For optimization of designs, CFD is ___________
a) Slow and expensive
b) Cost-effective but slow
c) Fast but expensive
d) Cost-effective and fast
Answer: d
Explanation: For design optimization, multiple tests should be done on different cases and many trials should be done. CFD is faster and cost-effective than experiments in this case.
7. An optimal prediction is ______________
a) Computational only
b) A combination of Computational and Experimental
c) Analytical only
d) Experimental only
Answer: b
Explanation: Experimental results are difficult to obtain in all the places, especially, in the locations of mounting. If we solely rely upon CFD, problem may arise out of wrong modelling. So, it is always better to use CFD models to get detailed results and using experiments to validate them.
8. Consider the following case. A teacher wants to show the effects of severe turbulence over an aircraft to the students. What is the best way to simulate it?
a) Wind tunnel with analytical backup
b) Wind tunnel with CFD backup
c) Flow visualization in wind tunnel
d) Turbulence modelling and animation using CFD tools
Answer: c
Explanation: As the intention is to just show the flow, CFD is not preferable. Solving turbulence model in CFD with animation needs complex solutions. So, in this case, choosing wind tunnel and visualizing the flow is the best way.
9. ____________ of a mathematical model is a limitation to CFD.
a) Reliability
b) Accuracy
c) Solvability
d) Validity
Answer: d
Explanation: Validity of a mathematical model is the relevancy of the model with the problem taken. If the mathematical model does not comply with the problem being solved, the entire prediction may go wrong.
10. I have to model a simple rectangular wing for my experimental aircraft testing a new technology on it. I have to choose an airfoil among the ten options given for my wing which would give the best aerodynamic performance. Suggest the best method to test and fix one airfoil.
a) CFD analysis
b) Literature
c) Wind tunnel models
d) Calculations
Answer: a
Explanation: This is an optimization case. CFD can be used to do trial and error analysis on the best fitting wing.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Simple CFD Techniques – CFD Softwares”.
1. Which one is a post-processing Software?
a) FieldView
b) CFD GEOM
c) CCM++
d) Typhon
Answer: a
Explanation: FieldView is a CFD post-processing software. FieldView is a data handling and visualizing tool that can be used for CFD post-processing.
2. Which of these plots are irrelevant to CFD post-processing?
a) Contour plots
b) Vector plots
c) xy plots
d) Bar plots
Answer: d
Explanation: Contour plots give a global view of the variation in flow properties. xy plots show the variation of a flow property with respect to an independent variable. Vector plots display vector quantities at discrete grid points. Bar plots cannot be used in CFD.
3. Is adaptive meshing possible in current CFD packages? If yes, which software offers it?
a) No
b) Yes, ANSYS Fluent 12.0
c) Yes, PHOENICS
d) Yes, ANSYS 8.0
Answer: b
Explanation: Adaptive meshing allows to change or refine mesh during the solution of the problem. This was not possible in earlier versions. From ANSYS Fluent 12.0, we get adaptive meshing.
4. The following plot represents flow velocity. Which of these points has the highest acceleration?
computational-fluid-dynamics-questions-answers-cfd-softwares-q4
a) 1
b) 2
c) 3
d) 4
Answer: b
Explanation: High acceleration means highly varying velocity. In a contour plot, the points with high gradient are represented by close lines. Point 2 has lines very close to each other. Therefore, point 2 has the highest acceleration.
5. Which one is an open source CFD solver tool?
a) OpenFOAM
b) ANSYS Fluent
c) TGrid
d) ParaView
Answer: a
Explanation: OpenFOAM is an open source CFD toolbox based on C++. It allows us to develop customized solvers. Some of the other open source software are CFL3D, Typhon, OVERFLOW, and Wind-US. ANSYS fluent is licensed software. TGrid and ParaView are for pre-processing and post-processing respectively.
6. Which of the following properties will need a plot like this?
computational-fluid-dynamics-questions-answers-cfd-softwares-q6
a) Pressure
b) Temperature
c) Velocity
d) Lift
Answer: c
Explanation: The given plot is a vector plot. Velocity is the directional flow property of a fluid. To represent its direction and magnitude, we can use the vector plot.
7. When was the first commercial CFD package released?
a) 1991
b) 1983
c) 1985
d) 1981
Answer: d
Explanation: PHOENICS is the first commercial CFD package released in the year 1981. This can simulate fluid flow, heat and mass transfer, chemical reactions, combustion, etc.
8. Full form of FOAM in OpenFOAM?
a) Field Optimization and Manipulation
b) Flow Optimization and Manipulation
c) Field Operation and Manipulation
d) Flow Operation and Manipulation
Answer: c
Explanation: OpenFOAM stands for open-source Field Operation and Manipulation. It was originally named as FOAM when created by Henry Weller in the late 1980s.
9. Which plot will you choose if you want your users to read quantitative data from your data visualization?
a) Contour plot
b) Vector plot
c) xy plot
d) Mesh plot
Answer: c
Explanation: xy plot shows the quantitative variation of a dependent variable with some independent variable. So, a user can use it readily to get quantitative data at a particular point.
10. Which is a pre-processing software?
a) ANSYS ICEM CFD
b) ANSYS Mechanical
c) ANSYS CFX
d) EnSight
Answer: a
Explanation: ICEM stands for Integrated Computer Engineering and Manufacturing. This is a software that allows us to create models and mesh it for generating grids.
This set of Computational Fluid Dynamics Assessment Questions and Answers focuses on “CFD Techniques – Alternating Direction Implicit Techniques”.
1. Which of these is important while solving a system with implicit methods?
a) Linearizing the difference equation
b) Linearizing the partial differential equation
c) Normalizing the difference equation
d) Normalizing the partial differential equation
Answer: a
Explanation: The nature of the original PDE should not be changed while solving the problem. If the problem is non-linear in PDE, to make it solvable by the implicit scheme, the difference equation should be linearized.
2. How many steps does the Alternating Direction Implicit scheme involve?
a) One step
b) Two steps
c) Three steps
d) Four steps
Answer: b
Explanation: The whole process is to initiate from the known values at the time-step t and move on to the required time-step t+Δ t. For this marching, the ADI scheme uses two steps. The solution does not directly move to the final step t+Δ t.
3. The intermediate step of the ADI scheme is at __________
a) t+\
t+\
t-\
t+\(\frac{\Delta t}{4}\)
Answer: b
Explanation: The first step of the ADI scheme is from t to t+\(\frac{\Delta t}{2}\). The second step of the ADI scheme is from t+\(\frac{\Delta t}{2}\) to t+Δt. Here, the intermediate step t+\(\frac{\Delta t}{2}\) is extra for the process and the results here are not actually needed.
4. Which of these statements is correct about the first step of the ADI scheme?
a) x-derivative is treated implicitly
b) y-derivative is treated implicitly
c) Time derivative is treated implicitly
d) Thomas algorithm is not used
Answer: a
Explanation: In the first step with the time interval \(\frac{\Delta t}{2}\), the spatial derivatives are replaced using the central difference scheme. Only the x-derivative is treated implicitly. Then the resulting equations are solved using the TDMA method.
5. The second step of the ADI scheme is swept over __________ direction.
a) both the x and y
b) the x
c) the y
d) the time
Answer: c
Explanation: The second step of the ADI scheme uses the time domain from t+\(\frac{\Delta t}{2}\) to t+Δt. Here, the y-derivative is treated implicitly after replacing the derivatives with central differences. The solutions are swept in the y-direction here.
6. If there are N grid points in both the x and y-directions, how many times does the ADI scheme use the Thomas algorithm?
a) N/2
b) 2
c) N 2
d) 2N
Answer: d
Explanation: In the first step of the ADI scheme, the Thomas algorithm is used N times to solve in the x-direction. Similarly, in the y-direction, again the Thomas algorithm is used N times to get the solutions. Totally, the Thomas algorithm is used 2N times.
7. The order of accuracy of the ADI scheme in the time direction is ___________
a) third-order
b) fourth-order
c) first-order
d) second-order
Answer: d
Explanation: The truncation error of the ADI scheme in the time direction is of order two. The ADI scheme is second-order accurate in the time direction. It uses two time-steps to move from the step t to t+Δt.
8. The order of accuracy of the ADI scheme in the x and y-directions are __________ and __________
a) second-order and first-order
b) first-order and second-order
c) second-order and second-order
d) second-order and third-order
Answer: c
Explanation: The truncation errors of the ADI scheme in the x and the y-directions are O(Δx 2 ) and O(Δy 2 ). Therefore, the order of accuracy of the ADI scheme in the x and the y-directions are two and two respectively.
9. The ADI scheme is particularly suitable for ____________ problems.
a) parabolic and elliptic
b) parabolic and hyperbolic
c) hyperbolic
d) parabolic
Answer: a
Explanation: The ADI scheme is useful to solve many fluid flow problems including the heat conduction and the mass diffusion problems. It is particularly suitable for the parabolic and elliptic problems.
10. Which of these is a popular version of the ADI scheme?
a) Operator splitting
b) Approximate factorization
c) ALU algorithm
d) SIMPLE algorithm
Answer: b
Explanation: The major objective of the ADI scheme is to make the Thomas algorithm for tri-diagonal matrices applicable to multi-dimensional problems. This scheme has other versions too. One of the popular versions of the ADI scheme is the approximate factorization scheme.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Governing Equations – Reynolds Transport Theorem”.
1. The Reynolds transport theorem establishes a relationship between __________ and ___________
a) Control mass system, Control volume system
b) Differential equation, Integral equation
c) Non-conservative equation, Conservative equation
d) Substantial derivative, Local derivative
Answer: a
Explanation: Equations formed by considering the control mass system and control volume system are not the same even if the same physical law is used. A relation between these equations is established by Reynolds transport theorem.
2. Let B denote any property of a fluid flow. The statement of Reynolds transport theorem is “The instantaneous total change of B inside the _____________ is equal to the instantaneous total change of B within the ______________ plus the net flow of B into and out of the _____________”
a) Control volume, Control mass, Control volume
b) Control volume, Control volume, Control mass
c) Control mass, Control mass, Control volume
d) Control mass, Control volume, Control volume
Answer: d
Explanation: Statement of Reynolds Transport Theorem: “The instantaneous total change of B inside the control mass is equal to the instantaneous total change of B within the control volume plus the net flow of B into and out of the control volume”.
3. Consider the following terms:
MV → Material Volume
V → Control Volume
S → Control Surface
B → Flow property
b → Intensive value of B in any small element of the fluid
ρ → Density of the flow
t → Instantaneous time
\
\
_{MV} = \frac{d}{dt}
+ \int_vb \rho \vec{v}.\vec{n} dV\)
b) \
_{MV} = \frac{d}{dt}
+ \int_sb \rho \vec{v}.\vec{n} dS\)
c) \
_V = \frac{d}{dt}
+ \int_sb \rho \vec{v}.\vec{n} dS\)
d) \
_{MV} = \int_vb \rho dV + \frac{d}{dt}
\)
Answer: b
Explanation:
\
_{MV} →\) Instantaneous total change of B in material volume
\
→ \) Instantaneous total change of ” B” within control volume
\
=
Therefore,
\
_{MV} = \frac{d}{dt}
+ \int_sb \rho \vec{v}.\vec{n}dS\).
4. Leibniz rule is applied to which of these terms in deriving Reynolds transport theorem?
a) Volume integral term of control volume
b) Differential term of material volume
c) Surface integral term of control volume
d) Volume integral term of material volume
Answer: a
Explanation: Using Leibniz rule, the differentiation of an integral term can be reduced. Here, differential of integral exists in the Volume integral term of Control Volume which is given by \
.\)
5. Why a surface integral is used to represent flow of B into and out of the control volume?
a) Control volume is moving
b) Flow of fluid is through the control surfaces
c) Fluid only on the control surfaces
d) Control volume is stationary
Answer: b
Explanation: Fluid can enter into or exit from the control volume through the control surface. If this flow velocity is integrated along the control surfaces, we can get the net inflow or outflow of fluid to the control volume.
6. When is Leibniz rule applicable to control volume?
a) When control volume is moving
b) When control volume is deforming
c) When control volume is fixed
d) In all conditions
Answer: c
Explanation: Leibniz rule is applicable to a system only if a system variable is independent. When control volume is fixed, position of the control volume becomes independent. So, Leibniz rule is applicable only to fixed control volumes.
7. Let,
V → Control Volume
B → Flow property
b → Intensive value of B in any small element of the fluid
ρ → Density of the flow
t → Instantaneous time
Which of these terms represent the ‘instantaneous total change of the flow property within the control volume’ after Leibniz rule is applied?
a) \
\)
b) \
dV\)
c) \
\(\rho \int_v \frac{\partial \rho}{\partial b} dV\)
Answer: b
Explanation: According to Leibniz rule, if the variation of f is independent of t,
\
dt = \int \frac{\partial}{\partial x}fdt\)
Instantaneous total change of the flow property within the control volume is given by,
\
\)
Applying Leibniz rule,
\
= \rho \int_v \frac{\partial b}{\partial T}dV\).
8. Gauss divergence theorem is used to convert a surface integral to volume integral. This is used in Reynolds Transport theorem. What is the purpose of this conversion?
a) Simplifying the term
b) Differentiating the flow property
c) Adding the flow property
d) Grouping terms related to control volume
Answer: d
Explanation: One term related to control volume is a volume integral. The other term is a surface integral. To group these two terms together, Gauss Divergence Theorem is used.
9. Gauss divergence is applied to which of these terms?
a) Instantaneous total change of B inside the control mass
b) Instantaneous total change of B within the control volume
c) Net flow of B into and out of the control volume
d) Net flow of B into and out of the control mass
Answer: c
Explanation: The term representing ‘net flow of B into and out of the control volume’ is a surface integral. This surface integral is converted into a volume integral using the Gauss divergence theorem.
10. Let,
V → Control Volume
b → Intensive value of B in any small element of the fluid
ρ → Density of the flow
\
\
dV\)
b) \
dS\)
c) \
dV\)
d) \
dS\)
Answer: a
Explanation: The term representing ‘net flow of B into and out of the control volume’ is
\
dV\).
11. The final equation of Reynolds transport theorem can be used to drive ____________ form of the conservation laws in fixed regions.
a) Eucledian
b) Lagrangian
c) Eulerian
d) Cartesian
Answer: c
Explanation: Reynolds transport theorem can be used to convert the material volume form of the conservation equations to Eulerian form.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Equations of State”.
1. How many equations are related to solving a flow field?
a) 2
b) 3
c) 5
d) 4
Answer: c
Explanation: There are 5 equations related to solving a flow field.
Mass conservation equation
x-momentum equation
y-momentum equation
z-momentum equation
Energy equation
2. Among the unknowns of a flow field, some of the properties are given below. Which set contains only thermodynamic properties?
a) Density, pressure, specific internal energy, temperature
b) Density, velocity, specific internal energy, temperature
c) Velocity, pressure, specific internal energy, temperature
d) Density, pressure, specific internal energy, Velocity
Answer: a
Explanation: Velocity is a property completely related to fluid flow. The other properties – density, pressure, specific internal energy and temperature are thermodynamic.
3. Relationship between thermodynamic variables of a flow field can be obtained through ___________
a) Momentum conservation
b) Thermodynamic equilibrium
c) Energy equations
d) Zeroth law of thermodynamics
Answer: b
Explanation: Thermodynamic equilibrium is a state of a matter where there is no transfer of energy. This condition can be used to relate the thermodynamic properties with one another.
4. Fluid velocity is very high. Will thermodynamic equilibrium be applicable to fluid flows?
a) Yes, the external conditions help them stay in thermodynamic equilibrium
b) No, their flow properties change abruptly
c) No, they are influenced by external conditions
d) Yes, the fluid can thermodynamically adjust itself quickly to be in thermodynamic equilibrium
Answer: d
Explanation: The velocity of fluid flow is very high. This may affect their thermodynamic equilibrium. But, the particles are small enough to thermodynamically adjust themselves to equilibrium so quickly.
5. We can describe the state of a substance in thermodynamic equilibrium using two state variables. What are these two variables?
a) Density and temperature
b) Density and pressure
c) Pressure and Temperature
d) Velocity and Temperature
Answer: a
Explanation: There are four thermodynamic variables – density, temperature, pressure and specific internal energy. Among these, pressure and specific internal energy can be represented using density and temperature.
6. Let,
ρ → Density
p → Pressure
T → Temperature
How can we represent p of a perfect gas in terms of ρ and/or T?
a) p=ρ RT
b) p=RT
c) p=ρ T
d) p=ρ R
Answer: a
Explanation: For perfect gases, pV=mRT. This can be written in the form p=ρ RT.
7. Let,
ρ → Density
T → Temperature
i → Specific internal energy
How can we represent i of a perfect gas in terms of ρ and/or T?
a) i=T
b) i ∝
c) i=
d) i ∝ T
Answer: d
Explanation: i=C v T is the relation between temperature and specific internal energy (where C v is the specific heat at constant velocity). This can be used to get the energy equation in terms of specific internal energy.
8. Which is/are the conservation laws that are enough to solve a complete fluid problem?
a) Energy and momentum conservation
b) Mass and energy conservation
c) Mass and momentum conservation
d) Mass equation
Answer: c
Explanation: A complete fluid flow problem can often be solved by only using the mass and momentum conservation equations. Energy conservation equation is not necessary.
9. Energy conservation equation is necessary to solve this property of fluid flow.
a) Pressure
b) Temperature
c) Density
d) Velocity
Answer: b
Explanation: Energy conservation should be solved for a fluid flow if we want the temperature distribution or if the system involves heat transfer.
10. Equations of state provide the linkage between ___________ and ____________
a) Conservative, non-conservative equation
b) Eulerian, Lagrangian equations
c) Energy equation, mass and momentum equations
d) Differential, Integral equations
Answer: c
Explanation: Equations of the state provide the linkage of Energy equation with mass and momentum equations. They give the thermodynamic properties in terms of the state variables.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Governing Equations – Flow Models”.
1. What does this diagram represent?
computational-fluid-dynamics-questions-answers-models-flow-q1
a) Finite control volume moving along with the flow
b) Stationary finite control volume
c) Infinitesimally small element with the fluid passing through it
d) Infinitesimally small element moving along with the flow
Answer: b
Explanation: The diagram represents a finite control volume stationary in position. Fluid flows into and out of this model.
2. A control volume based model gives ___________ equation.
a) Integral
b) Differential
c) Conservative
d) Non-conservative
Answer: a
Explanation: Control volumes are big enough that the flow properties can be integrated along it. Thus, a control volume yields an integral equation.
3. What is the need of constructing a model for analysing fluids?
a) Fluids are not stationary but they have the same velocity in different parts
b) Fluids are stationary and they have the same velocity in different parts
c) Fluids are not stationary and they have different velocities in different parts
d) Fluids are not stationary but they have the same velocity in different parts
Answer: c
Explanation: A solid body can be easily divided into different parts. A fluid is squishy and it has different velocities in different parts. This creates the need for a model to analyse it.
4. If this model is used to get the governing equations, what type of equation will be obtained?
computational-fluid-dynamics-questions-answers-models-flow-q4
a) Non-conservative differential
b) Conservative differential
c) Conservative integral
d) Non-conservative integral
Answer: d
Explanation: The given diagram represents a finite control volume moving along with the flow. This will give a non-conservative integral equation.
5. An equation modelled using infinitesimally small element leads to ____________
a) Partial differential equation
b) Integral equation
c) Differential equation
d) Linear differential equation
Answer: a
Explanation: Infinitesimally small element models directly leads to a partial differential equation. The element is infinitesimal in the sense of differential calculus.
6. A finite control volume moving along with the flow __________
a) Has its position coordinates stationary
b) Has its particles moving into and out of it
c) Has the properties differentiated
d) Has the same particles always inside it
Answer: d
Explanation: As the finite control volume moves along with the flow, the same particles are always concerned. The mass inside this model also does not change.
7. If this model is used to get the governing equations, what type of equation will be obtained?
computational-fluid-dynamics-questions-answers-models-flow-q7
a) Non-conservative differential
b) Conservative differential
c) Conservative integral
d) Non-conservative integral
Answer: b
Explanation: The given diagram represents an infinitesimally small element with the flow moving past it. This will give a conservative differential equation.
8. Other than finite control volume and infinitesimal small element, what is the third possible modelling of fluid flow?
a) Discrete approach
b) Quantum approach
c) Microscopic approach
d) Macroscopic approaches
Answer: c
Explanation: A microscopic approach is possible where the laws of nature are applied to the atoms and molecules. This is regarding kinetic theory.
9. What does this diagram represent?
computational-fluid-dynamics-questions-answers-models-flow-q9
a) Finite control volume moving along with the flow
b) Infinitesimally small element moving along with the flow
c) Infinitesimally small element with the fluid passing through it
d) Stationary finite control volume
Answer: b
Explanation: The small cube represents an infinitesimally small fluid element. This cube is moving. It represents an infinitesimally small element moving along with the flow.
10. A stationary model will result in ____________
a) Differential equation
b) Non-conservative equation
c) Conservative equation
d) Integral equation
Answer: c
Explanation: A stationary model has its position coordinates independent of time. This gives a conservative equation without any substantial derivative.
11. A model of fluid moving along with the flow gives non-conservative equation. What is the reason?
a) Position coordinates are dependent on time
b) Position coordinates are independent on time
c) Velocity is dependent on time
d) Velocity is independent of time
Answer: a
Explanation: As the flow model is not stationary, its position coordinates vary along with time. So, it results in a non-conservative equation.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Governing Equations – Eulerian and Lagrangian Conservation Laws”.
1. The principle of conservation is applicable to _______ systems.
a) isolated system
b) closed system
c) open system
d) all the systems irrespective of its type
Answer: a
Explanation: The principle of conservation is applicable only to the systems where there will not be any transfer of matter or energy. Isolated system will have these characters.
2. The fluid is subdivided into fluid parcels and every fluid parcel is followed as it moves through space and time. Which kind of formulation is this?
a) Cartesian
b) Eulerian
c) Lagrangian
d) Euclidian
Answer: c
Explanation: In Lagrangian fluid flow specification, the fluid is subdivided into many parts and each of this part is followed. These parts are called fluid parcels.
3. Each parcel in the Lagrangian formulation is tagged using __________
a) time-dependent position vector
b) time-independent position vector
c) time-dependent velocity vector
d) time-independent velocity vector
Answer: b
Explanation: The tag is time-independent so that the tag does not vary along with the flow. Position vector is chosen to follow the parcel along its position.
4. Which of these will best define positions of the parcel in increasing time?
a) Streamline
b) Streakline
c) Boundary line
d) Pathline
Answer: d
Explanation: Pathline is the one which represents the path of a fluid element along its way. So, to define the positions of the parcels, pathline is the best.
5. Which of these is an acceptable tag for Lagrangian parcels?
a) Parcel’s centre of mass at instantaneous time
b) Parcel’s centre of pressure at instantaneous time
c) Parcel’s centre of mass at initial time
d) Parcel’s centre of pressure at initial time
Answer: c
Explanation: Parcel’s centre of pressure is not suitable as it depends not only on position but on many variables too. Instantaneous time continuously varies along with the path. So, it cannot be a tag.
6. According to Eulerian approach, which of these is correct?
a) Both location and fluid move
b) Location moves and fluid is stationary
c) Both location and fluid are stationary
d) Location is stationary and fluid moves
Answer: d
Explanation: According to Eulerian approach, a particular location is chosen and the fluid flows past this location. This flowing fluid is analysed.
7. The independent variables in Eulerian approach are __________ and __________
a) instantaneous time and instantaneous position
b) initial time and instantaneous position
c) instantaneous time and Initial position
d) initial time and initial position
Answer: a
Explanation: Current position and time are the variables on which other variables depend on in Eulerian approach. These are the independent variables.
8. In Lagrangian approach, the flow parcels follow __________
a) pressure field
b) velocity field
c) temperature field
d) density field
Answer: b
Explanation: Velocity field defines the motion of fluid. Parcels also follow the same path of the fluid and therefore, parcels can be said to follow the velocity field.
9. Let,
t → Instantaneous time
\Missing open brace for subscript \
) = \frac{\partial \vec{x}
}{\partial t} \)
b) \
) = \frac{\partial \vec{x_0}
}{\partial t} \)
c) \
) = \frac{\partial \vec{x}
}{\partial t} \)
d) \
) = \frac{\partial \vec{x_0}
}{\partial t} \)
Answer: a
Explanation: Location at any instantaneous time is
\
\)
Velocity of fluid flow in Lagrangian approach is
\
) \)
The relationship between the approaches is
\
) = \frac{\partial \vec{x}
}{\partial t} \).
10. Which of the following frame of references does this diagram represent?
computational-fluid-dynamics-questions-answers-eulerian-lagrangian-conservation-laws-q10
a) Polar
b) Cartesian
c) Lagrangian
d) Eulerian
Answer: d
Explanation: In diagram, fluid particles move and the field of observation remains in the same position. This represents Eulerian Approach.
This set of Computational Fluid Dynamics Puzzles focuses on “Governing Equations – Velocity Divergence”.
1. In mathematical terms, how can the divergence of a velocity vector \
\) be represented?
a) \
\
\
\(\vec{V} \times\nabla\)
Answer: a
Explanation: \(\nabla.\vec{V}\)represents the divergence of a vector. \(\nabla\vec{V}\) is gradient which is not possible for a vector. \(\nabla \times\vec{V}\) is the curl of a vector. \(\vec{V} \times\nabla\) does not represent any property.
2. For a control volume moving along with the flow, which of these properties is a constant?
a) Volume
b) Shape
c) Mass
d) Velocity
Answer: c
Explanation: Mass of a control volume moving along with the flow will not vary. It is constant with time. The volume, shape and velocity of the control volume may not be the same at all points of time.
3. The time rate of change of a control volume moving along with the flow is represented by substantial derivative. Why?
a) Because the change is substantial
b) Because the change is more
c) Because of control volume
d) Because it is moving with the flow
Answer: d
Explanation: As the control volume moves along with the flow, their position coordinates continuously vary with time. This needs a substantial derivative.
4. What is the physical meaning of divergence of velocity?
a) Time rate of change of the volume per unit volume
b) Time rate of change of the volume of a moving fluid element per unit volume
c) Time rate of change of the volume
d) Time rate of change of the volume of a moving fluid element
Answer: b
Explanation: Divergence of velocity of a moving fluid model physically means that “time rate of change of the volume of a moving fluid element per unit volume”.
5. Divergence of velocity appears in the governing equations for _____________
a) infinitesimally small elements
b) stationary models
c) moving models
d) finite control volumes
Answer: c
Explanation: Divergence of velocity involves the change of volume of the fluid model. For stationary models, volume does not change. So, this applies for only moving fluid elements.
6. Let \
.\vec{ds}\) be the change in volume of elemental control volume in time Δt. Over the same time Δt, what is the change in volume of the whole control volume V with control surface S?
a) \
.\vec{ds}\)
b) \
\
.\vec{ds}\)
d) \
.\vec{ds}\)
Answer: d
Explanation: The change in volume of the whole control volume is the summation of \
.\vec{ds}\) over the total control surface S. This summation becomes integral as \
.\vec{ds}\) is the total change.
7. Consider a small control volume V with the surface dS with a normal vector \
. What is the change in volume of this small control volume ΔV?
computational-fluid-dynamics-puzzles-q7
a) \
.\vec{n}]\)
b) \
.\vec{n}]dS\)
c) \
]dS\)
d) \
.\vec{n}]dS\)
Answer: b
Explanation: The change in control volume can be calculated as the volume of a cylinder with altitude \
.\vec{n}\) and base area dS.
8. The time rate of change of volume represented by the following equation corresponds to _________
\
dV\) Where
V → Control volume
\
Control volume moving along with the flow
b) Infinitesimal element moving along with the flow
c) Control volume stationary
d) Infinitesimal element stationary
Answer: a
Explanation: Stationary models do not have a substantial derivative. If it is not control volume, volume integral is unnecessary. So, the model for this equation is ‘Control volume moving along with the flow’.
9. For infinitesimally small element moving along with the flow with velocity \
\
\
\
\(\nabla.\vec{V} = \frac{d}{dt} \)
Answer: c
Explanation: For control volumes,
\
dV\)
For infinitesimal elements, V can be represented as δ V. Therefore,
\
dV\)
As the elements are infinitesimally small, ∇.\
\delta V\)
\(\nabla \vec{V}=\frac{1}{\delta V} \frac{D}{Dt}\).
10. Expand divergence of velocity ∇.\
\
\
\
\(\frac{D\vec{V}}{Dt}\)
Answer: b
Explanation: Here,
\(\frac{\partial}{\partial x}\vec{i}+\frac{\partial}{\partial y}\vec{j}+\frac{\partial}{\partial z}\vec{k}\)
\(\vec{V}= u\vec{i} +v\vec{j}+w\vec{k}\)
∇.\(\vec{V}=\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}\)
For one-dimensional flow,
∇.\(\vec{V}=\frac{\partial u}{\partial x}\).
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Governing Equations – Substantial Derivative”.
1. How is the substantial derivative of velocity vector denoted?
a) \
\
\
\(\frac{D\vec{V}}{Dx}\)
Answer: a
Explanation: \(\frac{D\vec{V}}{Dt}\) is the substantial derivative. \(\frac{d\vec{V}}{dt}\) is the local derivative. \(\frac{\partial \vec{V}}{\partial t}\) is the partial derivative.
2. Expand the substantial derivative Dρ/Dt.
a) \
\
\
\(\frac{D\rho}{Dt}=\frac{\partial \rho}{\partial t}+u\frac{\partial \rho}{\partial x}+v\frac{\partial \rho}{\partial y}+w \frac{\partial \rho}{\partial z}\)
Answer: d
Explanation: As the location coordinates vary with time,
\(\frac{D\rho}{Dt}=\frac{\partial\rho}{\partial t}+\frac{\partial\rho}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial \rho}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial \rho}{\partial z}\frac{\partial z}{\partial t}\)
\(\frac{D\rho}{Dt}=\frac{\partial \rho}{\partial t}+u\frac{\partial \rho}{\partial x}+v\frac{\partial\rho}{\partial y}+\frac{\partial \rho}{\partial t}\)
3. Substantial derivative applies to ____________
a) Both stationary and moving models
b) Only moving models
c) Only stationary models
d) Neither stationary nor moving models
Answer: b
Explanation: Substantial derivatives arise as the coordinates move and they vary with time. So, they are applicable only to moving models.
4. The simplified form of substantial derivative can be given by __________
a) \
\
\
\(\frac{DT}{Dt}=\frac{\partial T}{\partial t}+\nabla \times T\)
Answer: c
Explanation:
\
.
\)
\(\frac{DT}{Dt}=\frac{\partial T}{\partial t}+\vec{V}.\nabla T.\)
5. Which of these statements best defines local derivative?
a) Time rate of change
b) Spatial rate of change
c) Time rate of change of a moving point
d) Time rate of change at a fixed point
Answer: d
Explanation: Local derivative is the term \(\frac{\partial}{\partial t}\) of a property. This defines the time rate of change of a property at a particular point with the assumption that the point is fixed.
6. A flow property has substantial derivative. What does this imply?
a) The property is a function of both time and space
b) The property is a function of time only
c) The property is a function of space only
d) The property is independent of time and space
Answer: a
Explanation: If a property has substantial derivative, it is differentiable by both time and space. This means that it is a function of time and space.
7. Substantial derivative = _____ + _____
a) Partial derivative, convective derivative
b) Local derivative, convective derivative
c) Local derivative, partial derivative
d) Total derivative, convective derivative
Answer: b
Explanation: Substantial derivative is the addition of local derivative and convective derivative .
8. Which of these terms represent the convective derivative of temperature ?
a) \
\
∇T
d) \(\frac{\partial T}{\partial t}\)
Answer: a
Explanation: \
is the convective derivative which is the time rate of change due to the movement of the fluid element.
9. Substantial derivative is the same as ________ of differential calculus.
a) Partial derivative
b) Instantaneous derivative
c) Total derivative
d) Local derivative
Answer: c
Explanation: Substantial derivative is the same as total derivative. However, total derivative is completely mathematical.
\(\frac{DT}{Dt}=\frac{\partial T}{\partial t}+u \frac{\partial T}{\partial x}+v \frac{\partial T}{\partial y}+w \frac{\partial T}{\partial z}\)
\(\frac{DT}{Dt}=\frac{\partial T}{\partial t}+u \frac{\partial T}{\partial x}+v \frac{\partial T}{\partial y}+w \frac{\partial T}{\partial z}\).
10. Which of these is not an equivalent to for substantial derivative?
a) Lagrangian derivative
b) Material derivative
c) Total derivative
d) Eulerian derivative
Answer: d
Explanation: Eulerian derivative means the local derivative
. Material and Lagrangian derivatives are the other names for substantial derivative. Total derivative is mathematical equivalent to substantial derivative.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “General Transport Equation”.
1. The general equation applicable to all the properties is called the general transport equation. What does this term ‘transport’ signify?
a) The equation is applicable to all properties
b) The equation can be transformed easily
c) The equation includes various transport processes
d) The equation is general
Answer: c
Explanation: The general transport equation involves all the transport processes which are responsible for the transfer of mass, energy or other properties.
2. What are the terms included in the transport equation?
a) Rate of change term, advective term, convective term, source term
b) Advective term, diffusive term, convective term, source term
c) Rate of change term, diffusive term, convective term, advective term
d) Rate of change term, diffusive term, convective term, source term
Answer: d
Explanation: Transport equation involves four terms which are rate of change, diffusion, convection and source of properties.
3. What does the term \
Rate of change
b) Convection
c) Diffusion
d) Source term
Answer: a
Explanation: This term represents the rate of change of fluid property inside the model of flow. It does not involve any kind of flow of the property.
4. Which of these terms represent the flow of fluid into and out of the observation model?
a) \
\
\)
c) div
d) ΓgradΦ
Answer: b
Explanation: \
\) is the convective term of the transport equation. Convection is the flow of fluid into and out of the model of observation.
5. Which term represents the diffusion of the property Φ?
a) div
b) div
c) curl
d) curl
Answer: b
Explanation: div represents diffusion. Diffusion is the movement of fluid from a high concentration to low concentration within the system.
6. What does this symbol Γ in the term div of the general transport equation mean?
a) Diffusion flux
b) Convection coefficient
c) Diffusion coefficient
d) Rate of diffusion
Answer: c
Explanation: Γ represents the diffusion coefficient. The diffusion coefficient is defined by Fick’s law of diffusion.
7. In terms of heat transfer, what does div mean?
a) Heat radiation
b) Heat convection
c) Thermal flow
d) Heat conduction
Answer: d
Explanation: Diffusive heat transfer is called heat conduction. This refers to the process of heat transfer without any movement of the particles.
8. When the general transport equation is written in the energy equation form, what does Γ become?
a) k
b) μ
c) σ
d) κ
Answer: a
Explanation: k represents thermal conductivity. As diffusion in heat transfer is heat conduction, diffusion coefficient becomes, thermal conductivity.
9. Which of these statements hold true?
a) Diffusive flux is always positive
b) Diffusive flux is positive in the direction of the positive gradient of fluid property
c) Diffusive flux is positive in the direction of the negative gradient of fluid property
d) Diffusive flux is always negative
Answer: c
Explanation: Diffusive flux is positive in the direction of the negative gradient of fluid property. Example, heat is conducted in the direction of decreasing temperature.
10. To get the mass conservation equation from the general transport equation given below, Φ=?
\
\)=div+S Φ
a) Mass of the fluid
b) 1
c) 0
d) Density of the fluid
Answer: b
Explanation: The mass conservation equation is obtained by replacing Φ with 1. As density is already present in the equation, 1 is enough to get mass conservation out of it.
11. The surface integral can be used to represent ____ and ____ terms of the transport equation.
a) Rate of change and diffusion
b) Rate of change and convection
c) Source and diffusion
d) Convection and diffusion
Answer: d
Explanation: Convection and diffusion terms are based on a transfer through the boundaries. The boundaries of a volume are its surfaces which makes surface integrals ideal for convection and diffusion.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Continuity Equation – Finite Control Volume”.
1. The physical principle behind the continuity equation is __________
a) Mass conservation
b) Zeroth law of thermodynamics
c) First law of thermodynamics
d) Energy conservation
Answer: a
Explanation: Continuity equation is derived from the mass conservation principle. It states that for an isolated system, the mass of the system must remain constant.
2. Which of these models directly gives this equation?
\
computational-fluid-dynamics-questions-answers-continuity-equation-finite-control-volume-q2a
b)
computational-fluid-dynamics-questions-answers-continuity-equation-finite-control-volume-q2b
c)
computational-fluid-dynamics-questions-answers-continuity-equation-finite-control-volume-q2c
d)
computational-fluid-dynamics-questions-answers-continuity-equation-finite-control-volume-q2d
Answer: b
Explanation: The equation is in conservative integral form. So, the model must be finite control volume fixed in space.
3. Which of these models directly gives this equation?
\
computational-fluid-dynamics-questions-answers-continuity-equation-finite-control-volume-q2d
b)
computational-fluid-dynamics-questions-answers-continuity-equation-finite-control-volume-q2b
c)
computational-fluid-dynamics-questions-answers-continuity-equation-finite-control-volume-q2a
d)
computational-fluid-dynamics-questions-answers-continuity-equation-finite-control-volume-q2c
Answer: c
Explanation: The equation is the non-conservative form of the integral continuity equation. This is obtained from a finite control volume moving along with the flow.
4. Consider a model of finite control volume fixed in space with elemental volume dV, vector elemental surface area d\
\
\
\
\(\iint_V\rho \vec{V}.d\vec{S}\)
Answer: d
Explanation: In general,
mass flow rate=density × velocity × area
For this case,
elemental mass flow rate = \(\rho \vec{V}.d \vec{S}\)
total mass flow rate=\(\iint_V\rho \vec{V}.d\vec{S}\)
5. Consider a model of finite control volume fixed in space with elemental volume dV, vector elemental surface area d\
\
\
ρdV
d) \(\frac{\partial}{\partial t} \iiint_V\rho dV\)
Answer: b
Explanation: Mass=density × volume
mass inside dV=ρdV
mass inside \( V=\iiint_V\rho dV\).
6. Consider a model of finite control volume moving along the flow with elemental volume dV, vector elemental surface area d\
\
\
\
ρdV
Answer: c
Explanation: Substantial derivative is used as the model is moving.
mass=density × volume
mass inside dV=ρdV
mass inside \( V=\iiint_V\rho dV\)
time rate of change of mass inside \(\frac{D}{Dt} \iiint_V\rho dV\).
7. To convert the non-conservative integral equation of mass conservation into the conservative integral form, which of these theorems is used?
a) Stokes theorem
b) Kelvin-Stokes theorem
c) Gauss-Siedel theorem
d) Gauss Divergence Theorem
Answer: d
Explanation: The expansion of non-conservative integral equation gives two volume integral terms. One of these terms representing the mass flow is converted into surface integral using the Gauss Divergence theorem.
8. Consider a model of finite control volume fixed in space with elemental volume dV, vector elemental surface area d\Missing open brace for subscript The mass flow is outward
b) The mass flow is inward
c) The mass flow is positive
d) The mass flow is negative
Answer: a
Explanation: d\(\vec{S}\) always points outwards to the control volume. So, the product \(\rho\vec{V}.d\vec{S}\) is positive when the mass flow is outwards.
9. What is the physical statement of mass conservation equation for a finite control volume fixed in space?
a) Net mass flow through the control surface = constant
b) Rate of change of mass inside the control volume = constant
c) Net mass flow through the control surface = Rate of change of mass inside the control volume
d) Net mass flow through the control surface≠Rate of change of mass inside the control volume
Answer: c
Explanation: Statement of mass conservation equation for a finite control volume fixed in space:
Net mass flow through the control surface is equal to the rate of change of mass inside the control volume.
10. What is the physical statement of mass conservation equation for a finite control volume moving along with the flow?
a) Rate of change of mass inside the control volume = 0
b) Rate of change of mass inside the control volume = constant
c) Net mass flow through the control surface = Rate of change of mass inside the control volume
d) Net mass flow through the control surface≠Rate of change of mass inside the control volume
Answer: b
Explanation: Statement of mass conservation equation for a finite control volume moving along with the flow:
Mass inside the control volume = constant
Rate of change of mass inside the control volume = 0.
This set of Computational Fluid Dynamics written test Questions & Answers focuses on “Continuity Equation – Infinitesimally Small Element”.
1. Which of the equations suit this model?
computational-fluid-dynamics-written-test-questions-answers-q1
a) \
\
\
=0\)
d) \(\frac{D\rho}{Dt}+\rho \nabla.\vec{V}=0\)
Answer: c
Explanation: The diagram represents infinitesimally small fluid element fixed in space. This model will give conservative differential equation which is \
=0\) from the options.
2. Consider an infinitesimally small fluid element with density ρ fixed in space and fluid is moving across this element with a velocity \Missing open brace for subscript \
\
[ρ]dx dy dz
d) \([\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}]dx \,dy \,dz\)
Answer: a
Explanation: Mass inflow in x direction=dy dz
mass outflow in x direction=\
dy \,dz\)
change in mass flow in x direction=\
dy \,dz-dy dz\)
change in mass flow in x direction=\
= \(\frac{\partial}{\partial x} dx \,dy \,dz + \frac{\partial}{\partial y} dx \,dy \,dz + \frac{\partial}{\partial z} dx \,dy \,dz\)
Net mass flow across the element = \(\Big[\frac{\partial}{\partial x}+\frac{\partial}{\partial y}+\frac{\partial}{\partial z}\Big]dx \,dy \,dz\)
3. Consider an infinitesimally small fluid element with density ρ fixed in space and fluid is moving across this element with a velocity \Missing open brace for subscript \
\
\
\)
d) \([\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}]dx \,dy \,dz\)
Answer: c
Explanation: Mass=density × volume
mass of fluid element=ρ×dx dy dz
rate of change in mass of fluid element=\(\frac{\partial\rho}{\partial t}dx \,dy \,dz \)
4. According to the conservation law, “Net mass flow across the fluid element is equal to the rate of change of mass inside the element”. But, stating the final equation, “Net mass flow across the fluid element + the rate of change of mass inside the element = 0”. Why is the operation not subtraction?
a) Irrespective of the law, the sum is always zero
b) The two terms are always opposite in sign
c) Change in sign is not considered
d) Rate of change may be increase or decrease
Answer: b
Explanation: The “net mass flow across the fluid element” being positive means that it is outward flow. If flow is outward, mass inside the fluid element decreases leading to a negative “rate of change of mass inside the element”. Thus, the two terms are always opposite in sign that they can be summed up to get zero.
5. Consider an infinitesimally small fluid element with density ρ fixed in space and fluid is moving across this element with a velocity \
\
\
\
\)
d) \
\)dx dy dz
Answer: d
Explanation:
Net mass flow across the element = change in mass flow in x direction + change in mass flow in y direction + change in mass flow in z direction
= \Extra \left or missing \rightdx \,dy \,dz.\vec{i} + \vec{j} + \vec{k})\right] \)
Net mass flow across the element = \
]dx \,dy \,dz\).
6. Which of the equations suit this model?
computational-fluid-dynamics-questions-answers-continuity-equation-infinitesimally-small-element - q6
a) \
\
\
\
=0\)
Answer: c
Explanation: \
=0\) is the non-conservative differential equation. Non-conservative differential equation is given by an infinitesimally small fluid element moving along with the flow.
7. Consider an infinitesimally small fluid element with density ρ moving along with the flow with a velocity \
\
\
\
\(\frac{D}{Dt}\)
Answer: a
Explanation: Substantial derivative is used as the model is moving.
mass = ρ δ V
time rate of change of mass=\(\frac{D}{Dt}\)
8. Consider an infinitesimally small fluid element with density ρ moving along with the flow with a velocity \
Integral
b) The rate of change of element’s volume
c) Elemental change in mass
d) Local derivative
Answer: b
Explanation: Applying mass conservation for this element,
time rate of change of mass = 0
\(\frac{D}{Dt}=0 \)
\(\delta V \frac{D}{Dt}+\rho\frac{D}{Dt}=0 \)
\(\frac{D}{Dt}+\rho\frac{1}{\delta V}\frac{D}{Dt}=0 \)
\(\frac{D}{Dt}+\rho\nabla.\rho=0 \)
Thus, the term arises from the rate of change of element’s volume.
9. Consider the continuity equation \
=0\). For an incompressible flow, this equation becomes ___________
a) \
=0\)
b) \
\
=0\)
d) \
=0\)
Answer: c
Explanation: Taking the continuity equation,
\
=0\)
For incompressible flow, ρ is constant
\
=0\)
The resulting equation is
\
=0\)
\
=0\)
\
=0\)
Thus, for incompressible flow, divergence of \(\vec{V}=0\).
10. Consider the continuity equation \
=0\). For a steady flow this equation becomes ___________
a) \
=0\)
b) \
=0\)
c) \
=0\)
d) \
=0\)
Answer: a
Explanation: Taking the continuity equation,
\
=0\)
For steady flow, flow variables do not vary with time.
\
=0\).
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Momentum Equation”.
1. What is the physical principle behind momentum equation?
a) Newton’s second law of motion
b) Newton’s first law of motion
c) Zeroth law of thermodynamics
d) First law of thermodynamics
Answer: a
Explanation: Momentum equation is derived using Newton’s second law of motion. This gives a relationship between force and acceleration. It also gives the condition for momentum conservation.
2. Which of these statements hold true?
a) Momentum conservation is applicable to neither individual directions nor the whole system
b) Momentum conservation is applicable to the whole system but not individually
c) Momentum conservation is applicable to both individual directions and the whole system
d) Momentum conservation is applicable only to the three directions individually
Answer: c
Explanation: According to newton’s second law, force is equal to the product of mass and acceleration, where force and acceleration are vector quantities
. Thus conservation is applicable to the whole system in vector terms and also to the individual directions.
3. Consider an element shown below. S is the source term.
computational-fluid-dynamics-questions-answers-momentum-equation-q3
If gravitational force is the only force acting on this element. Which of these following is correct?
a) S x =mg, S y =0, S z =mg
b) S x =0, S y =0, S z =mg
c) S y =0, S z =0, S x =mg
d) S z =0, S x =0, S y =mg
Answer: d
Explanation: For momentum equation, source is the body force of the element. If gravity is the only force acting on the element, then source=mass×acceleration due to gravity. This will act only in the y-direction.
4. Consider the element shown in the diagram. In diagram,
p → pressure
\(p+\frac{\Delta p}{\Delta x}dx=p+\frac{\partial p}{\partial x}dx\)
dx, day, dz → dimensions in x, y and z directions
computational-fluid-dynamics-questions-answers-momentum-equation-q4
What is the net pressure force in x direction?
a) \
\
\
\(-\frac{\partial p}{\partial x}dx \,dy \,dz\)
Answer: d
Explanation: From the diagram,
pressure force= pressure×area
inlet pressure force in x-direction = p×dy dz
outlet pressure force in x-direction = -
dydz
net pressure force in x-direction = -
dydz
net pressure force in x-direction=\(-\frac{\partial p}{\partial x}dx \,dy \,dz\)
5. The physical property Φ of the general transport equation is replaced by ________ to get momentum equation.
a) Velocity vector
b) Mass
c) Force vector
d) Acceleration vector
Answer: a
Explanation: The physical property in general transport equation should be replaced by velocity vector. This will result in momentum equation. For mass conservation, Φ = 1. For momentum conservation, Φ = 1×\(\vec{V}=\vec{V}\).
6. The source term in the momentum equation is ________
a) Pressure force
b) Body forces
c) Viscous force
d) Acceleration
Answer: b
Explanation: The effect of surface forces are external. They are not produced inside the body. Body forces are created inside the body of conservation. So, the body force term is the source term.
7. I am using an infinitesimally small element of fluid moving along with the flow as my model. What is the acceleration of this model in x-direction?
a) \
\
\
\(a_x=\frac{\partial u}{\partial t}\)
Answer: c
Explanation: Acceleration is the rate of change of velocity. Acceleration in the x-direction is the rate of change of velocity in x-direction . As the model is moving, rate of change is given by substantial derivative. Thus the acceleration in x-direction is \(a_x=\frac{Du}{Dt}\).
8. When I derived the momentum equation in x-direction, I got this equation \
computational-fluid-dynamics-questions-answers-momentum-equation-q8a
b) computational-fluid-dynamics-questions-answers-momentum-equation-q8b
c) computational-fluid-dynamics-questions-answers-momentum-equation-q8c
d) computational-fluid-dynamics-questions-answers-momentum-equation-q8d
Answer: b
Explanation: My equation is differential non-conservative. So, I must have used an infinitesimally small element moving along with the flow.
9. The momentum equation is \
Pressure force is taken as negative
b) Tensile force is taken as negative
c) Pressure force is taken as positive
d) Pressure force is unsigned
Answer: a
Explanation: When all the other forces and accelerations are positive, gradient of pressure force is negative
. This is possible only when pressure force is signed posititve.
10. Which of these terms represent convection in the momentum equation?
a) \
\)
b) \
\)
c) \
\)
d) \
\)
Answer: d
Explanation: : From general conservation equation, the convection term is div. For momentum equation, \
.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Governing Equations – Stress and Strain Tensor”.
1. Which among these forces used in momentum equation is a tensor?
a) Gravitational forces
b) Pressure forces
c) Viscous forces
d) Electromagnetic forces
Answer: c
Explanation: Viscous forces are tensors. The other forces given here are vectors.
2. What do the two subscripts of stress tensors represent?
a) Directions of stress and strain
b) Directions of stress and normal to the surface on which they are acting
c) Directions of strain and normal to the surface on which they are acting
d) Direction of stress and the flow direction
Answer: b
Explanation: The two subscripts of stress tensors indicate the direction of the stress and that of the normal to the surface on which they act. So, stress tensors give the location and direction of the stresses.
3. Which of these fluids have their stress tensor linearly varying to the strain rate?
a) computational-fluid-dynamics-questions-answers-stress-strain-tensor-q3a
b) computational-fluid-dynamics-questions-answers-stress-strain-tensor-q3b
c) computational-fluid-dynamics-questions-answers-stress-strain-tensor-q3c
d) computational-fluid-dynamics-questions-answers-stress-strain-tensor-q3d
Answer: d
Explanation: S-tress tensor linearly varies with the strain rate only for Newtonian fluids. For Newtonian fluid shear stress is proportional to du/dy. In the other cases, shear stress varies non-linearly with du/dy.
4. Which of the stress tensors from the diagram is represented by Τ xy ?
computational-fluid-dynamics-questions-answers-stress-strain-tensor-q4
a) 3
b) 2
c) 1
d) 4
Answer: a
Explanation: Τ xy indicate that the stress component acts in the y-direction on a surface normal to the x-direction. Representing this in the diagram, 3 is the corresponding tensor.
5. The divergence of the stress tensor is _____
a) Scalar
b) Vector
c) 0
d) 1
Answer: b
Explanation: Stress tensor is a square matrix given by
Τ xy = \(
\)
The divergence of this will result in a vector
∇. Τ= \(
\)
6. What are the two viscosity coefficients involved in the relationship between stress tensor and strain rate of fluids?
a) Kinematic viscosity and bulk viscosity
b) Dynamic viscosity and kinematic viscosity
c) Dynamic viscosity and bulk viscosity
d) Kinematic viscosity and volume viscosity
Answer: c
Explanation: The two viscosities involved in stress train relationship of fluids is dynamic viscosity coefficient and bulk viscosity coefficient. Bulk viscosity coefficient for diagonal elements respectively.
7. What is the relationship between bulk viscosity coefficient and the dynamic viscosity coefficient ?
a) λ=\
λ=\
λ=\
λ=\(-\frac{1}{2}\) μ
Answer: a
Explanation: The bulk viscosity coefficient represents fluid compressibility effects. λ=\(-\frac{2}{3}\) μ is the relationship between the bulk viscosity coefficient and the dynamic viscosity coefficient.
8. Express the shear stress tensor of a three-dimensional fluid flow element in terms of the velocity vector.
a) \Extra \left or missing \right^T\right\}+\lambda
I\)
b) \Extra \left or missing \right\right\}+\lambda
I\)
c) \Extra \left or missing \right^T+
^T\right\}\)
d) \Extra \left or missing \right^T+
^T\right\}+\lambda
I\)
Answer: d
Explanation: The shear stress tensor of a fluid element can be given by \Extra \left or missing \right^T+
^T\right\}+\lambda
I\). This is not applicable for practical cases. However, common fluids like air and water are assumed to be Newtonian for using this relationship.
9. Express \
\
\)
b) \
\)
c) \
\)
d) \
\)
Answer: a
Explanation: For non-diagonal elements,
\Extra \left or missing \right^T\right\}\)
\
\).
10. Viscous forces fall into which kind of the following forces acting on a body?
a) Pressure force
b) Tensile force
c) Body forces
d) Surface forces
Answer: c
Explanation: The two types of forces acting on a fluid are body forces and surface forces. Body forces are the forces produced by the fluid element itself. Surface forces are the one acting on the fluid elements. Viscous forces act on the element and it comes under surface forces.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Navier Stokes Equation”.
1. What are the independent variables in the Navier-Stokes equations?
a) x, y, z, ρ
b) x, y, z, τ
c) x, y, z, t, ρ
d) x, y, z, t
Answer: d
Explanation: There are four independent variables in Navier-Stokes equations. Three spatial variables and one time variable .
2. What are the dependent variables in the Navier-Stokes equations?
a) τ,T,p,ρ
b) p,ρ,T
c) u,v,w,T,p,ρ
d) u,v,w,T,p
Answer: c
Explanation: There are six dependent variables in the Navier-Stokes equations. They are pressure , temperature , density and three components of the velocity vector .
3. The Navier-Stokes equations are all partial differential equations. What will be the best reason behind this?
a) Ordinary differentials are not present in the Navier-Stokes equations
b) The dependent variables are functions of all of the independent variables
c) Each dependent variable depends on only one of the independent variables
d) Partial differentials are only present in the Navier-Stokes equations
Answer: b
Explanation: Each dependent variable in the Navier-Stokes equations depends on all of the independent variables. So, partial differentials are used to indicate that the other independent variables should be kept fixed while differentiating.
4. Turbulence problems particularly depend on this term of the Navier-Stokes equations. Which is that term?
a) Rate of change term
b) Convection term
c) Diffusion term
d) Source term
Answer: c
Explanation: Turbulence is caused by abrupt changes in velocities perpendicular to the flow. This, in turn, can be given in viscosity terms. Diffusion term of the Navier-Stokes equations holds the viscosity terms. So, without diffusion terms, we cannot model turbulence.
5. The Navier-Stokes equations are ____ system of equations.
a) coupled
b) uncoupled
c) exponential
d) radical
Answer: a
Explanation: Navier-Stokes equations are called a coupled system of equations because all of the equations should be solved to get the dependent variables. Equations cannot be solved separately to get the unknowns.
6. The diffusion term in the general transport equation is div. While equating this with the Navier-Stokes equations, what is Γ?
a) k
b) λ
c) μ
d) \(\vec{V}\)
Answer: c
Explanation: The diffusion term in the Navier-stokes equations is div for the x-momentum equation. Comparing this with the general transport equation, Γ is μ- dynamic viscosity coefficient.
7. The viscosity terms in x-momentum equation is \
. Which of these relations is used for this transformation?
a) Thermodynamic relations
b) Stress-strain relations
c) Fluid flow relations
d) Geometric relations
Answer: b
Explanation: Stress-strain relationship states that “the shear stresses are proportional to the gradient of velocities”. This relationship is used for the transformation.
8. Which is the diffusion terms of the y-momentum equation?
a) \
\
\
\(\frac{\partial\tau_{xy}}{\partial y}+\frac{\partial\tau_{yy}}{\partial y}+\frac{\partial\tau_{zy}}{\partial y}\)
Answer: a
Explanation: \(\frac{\partial\tau_{xy}}{\partial x}+\frac{\partial\tau_{yy}}{\partial y}+\frac{\partial\tau_{zy}}{\partial z}\) are the diffusion terms of the y-momentum equation. It involves all the shear stress tensors which are in the y-direction.
9. The major difference between the Navier-Stokes equations and the Euler equations is the dissipative transport phenomena. The impact of this phenomena in a system is ____
a) They decrease entropy
b) They increase entropy
c) They increase internal energy
d) They decrease internal energy
Answer: b
Explanation: Dissipation is the process where energy is transformed from one form into another. This transformation increases the entropy of the system.
10. Diffusion terms are not included in ____ of the Navier-Stokes equations.
a) continuity equation
b) y-momentum equation
c) z-momentum equation
d) energy equation
Answer: c
Explanation: Mass diffusion of the continuity equations are in general not included in the Navier-Stokes equations. This is because most of the fluid flow and thermodynamic processes do not include any change in concentration which is mass diffusion.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Euler Equation”.
1. The general transport equation is \
+div+S\). For Eulerian equations, which of the variables in the equation becomes zero?
a) Γ
b) ρ
c) Φ
d) \(\vec{u}\)
Answer: a
Explanation: Γ is the diffusion coefficient in the general transport equation. Diffusion of any property is not included in Eulerian equations. So, Γ=0.
2. Euler equations govern ____________ flows.
a) Viscous adiabatic flows
b) Inviscid flows
c) Adiabatic and inviscid flows
d) Adiabatic flows
Answer: c
Explanation: Euler equations constitute the governing equations of flow for adiabatic and inviscid flows. Here, the dissipative transport of flow properties is neglected.
3. Which of these is the non-conservative differential form of Eulerian x-momentum equation?
a) \
=-\frac{\partial p}{\partial x}+\rho f_x\)
b) \
\
\(\rho \frac{\partial u}{\partial t}=-\frac{\partial p}{\partial x}+\rho f_x\)
Answer: b
Explanation: Momentum equation excluding the viscous terms gives the Eulerian momentum equation. This can be given by \(\rho\frac{Du}{Dt}=-\frac{\partial p}{\partial x}+\rho f_x\).
4. Eulerian equations are suitable for which of these cases?
a) Compressible flows
b) Incompressible flows
c) Compressible flows at high Mach number
d) Incompressible flows at high Mach number
Answer: d
Explanation: Eulerian equations are best suited for examining incompressible flows at high Mach number. They are used to study flow over the whole aircraft.
5. Euler form of momentum equations does not involve this property.
a) Stress
b) Friction
c) Strain
d) Temperature
Answer: b
Explanation: Euler form of equations is for inviscid flows. For inviscid flows, viscosity is zero. So, there are no friction terms involved.
6. There is no difference between Navier-Stokes and Euler equations with respect to the continuity equation. Why?
a) Convection term plays the diffusion term’s role
b) Diffusion cannot be removed from the continuity equation
c) Its source term balances the difference
d) The continuity equation by itself has no diffusion term
Answer: d
Explanation: Diffusion term, in general, is given by div. For the continuity equation, Φ=1. And grad Φ=0. So, the continuity equation by itself has no diffusion term.
7. Which of these equations represent a Euler equation?
a) \
\
∇p=μ∇ 2 v+ρg
d) 0=μ∇ 2 v+ρg
Answer: a
Explanation: \(\frac{\rho Dv}{Dt}=-\nabla p+\rho g\) represents a Euler equation. All other equations have this term μ∇ 2 v representing diffusion.
8. Which of the variables in the equation \
f x , τ yx , τ zx
b) τ xx , τ yx , u
c) τ xx , τ yx , τ zx
d) τ xx , p, τ zx
Answer: c
Explanation: τ xx , τ yx , τ zx represent shear stresses due to viscous effects; u is the x-velocity; f x is the body force and p is the pressure. τ xx ,τ yx ,τ zx should become zero for the flow to be in-viscid and the equations to be Eulerian.
9. In Euler form of energy equations, which of these terms is not present?
a) Rate of change of energy
b) Heat radiation
c) Heat source
d) Thermal conductivity
Answer: d
Explanation: As the flow considered by Euler equations is adiabatic, heat cannot enter or exit the system. So, the thermal conduction is omitted.
10. To which of these flows, the Euler equation is applicable?
a) Couette flow
b) Potential flow
c) Stokes Flow
d) Poiseuille’s flow
Answer: b
Explanation: Among the given flows, only potential flows are in-viscid. So, the Euler equation is applicable to only potential flows among the above.
This set of Computational Fluid Dynamics Questions and Answers for Freshers focuses on “Energy Equation – Based on Thermal Properties”.
1. The energy equation in terms of total energy is
\
=-\nabla.\dot{q}_s-\nabla.
+\nabla.
+\vec{f_b}.\vec{V}+\dot{q}_v.\)
Where,
t → Time
ρ → Density
e → Specific total energy
\
Pressure term
b) Shear stress term
c) Body force term
d) Heat transfer term
Answer: c
Explanation: While converting the energy equation from total energy terms to internal energy terms, an equation with kinetic energy is subtracted from total energy. In this process, the body force term loses its explicit presence.
2. While converting the energy equation from one form to another, which of the following happens?
a) Either the left-hand side or the right-hand side of the equation changes
b) Both the left-hand side and the right-hand side of the equation change
c) The right-hand side of the equation changes
d) The left-hand side of the equation changes
Answer: b
Explanation: Changes are applied to only left-hand side terms in the equations, but they affect both the left and right-hand sides of the equation.
3. Expressing \
\
^2 \,and\, \psi=
^2+
^2+
^2\)
b) \
^2 \,and\, \phi=
^2+
^2+
^2\)
c) \
^2 \,and\, \phi=2
^2 + 2
^2 +2
^2+
^2 +\)
\
^2 +
^2\)
d) \
^2 \,and\, \psi=2
^2+2
^2+2
^2+
^2+\)
\
^2+
^2\)
Answer: d
Explanation: \
^2 \mu
^2+2
^2+2
^2+
^2+\)
\
^2+
^2)\).
4. If p and τ are the net pressure and net shear stress acting on an infinitesimally small element moving along with the flow
, what is the net work done on the system?
a) \
+\nabla .
)\)
b) \
+
)dx \,dy \,dz\)
c) \
+\nabla.
) dx \,dy \,dz\)
d) \+\nabla.)dx \,dy \,dz\)
Answer: c
Explanation: The rate of work done is power which is the product of force and velocity. This can be represented by \
+\nabla.
) dx \,dy \,dz\).
5. The energy equation which is in terms of temperature can be changed to terms of internal energy using ___________
a) momentum equation
b) stress-strain relations
c) equations of state
d) continuity equation
Answer: c
Explanation: Equations of state give the relationship between temperature and internal energy which is i=C v T. Using this relation, one can obtain the energy equation in internal energy terms.
6. The energy equation which is in terms of total energy can be changed to terms of internal energy using ___________
a) momentum equation
b) stress-strain relations
c) equations of state
d) continuity equation
Answer: a
Explanation: Total energy is the sum of internal energy and kinetic energy. By some manipulation in the momentum equation, we can get the kinetic energy terms. If this is subtracted from the energy equation written in total energy terms, we can get the same in terms of internal energy.
7. If \
moving along with the flow
, Which term is the work done by the body force?
a) \
\
\
\(\rho\vec{f}\)dx dy dz
Answer: c
Explanation: mass = density×volume
mass = ρdx dy dz
rate of work done = force×velocity
rate of work done = \(\rho\vec{f}.\vec{V}\)dx dy dz
\(\rho\vec{f}.\vec{V}\)dx dy dz is the work done by the body force.
8. Energy equation in terms of specific internal energy is
\
=-\nabla.\dot{q_s} – p\nabla .\vec{V}-\tau:\nabla \vec{V}+\dot{q_v}\)
Where,
t → Time
ρ → Density
\
\
=-\nabla .\dot{q_s}+\frac{Dp}{Dt} -\tau:\nabla \vec{V}+\dot{q_v}\)
b) \
=-\nabla .\dot{q_s}-p\nabla .\vec{V}-\tau:\nabla \vec{V}+\dot{q_v}\)
c) \
=-\nabla .\dot{q_s}-p\nabla .\vec{V}+\nabla .
-\tau:\nabla \vec{V}+\dot{q_v}\)
d) \
=-\nabla .\dot{q_s}+\vec{V}.\nabla p-\tau:\nabla \vec{V}+\dot{q_v}\)
Answer: a
Explanation: Take the given equation.
\
=-\nabla.\dot{q_s}-p\nabla.\vec{V}-\tau:\nabla \vec{V}+\dot{q_v} \)
The relation between internal energy and enthalpy is
\
-\nabla.
=-\nabla.\dot{q_s}-p\nabla .\vec{V}-\tau:\nabla \vec{V}+\dot{q_v} \)
\
= -\nabla.\dot{q}_s-\frac{\partial}{\partial t}-\nabla.
-p\nabla.\vec{V}-\tau:\nabla \vec{V}+\dot{q_v}\)
\
= -\nabla.\dot{q_s} +\frac{\partial}{\partial t}+\nabla.
-p\nabla.\vec{V}-\tau:\nabla \vec{V}+\dot{q_v}\)
\
= -\nabla.\dot{q_s} +\frac{\partial}{\partial t}+\vec{V}.\nabla p-\tau:\nabla \vec{V}+\dot{q_v}\)
\
= -\nabla.\dot{q_s} +\frac{D}{Dt}-\tau:\nabla \vec{V}+\dot{q_v}\)
This is the energy equation in specific enthalpy terms.
9. Let \
\
\
\)
c) \
\)
d) \
\)
Answer: b
Explanation: The total energy of the system is
\
\).
10. \
= -\nabla.\dot{q_s}-p\nabla.\vec{V}-\tau:\nabla\vec{V}+\dot{q_v}\). This form of the energy equation is applicable to _________
a) Both Newtonian and non-Newtonian fluids
b) Newtonian fluids
c) Non-Newtonian fluids
d) Pseudo-plastics
Answer: a
Explanation: The energy equation given here is in terms of shear stresses. So, no restrictions based on the viscosity of the fluid. It is applicable to both Newtonian and Non-Newtonian fluids.
This set of Computational Fluid Dynamics online test focuses on “Energy Equation – Temperature Terms”.
1. The physical principle behind the energy equation is _____________
a) Newton’s second law of motion
b) Zeroth law of thermodynamics
c) First law of thermodynamics
d) Newton’s first law of motion
Answer: c
Explanation: First law of thermodynamics is the physical principle behind the energy equation. This law states that “Energy can neither be produced nor be destroyed but can be converted from one form into another”.
2. Consider an infinitesimally small fluid element moving along with the flow. Apply the first law of thermodynamics to this model. Which of these statements is correct?
a) The rate of change of the total energy is equal to the rate of heat addition and work extraction
b) The rate of work extraction is equal to the rate of heat addition and the rate of change of the total energy
c) The rate of heat addition is equal to the rate of work extraction and the rate of change of the total energy
d) The rate of change of the total energy is equal to the rate of work extraction
Answer: a
Explanation: The first law of thermodynamics applied to a system states that “The rate of change of the total energy is equal to the rate of heat addition and work extraction”.
3. The rate of heat increase in a system depends on __________
a) the rate of heat transferred to the system
b) the rate of heat generated by the system
c) neither the rate of heat generated by the system nor the rate of heat transferred to the system
d) both the rate of heat transferred to the system and the rate of heat generated by the system
Answer: d
Explanation: Heat can be added to a system in two ways:
Transfer of heat across the surface of the element by surface forces.
The heat generated by the system itself.
4. To get the energy equation in terms of temperature, this law is used.
a) Newton’s third law of motion
b) Zeroth law of thermodynamics
c) Fick’s law
d) Fourier’s law of heat conduction
Answer: d
Explanation: Fourier’s law of heat conduction gives the relationship between heat energy and temperature. This is used in the energy equation to convert heat terms to temperature terms.
5. The rate of change of energy in a moving model is \
\). Which of these equations is used for this reduction?
a) Equations of state
b) Stress-strain equation
c) Momentum equation
d) Continuity equation
Answer: d
Explanation: Continuity equation is used as given below.
\
-e\nabla.
\)
Therefore,
\
-e\nabla.
\)
\
)+\nabla.
\)
Applying the continuity equation, \
=0\), and hence
\
\).
6. The relationship between the rate of heat transfer per unit area \
. Where, k is a scalar value of thermal conductivity and ∇T is the gradient of temperature. Which of these following is wrong according to the above equation?
a) Heat transfer is different in different directions
b) The rate of heat transfer depends upon the temperature gradient
c) Heat transfer is in the opposite direction of increasing temperature
d) k is the proportionality constant
Answer: a
Explanation: The statement “Heat transfer is different in different directions” is wrong. k is a scalar means that the heat transfer is the same in all directions and the material is isotropic. For different heat transfers in different directions , the proportionality constant should be a tensor.
7. The energy equation should be solved to get this variable of the flow.
a) Velocity
b) Temperature
c) Density
d) Pressure
Answer: b
Explanation: It is a must to solve the energy equation when we want the temperature distribution of the system. The energy equation can give information about temperature.
8. Which of these terms represent the rate of heat addition to the system due to heat transfer?
a) -k gradT
b) k gradT
c) ∇.
d) ∇.
Answer: c
Explanation: Rate of heat addition is
q=-k gradT
The total rate of heat added to the system is
-div=∇..
9. Consider the following diagram.
computational-fluid-dynamics-questions-answers-online-test-q9
In the diagram,
q → Time rate of heat energy per unit area
q+\
–\
–\
–\
\(\frac{\partial q}{\partial x}dx \,dy \,dz\)
Answer: b
Explanation:
heat transfer into the element in x-direction = q dy dz
heat transfer out of the element in x-direction = \
dy \,dz\)
net heat transfer in x-direction = q dy dz-\
dy \,dz\)
net heat transfer in x-direction = –\(\frac{\partial q}{\partial x}dx\,dy\,dz\).
10. In the general transport equation, ϕ is the flow property. To get the energy equation out of this general equation, which of these variables cannot be used?
a) v
b) T
c) h 0
d) i
Answer: a
Explanation: T, h 0 and I represent temperature, total enthalpy and internal energy. These can be used to get the energy equation. Y-velocity component cannot be used. It will result in the y-momentum equation.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Boundary Conditions”.
1. Which among these is used to specify a particular problem which we consider for solving in CFD?
a) Boundary conditions
b) Governing equations
c) Governing laws
d) Solution method
Answer: a
Explanation: Boundary conditions can define a particular problem. When the same governing equations are used to solve different problems, boundary conditions are the one which makes the problem unique.
2. What are the two major types of boundary conditions?
a) Wall and symmetry
b) Inlet and outlet
c) Dirichlet and Neumann
d) Initial and physical
Answer: c
Explanation: Dirichlet and Neumann boundary conditions are the two boundary conditions. They are used to define the conditions in the physical boundary of a problem.
3. Which of these best define the Dirichlet boundary conditions for the property Φ at a point ‘b’ in a boundary?
a) \
\
\
Φ b
Answer: d
Explanation: Dirichlet boundary conditions are where the values of the properties are directly given. So, among the options given, Φ b represents the Dirichlet boundary condition.
4. The boundary condition represented in the following diagram is ___________
computational-fluid-dynamics-questions-answers-boundary-conditions-q4
a) Dirichlet
b) Periodic
c) Neumann
d) Axisymmetric
Answer: c
Explanation: In the diagram, the heat flux value is given. This kind of boundary condition where the flux of a property is given is called the Neumann boundary condition.
5. The Neumann and Dirichlet boundary conditions are _________ and _________ in mathematical terms.
a) value specified, flux specified
b) flux specified, value specified
c) flux specified, gradient specified
d) value specified, time specified
Answer: b
Explanation: In mathematical terms, Neumann Boundary conditions are called flux specified and Dirichlet boundary conditions are called value specified boundary conditions. These names are based on their characteristics.
6. In Dirichlet boundary conditions, the flux values _________
a) can be calculated
b) are unknowns
c) are known
d) are 0
Answer: a
Explanation: In Dirichlet boundary conditions, the value of the flow property will be given. By using this, the flux values can be created. They are not unknowns.
7. Initial conditions are used for __________ problems.
a) time-dependent problems
b) boundary value problems
c) control volume problems
d) finite difference problems
Answer: b
Explanation: For time-dependent problems, initial conditions are used to specify the conditions at t=0. This will be helpful to find the solutions at the time points.
8. Which among these is a combination of value specified and flux specified boundary conditions?
a) Dirichlet
b) Mixed
c) Neumann
d) Symmetry
Answer: b
Explanation: Type 3 boundary conditions are the mixed boundary conditions. Mixed boundary condition combines Neumann and Dirichlet boundary conditions.
9. Which of these is not a combination of Neumann and Dirichlet Boundary conditions?
a) Cauchy boundary conditions
b) Wall boundary conditions
c) Mixed boundary conditions
d) Robin boundary conditions
Answer: b
Explanation: Wall boundary conditions are the boundary conditions specified at walls. The other types are different combinations of Dirichlet and Neumann Boundary conditions.
10. Which is not a type of boundary conditions in CFD?
a) Cyclic boundary conditions
b) Symmetry boundary conditions
c) Wall boundary conditions
d) Nodal boundary conditions
Answer: d
Explanation: Boundary conditions are not defined at nodes and nodal boundary conditions do not exist. There are many types of boundary conditions in CFD. Some of them are
Wall boundary conditions
Symmetry boundary conditions
Inlet boundary conditions
Outlet boundary conditions
Constant pressure boundary conditions
Periodic or cyclic boundary conditions
This set of Computational Fluid Dynamics Objective Questions & Answers focuses on “Boundary Conditions – Inlet and Outlet”.
1. Usually, in the inlet boundary conditions ___________ are known.
a) gradients of flow properties
b) diffusive fluxes of properties
c) convective fluxes of properties
d) flow properties
Answer: d
Explanation: Flow properties in the inlet should be known. If the flow properties are not known, it is good to move the boundary as far as possible from the region of interest.
2. The convective fluxes in the inlet boundary layer _________
a) are specified
b) can be calculated
c) should be approximated
d) are not necessary
Answer: b
Explanation: The convective flux of a property is a product of mass flow rate and the flow property. So, it can be easily obtained from the known values.
3. The diffusive fluxes in the inlet boundary conditions _________
a) are specified
b) can be calculated
c) are not necessary
d) should be approximated
Answer: d
Explanation: The diffusive fluxes are usually not known. They should be approximated using the boundary values of the variables and one-sided finite difference approximations for the gradients.
4. An extra grid is used before the physical boundary. What is the use of this grid?
a) To store the inlet boundary values
b) To solve the discretized equation
c) To calculate the inlet boundary values
d) To solve the system analytically
Answer: a
Explanation: The grid extends outside the physical boundary and these nodes are used to store the inlet variables. The discretized equation is solved starting from the first internal cell.
5. _________ is fixed at one inlet node and pressure correction is set to _________ in that node.
a) Gauge pressure, one
b) Absolute pressure, one
c) Gauge pressure, zero
d) Absolute pressure, zero
Answer: d
Explanation: While calculating, the pressure field does not give absolute pressures. A reference pressure should be set before to resolve this problem. To set a reference pressure, absolute pressure is fixed at inlet node and pressure correction is set to zero there.
6. If we are solving a k-ε model, the values of k and ε _________
a) are either specified or estimated at the inlet
b) are either specified or estimated at the outlet
c) need not be specified or estimated anywhere
d) are not estimated
Answer: a
Explanation: The most accurate solutions can be achieved only if k and ε are specified at the inlet boundary. However, if it is not possible to provide, CFD codes can estimate these values using a formula.
7. Which of these is the best practice regarding outlet boundaries?
a) Outlet boundaries should be at the exact outlet of the geometry
b) Outlet boundaries should be set as close as possible to the inlet boundaries
c) Outlet boundaries should be set as far as possible to the physical geometry
d) Outlet boundaries should be set as close as possible to the physical geometry
Answer: c
Explanation: At the outlet, usually we know only little about the flow properties. For this reason, we should set the outlet boundary conditions as far as possible from the region of interest. Otherwise, error may propagate upstream.
8. Which of these statements is false regarding the pressure correction at boundaries?
a) Pressure reference is set at the inlet boundaries
b) Pressure reference is set at the outlet boundaries
c) The link to the inlet boundary side is suppressed
d) The link to the outlet boundary side is suppressed
Answer: b
Explanation: Pressure reference is set at the inlet boundaries and the link to the inlet side is suppressed. In the outlet boundaries, there is no need for setting another reference pressure but the link to the outlet side is suppressed.
9. The mass flux out should be equal to the mass flux in. This is in accordance with _________
a) energy conservation
b) momentum conservation
c) continuity condition
d) flux conservation
Answer: c
Explanation: The theory of continuity states that the inlet and outlet mass fluxes should be the same. So, they are always matched by some external calculations.
10. Which of these following is used to match the outlet and inlet flow velocities?
a) u N,j = u N-1,j ×\
u N-1,j = u N,j ×\
u N-1,j = u N,j ×\
u N,j = u N-1,j ×\(\frac{M_{out}}{M_{in}}\)
Answer: a
Explanation: To match the flow velocities, velocity in the node immediate west of the outlet is multiplied by the mass ratio \(\frac{M_{in}}{M_{out}}\). This will help us to maintain continuity.
This set of Computational Fluid Dynamics Interview Questions and Answers for freshers focuses on “Boundary Conditions – Wall and Symmetry”.
1. For a no-slip condition which of these about velocity components is true near the wall boundary?
a) u=1, v=0, w=0
b) u=0, v=0, w=0
c) u=0, v=1, w=0
d) u=0, v=0, w=1
Answer: b
Explanation: No-slip is the ideal condition with the highest viscosity. Here, the fluid is relatively stationary with the wall and all of the velocity components are zero. Therefore, u=0, v=0, w=0.
2. Which of these represents the temperature of the fluid layer immediately near the wall at a condition analogous to no-slip? Note: T w is the temperature at the wall.
a) T=-1
b) T=1
c) T=T w
d) T=0
Answer: c
Explanation: For a condition analogous to no-slip in temperature, the temperature of the fluid layer near the wall is the same as the wall temperature. Therefore, T=T w .
3. Which of these is true for an impermeable wall?
a) \
\
\
\(\vec{V}.\vec{n}=0\) above the surface
Answer: c
Explanation: For an impermeable wall, there can be no mass flow into or out of the wall. Therefore, the velocity at the surface must be completely tangential and its normal component will be zero
.
4. For inviscid flows, which is correct immediately near the wall?
a) \
\
\
\(\vec{V} \lt 0\)
Answer: a
Explanation: \(\vec{V}\)=0 only for the no-slip condition. For inviscid flows, the flow velocity immediately near the wall is a finite non-zero value. We cannot specify its sign as the sign depends upon flow direction. We can only say \(\vec{V}≠0\).
5. For no-slip condition, which of these is true regarding the pressure correction equation if the wall is at the bottom?
a) a n =0
b) a w =0
c) a e =0
d) a s =0
Answer: d
Explanation: When velocities are known near the wall for no-slip condition, pressure correction is unnecessary. As the wall is at the bottom here, we set a s =0 to omit pressure correction in the southern side.
6. What is the shear force of a fluid
near the wall for a moving wall
)?
a) \
\
\
\(\vec{F}=-\mu\frac{\vec{u}}{\Delta y}\)
Answer: b
Explanation: Shear force is a product of shear stress and area. Shear stress is defined by Newton’s law of viscosity. For moving walls, the relative velocity \(\vec{u}-\overrightarrow{u_{wall}}\) should be taken. Therefore, the shear force is given by \(\vec{F}=-\mu\frac{\vec{u}-\overrightarrow{u_{wall}}}{\Delta y}\times area\).
7. Which of the following applies to a symmetry boundary?
a) There is no flow and no scalar flux across the boundary
b) There are flow and scalar fluxes across the boundary
c) There is no scalar flux but flow is possible across the boundary
d) There is no flow but scalar flux is possible across the boundary
Answer: a
Explanation: There are 2 conditions for a symmetry boundary: There is no flow across the boundary and there is no scalar flux across the boundary.
8. For a symmetry boundary, which is correct?
a) V n ≠0, τ nn =0
b) V n ≠0, τ nn ≠0
c) V n =0, τ nn ≠0
d) V n =0, τ nn =0
Answer: c
Explanation: For a symmetry boundary, there is no flow across the boundary. Therefore, V n =0. But, the gradients of normal flow are non-zero. So, τ nn ≠0.
9. A symmetry boundary is treated the same as a wall boundary for this reason.
a) There is flow across this boundary
b) No convection flux across this boundary
c) There is convection flux across this boundary
d) No flow across this boundary
Answer: d
Explanation: A symmetry boundary is treated as a wall boundary with an additional condition that there is no scalar flux across this boundary.
10. Which is true for a symmetry boundary?
a) Diffusive flux is non-zero
b) Diffusive flux is zero
c) Convective flux is zero
d) Convective flux is non-zero
Answer: c
Explanation: The convective flux at a symmetry boundary is always zero. The diffusive flux may or may not be zero depending upon the coincidence of this boundary with the Cartesian coordinates.
This set of Computational Fluid Dynamics Questions and Answers for Experienced people focuses on “Boundary Conditions – Constant Pressure and Periodic”.
1. When do we use a constant pressure boundary condition?
a) When there is an impermeable boundary
b) When there is constant pressure
c) When we do not know the flow distribution but we know the pressure at the boundaries
d) When we do not know the pressure at the boundaries
Answer: c
Explanation: When the flow distribution including velocity are not properly known but the boundary pressures are known, a constant pressure boundary condition is used.
2. While applying the constant pressure boundary condition, which of these is done?
a) Pressure is set to 0
b) Pressure correction is set to 1
c) Pressure correction is set to zero
d) Pressure is set to 1
Answer: c
Explanation: When constant pressure boundary is used, pressure is set to a fixed value which is the boundary pressure value and pressure correction is set to zero.
3. Which of these pose a problem in constant pressure boundaries?
a) Velocity
b) Flow direction
c) Density
d) Heat flux direction
Answer: b
Explanation: Constant pressure boundaries are used when the velocity is unknown. The main problem in this is the flow direction. It has to be found while solving the flow properties.
4. Which of these conditions is used to find the velocity component of the boundary while using constant pressure boundary condition?
a) Conservation of angular momentum
b) Conservation of linear momentum
c) Energy conservation
d) Continuity
Answer: d
Explanation: Velocity components are obtained as part of the solution procedure. The flow directions are obtained by ensuring that continuity is satisfied at every cell.
5. Constant pressure boundary condition makes the pressure correction cell nearest to the boundaries to act as ____________
a) Source or sink of mass
b) Source or sink of heat
c) Convection cell
d) Diffusion cell
Answer: a
Explanation: While implementing constant pressure boundary conditions, the nearest pressure correction cell acts as a source or sink of mass. This happens for all pressure boundary cells.
6. The velocity components in the nodes which are not at the boundary are found using ____________
a) energy equation
b) continuity equation
c) equations of state
d) momentum equation
Answer: d
Explanation: The velocities in the non-boundary nodes are found using u and v-momentum equations. After this, the boundary side velocity component is obtained using these values.
7. Periodic or cyclic boundary conditions are a type of ____________
a) wall boundary conditions
b) constant pressure boundary conditions
c) inlet boundary conditions
d) symmetry boundary conditions
Answer: d
Explanation: Periodic or cyclic boundary conditions are repeated symmetry boundary conditions. The matching of variables in Periodic or cyclic boundary conditions is done in the same way how it is done for symmetry boundary conditions.
8. For cyclic boundary conditions, which of these should be equated?
a) The flow variables
b) Gradient of the flow variables
c) Flux of the flow variables
d) Second derivative of the flow variables
Answer: c
Explanation: The flux of the flow variables in the inlet cyclic boundary should be matched with that of the outlet cyclic boundaries to apply a cyclic boundary. For example the flux of Φ 1,j and Φ n-1,j
9. The pairing operation can be done for all properties except ___________ of a flow.
a) pressure
b) density
c) velocity components
d) temperature
Answer: c
Explanation: Pairing can be done for all properties except the velocity components. Velocity components cannot be paired for the inlet and outlet boundary pairs.
10. For which of these flows, periodic or cyclic boundary conditions are applicable?
a) External flow over objects
b) Swirling flow inside a cylindrical furnace
c) Free surface flows
d) Buoyancy driven flows
Answer: b
Explanation: While solving swirling flow inside a cylindrical furnace, if cylindrical coordinates are used, we can pair the boundaries and hence periodic or cyclic boundary conditions can be used.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Partial Differential Equation”.
1. Which of these models of fluid flow give complete partial differential equations directly?
a) Finite control volume moving along with the flow
b) Finite control volume fixed in space
c) Infinitesimally small fluid element fixed in space
d) Infinitesimally small fluid moving along with the flow
Answer: c
Explanation: Infinitesimally small fluid element gives partial differential equations. When they are fixed in space, the equations are directly in partial differential form and there will be no need for changing a substantial derivative into partial differentials.
2. Where do we encounter partial differential equations in CFD?
a) Physical models
b) Assumptions
c) Governing equations
d) Discretized equations
Answer: c
Explanation: The governing equations of CFD are in partial differential form. This is because the flow variables depend upon four independent variables . When a flow variable is differentiated with respect to one of the independent variables, the others are kept constant.
3. What is the method used in CFD to solve partial differential equations?
a) Variable separation
b) Method of characteristics
c) Change of variables
d) Discretization
Answer: d
Explanation: In CFD, partial differential equations are discretized using Finite difference or Finite volume methods. These discretized equations are coupled and they are solved simultaneously to get the flow variables.
4. After discretizing the partial differential equations take which if these forms?
a) Exponential equations
b) Trigonometric equations
c) Logarithmic equations
d) Algebraic equations
Answer: d
Explanation: After discretization, the partial differential equations become algebraic equations with the flow variables as the unknowns which are then solved using some iterative method.
5. These are essential for solving partial differential equations.
a) Boundary conditions
b) Physical principle
c) Mathematical model
d) Algebraic equations
Answer: a
Explanation: The analytical solutions of partial differential equations depend upon boundary conditions and this is employed in CFD to some extent. Though the same PDE is solved, the solutions may differ based on the boundary conditions.
6. Find the order of the continuity equation for steady two-dimensional flow.
a) 1
b) 0
c) 2
d) 3
Answer: a
Explanation: Continuity equation for steady two-dimensional flow is given by \(\frac{\partial\rho u}{\partial x}+\frac{\partial \rho v}{\partial y}\). This is a first order partial differential equation.
7. The y-momentum equation falls into which of these types of PDEs?
a) 1-D first order equation
b) 2-D second order equation
c) 2-D first order equation
d) 1-D first order equation
Answer: b
Explanation: The y-momentum equation is
\
+2\mu\frac{\partial u}{\partial x}]+\frac{\partial}{\partial y}[\mu
]+\rho f_x\)
When expanded it gets second derivatives and hence it is a second order equation of x and y dimensions.
8. Which of these does not come under partial differential equations?
a) Laplace’s equation
b) Equations of motion
c) 1-D wave equation
d) Heat equation
Answer: b
Explanation: Equations of motion comes under ordinary differential equations. Laplace’s equation, wave equation and heat equations are all partial differential equations.
9. Which of these is not an analytical method to solve partial differential equations?
a) Change of variables
b) Superposition principle
c) Finite Element method
d) Integral transform
Answer: c
Explanation: Change of variables, Superposition principle, and Integral transform are all analytical methods. It is difficult to solve partial differential equations using analytical methods. Finite Element method is a numerical method to solve partial differential equations.
10. Linear partial differential equations are reduced to ordinary differential equations in which of these methods?
a) Change of variables
b) Fundamental equations
c) Superposition principle
d) Separation of variables
Answer: d
Explanation: In the separation of variables method, linear partial differential equations are reduced to ordinary differential equations and then these ODEs are solved.
11. The governing equations of CFD are ____________ partial differential equations.
a) Linear
b) Quasi-linear
c) Non-linear
d) Non-homogeneous
Answer: b
Explanation: The governing equations of CFD are quasi-linear partial differential equations. They have their highest order terms linearly and the coefficients are functions of the dependent variables itself.
12. Which of these is a quasi-linear partial differential equation?
a) \
\
\frac{\partial^2 u}{\partial y^2}=0\)
c) \
\
^2+\frac{\partial^2 u}{\partial y^2}=0\)
Answer: c
Explanation: \
\frac{\partial^2 u}{\partial y^2}=0\) are linear partial differential equations. \
^2+\frac{\partial^2 u}{\partial y^2}=0\) is non-linear.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Classification of PDE – 1”.
1. Which of these is not a type of flows based on their mathematical behaviour?
a) Circular
b) Elliptic
c) Parabolic
d) Hyperbolic
Answer: a
Explanation: The three types of flows based on the mathematical behaviour are Elliptic, Parabolic and Hyperbolic. These behaviours decide the method used to solve the mathematical model of the flows.
2. Which type of flow does the Laplace’s equation \
\) belong to?
a) Hyperbolic/ Parabolic
b) Hyperbolic
c) Parabolic
d) Elliptic
Answer: d
Explanation: The general equation is in this form.
\(A\frac{\partial ^2 \Phi}{\partial x^2}+B\frac{\partial ^2 \Phi}{\partial x\partial y}+C\frac{\partial^2\Phi}{\partial y^2}+D\frac{\partial\Phi}{\partial x}+E\frac{\partial \Phi}{\partial y}+F\Phi +G=0\)
Comparing \(\frac{\partial^2 \Phi}{\partial x^2}+\frac{\partial ^2 \Phi}{\partial y^2}=0\) with the above equation,
A=1
B=0
C=1
To find the type,
d=B 2 -4AC
d=-4
As d is negative, Laplace’s equation is elliptical.
3. The lines along which the derivatives of the dependent variables are indeterminate are called ___________
a) parabolic lines
b) characteristic lines
c) hyperbolic lines
d) transition lines
Answer: b
Explanation: The characteristic lines determine the type of the equation. These are defined as the lines along which the derivatives of the dependent variables do not exist.
4. Find the nature of the second-order wave equation.
a) Hyperbolic/elliptic
b) Parabolic
c) Hyperbolic
d) Elliptic
Answer: c
Explanation: The second-order wave equation is
\
.
A=-c 2
B=0
C=1
To find the type,
d=B 2 -4AC
d=4c 2
As d is positive, the second order wave equation is hyperbolic.
5. The classification of PDEs are governed by ________
a) Their highest order derivatives
b) Their least order derivatives
c) The number of terms
d) The constants
Answer: a
Explanation: The highest order derivatives of a PDE determines its type. Only the coefficients A, B and C are used to find the type of A\(\frac{\partial^2 \Phi}{\partial x^2}+B\frac{\partial ^2 \Phi}{\partial x\partial y}+C\frac{\partial^2 \Phi}{\partial y^2}+D\frac{\partial \Phi}{\partial x}+E\frac{\partial\Phi}{\partial y}+F\Phi +G=0\).
6. Which of these equations are used to classify PDEs?
a) \
-c=0\)
b) \
^2-b
=0\)
c) \
^2-
+1=0\)
d) \
^2-b
+c=0\)
Answer: d
Explanation: \
^2-b
+c=0\) is the characteristic equation for searching simple wave solutions. This is used to find the type of PDEs by substituting a, b and c by the coefficients of the second order derivatives of the given PDE.
7. Find the nature of the one-dimensional heat equation.
a) Circular
b) Elliptic
c) Hyperbolic
d) Parabolic
Answer: d
Explanation: The one-dimensional heat equation is
\
.
A=α
B=0
C=0
To find the type,
d=B 2 -4AC
d=0
As d is zero, the one-dimensional heat equation is parabolic.
8. The mathematical classification of inviscid flow equations are different from that of the viscous flow equations because of __________
a) absence of viscosity coefficients
b) absence of higher order terms
c) absence of convective terms
d) absence of diffusive terms
Answer: b
Explanation: The type of PDE is determined by the higher order terms. In the inviscid flow equations, the higher order terms are not present. So, the mathematical classification of inviscid and viscous flows is not the same.
9. Type of compressible flows depend upon _________
a) free stream pressure
b) free stream density
c) free stream velocity
d) free stream temperature
Answer: c
Explanation: Classification of compressible flows depend on M ∞ . This, in turn, depends on the free stream velocity. So, the type is determined by free stream velocity.
10. Find the nature of this system.
\
\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0, \frac{\partial u}{\partial y}-\frac{\partial v}{\partial x}=0. \)
a) Hyperbolic/elliptic
b) Elliptic
c) Hyperbolic
d) Parabolic
Answer: a
Explanation: For a system of equations, the general form is
\(a_1\frac{\partial u}{\partial x}+b_1\frac{\partial u}{\partial y}+c_1\frac{\partial v}{\partial x}+d_1\frac{\partial v}{\partial y}=0\)
\(a_2\frac{\partial u}{\partial x}+b_2\frac{\partial u}{\partial y}+c_2\frac{\partial v}{\partial x}+d_2\frac{\partial v}{\partial y}=0\)
Comparing the given equations with these two equations,
a 1 =1-M ∞ 2 , b 1 =0, c 1 =0, d 1 =1
a 2 =0, b 2 =1, c 2 =-1, d 2 =0
Here,
A=a 1 c 2 -a 2 c 1 =M ∞ 2 -1
B=-a 1 d 2 +a 2 d 1 -b 1 c 2 +b 2 c 1 =0
C=b 1 d 2 -b 2 d 1 =-1
Now,
d=B 2 -4AC = 4(M ∞ 2 -1)
Here,
When the flow is subsonic (M ∞ <1), d<1 and the equations are elliptic.
When the flow is supersonic (M ∞ >1), d>1 and the equations are hyperbolic.
This set of Computational Fluid Dynamics Interview Questions and Answers focuses on “Classification of PDE – 2”.
1. What are the two methods used to find the type of PDEs?
a) Lagrangian Method and Eulerian method
b) Cramer’s method and Eulerian method
c) Cramer’s method and Lagrangian Method
d) Cramer’s method and Eigenvalue method
Answer: d
Explanation: Partial differential equations can be classified using their characteristic lines. These are located using either the Cramer’s method or the Eigenvalue method.
2. Let u be a variable dependent on x and y. In the diagram, \(\frac{du}{dy}\, represents\, \frac{\partial u}{\partial y}\).
computational-fluid-dynamics-interview-questions-answers-q2
What does this line in the diagram represent?
a) Characteristic line
b) Eigenvalue line
c) Lagrange line
d) Cramer line
Answer: a
Explanation: Characteristic lines are those where the derivatives of the dependent variable are indeterminate. In the diagram, is indeterminate and hence the line represents a characteristic line.
3. How the type of PDE is identified using Cramer’s rule?
a) By equating the Cramer’s denominator to 1
b) By equating the Cramer’s numerator to 1
c) By equating the Cramer’s denominator to 0
d) By equating the Cramer’s numerator to 0
Answer: c
Explanation: The denominator of Cramer’s solution is equated to zero to find the type of PDE. The denominator is equated to zero to make the solution indeterminate.
4. What are the Cramer’s solutions equated to while using Cramer’s method of classifying a PDE?
a) The dependent variables
b) The derivatives of dependent variables
c) The second derivatives of dependent variables
d) The highest derivatives of dependent variables
Answer: b
Explanation: For characteristic lines, the derivatives of dependent variables are zero. Cramer’s rule is used to find these derivatives and then it is made indeterminate to find the type of PDE.
5. _________ of the characteristic curves is used to find the type of PDE.
a) Starting point
b) Centre
c) Length
d) Slope
Answer: d
Explanation: The nature of the slope of the characteristic curves gives the nature of the characteristic curves. This directly matches with the type of the PDE too.
6. What is the Cramer’s numerator when the solution is the derivative of dependent variables?
a) any negative value
b) any positive value
c) 1
d) 0
Answer: d
Explanation: The solution should be indeterminate. Therefore, if the denominator is 0, the numerator must also be zero. .
7. Consider the following system of PDEs.
\
\
\
\
\(
\) \(
\)
Answer: b
Explanation: From the given system of PDEs,
\(a_1\frac{\partial u}{\partial x}+b_1\frac{\partial u}{\partial y}+c_1\frac{\partial v}{\partial x}+d_1\frac{\partial v}{\partial y}=0\)
\(a_2\frac{\partial u}{\partial x}+b_2\frac{\partial u}{\partial y}+c_2\frac{\partial v}{\partial x}+d_2\frac{\partial v}{\partial y}=0\)
Let, \(W=
\) then
\(
\frac{\partial W}{\partial x}+
\frac{\partial W}{\partial y}=0 \)
\(\frac{\partial W}{\partial x}+
^{-1}
\frac{\partial W}{\partial y}=0\)
The Eigenvalues of \(
^{-1}\) \(
\) determines the class of PDE.
8. The Eigenvalues in the Eigenvalue method are ____________
a) the type of the characteristic lines
b) the type of PDE
c) the slope of the characteristic lines
d) the slope of PDE
Answer: c
Explanation: The Eigenvalues will give you the slope of the characteristic lines. Using this slope, we determine the type of the characteristic line and the type of PDE.
9. When the Eigenvalues are a mixture of real and imaginary values, the PDE is ___________
a) elliptic-hyperbolic
b) parabolic
c) elliptic
d) hyperbolic
Answer: a
Explanation: When we get a mixed type of Eigenvalues, the type of PDE is also mixed. Many practical equations have mixed behaviour also.
10. Solutions of a system of PDEs can be obtained by equating the numerator of Cramer’s solution while using Cramer’s rule. This method is used by __________
a) Integral transform
b) Change of variables
c) Separation of variables
d) Method of characteristics
Answer: d
Explanation: The Cramer’s numerator is equated to zero and compatibility equations are obtained. These are then solved to get the solution of PDEs.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Mathematical Behaviour of PDE – Elliptic Equations”.
1. The characteristic curves for an elliptic system are ___________
a) real and imaginary
b) both real
c) both imaginary
d) both zeros
Answer: c
Explanation: Elliptic equations have their determinant less than zero and the roots of this determinant are both imaginary. So, the curves are also imaginary.
2. Under which condition does the inviscid steady flow become elliptic?
a) M=1
b) M<1
c) M>1
d) M>5
Answer: b
Explanation: An inviscid steady flow is elliptic when the flow is subsonic. When the flow becomes supersonic, it changes its behaviour. Hence, it comes under the elliptic category when M<1.
3. Which of these are correct for an elliptic equation?
a) There is no limited region of influence or domain of dependence
b) There is no region of influence or domain of dependence
c) There is no region of influence, but there exists a domain of dependence
d) There is no domain of dependence, but there exists a region of influence
Answer: a
Explanation: For elliptic equations, the information is spread everywhere in all directions. They have an infinite region of influence or a limited domain of dependence for these equations. But, it cannot be stated that they do not have a region of influence or domain of dependence.
4. The solution of elliptic equations depends on ___________
a) one of its boundaries
b) all its boundaries
c) its opposite boundaries
d) its adjacent boundaries
Answer: d
Explanation: Any point inside the region of elliptic equations is influenced by the whole closed boundary. So, its solution depends on all of its boundaries.
5. Which of these statements is true for elliptic equations?
a) The solution can be approximated in some of the points
b) The solution can be marched from some initial conditions
c) The solution at all points must be carried out simultaneously
d) The solution process should be carried out simultaneously for some region and then marching can be done
Answer: c
Explanation: Any change at any point in the domain of elliptic equation influences all other points. So, the solution process should be carried out simultaneously and it cannot be marched.
6. The solution technique used to solve elliptic equations should ___________
a) allow each point to be influenced by its boundary-side neighbours
b) allow each point to be influenced by its west neighbour
c) allow each point to be influenced by its east neighbour
d) allow each point to be influenced by all its neighbours
Answer: d
Explanation: The solution technique of any problem should depend upon the type of the problem. Elliptic problems are influenced by all points in its domain. So, the solution technique should also allow all of its neighbours to influence each point.
7. Which of these flows is mathematically elliptic?
a) Transient viscous flow
b) Steady viscous flow
c) Steady inviscid flow
d) Transient inviscid flow
Answer: b
Explanation: Steady viscous flow exhibits elliptic behaviour. Steady inviscid flow behaves elliptic optionally. Transient flows are not elliptic.
8. Which of these is the prototype elliptic equation?
a) Incompressible irrotational flow
b) Incompressible rotational flow
c) Compressible irrotational flow
d) Compressible rotational flow
Answer: a
Explanation: The prototype elliptic equation is Laplace’s equation. This represents an incompressible irrotational fluid flow.
9. Which is an elliptic equation?
a) Wave equation
b) Transient heat conduction
c) Transient heat convection
d) Steady state heat conduction
Answer: d
Explanation: Transient problems and wave equations need marching solutions and cannot be elliptic. Steady state heat conduction is elliptic.
10. The below diagram represents elliptic equations.
computational-fluid-dynamics-questions-answers-behaviour-elliptic-equations-q10
The point ‘p’ is influenced by ___________
a) ab and bc
b) ab and cd
c) closed boundary abcd
d) ab
Answer: c
Explanation: The whole closed boundary influences the point ‘p’ along with its domain for elliptic equations. Thus ‘p’ is influenced by the closed boundary abcd.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Mathematical Behaviour of PDE – Parabolic Equations”.
1. Which of these are associated with a parabolic equation?
a) Initial and boundary conditions
b) Initial conditions only
c) Boundary conditions only
d) Neither initial conditions nor boundary conditions
Answer: a
Explanation: Initial conditions come into the picture as parabolic equations are marching. This marching is done along the boundary conditions. Thus, both the initial conditions and boundary conditions are necessary for parabolic equations.
2. Consider the flowing diagram.
computational-fluid-dynamics-questions-answers-behaviour-parabolic-equations-q2
Which of these flows can be represented by this diagram?
a) Steady viscous flow
b) Transient viscous flow
c) Transient inviscid flow
d) Steady inviscid flow
Answer: b
Explanation: The diagram represents parabolic equations. Transient viscous flows such as boundary layer flows are parabolic. So, the diagram can represent transient viscous flows and not the others.
3. Which among these problems is parabolic?
a) Steady inviscid flow
b) Steady state heat conduction
c) Unsteady heat conduction
d) Unsteady inviscid flow
Answer: c
Explanation: Unsteady heat conduction is a parabolic equation which marches through the time axis. When the problem reaches equilibrium, the rod gets a uniform distribution of temperature.
4. Imagine the point ‘p’ in the diagram is disturbed by some external conditions.
computational-fluid-dynamics-questions-answers-behaviour-parabolic-equations-q2
Which of these regions will be affected by this disturbance?
a) Neither 1 nor 2
b) Both 1 and 2
c) 1 only
d) 2 only
Answer: d
Explanation: Parabolic equations have marching solution. So, any disturbance at any point in the flow will affect only the flow behind it and not the flow ahead of it. So, region 2 will only be influenced by the disturbance.
5. Imagine a parabolic flow represented by this diagram.
computational-fluid-dynamics-questions-answers-behaviour-parabolic-equations-q2
Which of these lines represent initial conditions?
a) ab
b) p
c) ab
d) ad
Answer: a
Explanation: Parabolic equations in this diagram march from line ‘ab’ and move along the positive x-direction. So, the initial conditions are given by line ‘ab’.
6. Supersonic viscous problems _________
a) are always circular
b) cannot be parabolic
c) are parabolic
d) can be parabolized
Answer: d
Explanation: Supersonic viscous problems are governed by Navier-Stokes equations. Some assumptions made in this Navier-Stokes equations will give parabolic equations called Parabolized Navier Stokes equations.
7. Which of these equations is not parabolic?
a) \
+\frac{\partial}{\partial y}
+\frac{\partial}{\partial z}
=\rho\frac{\partial e}{\partial t}\)
b) \
=\rho\frac{\partial e}{\partial t}\)
c) \
+\frac{\partial}{\partial y}
+\frac{\partial}{\partial z}
]=\frac{\partial T}{\partial t} \)
d) \
+\frac{\partial}{\partial y}
+\frac{\partial}{\partial z}
=0\)
Answer: d
Explanation: \
+\frac{\partial}{\partial y}
+\frac{\partial}{\partial z}
=0\) represents steady heat conduction and it is elliptic. The other three equations represent unsteady heat conduction and they are parabolic.
8. Imagine a parabolic flow represented by this diagram.
computational-fluid-dynamics-questions-answers-behaviour-parabolic-equations-q2
What does the line ‘ad’ represent?
a) Right characteristic line
b) Left characteristic line
c) Boundary condition
d) Initial condition
Answer: c
Explanation: Lines ‘bc’ and ‘ad’ form the boundary of the flow. They represent the boundary conditions of the parabolic problem. The solutions march along these boundary lines.
9. Which of these assumptions are made for parabolizing Navier-Stokes equations?
a) Viscous terms are neglected
b) Viscous terms with derivatives in the stream-wise direction are neglected
c) Derivatives in the stream-wise direction are neglected
d) Viscous terms in the stream-wise direction are neglected
Answer: b
Explanation: Parabolized Navier-Stokes equations can be used for supersonic flows over sharp-nosed cones. In this case, the flow field does not involve any reversed or separated flows. So, the derivatives of the viscous terms in the stream-wise direction are neglected and the equations are parabolized.
10. Which of these apply to parabolic equations?
a) They have one real characteristic line
b) They have two real characteristic lines
c) They have two imaginary characteristic lines
d) They do not have characteristic lines
Answer: a
Explanation: Parabolic equations have their determinant equal to zero. So, they get only one real characteristic line through the point considered.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Mathematical Behaviour of PDE – Hyperbolic Equations”.
1. Consider the following diagram.
computational-fluid-dynamics-questions-answers-behaviour-hyperbolic-equations-q1
What does line ‘ab’ represent?
a) Left characteristic line
b) Right characteristic line
c) Data which depends on P
d) Initial data on which P is dependent
Answer: d
Explanation: Line ‘ab’ is the initial data which P depends on. They influence the values of dependent variables at point P. Solutions march from this line.
2. When the hyperbolic equations are represented by a diagram, what is the region dependent on the point of consideration called?
a) Boundary region
b) Domain of dependence
c) Region of influence
d) Initial region
Answer: c
Explanation: The area which is affected by point P is called the region of influence. Any disturbance at that point will disturb the solution in the region of influence.
3. Consider the diagram.
computational-fluid-dynamics-questions-answers-behaviour-hyperbolic-equations-q1
In the diagram what does region 1 mean?
a) Region of influence
b) Domain of dependence
c) Characteristic region
d) Boundary region
Answer: b
Explanation: The values at point P depend upon the region ahead of P. So, it is called the domain of dependence. This starts from the line ‘ab’.
4. When is the steady inviscid flow hyperbolic?
a) In supersonic flow regime
b) Never
c) Always
d) In subsonic flow regime
Answer: a
Explanation: In subsonic speeds, steady inviscid flow is elliptic. When the Mach number is more than 1, they become hyperbolic.
5. Which of these is true for hyperbolic equations?
a) They have 2 imaginary characteristic lines
b) They have 1 imaginary characteristic line
c) They have 1 real characteristic line
d) They have 2 real characteristic lines
Answer: d
Explanation: Hyperbolic equations get a positive determinant. Taking the square root of this positive determinant, we will get two real characteristic lines.
6. From the diagram,
computational-fluid-dynamics-questions-answers-behaviour-hyperbolic-equations-q1
What does, line Pc represent?
a) Left running boundary line
b) Right running characteristic line
c) Left running characteristic line
d) Left running boundary line
Answer: c
Explanation: Lines ‘ac’ and ‘bd’ represent the characteristic lines of the hyperbolic problem. From point P if we face the direction of marching, the left and right side characteristic lines are called the left and right running characteristic lines.
7. Which of these flows can be represented by the following diagram?
computational-fluid-dynamics-questions-answers-behaviour-hyperbolic-equations-q7
a) Unsteady viscous supersonic flow
b) Unsteady inviscid supersonic flow
c) Steady inviscid supersonic flow
d) Steady inviscid subsonic flow
Answer: b
Explanation: From the diagram, as the solutions are time marching, the flow must be unsteady. Unsteady viscous flow is parabolic. But the diagram is for hyperbolic problems. So, unsteady inviscid supersonic flow suits this diagram.
8. To solve a hyperbolic equation, which of these is/are necessary?
a) Initial conditions and boundary conditions
b) Initial conditions
c) Boundary conditions
d) Neither initial nor boundary conditions
Answer: a
Explanation: For solving any hyperbolic problem, initial conditions are needed to march from and boundary conditions are needed additionally.
9. Which of these equations is hyperbolic?
a) Unsteady Navier-Stokes equation
b) Steady Navier-Stokes equation
c) Steady Euler equation
d) Unsteady Euler equation
Answer: d
Explanation: Euler equations represent inviscid flows. Unsteady inviscid flows are always hyperbolic. Steady inviscid flows are conditionally hyperbolic.
10. The speed of propagation of the disturbance in hyperbolic problems is ____________
a) finite
b) infinite
c) 0
d) 1
Answer: a
Explanation: Disturbances propagate at a finite speed defined by some flow properties. In the elliptic and parabolic equations, this speed is infinite.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Mathematical Behaviour of PDE – Well Posed Problems”.
1. When can we say that a problem is suitable to be solved using CFD?
a) The PDE has no solution
b) The solution to PDE is unique and it depends continuously on the initial and boundary conditions
c) The PDE has more than one solution
d) The solution to PDE is unique and independent of the initial and boundary conditions
Answer: b
Explanation: We say a problem to be well suited for CFD when the partial differential equation representing the problem has a unique solution and that solution depends on the specified initial and boundary conditions.
2. What is the difficulty in modelling supersonic blunt body problem?
a) The PDE cannot be solved
b) A PDE cannot be formed
c) Mixed flow behaviour
d) Boundary conditions cannot be formed
Answer: c
Explanation: Supersonic flow over a blunt body creates a bow shock. This behaves as a normal shock near the leading edge and as an oblique shock downstream. So, the flow becomes a mixture of subsonic-supersonic flow based on spatial coordinates and makes the mathematical behaviour also mixed.
3. Which of these mathematical models suit unsteady Navier-Stokes equations?
a) Elliptic
b) Hyperbolic
c) Parabolic
d) Mixed
Answer: d
Explanation: The equations reduced from the Navier-Stokes equations have a particular behaviour. But, the pure unsteady Navier-Stokes equations do not have a particular behaviour. They exhibit mixed behaviour.
4. Which kind of flows need an initial condition?
a) Parabolic and hyperbolic
b) Hyperbolic and elliptic
c) Elliptic and Parabolic
d) Elliptic
Answer: a
Explanation: Parabolic and hyperbolic equations are marching problems. So, they need initial conditions. Elliptic equations need only boundary conditions.
5. The flow variables for a/ an _____________ should be solved simultaneously with the flow variables at all other points in the domain.
a) Hyperbolic flow
b) Elliptic flow
c) Parabolic flow
d) Mixed flow
Answer: b
Explanation: Elliptic equations do not march from any condition. They completely rely on the boundary conditions and to solve an elliptic problem, the whole domain should be solved simultaneously.
6. CFD solutions are excellent for flows with _____________
Note: ‘M’ is the Mach number of the flow.
a) M > 1
b) M < 1
c) M = 1
d) M > 5
Answer: b
Explanation: CFD solutions suit problems with Mach numbers well less than 1. When the Mach number is near or more than 1, they pose difficulties for the solution.
7. What is the problem in modelling flows with high Reynolds number?
a) A mixture of viscous and inviscid flow regions
b) A mixture of subsonic and supersonic flow regions
c) A mixture of elliptic and parabolic flows
d) A mixture of elliptic and hyperbolic flows
Answer: a
Explanation: At high Reynolds numbers, the viscous regions are very thin. The part of the flow where the boundary conditions are specified behaves different from the part near the body.
8. Under-specification of boundary conditions gives rise to _____________
a) solution equal to zero
b) no solution
c) a unique solution
d) n-number of solutions
Answer: d
Explanation: If enough boundary conditions are not specified, the number of solutions becomes more. A unique solution to the problem cannot be found.
9. What happens with the over-specification of boundary conditions?
a) Infinite solutions
b) Unique solutions
c) Unphysical solutions
d) No solution
Answer: c
Explanation: When boundary conditions are over-specified, it leads to severe unphysical solutions near the boundaries. It will not reinforce the solution.
10. What is the problem in solving flows with Mach numbers around and above 1?
a) Mach number
b) Compressibility
c) Continuity
d) Pressure
Answer: b
Explanation: Flows with Mach numbers around and above 1 has the additional issue of compressibility. Air becomes compressible at higher speeds and the effects of compressibility also should be included while solving the problem.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Discretization”.
1. Discretization of the physical domain of interest results in __________
a) Boundaries
b) Discretized equations
c) Discrete cells
d) Exponential equations
Answer: c
Explanation: Discretization of the physical domain means dividing the region of interest into discrete elements. This will result in discrete cells on which the rest of the solution process depends.
2. The number of discretized equations is equal to the number of __________
a) Discretized cells
b) Boundary conditions
c) Unknowns
d) Boundary-side elements
Answer: a
Explanation: Discretized equations are formed for each element obtained after discretizing the domain. So, the number of discretized cells and that of the discretized equations will be the same.
3. Equation discretization is the process of converting ___________ PDEs into discretized equations.
a) Open-form
b) Closed-form
c) Linear
d) Quasi-linear
Answer: b
Explanation: Closed form continuous PDEs are converted into discretized equations suitable for numerical solutions. Analytical solution of this closed form PDEs will give continuous equations which can be used to get the flow variable at any desired point in the domain.
4. Which of these methods is not a method of discretization?
a) Finite volume method
b) Finite difference method
c) Gauss-Seidel method
d) Spectral element method
Answer: c
Explanation: Gauss-Seidel method is a method of solving the discretized equations. Finite difference method, finite volume method and spectral element method are all methods of discretization.
5. Discretization of the governing equations result in ___________
a) Integral equations
b) Quasi-linear partial differential equations
c) Partial differential equations
d) Algebraic equations
Answer: d
Explanation: Discretization of the governing equations means converting the partial differential or the integral equations into algebraic equations.
6. The discretized equation connects each element with __________
a) the northern and southern elements
b) the boundary elements
c) the neighbouring elements
d) the eastern and western elements
Answer: c
Explanation: The discretized equation for a particular element connects it with its neighbouring elements in all the sides in general. The particular sides depend upon the problem taken.
7. If the domain and equations are not discretized, which of these will become true?
a) Numerical solution cannot be obtained
b) Analytical solution cannot be obtained
c) Initial conditions cannot be applied
d) Mathematical model cannot be obtained
Answer: a
Explanation: Discretization process is done to make the equations suitable for numerical solution. If not discretized, numerical methods could not be applied.
8. The discretized flow field properties exhibit ___________
a) continuous nature
b) piece-wise nature
c) quadratic nature
d) cubic nature
Answer: b
Explanation: The discretized flow field properties are discontinuous and they have a piece-wise nature. Until discretized, the variables are continuous.
9. The finite volume method ____________ the governing equations in each cell.
a) discretizes
b) sums up
c) integrates
d) multiplies
Answer: c
Explanation: In the finite volume method, the governing equations are integrated over each cell to form a semi-discretized equation. Then, the variation of flow variables is approximated.
10. In which of these methods, the differential equation is multiplied by a test function?
a) Finite difference method
b) Finite volume method
c) Finite element method
d) Spectral element method
Answer: d
Explanation: Spectral element method multiplies the differential equation by a random test function and then integrates it over the entire domain.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Discretization Aspects – Thomas Algorithm”.
1. Thomas algorithm is a ___________
a) Linear equations solver
b) Quadratic equations solver
c) Discretization method
d) Linear least square system
Answer: a
Explanation: Using a discretization method, the governing partial differential equation are converted into a system of algebraic equations. These discretized equations are solved using the Thomas algorithm.
2. Thomas algorithm can be used to solve __________
a) any matrix
b) all square matrices
c) only penta-diagonal matrices
d) only tri-diagonal matrices
Answer: d
Explanation: The other name of the Thomas algorithm is Tri-diagonal matrix algorithm. Tri-diagonal matrices are matrices with non-zero elements in the main diagonal and the diagonals above and below it.
3. Thomas algorithm is _________
a) an analytical method
b) a direct method
c) an iterative method
d) a least squares method
Answer: c
Explanation: Thomas algorithm solves a system of equations with non-repeated sequence of operations. It is a direct method to solve the system without involving repeated iterations and converging solutions.
4. Consider a system of equations where the i th equation is a i Φ i =b i Φ +c i Φ +d i . While solving this system using Thomas algorithm, we get Φ i =P i Φ +Q i . What are P i and Q i ?
a) \
\
\
\(P_i=\frac{d_i}{a_i-c_i P_{i-1}};Q_i=\frac{c_i Q_{i-1}+b_i}{a_i-c_i P_{i-1}}\)
Answer: b
Explanation: As given,
Φ i = P i Φ i+1 +Q i
Φ i-1 = P i-1 Φ i +Q i-1
The i th equation is,
a i Φ i = b i Φ i+1 + c i Φ i-1 + d i
a i Φ i = b i Φ i+1 + c i (P i-1 Φ i + Q i-1 ) + d i
a i Φ i – c i P i-1 Φ i = b i Φ i+1 +c i Q i-1 +d i
Φ i (a i -c i P i-1 ) = b i Φ i+1 +c i Q i-1 +d i
\(\Phi_i = \frac{b_i}{a_i-c_i P_{i-1}}\Phi_{i+1} + \frac{c_i Q_{i-1}+d_i}{a_i-c_i P_{i-1}} \)
Therefore,
\(P_i = \frac{b_i}{a_i-c_i P_{i-1}};Q_i=\frac{C_i Q_{i-1}+d_i}{a_i-c_i P_{i-1}}\).
5. Let the i th equation of a system of n equations be a i Φ i =b i Φ i+1 +c i Φ i-1 +d i . Which of these is correct?
a) c N =0; b N =0
b) c N =0; b 1 =0
c) c 1 =0; b N =0
d) c 1 =0; b 1 =0
Answer: c
Explanation: Node 1 will not have a previous node (c 1 =0). The last node will not have the next node (b N =0).
6. Using the Thomas algorithm, if the i th unknown is Φ i =P i Φ i+1 +Q i . what is the last unknown value Φ N equal to?
a) 0
b) P N
c) Q N
d) 1
Answer: c
Explanation: For the last equation, b N =0. So, P N =0. Φ N =P N Φ N+1 +Q N =Q N .
7. While solving a system of equations with the Thomas algorithm, in which order are the values of P i and Q i found?
a) Backwards
b) Forward
c) Simultaneously
d) Depends on the problem
Answer: b
Explanation: To find the current values of P i and Q i , the previous values P i-1 and Q i-1 should be known. So, this is started from the first equation and done in forward order.
8. After finding all the values of P i and Q i , in which order are the values of Φ i found?
a) Forward
b) Simultaneously
c) Backwards
d) Depends on the problem
Answer: c
Explanation: The last value of Φ N can be found using Q N . Using this, the previous value is found using the formula Φ i =P i Φ i+1 +Q i . So, it is done backwards.
9. Consider a system of equations where the i th equation is a i Φ i =b i Φ i+1 +c i Φ i-1 +d i . While solving this system using Thomas algorithm, we get Φ i =P i Φ i+1 +Q i . What are P 1 and Q 1 ?
a) \
\
\
\(P_1=\frac{b_1}{a_1};Q_1=\frac{d_1}{a_1}\)
Answer: d
Explanation: In general,
\(P_i=\frac{b_i}{a_i-c_i P_{i-1}};Q_i=\frac{c_i Q_{i-1}+d_i}{a_i-c_i P_{i-1}}\)
As c 1 =0,
\(P_1=\frac{b_1}{a_1};Q_1=\frac{d_1}{a_1}\).
10. A system of equations with which of these coefficient matrices can be solved using the Thomas algorithm?
a) \
\
\
\(
\)
Answer: a
Explanation: Thomas algorithm can be used to solve tri-diagonal matrices only.
\(
\)is a tri-diagonal matrix. \(
\) is an upper triangular matrix.\(
\) is a lower triangular matrix. \(
\) is a square matrix.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Discretization Aspects – Consistency”.
1. A solution to some algebraic equation is said to be consistent if _________
a) the error is bounded
b) the computation time is not prohibitive
c) the numerical solution approaches the exact solution when time step and grid spacing tends to zero
d) the solution does not change with further iterations
Answer: c
Explanation: A solution to some discretized algebraic equation is said to be consistent if that solution approaches the exact solution of the partial differential equation when time step and grid spacing are very small.
2. Consistency of a numerical solution is directly associated with __________
a) convergence
b) stability
c) iterative error
d) discretization error
Answer: d
Explanation: A numerical solution’s consistency has a direct dependence on discretization error. Discretization error occurs because of the discretization of the continuous solution. If this error is big, the solution will not match with the exact continuous solution.
3. What is the other name for Courant number?
a) CFL number
b) Peclet number
c) Nusselt number
d) Scarborough number
Answer: a
Explanation: Courant number is otherwise called CFL number. It is expanded as Courant-Friedrichs-Lewy number. It is named after Richard Courant, Kurt Friedrichs and Hans Lewy who first formed this number.
4. Courant number is applicable for __________
a) implicit transient schemes
b) explicit transient schemes
c) quadratic schemes
d) high-resolution schemes
Answer: b
Explanation: Courant number is used in explicit time integration methods involving numerical methods. It relates grid size and time steps of the explicit transient schemes. It cannot be applied for implicit transient schemes.
5. Consistency is defined when the discretization error approaches __________
a) infinity
b) 1
c) 0
d) -1
Answer: c
Explanation: Consistency is defined when the time step or grid spacing approaches 0. When this happens the discretization error becomes zero. When discretization error is zero, it means that the solution matches with the analytical solution.
6. Consistency comes into the picture because of _________
a) McLaurin series expansion
b) Power series expansion
c) Fourier series expansion
d) Taylor series expansion
Answer: d
Explanation: Consistency wants the numerical solution to be the same as the analytical solution. The numerical approximations of partial differential equations are done using the Taylor series expansion. The higher order terms in this series are neglected. This causes a difference between the numerical and the analytical solution.
7. For consistency to have some relationship with discretization error, the discretization error should be ____________
a) some powers of time-step
b) some powers of grid spacing
c) some powers of time-step and/ or grid spacing
d) some function of time-step and/ or grid spacing
Answer: a
Explanation: When time-step and grid spacing approaches zero, discretization error should approach zero. For this, it should be some powers of time-step and/ or grid spacing. If it is a function, it may or may not become zero.
8. For the solution of a system of discretized equations with consistent approximations to be consistent, which of these conditions is necessary?
a) Boundedness
b) Stability
c) Convergence
d) Accuracy
Answer: b
Explanation: Inconsistency problems arise when we truncate higher order terms. These approximations are consistent is the same order terms are truncated always. Though this condition is satisfied, it is a must for the system of equations to be stable to satisfy consistency.
9. If the discretization error is the ratio of grid spacing to time step, then for the system to be consistent, which of these is correct?
a) the ratio should be equal to one
b) the ratio should be equal to zero
c) the ratio should tend to zero
d) the ratio should be equal to negative one
Answer: c
Explanation: We know that consistency is satisfied if discretization error is zero. When discretization error is the ratio of grid spacing to time step, then actually the ratio should be zero. But, this is not practically possible as the grid spacing cannot be zero. So, grid spacing and time step must be reduced in a way that the ratio tends to zero.
10. Consistency should be ensured ___________
a) at the interior nodes
b) in the global domain
c) at the boundary nodes
d) at each node
Answer: d
Explanation: The algebraic equations for the partial differential governing equations are formed at each node of the domain. For each of these algebraic equations, consistency should be defined. There is no exception for the boundary nodes or the interior nodes.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Discretization Aspects – Stability”.
1. Stability is defined _________
a) only for iterative solvers
b) only for direct solvers
c) for all numerical solvers
d) for all discretization processes
Answer: a
Explanation: Only for iterative solvers, stability can be defined. It describes how well the equations can be solved using iterative solvers. It needs the solver not to diverge the solution from the exact answers.
2. Which of these is used to analyse the stability of a system?
a) Nusselt number
b) Courant number
c) Peclet number
d) Von Neumann’s method
Answer: d
Explanation: Von Neumann’s method is a widely used method of analysing the stability of any mathematical system. Since CFD uses numerical methods to solve problems, Von Neumann’s method is applicable to the CFD schemes also.
3. Stability is the property of a _________
a) partial differential equation
b) discretized equation
c) discretization process
d) mathematical model
Answer: b
Explanation: Stability is not the property of a discretization process; it is the property of the resulting system of discretized equation. This can be analysed mathematically for the system. Stability does not exist for a partial differential equation or a mathematical model.
4. For which of these problems, the error will be bounded if the system is stable?
a) Transient problems
b) Subsonic problems
c) Supersonic problems
d) Inviscid problems
Answer: a
Explanation: For transient problems, a stable system keeps the error bounded when time increases. Stability for a transient problem has special characteristics. Here, stability guarantees that the method gives a bounded solution if the exact solution is bounded.
5. A system is said to be stable if _________
a) the results for different boundary and initial conditions are different
b) the results for different boundary and initial conditions are the same
c) the system can be solved for different initial and boundary conditions
d) the result of two consequent iterations are the same
Answer: c
Explanation: A stable system of algebraic equation means that the system can be solved for different boundary conditions and initial conditions to get the flow properties. While varying the boundary conditions, the system should not become unsolvable.
6. Stability of explicit transient schemes is related to _________
a) over-relaxation
b) the time-step
c) the grid size
d) under-relaxation
Answer: b
Explanation: Stability of transient schemes vary for implicit and explicit problems. Stability of explicit transient scheme is ensured by limiting the time-step size. Stability of implicit transient scheme is improved by under-relaxing the equations.
7. Which of the following is a sufficient condition for a system to be stable?
a) Gauss criterion
b) Convergence criterion
c) Stopping criterion
d) Scarborough criterion
Answer: d
Explanation: A system of linear equations can be taken as stable if it satisfies the Scarborough criterion. Scarborough criterion gives a condition about the coefficient matrix of the algebraic system. This criterion also gives information about the boundedness of the problem.
8. What is the Scarborough criterion?
a) The coefficient matrix has larger values in the main diagonal
b) The coefficient matrix has larger values above the main diagonal
c) The coefficient matrix has larger values below the main diagonal
d) The coefficient matrix has larger values except the main diagonal
Answer: a
Explanation: Scarborough criterion needs the coefficient matrix to be diagonally dominant. This means that the diagonal elements should be larger than the non-diagonal elements for a particular row in the coefficient matrix.
9. A system is said to be stable if _________
a) it does not magnify the errors occurring in the course of solution
b) it does not converge
c) it reduces the errors occurring in the course of solution
d) it does not maintain the errors occurring in the course of solution
Answer: a
Explanation: A mathematical system is said to be stable if the system does not go on increasing the errors produced while solving the problem. So, the solution should not diverge from the exact answer though errors are present.
10. It is difficult to analyse the stability of _________
a) non-linear systems without boundary conditions
b) linear systems with boundary conditions
c) non-linear systems with boundary conditions
d) linear systems without boundary conditions
Answer: c
Explanation: While solving non-linear coupled equations with boundary conditions, analysing stability is difficult. So, the stability of a system is usually analysed for linear problems without boundary conditions. For non-linear systems, we rely on experience to know its stability.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Discretization Aspects – Convergence”.
1. Convergence is defined for _________
a) Elimination method
b) Iterative solvers
c) Direct solvers
d) Cramer’s method
Answer: a
Explanation: Convergence is a property of the iterative solvers used for solving the discretized system of equations. It cannot be defined for direct solvers as they do not have repeated steps and similar answers.
2. A solution is ideally converged if _________
a) the results match with the exact solution
b) the results for two consecutive iterations are the same
c) the results for two schemes are the same
d) the results for different boundary conditions are the same
Answer: b
Explanation: Ideally, a solution of a system of equations is said to be converged if the results of two consecutive iterations are exactly the same without any variation. No more iterations are required after this.
3. In real, how is convergence defined?
a) Variations are accepted
b) When the variation is less than the result
c) When the variation falls below a certain acceptable range
d) When the variation is the same as the result
Answer: c
Explanation: In real, the variation between two consecutive iterations cannot be exactly the same. The value of variation will be constantly decreasing. So, the solution is said to be converged when the range of variation is acceptable.
4. Convergence decides _________
a) the result of the numerical method
b) the method of iteration
c) the stability of the system
d) when to stop the iterations
Answer: d
Explanation: Iterative processes start with an initial guessed answer. This converges into the correct result as the number of iterations increases. Convergence criterion says when to stop this repeated process with acceptable error.
5. How is the tolerance of convergence decided?
a) Based on stability and consistency
b) Based on efficiency and accuracy
c) Based on efficiency and consistency
d) Based on consistency and accuracy
Answer: b
Explanation: In practical, the results of two iterations does not exactly match with each other. The iterations are stopped when the solution reaches some acceptable tolerance. This tolerance is decided in a way that it affects neither the accuracy of the solution nor its efficiency.
6. If the tolerance value to stop the iteration is too big, which of these properties will be affected?
a) Accuracy
b) Efficiency
c) Stability
d) Conservativeness
Answer: b
Explanation: The tolerance should be a balance between both accuracy and efficiency. If the tolerance is too big, iterations will stop soon but the answers will not be accurate. On the other hand, if the tolerance is too small, the number of iterations will be more. This will make the solution inefficient.
7. Which of these statements is wrong?
a) Convergence is applicable for iteration processes
b) Convergence is affected by accuracy and efficiency
c) Converged solutions do not vary much with further iterations
d) Converged solutions are exact
Answer: d
Explanation: Converged solutions are just correct in respect to the iteration. We cannot say that all the solutions which converge are correct. The converged solutions may be wrong I respect to other properties. Convergence does not ensure correct solutions.
8. Which of these is related to convergence?
a) Stopping criteria
b) Peclet number
c) Lax Equivalence Theorem
d) Scarborough criteria
Answer: c
Explanation: Lax Equivalence Theorem gives the condition for stability. It is applicable only for the finite difference methods applied to linear initial value problems. This is not applicable to non-linear systems.
9. Which of these properties is not included in the Lax Equivalence Theorem?
a) Stability
b) Boundedness
c) Consistency
d) Convergence
Answer: b
Explanation: Lax Equivalence Theorem states that “For a well-posed linear initial value problem solved by the finite difference approximation which satisfies consistency condition, stability is the necessary and sufficient condition for convergence”.
10. For small grid sizes, convergence is related to _________
a) truncation error
b) stability
c) consistency
d) boundedness
Answer: a
Explanation: When the grid sizes are sufficiently small, convergence is related to truncation error. The rate of convergence is governed by the order of principal truncation error component which is used to approximate the partial differential equations.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Discretization Aspects – Conservativeness”.
1. Conservation of the flow properties should be ensured in __________
a) both the global and local domains
b) the global domain only
c) the local domain only
d) the global domain and optionally in the local domain
Answer: a
Explanation: We know that the conservation laws govern the flows and the properties like energy and mass are conserved in the global domain. This must be applicable to the local domain also after discretization. Otherwise, the solution will be unrealistic.
2. The flux of one element must have _________ to the flux of the neighbouring element.
a) different magnitude and equal sign
b) different magnitude and opposite sign
c) equal magnitude and equal sign
d) equal magnitude and opposite sign
Answer: d
Explanation: Flux leaving the face of one element should be the flux entering the neighbouring element. So, the fluxes of two near-by elements should have equal magnitude and opposite sign. This ensures the conservativeness of the system.
3. Conservation of the system leaves a limitation to the __________
a) stability
b) solution error
c) convergence
d) boundedness
Answer: b
Explanation: If the conservation of mass, momentum and energy are ensured, the error can only interchange the values at different nodes. The overall system will not be erroneous. So, the solution error is a way decreased by the conservativeness of a system.
4. Non-conservative schemes can be consistent and stable if ____________
a) grid is fine
b) grid is coarse
c) solution converges
d) solution is bounded
Answer: a
Explanation: For fine rids, non-conservative schemes can also give a consistent and stable solution. The errors due to non-conservation are negligible. These errors become appreciable only if the grid is coarse and the grid size is more.
5. Which of these methods is usually conservative?
a) Finite Difference Method
b) Finite Element Method
c) Finite Volume Method
d) Iterative Method
Answer: c
Explanation: Conservativeness is defined for the discretization schemes only. Finite volume methods often guarantee conservation. They integrate the flow variables over each elemental domain. They ensure that flux leaving one domain is equal to that entering the neighbouring domain.
6. For a system with a source or sink to be conservative, which of these is correct?
a) The total source or sink in the domain is divided equally between the elements
b) The total source or sink in the domain is equal to the net flux through the boundaries
c) The total source or sink in the domain is not considered for checking conservativeness
d) Flux leaving the face of one element is equal to the flux entering the neighbouring element
Answer: b
Explanation: The general condition of equal and opposite fluxes cannot be applied to a system with sources or sinks. For this case, the total source or sink in the domain should be equal to the net flux through the boundaries of the whole domain.
7. Non-conservative schemes produce ___________
a) artificial radiation
b) artificial diffusion
c) artificial convection
d) artificial source or sink
Answer: d
Explanation: Non-conservative schemes do not have the net fluxes conserved. In the global domain, the fluxes entering the domain and that leaving the domain are not the same. This will create artificial sources and sinks which do not actually exist in the physical problem.
8. Which of these schemes ensure conservativeness?
a) Central differencing
b) Upwind differencing
c) TVD scheme
d) Quadratic schemes
Answer: a
Explanation: Conservation of flow property is ensured for the central differencing scheme over the entire domain. Here, the flux interpolation formula is consistent. So, only the two boundary fluxes remain when the global domain is considered.
9. Which of these higher-order schemes is conservative?
a) Upwind
b) TVD
c) QUICK
d) Power law scheme
Answer: c
Explanation: QUICK scheme is one of the higher-order schemes involving quadratic interpolation. While the other quadratic interpolation schemes give rise to conservation problems, the QUICK scheme ensures the conservation of the quantities. But, it has boundedness problems.
10. Quadratic interpolation results in conservation problems. Why?
a) Their physical problem is non-conservative
b) They involve quadratic equations
c) They are higher order schemes
d) Interpolation curves vary at the face
Answer: d
Explanation: Quadratic interpolation models the physical problems using quadratic equations. At the intersection faces, we get two different quadratic equations and their values do not cancel out there. This gives a problem to conservation in the Quadratic interpolation schemes.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Discretization Aspects – Boundedness”.
1. A flow property Φ is bounded when ___________
a) the value of flow properties at a node is bounded by its boundary values
b) the error is bounded
c) the numerical solution approaches the exact solution
d) the solution does not change with further iterations
Answer: a
Explanation: The flow property is said to be bounded if the internal nodal values of the flow property do not cross the minimum and maximum values of the flow properties in the boundaries. Physically the flow properties will not go beyond the boundary values. This should be guaranteed by the numerical approximations also.
2. For a system to be bounded, the coefficients of the nodes should ___________
a) have the same sign
b) be the same
c) be different everywhere
d) have different signs everywhere
Answer: a
Explanation: All the coefficients of the discretized form of the governing equations should have the same sign for the system to be bounded. In most of the cases, the signs should be all positive. This means that an increase in the flow variable at one node increases the variables at the neighbouring nodes too.
3. ___________ of the coefficient matrix is a desirable feature for boundedness.
a) Non-diagonal dominance
b) Singularity
c) Sparsity
d) Diagonal dominance
Answer: d
Explanation: A system is bounded if the Scarborough criterion is satisfied. This needs the coefficient matrix to be diagonally dominant. This is a desirable feature for boundedness of solutions. This depends upon the coefficients of the neighbouring nodes and the central nodes.
4. Boundedness of a system has a direct impact on ______________
a) stability
b) convergence
c) conservativeness
d) transportiveness
Answer: b
Explanation: Convergence of an iterative solution is when two consecutive iterations result in the same solution. If the discretized system does not satisfy boundedness, it is possible that the solution wiggles continuously without converging.
5. A coefficient matrix is diagonally dominant if ________
Note: a nb → coefficients of the neighbouring nodes
a p → coefficient of the central node
a) \
\
\
\(\frac{\big|a_p \big|}{\big|a_{nb} \big|}\) < 1
Answer: c
Explanation: For a coefficient matrix to be diagonally dominant, the diagonal elements should be larger than the non-diagonal elements of the row. So,\(\frac{\sum\big|a_{nb} \big|}{\big|a_p \big|}\) should be less than 1. This is the condition given by Scarborough criterion.
6. Which of these schemes often result in unbounded solutions?
a) Central difference schemes.
b) First order schemes
c) QUICK scheme
d) High-resolution schemes
Answer: c
Explanation: Higher order schemes result in unbounded solutions. This means that the solution is erroneous. This happens only when the grid is so coarse. The problem of boundedness can be overcome by refining the grids. The QUICK scheme is a higher order scheme.
7. Which of these schemes guarantee boundedness?
a) Central difference schemes.
b) Forward difference schemes
c) High-resolution schemes
d) First order schemes
Answer: d
Explanation: Irrespective of coarse or fine grids first order schemes always guarantee boundedness without any overshoots or undershoots. So, while choosing a coarse grid, it is better to use the first order schemes to get rid of errors.
8. Which of these conditions define boundedness?
Φ F → Flow property at boundary
Φ i → Flow property at nodes
a) Φ i ≤max(Φ F )
b) min(Φ F )≤Φ i ≤max(Φ F )
c) Φ i ≥min(Φ F )
d) Φ i ≥max(Φ F )
Answer: b
Explanation: The value of flow property Φ at any node inside the domain must be smaller than the minimum value of ϕ at the boundaries. Similarly, Φ at any node must be greater than the maximum value of Φ at the boundaries. Representing this mathematically, min(Φ F )≤ Φ i ≤max(Φ F ) holds true.
9. Applying the boundedness condition, which is correct?
ρ i → Density at nodes
a) ρ i < 1
b) ρ i < 0
c) ρ i > 0
d) ρ i > 1
Answer: c
Explanation: According to boundedness condition, flow properties should lie in the range. So, the non-negative properties like density, temperature and kinetic energy should always be positive. Therefore, ρ i > 0.
10. Flow properties are not bounded by the boundary values __________
a) in the absence of convection term
b) in the absence of source term
c) in the presence of source term
d) in the absence of convection term
Answer: c
Explanation: When there is a source or sink inside the domain of interest, the flow properties may increase or decrease drastically. This may lead to a value which does not lie in the range of the boundary values. So, a system with a source or sink will not be bounded.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Discretization Aspects – Transportiveness”.
1. Transportiveness has an influence on _________
a) the discretization scheme
b) the solution method for an algebraic system of equations
c) the mathematical model
d) the iterative scheme
Answer: a
Explanation: Transportiveness is borne by the discretization scheme. An appropriate method of discretization should be chosen based on the transportiveness of the system.
2. Which of these is related to the transportiveness?
a) Courant number
b) Reynolds number
c) Nusselt number
d) Peclet number
Answer: d
Explanation: Peclet number defines the transportiveness of a fluid flow property. The Peclet number has a direct impact on the isolines of a flow property around a specific node.
3. Peclet number is a ratio of _________ strength to the _________ strength.
a) Diffusive, convective
b) Convective, diffusive
c) Radiative, diffusive
d) Diffusive, radiative
Answer: b
Explanation: Peclet number defines how much convection flux dominates the diffusion flux of fluid flow. It will decide if the transport of flow property is because of convection or diffusion.
\(Peclet\, number=\frac{Convective\, strength}{Diffusive\, strength}\).
4. When the Peclet number is zero, the isolines are ___________
a) hyperbolic
b) elliptic
c) circular
d) parabolic
Answer: c
Explanation: When the Peclet number is zero, the isolines of flow property are circular with the current node at its centre. Here, diffusive strength dominates.
5. When the Peclet number is large, the isolines are ___________
a) hyperbolic
b) circular
c) elliptic
d) parabolic
Answer: c
Explanation: When the Peclet number is large, convection dominates the flow of a property. Here, the isolines will be elliptic with the current node at its focus.
6. When the Peclet number is zero, the value of flow property at the current node is influenced by ___________
a) the upstream node more
b) both the upstream and downstream nodes equally
c) the downstream node more
d) neither the upstream nor the downstream nodes
Answer: b
Explanation: When the Peclet number is zero, flow direction does not affect the value of flow property. Both the upstream and the downstream nodes have equal influence on the value at the current node.
7. When the Peclet number is large, the value of flow property at the current node influences ___________
a) the upstream node more
b) both the upstream and downstream nodes equally
c) neither the upstream nor the downstream nodes
d) the downstream node more
Answer: d
Explanation: The impact of flow property is affected by the flow direction when the Peclet number is large. The current node is affected by the upstream node and it affects the downstream node.
8. When transportiveness is not accounted in the discretization scheme, the solution becomes ___________
a) unstable
b) non-converging
c) inaccurate
d) non-conservative
Answer: a
Explanation: Transportiveness must be taken care of while choosing the discretization scheme. If not, solutions will have unphysical oscillations and become unstable.
9. Why are isolines circular when the Peclet number is zero?
a) Fluid is flowing and diffusion spreads equally
b) Fluid is stagnant and diffusion spreads equally
c) Fluid is stagnant and diffusion spreads directionally
d) Fluid is flowing and diffusion spreads directionally
Answer: b
Explanation: When the Peclet number is zero, the problem is dominated by diffusion. This means that the fluid is stagnant. Diffusion allows the flow of property to spread in the domain equally in all the directions.
10. Which of these schemes will suit a flow with a low Peclet number?
a) Iterative schemes
b) Backward differencing scheme
c) Central differencing scheme
d) Forward differencing scheme
Answer: c
Explanation: When the Peclet number is low, the central differencing scheme can be used. This is because, in this case, the flow variable is affected by both of the neighbouring nodes equally.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Discretization Aspects – Errors and Stability Analysis”.
1. Which of these conditions is unstable?
a) Error is amplified in increasing iterations
b) Error is decreasing in increasing iterations
c) Error is amplified in decreasing iterations
d) Error is maintained in increasing iterations
Answer: a
Explanation: A system is said to be unstable if the error increases with the increasing iterations. So, for a system to be stable, the error should decrease or at least should be maintained in further iterations.
2. The difference between the exact analytical solution of a partial differential equation and its numerical solution is as ___________
a) round-off error
b) discretization error
c) iteration error
d) modelling error
Answer: b
Explanation: Discretization error is the error arising due to the numerical solution of a partial differential equation. It includes the error due to the numerical approximation of the governing equations and the boundary conditions.
3. The error due to the discretization of the partial differential equation is called as ______________
a) round-off error
b) discretization error
c) truncation error
d) iteration error
Answer: c
Explanation: Truncation error arises when partial differential equations are approximated. Usually, the partial differential equations are approximated using a series expansion of infinite terms. The higher order terms are cut-off while approximating this series. So, it is called truncation error.
4. If the order of a discretized equation is ‘k’, what does it mean?
a) The last term of the equation is of th power
b) The last term of the equation is of k th power
c) Truncation error is proportional to th power
d) Truncation error is proportional to k th power
Answer: d
Explanation: When a discretized equation is said to be of the order ‘k’, it means that the last term has th power and the first term of the truncated equation has k th power. For example, in spatial derivative, the first term of the truncated equation will be proportional to k .
5. ___________ become significant after a repeated number of calculations.
a) Round-off errors
b) Discretization errors
c) Truncation errors
d) Modelling errors
Answer: a
Explanation: Round-off errors are introduced because of the round-off results produced by computers for a particular problem. When these round-off values are used for further calculations, they become significant after a certain time.
6. Round-off errors are important in ____________
a) modelling
b) iterations
c) discretization
d) truncation
Answer: b
Explanation: Iterations are where the same process is repeated with the last generated value. This way, while iterating, rounding off the results become significant as they affect further iterations. Round-off errors are aggregated here.
7. Which of these iterative processes is unstable? (Note: \(\epsilon_n\) is the error in the n th iteration).
a) \
\
\
\(\frac{\epsilon_{n+1}}{\epsilon_n} = 1\)
Answer: c
Explanation: For a system to be stable, the error should be decreasing or at least maintained.
Representing this mathematically,
ε n+1 ≤ ε n
\(\frac{\epsilon_{n+1}}{\epsilon_n} ≤ 1\)
For an unstable system,
\(\frac{\epsilon_{n+1}}{\epsilon_n} > 1\)
This happens in \(\frac{\epsilon_{n+1}}{\epsilon_n} = 1.25\). So, this system is unstable.
8. In Von Neumann analysis, the solution is expanded using ____________
a) Laurent series
b) McLaurin series
c) Taylor series
d) Fourier series
Answer: d
Explanation: Von Neumann stability analysis is otherwise called as Fourier stability analysis. It is a technique used to analyse the stability of linear partial differential equations. It is named as Fourier stability analysis as it is based on Fourier decomposition of numerical error.
9. The error occurring while approximating the physical problem is called as ____________
a) Modelling error
b) Physical error
c) Mathematical order
d) Iteration error
Answer: a
Explanation: The difference between the physical flow and the exact solution of the mathematical model is termed modelling error. This occurs because of approximating the physical model. It is not possible to model the exact physical scenario happening.
10. Tolerance for iteration errors is usually based on _______________
a) convergence
b) residuals
c) stability
d) round-off error
Answer: b
Explanation: A tolerance should be set for the convergence of iterations to say when to stop the iterations. To set this, tolerance residuals are analysed. Residual directly quantifies the error in the solution of a system.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Discretization Aspects – Grid Generation”.
1. Which of these analyses needs a stretched grid?
a) Transient flow over a flat plate
b) Incompressible flow over a flat plate
c) Viscous flow over a flat plate
d) Subsonic flow over a flat plate
Answer: c
Explanation: A stretched grid has less grid spacing in one side of the domain and stretched in the other side. For viscous flow over a flat plate, the boundary layer will have much variations of the flow properties than the far regions. So, for modelling this, stretched grids can be used.
2. Let x, y be the coordinates in the physical domain and ξ, η be the coordinates in the computational domain. In which of these cases, the horizontal lines are stretched and the vertical lines are equally spaced?
a) ξ=x; η=ln
b) ξ=ln; η=y
c) ξ=x; η=y
d) ξ=ln; η=ln
Answer: a
Explanation: The horizontal coordinate lines are stretched means that the grids are stretched in the y-direction. Coordinates are stretched when logarithmically spaced. So, to get vertically stretched grid, η=ln. Horizontal coordinates are undisturbed.
3. Form the continuity equation for steady 2-dimensional flow when the x-direction grids are stretched.
Density → ρ
x and y-velocities → u,v
Coordinates in physical domain → x,y
Coordinates in computational domain → ξ, η.
a) \
\
\
\(e^\xi\frac{\partial}{\partial\xi}+\frac{\partial}{\partial\eta}\)
Answer: b
Explanation: The continuity equation for steady 2-dimensional flow is:
\
⇒ x = e ξ -1 ⇒∂x = e ξ ∂ξ
η = y ⇒ y = η ⇒ ∂y = ∂η
Converting the continuity equation to computational domain terms,
\(\frac{\partial}{\partial\xi}\frac{\partial\xi}{\partial x} + \frac{\partial}{\partial \eta} \frac{\partial\eta}{\partial x} + \frac{\partial}{\partial\xi}\frac{\partial\xi}{\partial y}+\frac{\partial}{\partial\eta}\frac{\partial \eta}{\partial y}=0\)
\(\frac{\partial}{\partial\xi}\frac{1}{e^\xi} + \frac{\partial}{\partial \eta}1=0\)
\(\frac{\partial}{\partial\xi} + e^ξ\frac{\partial}{\partial \eta} =0\).
4. Consider a divergent nozzle as shown in the figure.
computational-fluid-dynamics-questions-answers-grid-generation-q4
Let x, y be the coordinates in the physical domain and ξ, η be the coordinates in the computational domain. Which of these equations can give the best-suited grid for this system?
a) ξ=x; η = y×y s
b) ξ=x×y s ; η=y×y s
c) \
\(\xi=x;\eta=\frac{y}{y_s}\)
Answer: d
Explanation: The given divergent nozzle needs a boundary-fitted grid to suit its divergent nature. The variation exists only in the y-direction. The x-directions grids can be uniformly spaced. So, the best-suited coordinates are \(\xi=x;\eta=\frac{y}{y_s}\).
5. Adaptive grids change automatically based on ______________
a) flow field gradients
b) time rate of change of the flow properties
c) grid gradients
d) time rate of change of the grid points
Answer: a
Explanation: Adaptive grids are the one which can automatically adapt their arrangement based on the needs of the problem. This is based on the flow-field gradients. High flow-field gradients indicate a sudden variation of properties based on spatial coordinates. So, more grids are automatically generated there.
6. Let x, y be the coordinates in the physical domain and ξ, η be the coordinates in the computational domain. Which of these is correct for adaptive grids?
a) \
\
\
\(\frac{\partial\xi}{\partial t}≠1 \)
Answer: c
Explanation: Adaptive grids change with varying time. So, the time rate of change of coordinates will never be equal to zero. This is given by\(\frac{\partial\xi}{\partial x}≠0 \). They may or may not vary from the physical coordinates.
7. Which of these properties are balanced by using adaptive grids?
a) Accuracy and convergence
b) Efficiency and stability
c) Accuracy and stability
d) Accuracy and efficiency
Answer: d
Explanation: By automatically generating the grid points, the places without much variation are given fewer grids and the places with high variation are given more grids. This helps in maintaining a balance between efficiency of the solution and accuracy of the answers.
8. What does elliptic grid generation mean?
a) Grids generated for elliptic equations
b) Grids transformed with elliptic equations
c) The computational domain is elliptic
d) The object under consideration is elliptic
Answer: b
Explanation: The one-to-one transformation of the coordinates in the physical domain to the coordinates in the computational domain is done using elliptic equations. It has nothing to do with elliptic nature of the governing equations.
9. What are zonal grids?
a) Grids generated for a particular zone of the domain of interest
b) Grids varying at different zones
c) Grids generated for a particular time in the flow
d) Grids varying with time
Answer: a
Explanation: While dealing with a complex problem, grids are generated separately for different zones which are called zonal grids. These are then attached together to form grids for the whole domain.
10. Which type of grids is the best for flow over an airfoil?
a) Stretched grids
b) Adaptive grids
c) Boundary-fitted grids
d) Elliptic grids
Answer: c
Explanation: The shape of the airfoil is complex. So, generating a grid that fits the boundary of this airfoil is best. Stretched grids are for flat surfaces. Adaptive grids are needed if the flow gradients are not known to the user. Elliptic grids are useful when the transformation is elliptic.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Numerical Methods – Components”.
1. Which is the first step in the numerical solution of a fluid flow problem?
a) Discretization
b) Physical model of the flow
c) Mathematical model of the flow
d) Iteration
Answer: c
Explanation: The first step of any numerical solution of a fluid problem is converting the physical flow into a mathematical model. Physical model of the flow is what we have to solve. After generating the mathematical model only steps like discretization and iterative solution follows.
2. What does the mathematical model of a fluid flow contain?
a) Partial differential equations
b) Discretized partial differential equations
c) Partial differential equations and boundary conditions
d) Discretized partial differential equations and boundary conditions
Answer: c
Explanation: After generating the mathematical model of the physical flow in a problem, we will have a set of partial differential equations along with its boundary conditions. The mathematical model is not complete without the boundary conditions which make the problem unique.
3. Choosing a particular type of discretization method is ineffective when ___________
a) mathematical model is complex
b) mathematical model is simple
c) grid is coarse
d) grid is very fine
Answer: d
Explanation: When the grid size is very small, whatever the type of discretization method is, the results will be the same. As very fine grids are not practically acceptable, we choose a particular type of discretization method which will be the best fit for the problem.
4. The mathematical model is based on ____________
a) physical principles and assumptions
b) physical principles
c) flow model
d) flow model and assumptions
Answer: a
Explanation: For generating the mathematical model, first of all, the physical principles which are applicable to the given flow should be taken. Along with these some assumptions also must be made to make the model suit the mathematical solution. These assumptions result in modelling errors.
5. Express the 2-dimensional continuity equation in cylindrical coordinates.
a) \
\
\
\(\frac{\partial
}{\partial r}+\frac{1}{r}\frac{\partial
}{\partial\theta}+\rho \frac{v_r}{r}+\frac{\partial\rho}{\partial t}=0\)
Answer: b
Explanation: In Cartesian coordinates, radial and angular velocities replace the x and y velocity components. Similarly, is the coordinate system used here. The continuity equation in this system can be given by \(\frac{\partial
}{\partial r}+\frac{1}{r}\frac{\partial
}{\partial\theta}+\rho \frac{v_r}{r}+\frac{\partial\rho}{\partial t}=0 \).
6. Each node has 4 nearest neighbours. This statement is correct for which of these grid types?
a) Structured 2-D grids
b) Unstructured 2-D grids
c) Structured 3-D grids
d) Unstructured 3-D grids
Answer: a
Explanation: Structured grids have 2 nearest neighbours in 1-D, 4 in 2-D and 6 in 3-D. There is no standard number of nearest neighbours in the unstructured grid type except 1-D case where there is no option for the grids to have more than two neighbours.
7. Which of these features of structured grids is a disadvantage?
a) Easy to solve
b) Suitable for simple geometries
c) Efficient in memory requirements
d) Less time requirement
Answer: b
Explanation: One of the major disadvantages of structured grids is that they are not suitable for complex geometries. Those can be modelled using unstructured grids only. The other disadvantage of structured grids is distribution.
8. Which of these grids are called Chimera grids?
a) Structured grids with overlapping blocks
b) Block-structured grids
c) Block-structured grids with overlapping blocks
d) Structured grids
Answer: c
Explanation: Block-structured grids with overlapping blocks are called composite or chimera grids. The disadvantage of these grids is that conservation is not ensured in boundaries. This is helpful to follow moving bodies.
9. While using a Finite Element Method, one has to approximate ____________
a) boundary conditions
b) integrals at grid faces
c) derivatives at grid points
d) shape functions and weighting functions
Answer: d
Explanation: While using the Finite Element Method for solving a problem, shape functions and weighting functions are approximated. Integrals at grid faces are approximated for Finite Volume Methods. Derivatives at grid points are approximated for Finite Difference Methods.
10. Which of these coordinates are not used in CFD?
a) Orthogonal coordinates
b) Cartesian coordinates
c) Spherical coordinates
d) Number line
Answer: d
Explanation: The type of coordinate system can be opted between Cartesian, Cylindrical, Spherical, Curvilinear orthogonal and non-orthogonal. A Number line cannot be used as a coordinate system in CFD.
This set of Computational Fluid Dynamics test focuses on “Numerical Methods – Mesh Topology”.
1. Which of these is related to the flux terms?
a) Element connectivity
b) Node connectivity
c) Face connectivity
d) Vertex connectivity
Answer: c
Explanation: For faces, information about elements sharing the face is stored for computational uses. Since the flux terms completely depend upon the faces, face connectivity should be in a way that it represents flux.
2. The direction of the normal vector to a face in mesh ____________
a) is from the owner element to the neighbour element
b) is from the neighbour element to the owner element
c) is in the direction of the flux
d) is in the direction opposite to the flux
Answer: a
Explanation: The orientation of the faces is such that the normal vector to the face points from the owner element to the neighbour element. Depending on this the sign of flux term changes. Orientation does not depend on fluxes.
3. In the boundary faces, the normal vector points _____________
a) to the owner element
b) outside the domain
c) in the direction of the flux
d) in the direction opposite to the flux
Answer: b
Explanation: The boundary faces bound only one element and they do not have any neighbouring elements. So, the owner elements in the boundaries have their faces with the normal vectors pointing outside the domain.
4. Which of these points is shared by the maximum number of elements?
a) Grid point
b) Cell centre
c) Face centre
d) Vertex
Answer: d
Explanation: A vertex is shared by the most number of elements. In structured grids, a vertex point is shared by eight elements. Gridpoint can be either the cell centre or the vertex depending upon the type of discretization.
5. Vertex connectivity is important while __________
a) solving the discretized equation
b) discretizing
c) post-processing
d) pre-processing
Answer: c
Explanation: Vertex connectivity is important for post-processing especially while computing gradients. Vertex connectivity generally contains information like the elements and faces sharing that vertex.
6. Which of these establish a one-to-one relationship between two elements?
a) Face
b) Node
c) Vertex
d) Element
Answer: a
Explanation: A one-to-one relationship is given by the faces of a mesh. One face is shared by two elements. It usually contains information about the fluxes flowing between those elements.
7. Element connectivity is responsible for ___________
a) consistency of the fluxes of different elements
b) consistency of the equations formed for different elements
c) flow field variables
d) gradient of the flow field variables
Answer: b
Explanation: Element connectivity relates the local matrix to the global matrix. This ensures that the equations formed for one element are consistent with those formed for the other elements in the computational domain.
8. In a two-dimensional flow, the algebraic equation of an element relates the element with ___________
a) its face centres
b) its vertices
c) its faces
d) its neighbours
Answer: d
Explanation: Each element in a mesh has its own algebraic equation. This algebraic equation has the coefficients of the neighbouring elements too. This way, the elements are connected to their neighbours by these algebraic equations.
9. The aspect ratio of each element should be ___________
a) less than one
b) equal to one
c) around one
d) greater than one
Answer: c
Explanation: Ideally, the aspect ratio of each element should be equal to 1. But, this cannot be practically ensured. So, in real, the elements have their aspect ratio around 1. If it is large, it will lead to errors.
10. The variation of the size of a cell from an optimal cell size is its __________
a) centroid
b) structure
c) orthogonality
d) skewness
Answer: d
Explanation: The skewness of a cell is its variation from the optimal cell size. This is an apt indicator of the quality and suitability of a mesh. Large skewness leads to less accuracy.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Numerical Methods – Discretization Approaches”.
1. The diagram represents a one-dimensional mesh.
computational-fluid-dynamics-questions-answers-discretization-approaches-q1
The conservation equations are applied to which of these points while discretizing the equation using the Finite Difference Method?
a) 5, 7
b) 2, 3
c) 5, 6, 7
d) 1, 2, 3, 4
Answer: d
Explanation: The Finite Difference Method applies conservation equation to the nodes of each cell. The vertices of the cells form the nodes. So, points 1, 2, 3 and 4 are used for discretizing.
2. Which of these is the oldest method for numerical solution of partial differential equations?
a) Finite Element Method
b) Finite Difference Method
c) Finite Volume method
d) Spectral Element Method
Answer: b
Explanation: The Finite Difference Method is the oldest method for solving partial differential equations numerically. It is believed that this method is developed by Euler in the 18th century. This is the easiest method too.
3. Which of these properties is not ensured in Finite Difference Methods?
a) Stability
b) Convergence
c) Conservativeness
d) Accuracy
Answer: c
Explanation: Though the Finite Difference method is the easiest, it is suitable only for simple problems involving structured grids. Another main disadvantage is that conservation is not enforced unless special care is taken.
4. To obtain the derivatives of the variables with respect to the coordinates, which of these approximations are used in the Finite Difference Method?
a) Taylor series and polynomial fitting
b) Fourier series and polynomial fitting
c) Taylor series and Fourier series
d) Taylor series and interpolation
Answer: a
Explanation: The conservation equation of the partial differential form is used by the Finite Difference Method. To approximate the derivatives in this equation, Taylor series and polynomial fitting are used.
5. Which of these is an advantage of the Finite Difference Method over the Finite Volume Method?
a) Conservativeness
b) Higher-order
c) Stability
d) Complex problems
Answer: b
Explanation: The Finite Volume Method cannot be applied to higher orders. The disadvantage of the Finite Volume Method, when compared to the Finite Difference Method, is that for orders higher than second order are more difficult to develop in 3-D.
6. The diagram represents a one-dimensional mesh.
computational-fluid-dynamics-questions-answers-discretization-approaches-q1
The conservation equations are applied to which of these points while discretizing the equation using the Finite Volume Method?
a) 5, 7
b) 2, 3
c) 5, 6, 7
d) 1, 2, 3, 4
Answer: c
Explanation: Governing equations are applied to each control volume at its centre in the Finite Volume Method. In the diagram, points 5, 6 and 7 represent cell centres. To get the nodal values, these answers are interpolated.
7. Conservation of Finite Volume Method depends on ___________
a) surface integrals
b) volume integrals
c) convection
d) diffusion
Answer: a
Explanation: The Finite Volume Methods are conservative as long as the surface integrals which represent the convective and diffusive fluxes are the same for two control volumes sharing the same boundary.
8. A hybrid method of which of these methods exists?
a) Finite Difference Method and Finite Element Method
b) Finite Volume Method and Finite Difference Method
c) Finite Volume Method and Finite Element Method
d) Finite Volume Method and Spectral Element Method
Answer: c
Explanation: A hybrid method integrating Finite Volume Method and Finite Element Method called the Control Volume based Finite Element Method is also used for solving PDEs. In this, shape functions are used to describe the variation of the variables over an element.
9. In which of these methods, the calculation for interior nodes is done at the post-processing stage?
a) Spectral element method
b) Boundary element method
c) Finite Element method
d) Finite Volume Method
Answer: b
Explanation: In Boundary element method, the given boundary conditions are used to fit the boundary values into the integral equations. After this, using the boundary solutions, the solution for the interior nodes is carried out at the post-processing stage.
10. Which of these statements are wrong considering Control Volume based Finite Element Method?
a) Control volumes are formed
b) Integral form of the conservation equation is used
c) The centroid of the elements is used
d) Conservation equations are applied to the centroids
Answer: d
Explanation: In CV-FEM, control volumes are formed by joining the centroids of the elements. The conservation equation in the integral form is applied to these control volumes as in the Finite Volume Method.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Numerical Methods – Variable Arrangements and Velocity Components”.
1. Which of these coordinate systems will best suit to model the flow in a pipe with a swirl?
a) Cylindrical coordinates
b) Cartesian coordinates
c) Spherical coordinates
d) Polar coordinates
Answer: a
Explanation: In the case of flow over a pipe, cylindrical coordinates do the best as the dependent spatial variables are just 2 which would be 3 if it is Cartesian coordinates. But, in this case, a swirl component is included which makes the flow dependent on three dimensions anyway. To keep the complexities less, Cartesian coordinates should be chosen for this problem.
2. While using grid oriented velocity components, conservation of which of these equations will be lost?
a) Energy equation
b) Euler equations
c) Momentum equation
d) Continuity equation
Answer: c
Explanation: If grid-oriented velocity components are used, non-conservative source or sink terms will appear in Momentum equations. This will affect the conservation of Momentum equation. Euler equations is a set including all three – continuity, momentum, energy equations.
3. In the polar-cylindrical momentum equation for the r-component, the term \Missing open brace for subscript Centrifugal force
b) Centripetal force
c) Tangential force
d) Coriolis force
Answer: a
Explanation: When the momentum equation is expressed in cylindrical terms, due to the transformation from Cartesian coordinates to polar coordinates, a centrifugal term arises. This is the term \(\frac{\rho v \theta^2}{r}\). It describes the transfer of θ-momentum into r-momentum due to the change of direction of angular velocity.
4. Which of these terms in the momentum equation for θ – component represents the Coriolis force?
a) \
\
\
\(\frac{\partial v_\theta}{\partial r}\)
Answer: c
Explanation: The term \(\frac{\rho v_r v_\theta}{r}\) represents Coriolis force in the θ – momentum equation. Coriolis force arises due to the motion of the coordinate system taken. The term \(\frac{\rho v_r v_\theta}{r}\) represents the source or sink of θ -momentum.
5. Curvilinear coordinates do not suit __________
a) Unstructured grids
b) Structured grids
c) Orthogonal grids
d) Non-orthogonal grids
Answer: a
Explanation: For curvilinear coordinates, the grid should be very smooth and the change of grid direction from point to point must be very small. Since the grid direction in the unstructured grids varies much, they do not suit unstructured grids.
6. While changing the coordinates from Cartesian to non-orthogonal in the Finite Difference Method, which of the following remains the same?
a) all the terms in the equation
b) conservation properties of the equation
c) number of terms in the equation
d) source terms of the equation
Answer: b
Explanation: In the Finite Difference Method, while changing from Cartesian to non-orthogonal grids, the terms change and the number of terms increases. But, the conservation properties of the equation remain the same.
7. Staggered grid arrangements are used to establish strong coupling between ____________ and __________
a) velocities and momentum components
b) velocities and pressure gradients
c) velocities and viscosity terms
d) viscosity and source terms
Answer: d
Explanation: Coupling problem between velocities and pressure gradients is the reason why staggered grids are formed. Staggered grids ensure the coupling between these two terms. The velocity components normal to the cell face should lie between the pressure coordinates on either side of that face.
8. Which of these arrangements of the velocity and pressure gradient components is suitable for non-orthogonal grids?
a) Staggered arrangement with covariant base
b) Collocated arrangement
c) Staggered arrangement
d) Covariant arrangement
Answer: b
Explanation: The collocated arrangement is the simplest as all the variables share the same control volume. But this requires more interpolation. When the grid is non-orthogonal, the collocated arrangement is the best as the other arrangements are more difficult.
9. Which of these flow properties does not suit covariant or contra-variant bases?
a) Velocity
b) Density
c) Stress
d) Viscous forces
Answer: b
Explanation: Covariant or contra-variant bases can be used only with vectors or tensors. They cannot be used with scalars. Here, Velocities stress and forces all come under either vectors or tensors. Density is a scalar property. So, it cannot be represented using covariant or contra-variant bases.
10. Stress components cannot be expressed in terms of __________
a) Cartesian tensors
b) Contra-variant tensors
c) Covariant tensors
d) Metric tensors
Answer: d
Explanation: In general, Stress components are expressed in the Cartesian tensors form. It is also possible to define them in covariant or contra-variant terms. But they cannot be expressed in metric tensor. Metric tensors are used to transform covariant tensors into contra-variant tensors.
This set of Computational Fluid Dynamics Interview Questions and Answers for Experienced people focuses on “Numerical Methods – Direct Solvers for Discretized Equations”.
1. The coefficient matrix in the global matrix format of the algebraic equations is a ______________
a) sparse matrix
b) dense matrix
c) diagonal matrix
d) symmetric matrix
Answer: a
Explanation: Each row in the coefficient matrix represents an equation corresponding to a single node. This equation contains coefficients of the neighbouring elements only depending on the element connectivity of the owner element. So, all other elements of that row become zero. This makes the coefficient matrix sparse.
2. Which of these direct methods is suitable only for banded matrices?
a) Gauss elimination
b) LU decomposition
c) PDMA
d) LU decomposition by Gauss elimination
Answer: c
Explanation: PDMA stands for Penta-Diagonal Matrix Algorithm. This is used for solving a Penta-diagonal matrix which has non-zero elements only in its main diagonal and two diagonals above and below it. A Penta-diagonal matrix is a sparse banded matrix.
3. After the forward elimination step of the Gauss elimination method, the coefficient matrix is reduced to ____________
a) a lower triangular matrix
b) an upper triangular matrix
c) a diagonal matrix
d) a banded matrix
Answer: b
Explanation: After the forward elimination step, the main diagonal elements and the elements above it are non-zeros. All the elements below the main diagonal are zeros. So, we can say that it gives an upper triangular matrix to be solved.
4. Which of these formulae is used in the backward substitution step of Gauss-elimination method?
Note: The global matrix is
AΦ=b
i → row number.
j → column number.
N→ Number of unknowns.
a) \
\
\
\(\Phi_i=b_i-\frac{\sum_{j=i+1}^NA_{ij}\phi_j}{A_{ii}} \)
Answer: a
Explanation: The system is solved backwards from the last step. Using the previous value of φ, the current value is found. The formula is
\(\Phi_i=\frac{b_i-\sum_{j=i+1}^N a_{ij}\phi_j}{A_{ii}} \).
5. If N is the number of unknowns, the number of operations required for backward substitution is ____________
a) \
\
\
\(\frac{N^2}{3}\)
Answer: c
Explanation: The number of operations required for the overall Gauss elimination method is \(\frac{N^3}{3}\). The number of operations for backward substitution is \(\frac{N^2}{2}\). This high computational cost is the disadvantage of the Gauss elimination method.
6. Which of these is true for the LU decomposition method?
a) LΦ=Ub
b) LUΦ=b
c) Φ=LUb
d) UΦ=Lb
Answer: b
Explanation: For the LU decomposition method,
A=LU
Where L and U stand for Lower and Upper triangular matrices respectively.
Substituting in the global matrix,
LUΦ=b.
7. The number of operations for LU decomposition method is ____________ the number of operations for the Gauss elimination method.
a) twice
b) half of
c) one-third of
d) thrice
Answer: a
Explanation: LU decomposition is computationally more expensive than the Gauss elimination method. The number of operations involved is \(\frac{2N^3}{3}\). This is because the same process of Gauss elimination is carried out twice in the LU decomposition case.
8. Consider the global matrix AΦ=b. If my coefficient matrix A is the same for different b vectors, which of these methods is economic?
a) Gauss elimination
b) TDMA
c) LU decomposition
d) PDMA
Answer: c
Explanation: Once matrix A is factorized in the LU decomposition method, the same factorized matrices can be used to solve different b vectors. Because decomposition does not depend upon the b vector. This is the major advantage of the LU decomposition method.
9. TDMA and PDMA are particularly suitable for _____________
a) Cartesian grid problems
b) Curvilinear grid problems
c) Unstructured grid problems
d) Structured grid problems
Answer: d
Explanation: When a structured grid is used for discretization, it results in a coefficient matrix with its non-zero elements aligning along a few diagonals. The number of non-zero diagonals depends on the discretization stencil and the dimension of the problem. So, TDMA and PDMA are suitable to solve this kind of banded matrix.
10. The general equation for PDMA is a i Φ i +b i Φ i+2 +c i Φ i+1 +d i Φ i-1 +e i Φ i-2 =f i . Which of the following is incorrect?
.
a) e 2 =0
b) c N-1 =0
c) e 1 =0
d) c N =0
Answer: b
Explanation: By analysing the equation for PDMA, for the first two equations,
d 1 =e 1 =e 2 =0
Similarly, for the last two equations,
b N-1 =b N =c N =0.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Numerical Methods – Iterative Solvers for Discretized Equations”.
1. Which of these statements is incorrect about iterative methods?
a) Low computational cost
b) Low computer storage
c) Not suitable for sparse matrices
d) Needs initial guess
Answer: c
Explanation: Iterative methods are chosen when the coefficient matrix is sparse . They need less storage and less computational cost. They start the solution from an initial guess and proceed to find the answer from this initial guess.
2. Let us divide the coefficient matrix into this form: A=D+L+U. Which of these matrices can be L in this equation?
a) \
\
\
\(
\)
Answer: b
Explanation: In A=D+L+U, A is the coefficient matrix; D is a diagonal matrix; L is a strict lower triangular matrix; U is a strict upper triangular matrix. A strict lower triangular matrix contains non-zero elements only below its main diagonal.
3. Consider the global matrix AΦ=b. Let the coefficient matrix A=M-N. At the i th iteration, the general form can be given as ______________ (Note: Φ i is the value of Φ in the i th iteration).
a) MΦ i =NΦ i-1 +b
b) MΦ i =NΦ i +b
c) MΦ i =NΦ i-1 -b
d) MΦ i =NΦ i -b
Answer: a
Explanation: The global matrix is
AΦ=b
Replacing A with M-N,
Φ=b
MΦ=NΦ+b
At a particular step, we only know the Φ values of the previous step only. So,
MΦ i =NΦ i-1 +b.
4. Consider the global matrix AΦ=b. What is the residual at the ith iteration? (Note: Φ i is the value of Φ in the i th iteration).
a) AΦ i-1 +b
b) AΦ i-1 -b
c) AΦ i +b
d) AΦ i -b
Answer: d
Explanation: While solving the system AΦ=b, AΦ-b should be equal to zero. Since, the iterative method will not produce such an answer, the decision when to stop the iteration relies upon a tolerance value. When the residual AΦ i -b becomes less than the tolerance, iterations are stopped.
5. When compared to the Jacobi method, the Gauss-Siedel method ____________
a) has better convergence and needs less memory
b) has less convergence and needs more memory
c) has less convergence and needs less memory
d) has better convergence and needs more memory
Answer: a
Explanation: Gauss-Seidel method uses the latest values at a particular iteration. So, it has better convergence. The same way, as there is no need for storing the values of previous iterations, they require less memory too.
6. Preconditioners improve ____________ of the coefficient matrix.
a) sparsity
b) rank
c) spectral properties
d) Singularity
Answer: c
Explanation: For a system of equations to have a better rate of convergence, the coefficient matrix should have a less spectral radius. So, Preconditioners are used to improve the spectral characteristics of a system to give the same solution.
7. What does the letter ‘I’ stand for in ILU decomposition problem?
a) Inbuilt
b) Incomplete
c) Iterative
d) Imbalanced
Answer: b
Explanation: ILU means Incomplete LU decomposition method. This does incomplete factorization of the coefficient matrix into upper and lower triangular matrices. The L and U matrices have the same number of non-zero elements as in the lower and upper parts of A.
8. In which of these methods, after factorization, the pattern of zero elements in the combined L and U matrices the same as the original coefficient matrix?
a) LU decomposition
b) ILU decomposition
c) ILU decomposition
d) ILU decomposition
Answer: a
Explanation: The extra non-zero elements in the factorized matrices are called fill-ins. In ILU, p indicates the order of fill-in allowed. So, where there is no fill-in, the pattern of combined L and U matrices and the coefficient matrix will be the same.
9. Which of these methods is not restricted to symmetric positive definite matrices?
a) The method of steepest descent
b) Conjugate gradient method
c) Bi-conjugate gradient method
d) Gauss-Seidel method
Answer: d
Explanation: The gradient iterative solvers are restricted to symmetric positive definite matrices. The gradient methods are:
The method of Steepest descent
Conjugate gradient method
Bi-conjugate gradient method
Preconditioned bi-conjugate gradient method.
10. In incomplete Cholesky decomposition, the preconditioner matrix produced is ____________ .
a) UL’
b) LU’
c) LL’
d) LL 2
Answer: c
Explanation: In incomplete Cholesky decomposition, factorization is done only for the lower triangular matrix and the preconditioner matrix is LL’. The coefficient matrix is also approximately equal to the preconditioner matrix.
This set of Computational Fluid Dynamics Quiz focuses on “Numerical Methods – Multi-grid Approach for Solving Discretized Equations”.
1. The multi-grid approach is used to assist __________
a) iterative solvers
b) direct solvers
c) gradient solvers
d) pre-conditioned solvers
Answer: a
Explanation: When the iterative solvers are used to solve medium to large system of equations, the error is big and accuracy becomes less. This poses a problem with iterative solvers. So, they are supplemented with the multi-grid approach.
2. Which of these properties are affected when the multi-grid approach is not used?
a) Conservativeness
b) Convergence
c) Consistency
d) Stability
Answer: b
Explanation: As the accuracy in the iterative solvers for large equations are not good, the rate of convergence is very less. A solution to this problem is given by the multi-grid approach.
3. Which of these errors need a multi-grid approach?
a) Low amplitude error
b) High amplitude error
c) Low frequency error
d) High frequency error
Answer: c
Explanation: High frequency oscillatory errors are easily eliminated using iterative methods like Jacobi and Gauss-Seidel. But, these techniques cannot handle smooth and low frequency errors without a multi-grid approach.
4. Multi-grid approach switches between ___________ and ____________ grids to meet the errors.
a) structured and unstructured
b) collocated and staggered
c) cylindrical to polar
d) fine and coarse
Answer: d
Explanation: The low frequency errors create problem in fine grids as the error in one cell is very small and negligible because of the high wavelength of the errors. So, multi-grid approach changes these fine grids into coarser one to make it considerable in one cell.
5. The multi-grid approach is a ___________ process.
a) direct
b) iterative
c) cyclic
d) periodic
Answer: c
Explanation: The multi-grid approach involves traversal from a fine grid into a coarse one in order to make the error considerable and again another traversal from a coarse grid into a fine one after the error correction.
6. Let r k be the residual in the k th level of multi-grid approach. Which of these give the restriction operator?
a) \
\
r k+1 -r k
d) r k -r k+1
Answer: a
Explanation: The first step in the multi-grid approach is the restriction step. Here, the process starts with a fine grid. After a few iterations, the error is transferred to a coarser grid level. Again, some iterations are performed in that step and the process continues. The restriction operator for the error transferred from one step to the higher step is \(\frac{r^{k+1}}{r^k} \).
7. For an algebraic multi-grid approach, the residual in the k th level is __________
a) residual in the th level
b) summation of the residuals in the th level
c) residual in the th level
d) summation of the residuals in the th level
Answer: d
Explanation: While transferring the errors from one level to another, the residual of the k th level is given by the summation of the residuals of all the terms in the previous th level.
8. Which of these traversal cycles are possible for an algebraic multi-grid approach?
a) W-cycle
b) V-cycle
c) U-cycle
d) F-cycle
Answer: c
Explanation: There are three possible traversal cycles for algebraic multi-grid approaches. They are W-cycle, V-cycle and F-cycle. V-cycle is a direct traverse without any nesting. W-cycle involves nesting. F-cycle is a hybrid cycle of W and V-cycles.
9. Errors are transferred from a fine grid to a coarser one. Similarly __________ is transferred from a coarse grid to a finer one.
a) residual
b) correction
c) restriction
d) prolongation
Answer: b
Explanation: Corrections are transferred from a coarse grid to a finer one. Correction is basically obtained from the solution of the system of equations at the coarse grid. This, in terms of ratios of errors, is transferred to the finer grid.
10. Which of these is the opposite step of restriction?
a) Prolongation
b) Traversal
c) Agglomeration
d) Coarsening
Answer: a
Explanation: Prolongation is the step where correction is transferred from a coarse grid to a finer one. This is done the same way how errors are transferred from a fine grid to a coarser one in restriction step.
This set of Computational Fluid Dynamics Question Paper focuses on “Numerical Methods – Coupled Equations and Non-Linear Equations Solution”.
1. When coupled equations are ___________ sequential solutions are used.
a) linear and highly coupled
b) non-linear and uncoupled
c) linear and uncoupled
d) non-linear and highly coupled
Answer: a
Explanation: There are two ways to solve coupled equations – simultaneous and sequential. In the simultaneous methods, equations are solved together for the unknowns. The sequential methods are used to solve a highly coupled system with linear equations.
2. In sequential methods for solving coupled equations, except the variable for which the equations are solved, the other variables are treated as ___________
a) zeros
b) unknowns
c) known values
d) ones
Answer: c
Explanation: In sequential methods of solving coupled equations, the variable for which the system is solved is treated as unknown. All other variables are treated as known values with some approximations.
3. For solving for a single unknown in sequential solvers ____________ is used.
a) Direct solver
b) LU decomposition
c) Elimination method
d) Iterative solver
Answer: d
Explanation: For each inner iteration, one variable is unknown and all other variables are treated as known values. It is ineffective to solve this system accurately for one unknown. So, the iterative solvers are preferred to direct solvers in this case.
4. In solving non-linear systems, there is a trade-off between ___________ and ___________
a) speed and stability
b) speed and security
c) stability and convergence
d) stability and error
Answer: b
Explanation: In solving the non-linear system, there are two methods – Newton’s method and Global method. Newton’s method is faster and the Global method is guaranteed not to diverge. So, there is always a trade-off between speed and security.
5. The master method for solving the non-linear system of equations is __________
a) Newton’s method
b) Global method
c) Jacobi method
d) Gradient method
Answer: a
Explanation: Newton Raphson is the most widely used method for solving a non-linear system of equations. It is preferred in most of the cases as the rate of convergence is more. It converges fast.
6. Newton’s method linearizes the function using ___________
a) McLaurin series
b) Laurent series
c) Taylor series
d) Fourier series
Answer: c
Explanation: Newton’s method uses the first two terms of Taylor’s series to linearize the non-linear system. This is further simplified to get the formula to be iterated and get the roots.
7. Which of these creates a problem in Newton’s method for solving non-linear system of equations?
a) Taylor series
b) Jacobian
c) Convergence
d) Speed
Answer: b
Explanation: At each iteration of Newton’s method, Jacobian has to be evaluated for the method to be effective. Evaluation of the Jacobian with n 2 elements at each step will be expensive. Moreover, a direct method of evaluating the Jacobian does not exist.
8. When evaluation of the derivative of the non-linear function is not possible, which method is used?
a) Newton’s method
b) Global method
c) Jacobi method
d) Secant method
Answer: d
Explanation: An alternative to Newton’s method is the Secant method. This is much slower than Newton’s method. However, when the derivative of the function cannot be evaluated, this method is chosen as it does not involve any derivative.
9. The non-linear terms like convection and source terms in a system are linearized using __________
a) Iterative gradient method
b) Jacobi method
c) Picard iteration
d) Incomplete LU decomposition
Answer: c
Explanation: The usual solution of non-linear coupled system is the sequential decoupled method. For this, the non-linear terms are linearized using the Picard iteration method. This is applied for convection and source terms of the equation.
10. While using the Picard iteration, how is the source term decomposed?
a) q Φ =b 0 +b 1 Φ
b) q Φ =b 0 Φ+b 1 Φ 2
c) q Φ =b 0 Φ
d) q Φ =b 0 +b 1 Φ 2
Answer: a
Explanation: Picard iteration is used with the source term to decompose and linearize it. It decomposes to q Φ =b 0 +b 1 Φ. The term b 0 is absorbed by the RHS of the system. The term b 1 Φ is added to the coefficient matrix.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Finite Difference Method”.
1. The truncation error in a finite difference expansion is \
_{i,j} \frac{\Delta x}{2}-
_{i,j} \frac{^3}{6}\). What is the order of accuracy of the finite difference equation?
a) 1
b) 2
c) -2
d) -1
Answer: a
Explanation: The lowest order term in the truncation error given is \(\frac{\Delta x}{2}\), which is of the order 1. This defines the order of accuracy of the equation. So, the order of accuracy here is 1.
2. Consider the equation \
_{i,j}=
\) formulated using the Taylor series expansion. Find the type of equation.
a) first-order forward difference
b) first-order rearward difference
c) second-order forward difference
d) second-order rearward difference
Answer: b
Explanation: The equation uses the terms u i,j and its rearward term u i,j-1 . So, it represents a rearward difference. The order of accuracy of the equation depends on its truncation error. The truncation error has the least order term \(\frac{-\Delta y}{2}\). So, it is first order accurate.
3. Find the second-order accurate finite difference approximation of the first derivative of the velocity component in the x-direction using the Taylor series expansion. .
a) \
\
\
\(\frac{u_{i,j+1}-u_{i,j-1}}{2\Delta x}\)
Answer: c
Explanation: The only second-order accurate finite difference approximation of the first derivative is the central difference. For getting the central difference term,
\
_{i,j}\Delta x+
_{i,j}\frac{^2}{2}+⋯\)
\
_{i,j}\Delta x+
_{i,j}\frac{^2}{2}+⋯\)
To get \
_{i,j^,}\)
\
_{i,j} \Delta x+⋯\)
\
_{i,j}=\frac{u_{i+1,j}-u_{i-1,j}}{2 \Delta x}\) .
4. Find the first-order forward difference approximation of \
_{i,j}\) using the Taylor series expansion.
a) \
\
\
\(\frac{u_{i+1,j}-u_{i,j}}{\Delta x}\)
Answer: d
Explanation: To get the first-order forward difference approximation,
The Taylor series expansion of u i+1,j is
\
_{i,j}\Delta x+
_{i,j}\frac{^2}{2}+⋯\)
\
_{i,j}=\frac{u_{i+1,j}-u_{i,j}}{\Delta x}\).
5. Using the Taylor series expansion, What is the first term of the truncation error of the finite difference equation \
_{i,j}=\frac{u_{i+1,j}-u_{i,j}}{\Delta y}\)?
a) \
_{i,j}\frac{\Delta x}{2}\)
b) \
_{i,j}\frac{\Delta x}{3}\)
c) \
_{i,j}\frac{\Delta x}{3}\)
d) \
_{i,j}\frac{\Delta x}{2}\)
Answer: a
Explanation: The Taylor series expansion of u i+1,j is
\
_{i,j} \Delta x+
_{i,j} \frac{^2}{2}+⋯\)
\
_{i,j}+
_{i,j}\frac{\Delta x}{2}+⋯\)
\
_{i,j}=\frac{u_{i+1,j}-u_{i,j}}{\Delta x}-
_{i,j}\frac{\Delta x}{2}-…\)
The term –\
_{i,j}\frac{\Delta x}{2}-… \)is truncated. So, the first term of truncation error is –\
_{i,j}\frac{\Delta x}{2}\).
6. Find the central second difference of u in y-direction using the Taylor series expansion.
a) \
\
\
\(\frac{u_{i,j+1}+2u_{i,j}-u_{i,j-1}}{^2}\)
Answer: b
Explanation: To get the second difference,
\
_{i,j}^2+⋯\)
\
_{i,j}=\frac{u_{i,j+1}-2 u_{i,j}+u_{i,j-1}}{^2} +⋯\)
After truncating,
\
_{i,j}=\frac{u_{i,j+1}-2 u_{i,j}+u_{i,j-1}}{^2}\).
7. What is the order of the central difference for the mixed derivative \
1
b) 2
c) 3
d) 4
Answer: b
Explanation: The first term in the truncation error of the central difference for the mixed derivative \
\frac{^2}{12}\). So, the order of accuracy is 2.
8. Find \(\frac{\partial u}{\partial r}\) at point 1 using forward difference method.
computational-fluid-dynamics-questions-answers-finite-difference-method-q8
a) 1000
b) 100
c) 500
d) 5000
Answer: a
Explanation: Using the forward difference method,
\(\frac{\partial u}{\partial r}=\frac{u_2-u_1}{\Delta r}=\frac{5-0}{0.5×10^{-2}}=\frac{5}{5×10^{-3}}=1000.\)
9. What is the least order of accuracy for the second derivatives?
a) first-order
b) third-order
c) fourth-order
d) second-order
Answer: d
Explanation: The least possible order of accuracy for the second derivatives is 2. There cannot be a first-order second derivative as the second derivatives need terms less than the second order for the approximation.
10. Order of accuracy m means _____________
a) as the grid size is reduced, the approximations converge to the exact solution with an error proportional to m powers of the grid size
b) as the grid size is reduced, the approximations converge to the exact solution with an error proportional to m times of the grid size
c) as the grid size is reduced, the approximations diverge from the exact solution with an error proportional to m powers of the grid size
d) as the grid size is reduced, the approximations diverge from the exact solution with an error proportional to m times of the grid size
Answer: a
Explanation: An order of accuracy m means that the truncation error starts with a term proportional to grid size m . So, the statement “as the grid size is reduced, the approximations converge to the exact solution with an error proportional to m powers of the grid size” is correct.
11. Which of these diagrams represent the central difference approximation of \
computational-fluid-dynamics-questions-answers-finite-difference-method-q11a
b) computational-fluid-dynamics-questions-answers-finite-difference-method-q11b
c) computational-fluid-dynamics-questions-answers-finite-difference-method-q11c
d) computational-fluid-dynamics-questions-answers-finite-difference-method-q11d
Answer: d
Explanation: The line connecting the previous and the next node represents the central difference. The line connecting the previous and the current node represents the rearward difference. The line connecting the current and the next node represents the forward difference. The tangent to the curve at the point ‘i’ is the exact solution.
This set of Computational Fluid Dynamics Questions and Answers for Aptitude test focuses on “Explicit and Implicit Finite Difference Methods”.
1. Which of these methods of solving a system of equations will be needed after using an explicit scheme?
a) Sequential
b) Simultaneous
c) Iterative
d) Direct
Answer: a
Explanation: Explicit schemes result in marching solutions. Each step is dependent on the previous step only for one variable. The rest of the variables are found using the first obtained one. So, a simultaneous solution will not be needed here.
2. What is the main disadvantage of explicit schemes in a time-dependent problem?
a) Marching solution
b) Simultaneous equations
c) Small time-step size
d) Small grid size
Answer: c
Explanation: Explicit time-based schemes have a limited time-step size. Big time steps cannot be used. So, the total time of computation required to solve the system is very large when compared to the implicit schemes.
3. Implicit time-based problems will result in __________
a) Coupled equations
b) Uncoupled equations
c) Linear equations
d) Non-linear equations
Answer: a
Explanation: Implicit time-dependent solutions do not have a single unknown in a new time step. All the variables at a time step are coupled. So, they must be solved simultaneously to get the variables.
4. Which of these properties limit the time-step size in the explicit schemes?
a) Convergence
b) Stability
c) Consistency
d) Error
Answer: b
Explanation: The time step-size of an explicit scheme cannot be big. They are limited by the stability criterion. If the time-step size is bigger than the limit given by this criterion, the results will go extremely unstable.
5. What is advantageous in implicit schemes?
a) Error
b) Consistency
c) Convergence
d) Stability
Answer: d
Explanation: Implicit schemes do not have any restriction for the time-step size. They are stable for large time-steps also. Some of the implicit schemes are even unconditionally stable. Stability problems do not arise in implicit schemes.
6. Which of these is correct regarding implicit schemes?
a) Truncation error is less
b) Computation time is more
c) Time-step size is small
d) Easy to set-up
Answer: a
Explanation: As the time-step size is very large, the truncation error may become large and the accuracy of results may be less when compared to that of the explicit schemes. The total time of computation is less. But the algorithm is difficult to set-up.
7. Which of these may cause a problem to implicit schemes?
a) Coupled equations
b) Partial differential equations
c) Non-linear equations
d) Linear equations
Answer: c
Explanation: Though the implicit scheme has a great advantage of larger time steps, each step in an implicit scheme is large and takes more computational time. If the equation is non-linear, solving the system simultaneously will become more difficult. Usually, for these cases, the equations are linearized.
8. The time-step size in explicit schemes depends upon _____________
a) Grid size
b) Number of iterations
c) Total time interval
d) Given mathematical equation
Answer: a
Explanation: There is a limit posed to time-step size in explicit schemes. This limit depends on the grid size chosen. Once, the grid size is chosen, from the formula given by stability criterion, the maximum possible time-step size can be calculated.
9. Which of these schemes will lead to an implicit problem?
a) Higher-order schemes
b) SIMPLE algorithm
c) High-resolution scheme
d) Crank-Nicolson scheme
Answer: d
Explanation: Crank-Nicolson scheme is used to solve problems governed by parabolic equations. They result in implicit time-dependent problems. In CFD, they are usually used for finite difference solutions of boundary layer problems.
10. Consider the one-dimensional heat conduction equation. Apply forward difference method to approximate time rate and central difference method to approximate x-derivative. The resulting equation is in _____________
a) Implicit linear form
b) Explicit linear form
c) Explicit non-linear form
d) Implicit non-linear form
Answer: b
Explanation: The one-dimensional heat conduction equation is
\(\frac{\partial T}{\partial t}=α \frac{\partial^2 T}{\partial t^2} \)
Applying the difference approximations,
\(\frac{T_i^{n+1}-T_i^n}{\Delta t}=\alpha \frac{T_{i+1}^n-2T_i^n+T_{i-1}^n}{
} \)
\(T_i^{n+1}=T_i^n+\alpha\frac{\Delta t
}{
} \)
The equation is in explicit linear form.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Finite Difference Methods – Spectral Methods”.
1. Spectral methods are particularly suitable for __________
a) Subsonic flows
b) Boundary layer flows
c) Compressible flows
d) Turbulence modelling
Answer: d
Explanation: Spectral methods are not so commonly used like the Finite Volume and Finite Difference methods in CFD. But, they are specifically very good methods for analyzing turbulence models, especially with uniform grids.
2. Spectral methods use ___________
a) Fourier series
b) Taylor series
c) McLaurin series
d) Laurent series
Answer: a
Explanation: Finite Difference methods use Taylor series. In spectral methods, spatial derivatives are evaluated using the Fourier series or one of their generalization. The simplest spectral methods deal with periodic functions specified by their values at a uniformly spaced set of points.
3. What causes aliasing in Spectral methods?
a) Small grid sizes
b) Fourier series
c) Arbitrary values for Fourier series
d) Periodic nature
Answer: c
Explanation: For Fourier expansion, there is a nearly arbitrary value to be assumed. The results depend on this arbitrary value. If this value is not chosen properly, it will lead to aliasing error. This means pointing the same element more than one time.
4. What is the advantage of using fourier series in the spectral method?
a) Less grid size
b) Large number of grid points
c) Continuous results
d) Flexibility
Answer: c
Explanation: The Fourier series can be interpolated to get the dependent function. This will help us to get the results at a continuous space instead of results at particular grid points. This is an advantage over the other discretization techniques.
5. For higher order derivatives, spectral methods ___________
a) can be easily generated
b) are not suitable
c) difficult to generate
d) become invalid
Answer: a
Explanation: Spectral method can easily be generated for higher derivatives. The Fourier coefficients will vary in higher order derivatives. Other than this, there are not much changes needed for higher orders.
6. Spectral methods are much more accurate than the Finite Difference methods ___________
a) unconditionally
b) conditionally depending on the problem taken
c) conditionally when the number of grid points is small
d) conditionally when grid size is small
Answer: d
Explanation: The error in the solution decreases exponentially with the number of grid points. The results are more precise when the number of grid points is more. So, the grid size should be small. This makes the spectral method with more grids accurate than the Finite Difference methods.
7. The cost of computing the Fourier coefficients is ___________
.
a) N 3
b) N 2
c) \
\(\frac{N^3}{3} \)
Answer: b
Explanation: The computational cost required for computing Fourier coefficients, if done in the most obvious manner, is N 2 . This is prohibitively expensive. It is twice that of the backward substitution for Gauss-Elimination method.
8. The cost of computation for Fourier coefficients can be reduced by ___________
a) FFT
b) DFT
c) IDFT
d) IFT
Answer: a
Explanation: FFT stands for Fast Fourier Transform which is an algorithm for finding Discrete Fourier Transform of a sequence or the Inverse Discrete Fourier Transform . This is used to reduce the computational cost.
9. What is the cost of computation of FFT?
.
a) N
b) log N
c) N log N
d) \(\frac{N^2}{2} \)
Answer: c
Explanation: The cost of computation is reduced by FFT. FFT has a cost of computation of N log N orders. This is much lower than N 2 , especially when N is large. This reduces the problem of computation cost in the Spectral method.
10. To make the spectral method advantageous _____________
a) Function must be periodic but grids can be non-uniform
b) Grids should be uniform and function must be periodic
c) Grids should be uniform but function can be non-periodic
d) Grids should be structured and function must be periodic
Answer: b
Explanation: To get the full advantages of this Spectral method, the function must be periodic of the dependent variable and the grids should be uniform. This makes the Spectral method inflexible when compared to the other discretization methods.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Finite Difference Methods – Lax-Wendroff Technique”.
1. The Lax-Wendroff technique is ____________
a) explicit, finite-difference method
b) implicit, finite-difference method
c) explicit, finite volume method
d) implicit, finite volume method
Answer: a
Explanation: Lax-Wendroff technique is particularly suitable for marching solutions of hyperbolic and parabolic partial differential equations. It is an explicit method which uses the finite difference scheme for marching solutions.
2. What is the order of accuracy of the Lax-Wendroff technique?
a) fourth-order
b) third-order
c) first-order
d) second-order
Answer: d
Explanation: The Lax-Wendroff technique is second order accurate in both space and time. The first term in the truncation error has an order 2. This order of accuracy makes the algebra behind the technique complex.
3. Which series expansion is used by the Lax-Wendroff Technique?
a) Taylor Series
b) Fourier series
c) McLaurin series
d) Laurent series
Answer: a
Explanation: Lax-Wendroff technique uses the Taylor series expansion to approximate its time derivatives. This makes the technique marching in time in an explicit way. The number of terms used for this expansion decides the accuracy of this system.
4. How many terms of the Taylor series expansion is used in the Lax-Wendroff technique?
a) 1 and 2
b) 0 , 1 and 2
c) 0 and 1
d) 0
Answer: b
Explanation: The first three terms of the Taylor series expansion for the time marching term is used in the Lax-Wendroff Technique. This leads to the second-order accuracy of the system. Known values at previous time-step are used to find the value at the current time-step.
5. Expand the term \
\
^t+
_{i,j}^t \Delta t+
_{i,j}^t \frac{^2}{2} \)
b) \
_{i,j}^{t+\Delta t} \Delta t+
_{i,j}^{t+\Delta t}\frac{^2}{2}\)
c) \
_{i,j}^{av} \Delta t+
_{i,j}^{av}\frac{^2}{2}\)
d) \
_{i,j}^{t-\Delta t} \Delta t+
_{i,j}^{t-\Delta t}\frac{^2}{2}\)
Answer: a
Explanation: Lax-Wendroff technique uses the previous time-step values to get the current time-step values using the Taylor series expansion. The first three terms of the Taylor’s series expansion is used.
\
^t+
_{i,j}^t \Delta t+
_{i,j}^t \frac{^2}{2} \)
.
6. What is the disadvantage of the Lax-Wendroff technique?
a) Stability
b) Explicit
c) Order of accuracy
d) \
_{i,j}^t\)
Answer: d
Explanation: The second order term in the Taylor series expansion of the Lax-Wendroff technique is its disadvantage. This term leads to a complex algebra while getting it using the difference schemes. The lengthy algebra here is the only considerable disadvantage of this technique.
7. Consider three-dimensional Euler equations. Which equation will you use to find the value \
_{i,j}^t\)?
a) Energy equation
b) y-momentum equation
c) x-momentum equation
d) Continuity equation
Answer: c
Explanation: The x-momentum equation gives the time derivative if the x-component of velocity at a particular time in terms of the other flow variables and their special derivatives. So, this can be used to get the time derivative \
_{i,j}^t\).
8. Consider three-dimensional Euler equations. What will you do to get the value of \
_{i,j}^t\)?
a) Differentiate \
Differentiate the continuity equation with respect to time
c) Differentiate the value of \
_{i,j}^t\) with respect to time
d) Differentiate the value of \(\rho_{i,j}^t\) with respect to time twice
Answer: b
Explanation: Differentiating the value of any variable or the value of its derivative have no sense as it will result in zero. To differentiateup \
\).
9. Which of these is wrong for the Lax-Wendroff technique?
a) Linearization is needed
b) Simultaneous equations are not required
c) It is simple to solve
d) It uses the finite difference method
Answer: c
Explanation: Lax-Wendroff technique uses the finite difference method to get time-dependent solutions. The need for linearization depends upon the equation to be solved. Simultaneous equations are not required as the resulting system is explicit. But the system is not simple to solve. It involves lengthy algebra to get the second order terms.
10. Which is the technique used to overcome the disadvantages of the Lax-Wendroff technique?
a) Upwind scheme
b) MacCormack’s technique
c) Downwind scheme
d) Richtmeyer method
Answer: b
Explanation: To overcome the lengthy algorithm of the Lax-Wendroff technique, MacCormack’s technique is used which can produce results of the same order of accuracy with a simpler method which does not want the second-order derivative.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Finite Difference Methods – MacCormack’s Technique”.
1. MacCormack’s technique is __________
a) explicit, finite-difference method
b) implicit, finite-difference method
c) explicit, finite volume method
d) implicit, finite volume method
Answer: a
Explanation: Like the Lax-Wendroff technique, MacCormack’s technique is also particularly suitable for marching solutions of hyperbolic and parabolic partial differential equations. It is also an explicit finite difference scheme for marching solutions.
2. Which series expansion is used by the MacCormack’s technique?
a) Taylor Series
b) Fourier series
c) McLaurin series
d) Laurent series
Answer: a
Explanation: The MacCormack’s technique uses the Taylor series expansion to approximate its time derivatives like the finite difference scheme. But the accuracy here is not dependent on the order of the derivative. It has improved accuracy.
3. What is the order of accuracy of the MacCormack’s technique?
a) Fourth-order
b) Third-order
c) First-order
d) Second-order
Answer: d
Explanation: MacCormack’s technique is second order accurate in both space and time. There is a special method used in MacCormack’s technique to make the order of accuracy two, even after reducing the lengthy algebra.
4. Which of these terms of the Taylor series expansion is used in the MacCormack’s technique?
a) 1 and 2
b) 1
c) 0 and 1
d) 0
Answer: c
Explanation: Only the first two terms in the Taylor series expansion is used in the MacCormack’s technique. The first two terms are 0 and 1 . All other higher-order terms are omitted. But, the order of accuracy is maintained here as two.
5. Expand the term \
\
_{i,j}^{av}\Delta t\)
b) \
_{i,j}^{av} \Delta t\)
c) \
_{i,j}^t \Delta t\)
d) \
_{i,j}^t \Delta t\)
Answer: a
Explanation: The MacCormack’s technique uses the previous time-step values of the dependent variable and its derivative is obtained at between times t and t+Δ t. Only the first two terms of the Taylor series expansion is used.
\
_{i,j}^{av}\Delta t\).
6. Which of these methods is used for finding the average time derivative in MacCormack’s technique?
a) Trial and error method
b) Predictor-corrector method
c) Genetic algorithm
d) Relaxation method
Answer: b
Explanation: To get the time derivative at the average time between t and t+Δ t, the MacCormack’s technique uses the Predictor-corrector method. This is used as we do not know the value for the time derivative at the time-step t+Δ t.
7. How is the value \
_{i,j}^{av}\) obtained in the MacCormack’s expansion to find \
Truncated mean of \
_{i,j}^t and
_{i,j}^{t+\Delta t}\)
b) Weighted average of \
_{i,j}^t and
_{i,j}^{t+\Delta t}\)
c) Geometric mean of \
_{i,j}^t and
_{i,j}^{t+\Delta t}\)
d) Arithmetic mean of \
_{i,j}^t and
_{i,j}^{t+\Delta t}\)
Answer: d
Explanation: The value of \
_{i,j}^{av}\) is the arithmetic mean of \
_{i,j}\) at t and \
_{i,j}\) at t+Δt.
\
_{i,j}^{av}=\frac{1}{2}[
_{i,j}^t+
_{i,j}^{t+\Delta t}]\).
8. Which value is predicted in the predictor step of the MacCormack’s technique?
a) Variable at the average time-step
b) Variable at the upcoming time-step
c) Time derivative of the variable at the upcoming time-step
d) Time derivative of the variable at the average time-step
Answer: b
Explanation: To get the derivative in the upcoming time-step for finding the time derivative of the variable at the average time, the variable at that time step is needed. This is the value which we intend to find using MacCormack’s technique. So, in the predictor step, the value of the variable at that time step is predicted.
9. Which of these values used to find \
_{i,j}^{av}\) is a predicted one?
a) \
_{i,j}^t\)
b) Neither \
_{i,j}^t nor
_{i,j}^{t+\Delta t}\)
c) \
_{i,j}^{t+\Delta t}\)
d) Both \
_{i,j}^t and
_{i,j}^{t+\Delta t}\)
Answer: c
Explanation: \
_{i,j}^{t+\Delta t}\) is predicted using the continuity equation and the value of \
_{i,j}^{av}\). The continuity equation is used as we need the time rate of change of density.
10. I am using forward differences in the predictor step. Which method would you suggest me to use in the corrector step?
a) Rearward differences
b) Central differences
c) Forward differences
d) Second-order differences
Answer: a
Explanation: If forward differences are used in the predictor step, rearward differences should be used in the corrector step and vice versa. At every time-step, this sequence should be changed while solving a time-marching problem.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Errors in Finite Difference Approximations”.
1. Which is the major error occurring due to the finite difference approximations?
a) Discretization error
b) Round-off error
c) Iteration error
d) Modelling errors
Answer: a
Explanation: The major error occurring in the finite difference method is the discretization error. This error occurs due to both temporal and spatial discretization using an approximation for the discretization. This is also called a numerical error.
2. What is the source of discretization error in the finite difference method?
a) Numerical error
b) Round-off error
c) Truncation error
d) Modelling error
Answer: c
Explanation: Discretization error occurs because of the truncation errors which arise while discretizing the PDEs. It is named truncation error as the root cause of it is the truncation of the higher order terms in the series expansion.
3. The exact solution of the partial differential equation varies from the exact solution of the discretized equations by ___________
a) truncation error
b) discretization error
c) iteration error
d) modelling error
Answer: b
Explanation: The difference between the exact solution of the partial differential equation and the solution of the algebraic equation is the discretization error. Mathematically
Φ=Φ num +∈ d
Where,
Φ → Exact solution
Φ num → Numerical solution
∈ d → Discretization error
4. Truncation error is the difference between __________
a) the exact solution of the partial differential equation and the discretized equations
b) the exact partial differential equation and the discretized equations
c) the exact solution and the numerical solution of the partial differential equations
d) the exact partial differential equation and its solution
Answer: b
Explanation: Truncation error is the difference between the exact partial differential equation and the discretized algebraic equation. This arises as we cut-off the higher order terms in the Taylor series expansion.
5. Information about the magnitude and distribution of the truncation error can be useful for ____________
a) correcting the error
b) increasing the stability
c) refining the grid
d) converging the solution
Answer: c
Explanation: It is not possible to decrease the discretization error as it will lead to increased algebra and computation. So, the information about the discretization error can be used only for refining the girds and achieve the same level of discretization everywhere in the solution.
6. How is the discretization error found?
a) Difference between the solutions obtained from systematically refined grids
b) Difference between the exact and the numerical solutions
c) Difference between the exact solution and the solution from the refined grid
d) Difference between the coarse grid solution and the exact solution
Answer: a
Explanation: Discretization error is originally the difference between the exact solution of the partial differential equation and the solution of the algebraic equation. But, as the exact solution is not known, we estimate the discretization error as the difference between the solutions obtained from systematically refined grids.
7. What is Richardson extrapolation used for?
a) To increase the accuracy
b) To decrease the error
c) To create convergence monotony
d) To increase the rate of convergence
Answer: d
Explanation: Richardson extrapolation is a sequence acceleration method. It is used to increase the rate of convergence of a system. In the finite difference method, it is used to find accurate results from the discretized results.
8. What does Richardson extrapolation do in finite difference schemes?
a) Add the error estimate to the results of the finest grid
b) Subtract the error estimate from the results of the finest grid
c) Add the error estimate to the results of the current grid
d) Subtract the error estimate from the results of the current grid
Answer: a
Explanation: When we have the results of many grid arrangements ranging from coarse to fine. A solution which is more accurate than the solution of the finest grid can be obtained by adding the error estimate to the results of the finest grid available.
9. When is the Richardson extrapolation accurate?
a) When the system is stable
b) When the convergence is monotonic
c) When the system is consistent
d) When the system is linear
Answer: b
Explanation: The Richardson extrapolation is very useful as it is a simple method. But, it can give accurate results only when the convergence is monotonic. This convergence represents the convergence of error while refining the grid.
10. What happens when the convergence is not monotonic?
a) Solutions will always converge
b) Solutions will not converge
c) Erroneous solutions may converge
d) Error will increase
Answer: c
Explanation: The order of convergence is valid only when the convergence is monotonic. This is because, for two consecutive grids, results may also converge even if the error is large. Then, a third grid arrangement must be used to assure the real convergence of the solution.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Finite Volume Method”.
1. Which of these models will directly give the conservative equations suitable for the finite volume method?
a) Finite control volume moving along with the flow
b) Finite control volume fixed in space
c) Infinitesimally small fluid element moving along with the flow
d) Infinitesimally small fluid element fixed in space
Answer: b
Explanation: Finite volume method uses the conservation equation in the integral form without any substantial derivative. This can be given by a finite control volume fixed in space. This directly results in the conservative integral form of the governing equations.
2. Which of these terms need a surface integral?
a) Diffusion and rate of change terms
b) Convection and source terms
c) Convection and diffusion terms
d) Diffusion and source terms
Answer: c
Explanation: The convection and diffusion terms have fluxes flowing along the surfaces of the control volume. So, they have to be integrated over the faces of the control volume. So, the convection and diffusion terms need a surface integral.
3. Which of these terms need a volume integral while modelling steady flows?
a) Convection term
b) Diffusion term
c) Source term
d) Rate of change term
Answer: c
Explanation: The convection and diffusion terms need surface integrals and not volume integrals. The source and rate of change terms need volume integral. Since the flow taken is a steady flow, the rate of change term will be equal to zero. So, in this case, only the source term needs a volume integral.
4. Consider a two-dimensional flow. If f is the component of the flux vector normal to the control volume faces, which of these terms represent ∫ S fd\
\
\
\
\(\Sigma_{k=1}^8 \int_{S_k} f d\vec{S}\)
Answer: a
Explanation: In a two-dimensional flow, the number of faces bounding a control volume is four. So, the summation of the integrals along these four faces will be equal to the total flux of the control volume. Here, f may be convective or diffusive flux.
5. Approximate the surface integral ∫ S n fd\(\vec{S}\) using the midpoint rule.
computational-fluid-dynamics-questions-answers-finite-volume-method-q5
a) f n S n
b) S n (f ne +f nw )
c) \(\frac{S_n}{2}\) (f ne +f nw )
d) \(\frac{S_n}{2}\) f n
Answer: a
Explanation: For surface integral, the value of the integrand at the face centre is used in the midpoint approximation. The neighbouring values are not used. So,
∫ S n fd\(\vec{S}\)=f n S n
6. Approximate the surface integral in the eastern face ∫ S e fd\(\vec{S}\) of a two-dimensional problem using the trapezoidal rule.
computational-fluid-dynamics-questions-answers-finite-volume-method-q5
a) \(\frac{3}{2}\)(f ne +f se )
b) 3 \(\frac{S_e}{2}\)(f ne +f se )
c) \(\frac{1}{2}\)(f ne +f se )
d) \(\frac{S_e}{2}\) (f ne +f se )
Answer: c
Explanation: The trapezoidal rule is a second-order accurate approximation. It needs the values of the integrand at two points. Here, as we need the surface integral in the eastern face, the value is approximated using the northern and the southern nodes of the eastern face.
∫ S e fd\(\vec{S}=\frac{1}{2}\) (f ne +f se ).
7. In a two dimensional flow, how many terms does Simpson’s rule need to approximate a surface integral?
a) four terms
b) one term
c) two terms
d) three terms
Answer: d
Explanation: For a two-dimensional flow and the surface integral, the midpoint rule needs only one term . The trapezoidal rule needs two terms . The Simpson’s rule needs three terms .
8. For three-dimensional flows, what is the approximation of the volume integral using the midpoint rule?
a) Product of the integrand at the face centre and the volume of the control volume
b) Product of the integrand at the control volume centre and the volume of the control volume
c) Product of the integrand at the control volume centre and the surface area of the control volume
d) Product of the integrand at the face centre and the surface area of the control volume
Answer: b
Explanation: Using the midpoint rule, the volume integral is approximated as the product of the integrand at the centre of the cell and the volume of the cell. Representing it mathematically,
∫ V qdV = q P ×ΔV.
9. In a one-dimensional flow, the volume integral becomes __________
a) a line integral
b) an area integral
c) a surface integral
d) a surface integral and the Gauss divergence theorem
Answer: a
Explanation: For a one-dimensional flow, the volume of the control volume is in a single dimension which is the length of that cell. So, there will be no need for a volume, area or surface integrals. It is enough to integrate over the length of that cell.
10. Approximate the surface integral ∫ S w f d\(\vec{S}\) using the Simpson’s rule.
computational-fluid-dynamics-questions-answers-finite-volume-method-q5
a) \(\frac{S_w}{6}\)(2f nw +2f w +2f sw )
b) \(\frac{S_w}{4}\)(2f nw +2f sw )
c) \(\frac{S_w}{6}\)(f nw +4f w +f sw )
d) \(\frac{S_w}{4}\)(f nw +2f w +f sw )
Answer: c
Explanation: The Simpson’s rule uses values of the integrand at three points – centre of the face and the two vertices in the same face. It is given by
\(\frac{S_w}{6}\)(f nw +4f w +f sw ).
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Finite Volume Methods – Order of Accuracy”.
1. For integrating the convective and diffusive fluxes using the mean value approximation, the value at the ___________ is used.
a) face centre
b) cell centre
c) node
d) vertex
Answer: a
Explanation: The convective and diffusive fluxes undergo a surface integral. To get the surface integral, they are integrated over the area of the face considered. While approximating using the mean value approximation, the central value is used. So, the value at the face centre is used.
2. For integrating the source term, if the value is approximated using the mean value theorem, the value at the _____________ is used.
a) face centre
b) cell centre
c) boundary face
d) vertex
Answer: b
Explanation: Source term integral is a volume integral. Approximating the volume integral, the value at the centre of the cell and the volume of the cell are multiplied. So, the value at the cell centre is used.
3. What is the order of accuracy of the midpoint rule approximation?
a) Fourth-order
b) Third-order
c) Second-order
d) First-order
Answer: c
Explanation: The midpoint rule is the method which uses the value of the function only at its midpoint to approximate the integration. It is second order accurate. This is the simplest method of approximation.
4. I know the value of flux at point x c . How will you find the value of this flux at point x away form x c ?
a) Simpson’s rule
b) Trapezoidal rule
c) Mean value theorem
d) Taylor series expansion
Answer: d
Explanation: Simpson’s rule, Trapezoidal rule and the Mean value theorem are all used to integrate a function numerically. To get a value at the point near another point where the value is known, we use the Taylor series.
5. Which is the order of accuracy used for?
a) To quantify the rate of convergence
b) To find the error
c) To find the stability
d) To find the consistency
Answer: a
Explanation: The order of accuracy of a numerical method is used to quantify the rate of convergence of the approximation. This can be applied to both the finite difference and the finite volume methods.
6. In which of these methods the function is assumed to vary linearly with the independent variable?
a) Trapezoidal rule and Simpson’s rule
b) Trapezoidal rule
c) Midpoint rule and Simpson’s rule
d) Only Simpson’s rule
Answer: b
Explanation: The trapezoidal rule assumes that the dependent variable varies linearly with the independent variable. So, the mean of the values at the endpoints is taken to calculate the integral value.
7. What is the order of accuracy for the method of numerical approximation represented by this diagram?
computational-fluid-dynamics-questions-answers-order-accuracy-q7
a) Fourth-order
b) Third-order
c) Second-order
d) First-order
Answer: c
Explanation: The given diagram represents the Trapezoidal rule for the approximation of an integral. It uses the values at the starting and the ending of the function. The order of accuracy of the Trapezoidal rule is two.
8. Simpson’s rule assumes the function to be ___________
a) constant
b) cubic
c) linear
d) quadratic
Answer: d
Explanation: Simpson’s rule assumes the function to be integrated as a quadratic function. The points chosen to generate the quadratic line are the two endpoints and the midpoint. So, this approximation is more accurate.
9. How is the order of accuracy of the mean value approach found?
a) Using spatial variation
b) Using trapezoidal rule
c) Using convergence
d) Using Fourier expansion
Answer: a
Explanation: The mean value theorem approximates the function to be linear. If the real spatial variation of the function is found and the two are subtracted, the order of accuracy can be obtained.
10. What is the order of accuracy for the method of numerical approximation represented by this diagram?
computational-fluid-dynamics-questions-answers-order-accuracy-q10
a) second-order
b) fourth-order
c) first-order
d) third-order
Answer: b
Explanation: The diagram represents Simpson’s rule of approximation. This rule uses quadratic interpolation of the two endpoints and the midpoint of the function. The order of accuracy of this approximation is four.
This set of Tough Computational Fluid Dynamics Questions and Answers focuses on “Variable Arrangement in FVM”.
1. What are the two possible variable arrangements for the finite volume method?
a) Cell-centred and Vertex-centred
b) Cell-centred and Face-centred
c) Face-centred and Vertex-centred
d) Face-centred and Boundary-centred
Answer: a
Explanation: Unlike FDM, in FVM, the grid points are taken inside the elements. There are two ways of arranging these elements for a finite volume method. They are cell-centred and Vertex centred arrangements.
2. The variables are calculated at the __________ in the vertex-centred arrangements.
a) element-edges
b) face-centres
c) centroids
d) vertices
Answer: d
Explanation: In the vertex-centred arrangements, all the flow variables and their related quantities are calculated and stored in the vertices. Elements are constructed around these vertices using different methods.
3. Dual cell or dual mesh method is used to __________
a) create boundary elements
b) create elements around a vertex
c) create cell-centred arrangement
d) create boundary faces
Answer: b
Explanation: In the vertex-centred arrangements, elements are created around the vertices by joining either their centroids or their centroids with the face-centroids. This method of forming the vertex-centred arrangement is called dual cell or dual mesh method.
4. Consider a 2-D finite volume problem. A vertex-centred arrangement is created by connecting the centroids of the elements sharing the vertex. What is the problem that may arise because of this arrangement?
a) Unstructured elements
b) Conjunctional elements
c) Orthogonal elements
d) Overlapping elements
Answer: d
Explanation: The above-mentioned method is a method of creating the vertex-centred arrangement. As mentioned, elements are created by joining the centroids of the surrounding elements. This way, the lines connecting the centroids may overlap and result in overlapping elements.
5. Consider a 3-D problem. Non-overlapping elements can be created by joining __________
a) face-centroids and edge-centroids
b) cell-centroids and edge -centroids
c) cell-centroids, face-centroids and edge-centroids
d) cell-centroids and face-centroids
Answer: c
Explanation: To overcome the problem of overlapping elements, in two-dimensional problem, the cell-centroids and the face-centroids are connected. Extending this to a three-dimensional problem, cell-centroids, face-centroids and edge-centroids are connected. Therefore, ensuring that the connecting lines do not overlap.
6. In vertex-centred arrangements, the variables at the vertices are only known. How is the variation of variables in these elements calculated?
a) Interpolation profiles or Taylor series expansion
b) Shape functions or Taylor series expansion
c) Shape functions or interpolation profiles
d) Shape functions or Fourier series expansion
Answer: c
Explanation: In vertex-centred arrangements, the flow variables are calculated and stored at the vertices only. The variation of flow properties through the element can be obtained by using either shape functions or interpolation profiles.
7. Vertex-centred approach gives accurate solution for ___________ but not for _________
a) diffusion term, convection term
b) source term, convection term
c) convection term, diffusion term
d) convection term, source term
Answer: d
Explanation: Vertex-centred arrangements give an accurate resolution of face fluxes such as convection and diffusion terms. But, they yield a lower order of accuracy for element based integrations .
8. Cell-centred FVM is __________ accurate.
a) first-order
b) second-order
c) third-order
d) fourth-order
Answer: b
Explanation: Cell-centred arrangements have predefined elements and their centroids are the grid points. Here, the elements are identical to discretization elements and they are second-order accurate.
9. Which of these is needed for a vertex-centred arrangement and not for cell-centred elements?
a) Pre-defined shape functions
b) Face-centroids
c) Edge-centroids
d) Cell-centroids
Answer: a
Explanation: Cell-centroids are must for cell-centred arrangements as they are the grid points here. Face-centroids and edge-centroids are also needed here during numerical integration. Pre-defined shape functions are not needed as they use general polygonal elements.
10. The cell-centred arrangements are disadvantageous when we have ___________
a) Non-conjunctional and non-orthogonal elements
b) Conjunctional and orthogonal elements
c) Unstructured and non-orthogonal elements
d) Structured and orthogonal elements
Answer: a
Explanation: The treatment of non-conjunctional elements and the manner the diffusion term is discretized for non-orthogonal grids are disadvantages of cell-centred arrangements. They cannot produce good results in these cases.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Structured Grids in FVM”.
1. For a regular structured grid, which of these statements is true?
a) Each interior cell is connected to the same number of neighbours
b) Each cell is connected to the same number of neighbours
c) Each boundary cell is connected to the same number of neighbours as an interior cell
d) Each direction has the same number of cells
Answer: a
Explanation: The boundary cells will have less number of neighbours when compared to the interior cells. As the statement telling “Each cell” includes the boundary cells also, it does not hold true. It is not necessary for the structured grids to have the same number of cells in every direction. So, the statement “Each interior cell is connected to the same number of neighbours” is only true.
2. Structured grids give ____________ and need ____________
a) no access to elements, less memory for storage
b) easy access to elements, less memory for storage
c) easy access to elements, more memory for storage
d) no access to elements, more memory for storage
Answer: b
Explanation: Structured grids have properly ordered cells. So, it is easy to access their elements or grid points. As access is easy, it is easy to store the values also in memory. So, less memory is required.
3. Topological information is embedded in a structured mesh through ___________
a) neighbours
b) boundaries
c) indices
d) discretization
Answer: c
Explanation: The topological information is embedded in the mesh structure through the indexing system. This also leads to greater efficiency in coding, cache utilization and vectorization.
4. In a three-dimensional structured grid, an element has ____________ faces and ____________ vertices.
a) 8, 6
b) 8, 8
c) 4, 4
d) 6, 8
Answer: d
Explanation: The elements of a structured grid are in a hexagonal shape. They have six faces and eight vertices. Each interior element is surrounded by six neighbours. Unlike the unstructured grids, these are fixed in a structured grid.
5. In a two-dimensional grid, the elements are ____________ in shape and have ___________ neighbours.
a) Quadrilateral, 4
b) Cube, 6
c) Cube, 4
d) Cuboid, 4
Answer: a
Explanation: For a two-dimensional grid, there are four faces and four vertices for every element. And, each element is surrounded by four elements. With this, we cannot say if they are cubes or cuboids. In general, we can say they will definitely be quadrilaterals.
6. Which is correct?
a) Local index has a single index and global index has multiple indices
b) Global index has a single index and local index has multiple indices
c) Both local and global indices have a single index
d) Both local and global indices have multiple indices
Answer: b
Explanation: Local indexing is the indexing of elements around a particular cell. It has to mention the direction also. So, it involves multiple indexing. Global indexing is just a single number representing a cell or element.
7. Consider the two 2-D cells given here.
computational-fluid-dynamics-questions-answers-structured-grids-fvm-q7
Which of these is correct? .
a) S1 i,j-1 =S2 i,j
b) S1 i,j-1 =-S2 i,j
c) S3 i,j-1 =-S4 i,j
d) S3 i,j-1 =S4 i,j
Answer: c
Explanation: Arranging these two grids according to the directions of i and j, we get
computational-fluid-dynamics-questions-answers-structured-grids-fvm-q7a
Each surface points outside. But, their magnitude is the same. Therefore,
S3 i,j-1 =-S4 i,j .
8. In structured grids, computer memory is saved by ____________
a) multi-dimensions
b) localization
c) linearization
d) vectorization
Answer: d
Explanation: As it is easy to index the elements in structured grids, the values are stored using global indexing which does not need three-dimensional indexing. So, it can be vectorized and saved as a single array of elements. This saves computer memory.
9. The values of the flow variables at the faces can be calculated by _____________ in a structured grid.
a) Interpolation
b) Taylor series
c) Fourier series
d) Shape functions
Answer: a
Explanation: As the cells of a structured grid have simple quadrilateral shapes, the values at a cell faces can be easily interpolated using the values of the flow variables at the cell centroids of the two cells sharing that particular face.
10. Which of these represent the discretization indexing?
a) Φ i,j+1 , Φ i,j-1
b) Φ e , Φ w
c) Φ n+1 , Φ n-1
d) Φ i+1,j , Φ i-1,j
Answer: b
Explanation: Discretization indexing is the type of indexing which is used locally while mentioning the neighbours of an element. Here, Φ e is the eastern node; Φ w is the western node; Φ n is the northern node; Φ s is the southern node.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “FVM – Unstructured Grids”.
1. The advantage of using unstructured grids is ___________
a) vectorization
b) flexibility
c) simple arrangement
d) less memory requirement
Answer: b
Explanation: Flexibility is the greatest advantage of using unstructured grids. It offers more flexibility while meshing in terms of the element types that can be used and in terms of where the elements can be concreted.
2. Which of these statements is true for unstructured grids?
a) It is simple to model and use
b) Element connectivity is implicitly defined
c) Element connectivity does not depend upon indices
d) Bounding faces can be easily found
Answer: c
Explanation: All the entities should be named separately in an unstructured grid. There is no way to directly link the various entities of an unstructured grid. Thus, local connectivity has to be defined explicitly.
3. What is the cost of flexibility in unstructured grids?
a) Consistency
b) Stability
c) Accuracy
d) Complexity
Answer: d
Explanation: All these properties such as consistency, stability and accuracy depends upon the method used for discretization. They do not depend on whether the grid is structured or unstructured. The cost of increased flexibility in unstructured grids is its complexity.
4. Which of these entities need the information to be explicitly stored?
a) Elements, faces, nodes and neighbours
b) Elements, faces and nodes
c) Elements, faces and neighbours
d) Elements, nodes and neighbours
Answer: a
Explanation: The data structure for an unstructured grid should contain information about the elements, faces, nodes and in addition to these about the neighbouring elements also. As they are not properly arranged, the neighbouring elements cannot be defined using indices.
5. The direction of the normal to a face in an unstructured mesh depends upon __________
a) Local indices
b) Global indices
c) Direction of flow
d) Direction of increasing indices
Answer: b
Explanation: The direction of the normal to a face depends on the indices in both structured and unstructured grids. For structured grids, it depends on the direction of increasing indices. For unstructured grids, the direction depends on the global index number.
6. Which of these is correct for the direction of the normal vectors to faces?
a) computational-fluid-dynamics-questions-answers-unstructured-grids-fvm-q6a
b) computational-fluid-dynamics-questions-answers-unstructured-grids-fvm-q6b
c) computational-fluid-dynamics-questions-answers-unstructured-grids-fvm-q6c
d) computational-fluid-dynamics-questions-answers-unstructured-grids-fvm-q6d
Answer: c
Explanation: The normal vectors to faces in an unstructured mesh depends upon the global indices. The vector always points in the direction of increasing global indices. For element 5, a normal vector connecting the elements 5 and 6 will point towards 6. All other vectors will point towards 5.
7. Which of these formulae is correct to find the gradient of the element ‘k’?
a) ∇Φ k = \
\)
b) ∇Φ k = \
\)
c) ∇Φ k = \
\)
d) ∇Φ k = \
\)
Answer: d
Explanation: The formula for the gradient, in general, is the summation of the product of the flow variables and the area of the faces. As the direction matters here, the summation will be negative till ‘n’ reaches ‘k’. Therefore,
∇Φ k = \
\).
8. The topology of the faces in an unstructured grid depends upon ___________
a) Straddling elements
b) Boundary elements
c) Interior elements
d) Neighbouring elements
Answer: a
Explanation: The information about the straddling elements is what decides the topology of the faces in an unstructured grid. Here, the flux at every element is the same. They do not vary in the direction.
9. When compared to the algorithm to calculate the gradient of a structured grid, the algorithm for unstructured grids ___________
a) need more computational cost
b) need less computational cost
c) is more accurate
d) is less accurate
Answer: b
Explanation: Accuracy wise, both the grids result in the same level of accuracy as they do not involve any approximation. But, the unstructured grid algorithm needs less computational cost as it is written in terms of the global indices. This can be adopted for structured grids also.
10. Gradient for an unstructured grid is calculated by ___________
a) looping over the nodes in the computational domain
b) looping over the elements in the computational domain
c) looping over the faces in the computational domain
d) looping over the gradients in the computational domain
Answer: c
Explanation: Computation of gradient for the overall domain is more efficient than calculating locally element by element. For this purpose, all the faces are looped over and the gradients are updated.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Types of FVM Elements”.
1. What is the minimum number of vertices that a 3-D element can have?
a) 6 vertices
b) 5 vertices
c) 3 vertices
d) 4 vertices
Answer: d
Explanation: At least four points are needed to make an element. A tetrahedron is a shape which has the least number of vertices and the least number of faces. It has 4 faces and 4 vertices. A 2-D element can be formed using three vertices.
2. What is the shape of a tetrahedral element’s face?
a) Cuboid
b) Cube
c) Triangle
d) Quadrilateral
Answer: c
Explanation: A tetrahedron has four triangular faces. Each face is formed by connecting three vertices of the tetrahedron. This is one of the most common shapes of element used in unstructured grids.
3. The general shape of a 3-D element is __________
a) Quadrilateral
b) Tetrahedral
c) Polyhedron
d) Polygon
Answer: c
Explanation: The number of sides in an element is not restricted. The general shape of a two-dimensional element can be named polygon which can have any number of sides. The same way, the general shape of a three-dimensional element can be named polygon which can have any number of faces.
4. Which of these three-dimensional elements has faces as a mixture of two shapes?
a) Tetrahedron
b) Prism
c) Hexahedron
d) Octahedron
Answer: b
Explanation: A prism is made up of two triangular faces and three rectangular faces. This is a regular shape which has faces of mixed shapes. It has six vertices. It can be used to make a structured grid.
5. Each face of a hexahedron is __________
a) Cuboid
b) Cube
c) Triangle
d) Quadrilateral
Answer: d
Explanation: A hexahedron is made up of six faces. Each of its faces may have the same dimensions or may not. In general, all hexahedrons will have faces in quadrilateral shape.
6. To find the volume and centroid of a polyhedral, it is divided into ____________
a) Quadrilaterals
b) Pyramids
c) Polygons
d) Hexahedrons
Answer: b
Explanation: The first step of calculating the volume and centroid of any three-dimensional element is to pyramids. The summation of the volumes of these pyramids is the volume of the whole polyhedron.
7. The apex of the sub-element while calculating the volume of a polyhedron is ____________
a) its centre of mass
b) its centre of gravity
c) its centroid
d) its geometric centre
Answer: d
Explanation: First, the geometric centre of the polyhedron is located. This is the apex of the pyramid. The distance of this apex from the face of the element gives the height of the pyramid. From the base and height of the pyramid, its volume is calculated.
8. The centroid of the polyhedron is ___________
a) the summation of the centroids of the pyramids
b) the arithmetic mean of the centroids of the pyramids
c) the weighted average of centroids of the pyramids
d) the Pythagorean mean of centroids of the pyramids
Answer: c
Explanation: The centroid of each of the sub-pyramids is calculated. Their volumes are also calculated. The weighted average of these centroids is calculated by using the volumes as the weights. This gives the centroid of the element.
9. Which of these formulae is used to calculate the centroid of a polyhedron?
a) \
\
\
\(\frac{\sum_{n=1}^{No. \,of \,pyramids}Centroid_n×Volume_n}{Centroid_n} \)
Answer: b
Explanation: The centroid of a polyhedron is the volume-weighted average of the centroid of its sub-elements. Giving mathematically,
\(\frac{\sum_{n=1}^{No. \,of \,pyramids}Centroid_n×Volume_n}{\sum_{n=1}^{No. of pyramids}Volume_n}\).
10. In case of a two-dimensional element, which of these will serve as the face area?
a) Distance of the connecting line
b) Area of the element
c) Unit area
d) Volume of the element
Answer: a
Explanation: For a two-dimensional element, the area of the element is equivalent to its volume. Similarly, the length of the connecting line is equivalent to the face area. However, the direction of flux is given by a normal vector.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “The Geometry of FVM Elements”.
1. How are the faces of a 3-D element divided to find the area?
a) Squares
b) Quadrilaterals
c) Rectangles
d) Triangles
Answer: d
Explanation: The faces of a 3-D element are 2-D polygons. With the intention of finding the area of these faces, they are divided into triangles of different type and sizes. Triangles give the advantage of having any length in all three of its sides.
2. Which of these points form the apex of the sub-elements of the faces?
a) Centre of mass of the face
b) Vertex of the face
c) Geometric centre of the face
d) Apex of the face
Answer: c
Explanation: The sub-elements are all triangular in shape. The apex of these triangles are all the same point. The geometric centre of the face is chosen as the apex of the triangles. The geometric centre is not the centroid of the polygon.
3. Which formula is suitable for finding the geometric centre of a polygonal face?
a) \(\frac{1}{No.of points}\sum_{i=1}^{No.of points}\) Point defining the polygon i
b) \(\sum_{i=1}^{No.of\, points}\) Point defining the polygon i
c) \(\frac{1}{Point\, defining\, the\, polygon}\sum_{i=1}^{No.of\, points}\) Point defining the polygon i
d) \(\frac{1}{No.of\, points}\sum_{i=1}^{No.of\, points}\)Centroid i
Answer: a
Explanation: The average of all the points that define the polygon is the geometric centre of the polygon. Therefore,
Geometric centre=\(\frac{1}{No.of points}\sum_{i=1}^{No.of points}\) Point defining the polygon i
4. Which of these formulae is used to find the area of the sub-elements in CFD?
a) Vector product of two sides
b) Half of the vector product of two sides
c) Half of the base times height
d) Half of the vector product of all three sides
Answer: b
Explanation: Sub-elements are in triangular shape. Half of the base times height is the formula generally used to find the area of a triangle. But, CFD uses the vector-based formula to make it algorithmically easier, which is given by half of the vector product of two sides.
5. The centroid of the faces of a 3-D element is obtained by _________
a) Area-weighted average of the sub-elements
b) Average of the sub-elements
c) Area-weighted average of the centroid of the sub-elements
d) Volume-weighted average of the centroid of the sub-elements
Answer: c
Explanation: The centroid of each of the sub-elements are first located. The weighted average of these centroids is the centroid of the whole face. Areas of the triangles are used as the weight for this average.
6. The surface area of the face of a 3-D element is a _____________
a) 3-D tensor
b) 2-D tensor
c) Scalar
d) Vector
Answer: d
Explanation: The surface area of the face of a 3-D element is a vector with the area as its magnitude and pointing in a direction. This direction decides if the face points outwards or inwards. Moreover, the area is the result of the cross product of two vectors which will again be a vector.
7. How is it identified whether a vector is pointing outwards or inwards?
a) Sign of the vector joining the element’s centroid with the face’s centroid is used
b) Sign of the surface vector is used
c) Cross product of the surface vector and the vector joining the element’s centroid with the face’s centroid
d) Dot product of the surface vector and the vector joining the element’s centroid with the face’s centroid
Answer: d
Explanation: The vector joining the centroid of the element with the centroid of the face always points outwards. So, the dot product of this vector with the surface vector is found. If the sign of the dot product is positive, the surface vector points outwards. Otherwise, it points inwards.
8. I know the value of the flow variable at two points. In which of these cases, is it easy for me to calculate the flow variable at a point between these two?
a) Three-dimensional FVM
b) Two-dimensional FVM
c) One-dimensional FVM
d) Two-dimensional FDM
Answer: c
Explanation: If the values at two points are known, in the one-dimensional case, it is easy to find the values at any other point. In the two-dimensional and three-dimensional cases, they pose complication to the calculation.
9. To find the value of a flow variable at a third point in between two points with known values, which of these methods can be used for the one-dimensional case?
a) Shape function
b) Interpolation
c) Taylor series
d) Fourier series
Answer: b
Explanation: If the value of the flow variable is known at two points and the value at the third point which lies in between these two is to be found, simple interpolation will be enough in the one-dimensional case. The value at any point in between these two points can be found using these two points.
10. Consider the diagram.
computational-fluid-dynamics-questions-answers-geometry-fvm-elements-q10
When is the formula Φ f =g f Φ F +(1-g f )Φ c ; g f =\
Only if Φ varies linearly
b) Only if Φ varies quadratically
c) Only if Φ varies cubically
d) Always
Answer: a
Explanation: The formula Φ f =g f Φ F +(1-g f )Φ c ; g f =\(\frac{distance_{cf}}{distance_{cf}+distance_{fF}}\) is the linear interpolation formula to find Φ f . If the variation of Φ is not linear (the three points Φ f , Φ c and Φ F are not collinear), the formula becomes invalid.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Turbulence Modelling”.
1. Which of these does not characterize a turbulent flow?
a) Time-independent
b) Rapid mixing
c) Three-dimensional fluctuation
d) Unstable
Answer: a
Explanation: Turbulent flows are chaotic, diffusive causing rapid mixing, time-dependent as they are unsteady, and involve vorticity fluctuation in all three-dimensions. Turbulence develops an instability in flows.
2. Which of these methods is not used for turbulence modelling?
a) RANS
b) SIMPLE
c) DNS
d) LES
Answer: b
Explanation: The SIMPLE algorithm is a widely used iterative solution technique for Navier-Stokes Equations for pressure linked systems. It is not used to solve turbulent models. The other techniques – RANS, DNS and LES – are all used to solve turbulent flows.
3. Which among these techniques is the first one invented for solving turbulent flows?
a) DNS
b) LES
c) RANS
d) FANS
Answer: a
Explanation: DNS is the first method used to solve the turbulent flow models. DNS stands for Direct Numerical Simulation. It uses the energy cascade concept developed by Kolmogorov. Its time-steps are limited by Courant number.
4. Which of these methods is invented to overcome the disadvantage of DNS technique?
a) Temporal discretization
b) FANS
c) RANS
d) LES
Answer: d
Explanation: The LES method is used to overcome the disadvantages of the DNS technique. LES stands for Large Eddy Simulation. This filters the Navier-Stokes equations to discard some scales and keep the rest.
5. What is the disadvantage of the DNS technique?
a) Time averaging
b) Spatial averaging
c) Computationally demanding
d) Large time-steps
Answer: c
Explanation: The time-step sizes in DNS technique is limited by courant number. So, it involves many steps to reach the actual interval needed to be crossed. This makes the technique computationally demanding. This is the disadvantage of DNS technique.
6. Which of these methods is used to overcome the high resolution of turbulent flows?
a) Weighted average
b) Statistical analyses
c) Data analysis
d) Analytical method
Answer: b
Explanation: Statistical analysis is used to simplify the resolution of turbulent flows. Statistical time averaging is used to reduce the random fluctuations in the time-dependent nature of turbulence flows.
7. Which of these correct about the LES method?
a) Navier-Stokes equation is not needed
b) Small turbulent models are directly simulated
c) Sub-grid scale models are used for small turbulent scales
d) Sub-grid scale models are used for large turbulent scales
Answer: c
Explanation: Large scale turbulent structures are directly simulated and the small scales are simulated using sub-grid models. A spatial statistical filter is used to filter the Navier-Stokes equation to determine which scale to keep and which to discard.
8. Which of these methods use time-based averaging?
a) Statistical averaging with DNS
b) DNS
c) LES
d) RANS
Answer: d
Explanation: The statistical averaging and LES methods use spatial statistical averaging. RANS stands for Reynolds Averaged Navier-Stokes equations where statistical averaging is based on time, unlike LES.
9. What is the expansion of FANS method?
a) Favre Averaged Navier-Stokes equations
b) Fully Averaged Navier-Stokes equations
c) Favre Averaged Numerical Simulation
d) Fully Averaged Numerical Simulation
Answer: a
Explanation: The FANS model is another method like RANS. FANS stands for Favre Averaged Navier-Stokes equations. This employs Favre averaging technique used for turbulent compressible flows. Both RANS and FANS use modified Navier-Stokes equations.
10. The weight used to average time in FANS method is _________
a) volume
b) mass
c) area
d) density
Answer: b
Explanation: The averaging in FANS method is a weighted average method which uses mass as the weight for it. Here, mass decides the importance of the terms and whether it can be filtered or not.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Turbulence Modelling – Turbulent Flows Characteristics”.
1. What is Reynolds stress?
a) Stress due to velocity fluctuations
b) Tangential component of pressure
c) Stress due to pressure fluctuations
d) Normal component of viscosity
Answer: a
Explanation: Turbulent flows are highly chaotic and unstable. The high fluctuation in the turbulent flows creates highly varying velocities. These velocities create additional stresses called Reynolds stress.
2. The mathematical technique used to represent the random nature of the turbulent flow.
a) RANS
b) Reynolds decomposition
c) Parallel decomposition
d) DNS
Answer: b
Explanation: The random nature of a turbulent flow needs a proper description of the motion of all the flow particles. Reynolds decomposition is used to represent this random nature. This decomposes the flow variables into two components.
3. Represent the velocity of turbulent flow using Reynolds decomposition.
a) Steady velocity + Mean velocity
b) Steady velocity + Fluctuating component of velocity
c) Variation in velocity + Fluctuating component of velocity
d) Mean variation + Fluctuating component of velocity
Answer: b
Explanation: Reynolds decomposition separates the steady mean component and some statistical properties of their fluctuations. Total velocity = Steady velocity + Fluctuating component of velocity.
4. Eddies in turbulent flows result in _________
a) high diffusion coefficients
b) less diffusion coefficients
c) high value of the source term
d) low value of the source term
Answer: a
Explanation: Because of the eddying motion of the turbulent flows, heat, mass and momentum are effectively exchanged. So, the values of diffusion coefficients of heat, mass and momentum are high.
5. Large turbulent eddies extract energy from the mean flow by this process.
a) Energy decomposition
b) Eddy extracting
c) Vortex stretching
d) Substantial variation
Answer: c
Explanation: Vortex stretching is the lengthening of vortices with a corresponding increase in the component of vorticity in the stretching direction. This vortex stretching is responsible for the largest turbulent eddies to interact with the mean flow and extract energy from them.
6. Which of these is correct for large eddies?
a) High viscosity and linear momentum are conserved
b) Low viscosity and linear momentum are conserved
c) High viscosity and angular momentum are conserved
d) Low viscosity and angular momentum are conserved
Answer: d
Explanation: Vortex stretching leads to conservation of angular momentum. As the large eddies are associated with vortex stretching, here angular momentum is conserved. They have a high Reynolds number which directly tells that they are relatively inviscid.
7. Which of these is highly energetic?
a) Kolmogorov micro-scale eddies
b) Small eddies
c) Medium eddies
d) Large eddies
Answer: d
Explanation: Energy content peaks at low wavenumber. At low wave numbers, eddies are large. The larger eddies are the most energetic as they acquire energy from the mean flow through direct interaction.
8. Large eddies are _________
a) two-dimensional and isotropic
b) two-dimensional and isotropic
c) three-dimensional and anisotropic
d) two-dimensional and anisotropic
Answer: c
Explanation: Turbulent flows are characterized by their three-dimensional fluctuation. So, the large eddies are also three-dimensional. The flow variables highly vary in all three directions. This leads to an anisotropic nature. Especially, large eddies are highly anisotropic.
9. Transfer of kinetic energy from large eddies to smaller eddies is called as _________
a) Energy cascade
b) Momentum cascade
c) Energy decomposition
d) Momentum decomposition
Answer: a
Explanation: Smaller eddies are stretched strongly by large eddies. Thereby, kinetic energy is transferred from larger eddies to smaller and smaller eddies. This is called energy cascade.
10. Reynolds number gives the relative importance of __________
a) viscous force and tangential force
b) inertia force and viscous force
c) inertia force and pressure force
d) pressure force and viscous force
Answer: b
Explanation: Reynolds number decides whether the flow is laminar or turbulent. Reynolds number of flow gives the relative importance of inertia force and viscous forces.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Turbulence Modelling – Turbulent Flow Structure”.
1. The rotational flow structure of turbulent flows is termed as ___________
a) Turbulent eddies
b) Haida eddies
c) Whirlpool
d) Wake turbulence
Answer: a
Explanation: The turbulent fluctuation has a three-dimensional spatial character. Visualizations of turbulent flows show that they have a highly rotational flow structure. These rotational flow structures are called Turbulent eddies.
2. According to Kolmogorov, the structure of the smallest eddies depends on ___________
a) rate of dissipation of kinetic energy
b) rate of dissipation of turbulent energy
c) rate of convection of kinetic energy
d) rate of convection of turbulent energy
Answer: b
Explanation: Kolmogorov stated that the structure spectral energy of the smallest eddies depends only on the rate of dissipation of turbulent energy. But later studies revealed that only the spectral energy of the smallest eddies depends only on the rate of dissipation of turbulent energy and not their structure.
3. The details about the structure of the fluctuations are contained in ___________
a) sum of different variables
b) sum of different pairs of variables
c) moments of different pairs of variables
d) moments of different variables
Answer: c
Explanation: Turbulent flows have complex rotating three-dimensional fluctuations. Details about the structure of these fluctuations are contained in moments constructed from pairs of different variables.
4. Consider the turbulent structure of thin shear layers. Which of these statements is correct?
a) The rates of change of the flow variables are very high in every direction
b) Flow variables do not vary much in any direction
c) The rates of change of the flow variables in the cross-sectional direction are smaller than that of the flow direction
d) The rates of change of the flow variables in the flow direction are smaller than that of the cross-sectional direction
Answer: d
Explanation: Considering turbulence in thin shear layers, large variations are concentrated in thin regions. The variables do not change much in the flow direction. But, the rates of change in the cross-sectional direction are more.
5. The structure of free turbulent flow is controlled by ___________
a) Only the local environment
b) The source
c) The sink
d) The source along with the local environment
Answer: a
Explanation: For a free turbulent flow, as the flow progresses, the effect of source degrades. Only the local environment tends to control the structure of the flow at any time. This is inferred from experiments.
6. For a turbulent flow, which of these is correct?
a) Fluctuating pressure is equal everywhere
b) Fluctuating velocity and pressure are not equal everywhere
c) Fluctuating velocity and pressure are equal everywhere
d) Fluctuating velocity is equal everywhere
Answer: b
Explanation: Turbulent flow structures are highly anisotropic. The flow variables are not the same in a particular direction. Therefore, the fluctuating velocities are also not the same everywhere. Instead, they show continuous variation.
7. Methods of time-averaging statistics are applicable only to ___________
a) Free turbulent structures
b) Boundary layer turbulent structures
c) Coherent turbulent structures
d) Incoherent turbulent structures
Answer: c
Explanation: For a time-averaged statistic to be applicable, the turbulent structure should have temporal coherence. Though turbulent flows are chaotic, they should be resolved into coherent structures to apply these methods.
8. The various fluid parcels come into contact in turbulent flow by ___________
a) Turbulent energy transfer
b) Turbulent vorticity
c) Turbulent dissipation
d) Turbulent diffusion
Answer: d
Explanation: Turbulence increases the rate at which conserved quantities are mixed. Here, different fluid parcels are brought into contact. As this mixing is accomplished by diffusion, the process is called turbulent diffusion.
9. Dissipation in turbulent flows converts ____________ energy to ____________ energy.
a) kinetic, internal
b) internal, kinetic
c) kinetic, viscous
d) viscous, kinetic
Answer: a
Explanation: Viscous effects reduce the velocity gradients and hence the kinetic energy of the flow. Thus, mixing is dissipative. This lost energy is converted into thermal internal energy of the flow. Thus, kinetic energy is converted into internal energy.
10. Direct Numerical Simulation of turbulent flows become difficult because of ____________
a) Viscosity scales
b) Time and length scales
c) Energy scales
d) Velocity scales
Answer: b
Explanation: High fluctuation in turbulent flows result in a broad range of length and time scales. This makes the Direct Numerical Simulation of turbulent flows difficult. It imposes a limitation to the DNS method.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Turbulence Modelling – Turbulent Scale”.
1. Small eddy scales are called as ___________
a) Batchelor scales
b) Taylor micro-scales
c) Kolmogorov micro-scales
d) Integral length scales
Answer: c
Explanation: The smallest eddies with dominating viscosity are called Kolmogorov micro-scales. It is named after the Russian scientist Andrey Kolmogorov who carried out works on the structure of turbulence in the 1940s.
2. What do the length, velocity and time-scale ratios mean?
a) The ratio of large-scale and characteristic properties
b) The ratio of small-scale and characteristic properties
c) The ratio of small and large scale eddies
d) The ratio of actual and characteristic properties
Answer: c
Explanation: The time-scale ratio is the ratio between small-scale time and large-scale time . Similarly, the length and velocity-scale ratios are also the ratios between small and large-scale lengths and velocities.
3. What is the relationship between Length-scale ratio and Reynolds number ?
a) Re 3/4
b) Re -3/4
c) Re 1/2
d) Re -1/2
Answer: b
Explanation: Length scale ratio is obtained by estimation of the dissipation rate in the large scale flow features.
Length-scale ratio= Re -3/4 .
4. Small-scale eddy motions have ___________
a) does not vary from the energy losses of the large-scale eddy motions
b) the same order of energy losses as the large-scale eddy motions
c) decreased energy losses
d) increased energy losses
Answer: d
Explanation: Small-scale eddy motions are dissipative. The energy associated with these motions is converted into thermal internal energy. This leads to increased energy losses associated with the turbulent flows.
5. The Reynolds number associated with the smallest scale of motion in the turbulent flow is ____________
a) 0.1
b) 1
c) 10
d) 100
Answer: a
Explanation: The smallest scales of motion in the turbulent flows are dominated by viscous flows. They are associated with a Reynolds number 1. The smallest scales are those for which inertia and viscous flows are the same.
6. Express the time-scale ratio in terms of Reynolds number .
a) Re -1/4
b) Re 3/4
c) Re -1/2
d) Re 1/2
Answer: c
Explanation: The ratio of time-scales is obtained using the length scale ratios. It is given by
Time-scale ratio= Re -1/2 .
7. To establish the relationship between turbulent scales and the Reynolds number, which of these methods is used?
a) Statistical averaging
b) Dimensional analysis
c) Weighted averaging
d) Geometric algebra
Answer: b
Explanation: The ratio of large and small scales of turbulent flows can be given in terms of the Reynolds number. These relations are established using dimensional analysis. Dimensional analysis is a method which compares the dimensions in both the sides of the equations to set the relationship.
8. The behaviour of large eddies depend on __________
a) Viscosity and time scales
b) Velocity and time scales
c) Time and length scales
d) Velocity and length scales
Answer: d
Explanation: Velocity and length of the large eddies will be large. The variation of flow with time is not much. Thus, they are not dependent much on the time scales. Instead, they are dependent on the velocity and length scales.
9. When the Reynolds number increases, the difference between the large and small scales ____________
a) increases
b) decreases
c) remains constant
d) cannot be defined
Answer: a
Explanation: As the Reynolds number increases, the flow becomes more turbulent. This increases the difference between the large and small eddies . It is called scale separation.
10. Which of these is equal to the velocity-scale ratio?
a) Re -3/4
b) Re -1/3
c) Re -1/4
d) Re -2/4
Answer: b
Explanation: Once the length and time-scale ratios are known, we can get the velocities scale ratios using these two. It is given by
Velocity-scale ratio = Re -1/4 .
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Turbulence Modelling – Kolmogorov Energy Spectrum”.
1. Kolmogorov micro-scales can be expressed in terms of ___________
a) Rate of dissipation of turbulent energy and fluid viscosity
b) Turbulent energy and fluid velocity
c) Fluid velocity and viscosity
d) Turbulent energy and fluid viscosity
Answer: a
Explanation: Kolmogorov micro-scales can be expressed in terms of the rate of energy dissipation of the turbulent flow and the fluid viscosity. It uses the statement that the rate of production of turbulent energy and the rate of dissipation should be balanced.
2. Kolmogorov spectral energy is a function of ____________
a) Velocity
b) Wavenumber
c) Kinematic viscosity
d) Dynamic viscosity
Answer: b
Explanation: Kolmogorov spectral energy is, in general, a function of the wavenumber. The wavenumber is, in turn, a function of wavelength given by
κ=\(\frac{2\pi}{\lambda}\).
3. Spectral energy is equal to ___________
a) Kinetic energy per unit mass per unit wavenumber
b) Rate of dissipation of turbulent energy per unit mass per unit wavenumber
c) Rate of dissipation of turbulent energy per unit wavenumber
d) Kinetic energy per unit wavenumber
Answer: a
Explanation: All the fluctuating properties of a turbulent flow contain some energy. The spectral energy can be given as the kinetic energy per unit mass per unit wavenumber of fluctuations around the wavenumber.
4. What is the unit of spectral energy?
a) \
\
\
\(\frac{m^2}{s^2}\)
Answer: b
Explanation: Spectral energy is the kinetic energy per unit mass per unit wavenumber. So, the unit of spectral energy is given by
\(\frac{kg m^2}{s^2}×\frac{1}{kg}×m=\frac{m^3}{s^2}\).
5. What is the range of length and frequency of the Kolmogorov micro-scale eddies respectively?
a) 0.1 to 1 mm, around 1 kHz
b) 1-10 mm, around 10 kHz
c) 0.01 to 0.1 mm, around 10 kHz
d) around 10 mm, 0.1 to 0.01 kHz
Answer: c
Explanation: The smallest scales of motion in a turbulent flow is called the Kolmogorov micro-scale. They have a wavelength of around 0.01 to 0.1 mm and frequencies around 10 kiloHertz. They have Reynolds number very near to one.
6. If ν is the kinematic viscosity and ε is the rate of dissipation of turbulent energy, to which of these terms is the spectral energy of Kolmogorov micro-scale eddies proportional to?
a) ν 5/3 ε 1/3
b) ν 3/2 ε 1/2
c) ν 1/2 ε 3/2
d) ν 5/4 ε 1/4
Answer: d
Explanation: Kolmogorov argued that the behaviour of the smallest turbulent eddies depends on the rate of dissipation of turbulent energy. But, later studies revealed that only the spectral energy of the smallest turbulent eddies depends on the rate of dissipation of turbulent energy and the relationship is given by
spectral energy ∝ ν 5/4 ε 1/4 .
7. The size of the eddies and the wavenumber are __________
a) inversely proportional
b) directly proportional
c) not related to each other
d) related but it varies according to the energy
Answer: a
Explanation: Wavenumber is the frequency of eddies and the size will be related to the wavelength. So, the size and wavenumber are inversely proportional. The large eddies have low wavenumber and the small eddies have high wavenumbers.
8. ____________ is associated with high wavenumbers.
a) Conduction
b) Dissipation
c) Kinetic energy
d) Potential energy
Answer: b
Explanation: High wavenumbers represent small eddies. Dissipation is associated with small eddies and kinetic energy is associated with large eddies. This is given by the relation between energy and wavenumber.
9. _____________ are called the inertial sub-range of turbulence.
a) Kolmogorov micro-scale
b) Small scale eddies
c) Intermediate scale eddies
d) Large scale eddies
Answer: c
Explanation: Transfer of energy from large eddies to small eddies is called the energy cascade. The transfer brings turbulent kinetic energy from large scales to small scales. The intermediate range in this transfer is called the inertial sub-range.
10. If κ is the wavenumber and ε is the rate of dissipation of turbulent energy, which of these is proportional to the spectral energy of the inertial sub-range of turbulence?
a) κ -5/3 ε -2/3
b) κ 5/3 ε -2/3
c) κ 5/3 ε 2/3
d) κ -5/3 ε 2/3
Answer: d
Explanation: The spectral energy in terms of the wavenumber and the rate of dissipation of turbulent energy is given by E ∝ κ -5/3 ε 2/3 . Where the proportionality constant is 1.5
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Free and Wall Turbulence”.
1. The process which is responsible for spreading of turbulent flows in the flow direction is __________
a) Pluming
b) Entrainment
c) Turbulent mixing
d) Turbulent generation
Answer: b
Explanation: While turbulent flows burst out of its region, fluid from the surrounding is drawn into the turbulent region. This is the process of entrainment. This is the cause of spreading of turbulent flows in the flow direction.
2. Which of these terms represent the burst of turbulent activity to the outer region?
a) Crisis
b) Intermittency
c) Turbulent burst
d) Turbulent jumps
Answer: b
Explanation: Intermittency is the irregular alteration of phases. In turbulent flows, it is seen in the irregular alteration between the turbulent and non-turbulent region of jet flow. Intermittency represents the burst of turbulent activity to the surrounding region.
3. A turbulent jet is formed because ___________
a) A region of high-speed flow is surrounded by a stationary fluid
b) An object disturbs the flow
c) Interaction between an object and a fast moving fluid
d) Interaction of fast and slow moving fluids
Answer: a
Explanation: In mixing layers, the interaction of fast and slow moving fluids create turbulence. A turbulent wake is created by an object which disturbs the flow. When a high-speed flow is surrounded by a stationary fluid, the turbulent jet is formed.
4. The velocity of a free turbulent flow at any particular distance in the cross-stream direction is a function of ___________
a) The ratio of the distance in the cross-stream direction from the centreline and half-width at that cross section
b) The source velocity
c) The cross-stream velocity of the source
d) The velocity in the flow direction of the source
Answer: a
Explanation: The velocity at any point in the cross-stream direction at a particular cross section depends on the ratio of the distance from the centreline and the half width of the cross-stream.
5. Turbulent entrainment leads to ___________
a) Increase in the magnitude of the velocity gradients in the flow direction
b) Increase in the magnitude of the velocity gradients in the cross-stream direction
c) Decrease in the magnitude of the velocity gradients in the flow direction
d) Decrease in the magnitude of the velocity gradients in the cross-stream direction
Answer: c
Explanation: Because of the entrainment of the surrounding fluid, the velocity gradients decrease in magnitude in the flow direction. This also decreases the difference between the speed of the wake fluid and its surroundings.
6. Which of these is correct for mixing layer turbulent flows?
Note:
U max → Maximum velocity at a particular cross-section
U min → Minimum velocity at a particular cross-section
y → Distance in the cross-stream direction from the centre line at the cross section
b → Cross-stream half width
U → Velocity at the distance ‘y’.
a) \
\)
b) \
\)
c) \
\)
d) \
\)
Answer: d
Explanation: For mixed flows, the velocity is dependent on minimum velocity. This corresponds to the velocity of the slow-moving fluid. Maximum velocity corresponds to the fast-moving fluid. The function is given by
\
\).
7. If y is the distance in the cross-stream direction from the centre line at a particular cross-section b → Cross-stream half width. The mean velocity gradients and all the velocity fluctuations become zero when the value \Missing open brace for subscript goes above unity
b) goes below unity
c) goes above zero
d) goes below zero
Answer: a
Explanation: The value \(\frac{y}{b}\) going above 1 means that y crosses b and goes out of the turbulent region. So, the velocity gradients and its fluctuations also will become zero when the value \(\frac{y}{b}\) goes above 1.
8. Which of these is correct for turbulent jets?
Note:
U max → Maximum velocity at a particular cross-section
U min → Minimum velocity at a particular cross-section
y → Distance in the cross-stream direction from the centre line at the cross section
b → Cross-stream half width
U → Velocity at the distance ‘y’.
a) \
\)
b) \
\)
c) \
\)
d) \
\)
Answer: d
Explanation: For a turbulent jet, the minimum velocity is zero corresponding to the stationary surrounding fluid. So, the equation becomes
\
\).
9. The mean velocity gradient is zero at the centreline for ___________
a) jets
b) mixing flows
c) mixing flows and wakes
d) jets and wakes
Answer: d
Explanation: For turbulent flows at jets and wakes, the sign must change at the symmetry line. The symmetry line is the centreline here. For the sign change to be possible, the velocity gradients should reach zero at this line.
10. The velocity at cross-stream of a turbulent wake is calculated using the formula
\
\)
Note:
U max → Maximum velocity at a particular cross-section
U min → Minimum velocity at a particular cross-section
y → Distance in the cross-stream direction from the centre line at the cross section
b → Cross-stream half width
U → Velocity at the distance ‘y’
The minimum velocity here corresponds to _____________
a) Velocities at the edges
b) Velocities just downstream of the object
c) Velocities of the surrounding free stream
d) Velocities at the centreline
Answer: b
Explanation: \
\) is the formula used to calculate the velocity at a distance from the centre-line in the cross-stream direction. The minimum velocities here corresponds to the starting of the wake, which is just downstream of the object.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Turbulent Viscosity”.
1. The value of turbulent viscosity is fairly close to that of __________
a) Turbulent diffusivity
b) Newton’s viscosity
c) Kinematic viscosity
d) Dynamic viscosity
Answer: a
Explanation: Transport of momentum is due to viscosity and transport of heat or mass is due to diffusivity. In turbulent flows, both of these are due to the same mechanism which is eddy mixing. So, the value of turbulent viscosity is considered to be close to that of turbulent diffusivity.
2. Which of these scientists introduced turbulent viscosity?
a) Kolmogorov
b) Smagorinsky
c) Prandtl
d) Boussinesq
Answer: d
Explanation: Boussinesq introduced the concept of eddy viscosity or turbulent viscosity in turbulent flows. Boussinesq equated the turbulent stresses to the mean flow. Here, the new constant of proportionality called the turbulent viscosity was introduced.
3. The kinematic turbulent viscosity is __________
a) equal to the product of turbulent length and time scales
b) equal to the product of turbulent length and velocity scales
c) proportional to the product of turbulent length and velocity scales
d) proportional to the product of turbulent length and time scales
Answer: c
Explanation: By using dimensional analysis, the relationship between kinematic turbulent viscosity, turbulent length and time scales can be established using the dimensional analysis. It can be expressed as
ν t ∝vl
Where,
ν t → Kinematic turbulent viscosity.
v → Turbulent velocity scale.
l → Turbulent length scale.
4. The dynamic turbulent viscosity is ___________
a) equal to the product of turbulent length and time scales
b) proportional to the product of turbulent length and time scales
c) proportional to the product of kinematic turbulent viscosity and density of the fluid
d) equal to the product of kinematic turbulent viscosity and density of the fluid
Answer: d
Explanation: Dynamic viscosity is the product of kinematic viscosity and density in general. This applies to the turbulent viscosities also.
μ t =ρν t
Where,
ν t → Kinematic turbulent viscosity.
μ t → Dynamic turbulent viscosity.
ρ → Density.
5. __________ relates turbulent viscosity and diffusivity.
a) Reynolds number
b) Reynolds analogy
c) Reynolds-Averaged Navier-Stokes equations
d) Favre-Averaged Navier-Stokes equations
Answer: b
Explanation: Reynolds analogy relates the turbulent momentum and heat transfer. It states that “both of these are due to the same mechanism called turbulent eddies and hence the values of turbulent viscosity and diffusivity will be close to each other”.
6. Which of these models solves a system for the turbulent kinematic viscosity?
a) DNS
b) LES
c) Spalart-Allmaras
d) RANS
Answer: c
Explanation: Spalart- Allmaras is a model for solving turbulent flow especially invented for aerospace problems. It gives good results for turbulent boundary layer models. It solves the transport equation for turbulent kinematic viscosity.
7. The units of kinematic and dynamic turbulent viscosities are ___________ respectively.
a) m/s 2 and kg m/s
b) m 2 /s 2 and kg/m s
c) m 3 /s and kg m/s
d) m 2 /s and kg/m s
Answer: d
Explanation: The units of turbulent viscosities are the same as the viscosities given by Newton’s law. The unit of kinematic turbulent viscosity is m 2 /s. The unit of dynamic viscosity is kg/m s.
8. Which of these is not a turbulent viscosity model?
a) k-ω
b) k-ε
c) DNS
d) SST
Answer: c
Explanation: The models k-ε, k-ω and SST are all models which include turbulent viscosity. DNS stands for Direct Numerical Simulation which is the basic model to solve turbulent flows. It does not involve turbulent viscosities.
9. In the k-ε model, the turbulent viscosity is given as ___________
a) μ t ∝k/ε
b) μ t ∝k 2 /ε
c) μ t =k/ε
d) μ t =k 2 /ε
Answer: b
Explanation: In the k-ε model, the turbulent dynamic viscosity is given in terms of k and ε. The relation is given as
μ t ∝k 2 /ε.
10. Which of these models is made different from its parent model by turbulent viscosity?
a) Realizable k-ε model
b) k-ε model
c) Spalart-Allmaras model
d) SST model
Answer: a
Explanation: The realizable k-ε model is a variant of the k-ε model. The realizable k-ε model is different from its parent that a new turbulent viscosity is formulated. Another variation is that a new transport equation for the dissipation rate is formed.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Turbulent Schmidt Number”.
1. Turbulent Schmidt number is the ratio of ____________
a) turbulent viscosity to turbulent diffusivity
b) turbulent diffusivity to turbulent viscosity
c) turbulent rate to turbulent diffusivity
d) turbulent diffusivity to turbulent rate
Answer: a
Explanation: Turbulent Schmidt number gives the ratio of turbulent transfer of momentum to the turbulent transfer of mass. It is given by
Turbulent Schmidt number=\(\frac{Turbulent\, viscosity}{Turbulent\, diffusivity}\).
2. Turbulent Prandtl number is the ratio of ____________
a) turbulent transport of heat to turbulent transport of momentum
b) turbulent transport of momentum to turbulent transport of heat
c) turbulent viscosity to turbulent transport of heat
d) turbulent transport of heat to turbulent viscosity
Answer: b
Explanation: Prandtl number is the ratio of transport of momentum to transport of heat. Turbulent Prandtl number is
Turbulent Prandtl number=\(\frac{Turbulent\, viscosity}{Turbulent\, diffusivity}\).
3. What is the unit of turbulent Schmidt/Prandtl number?
a) m 2 /s
b) It is dimensionless
c) m 2 /s 2
d) m/s 2
Answer: b
Explanation: Both viscosity and diffusivity are in the same units. As the turbulent Schmidt or Prandtl number is the ratio of viscosity to conductivity, the number becomes dimensionless. So, it does not have units.
4. According to Reynolds analogy, what is the value of turbulent Schmidt number?
a) -1
b) 0
c) 1
d) ∞
Answer: c
Explanation: Reynolds analogy states that “because of eddy mixing, the values of turbulent viscosity and turbulent diffusivity will be fairly close to each other”. So, dividing both we will get unity. Therefore, turbulent Schmidt or Prandtl number becomes unity.
5. Which of these is correct when the turbulent Prandtl number is unity?
a) Turbulent diffusivity is zero
b) Turbulent viscosity is zero
c) The flow becomes laminar
d) The velocity and temperature profiles are identical
Answer: d
Explanation: When the turbulent Prandtl number is one, both turbulent viscosity and turbulent diffusivity are the same. So, the velocity and temperature profiles near the solid boundary or wall become identical.
6. Turbulent Prandtl number is used in CFD to ____________
a) change viscosity profiles
b) change temperature profiles
c) modulate heat transfer results
d) modulate velocity profiles
Answer: c
Explanation: Prandtl number relates momentum transfer and heat transfer in a turbulent flow. So, it is used in turbulent flows with heat transfer. In general CFD problems, it is used to tune the heat transfer results.
7. Turbulent Prandtl number is useful in __________
a) SST
b) DNS
c) LES
d) RANS
Answer: d
Explanation: RANS is Reynolds-Averaged Navier-Stokes equations. RANS is a method used to solve the turbulent flows with modified Navier-Stokes equations. The turbulent Prandtl number is used here to solve the problem.
8. Turbulent Schmidt number is used to solve ____________
a) mass transfer in a turbulent boundary layer
b) compressible flows
c) boundary layer flows
d) heat transfer in a turbulent boundary layer
Answer: a
Explanation: As the turbulent Schmidt number relates momentum and mass transports, it is used to solve mass transfer in turbulent boundary layers. In general, it is used along with the Reynolds analogy.
9. The range of turbulent Schmidt number is ___________
a) 0.2 to 3.5
b) 0.2 to 1.5
c) 1 to 3.5
d) 0 to 0.2
Answer: a
Explanation: According to the experiments, the turbulent Schmidt number ranges from 0.2 to 3.5. Only when using the Reynolds analogy, the approximation becomes a value near unity. Otherwise, this value varies.
10. The range of turbulent Prandtl number is ___________
a) 0.5 to 0.7
b) 0.7 to 0.9
c) 0.9 to 1.5
d) 1.5 to 1.7
Answer: b
Explanation: The average value of the turbulent Prandtl number is 0.85. The experimental turbulent Prandtl number value varies from 0.7 to 0.9. It stays below unity.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Turbulent Boundary Layer”.
1. For flows over a flat plate, at length scales near to the length of the flat plate, which of these is correct?
a) Inertial force is zero
b) Inertial force is large
c) Inertial force is equal to viscous force
d) Viscous force is large
Answer: b
Explanation: Reynolds number depends on the length scale taken for the calculation. At the length scales near to that of the length of the flat plate, the Reynolds number will be high. Therefore, the inertial forces will be large.
2. Which of these statements is correct?
a) Inertia forces dominate in the flow far from the wall
b) Viscous forces dominate in the flow far from the wall
c) Inertia forces are small in the flow far from the wall
d) Viscous forces are large in the flow near the wall
Answer: a
Explanation: As the flow of a fluid near the wall moves away from the wall, the Reynolds number of the flow increases accounting to the increase in the distance. This leads to high inertial forces and low viscous forces.
3. Which of these laws define the dimensionless quantities u + and y + ?
a) Velocity-defect law
b) Log-law
c) Newton’s law of viscosity
d) Law of the wall
Answer: d
Explanation: The law of the wall is the relationship between the mean flow velocity and the distance from the wall derived using dimensional analysis. This gives the relationship between two important dimensionless quantities u + and y + .
4. What is u + ?
a) The ratio of velocity parallel to the wall to the friction velocity
b) The ratio of the friction velocity to velocity parallel to the wall
c) The ratio of free-stream velocity and friction velocity
d) The ratio of friction velocity and free-stream velocity
Answer: a
Explanation: u + is the dimensionless velocity. It is defined as the ratio of the velocity of fluid particle parallel to the wall at a particular distance from the wall to the friction velocity. Friction velocity is the square root of the ratio of shear stress to the density of fluid.
5. The velocity at a point far away from the wall is defined by ____________
a) Power law
b) Log-law
c) Velocity-defect law
d) Newton’s law of viscosity
Answer: c
Explanation: Far away from the wall the fluid flow is retarded by the wall shear stress. The velocity at such a point is defined as
\
\)
This is called velocity-defect law.
6. The fluid layer which is in contact with a smooth wall is called ____________
a) Inviscid layer
b) Linear sub-layer
c) Log-law layer
d) Wake-law layer
Answer: b
Explanation: In the fluid layer which is in contact with a smooth wall, the value of dimensionless velocity and dimensionless cross-stream distance tend to be the same. Because of this linear relationship, the layer is named linear sub-layer.
7. What is the range of y + in the viscous sub-layer?
a) 0<y + <20
b) 0<y + <5
c) 0<y + <10
d) 0<y + <15
Answer: b
Explanation: This is the layer which is in immediate contact with the smooth wall. It obeys Newton’s law of viscosity. The shear stress in this layer is constant and approximately equal to that of the wall. It extends from the wall till y + reaches 5.
8. The layer with viscous and turbulent stresses in equal magnitude is called _____________
a) Viscous sub-layer
b) Log-law layer
c) Buffer layer
d) Velocity-defect layer
Answer: c
Explanation: The layer above the linear sub-layer has equally important turbulent and viscous stresses. Neither of these is dominating nor inconsiderable. A layer in this area where both the viscous and turbulent stresses are of equal magnitude is called the buffer layer.
9. What is the other name of the velocity-defect law?
a) Linear law
b) Log law
c) Law of the wall
d) Law of the wake
Answer: d
Explanation: Velocity defect law is applicable to the layer far away from the wall. This layer has less viscous effects and inertia forces are dominating here. The velocity-defect law is otherwise called the law of the wake.
10. What is the range of y + in the log-law layer?
a) 30<y + <500
b) 20<y + <500
c) 30<y + <400
d) 20<y + <400
Answer: a
Explanation: The log-law layer extends outside the viscous or linear sub-layer. Here, both viscous and turbulent effects are important. It ranges between 30<y + <500. It is called log-law layer because of the logarithmic relationship between u + and y + .
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Turbulence Modelling – Averaging Methods”.
1. The methods of averaging are collectively called as ______________
a) Reynolds averaging
b) Boussinesq averaging
c) Kolmogorov averaging
d) Schmidt averaging
Answer: a
Explanation: The averaging techniques include time averaging, Spatial averaging and ensemble averaging. These are collectively called the methods of averaging. They are used to simplify the algebra without actually disturbing the accuracy much.
2. What are the methods of averaging used to?
a) To decompose the flow variable
b) To get the mean component of the flow variable
c) To get the remove the fluctuating component
d) To solve the flow variables
Answer: b
Explanation: By Reynolds decomposition, the flow variables are decomposed into mean and fluctuating components. These methods of averaging are used to get the mean component during this decomposition.
3. Which of these averaging methods is useful for any kind of turbulent flows?
a) Ensemble averaging
b) Time averaging
c) Volume averaging
d) Spatial averaging
Answer: a
Explanation: Ensemble averaging is a method used in statistical mechanics. Here, it is used as one of the methods of averaging. This is suitable for any type of turbulent flows including unsteady turbulent flows.
4. Which of these represent time averaging?
a) \
\
\
\(lim_{T→∞}\frac{1}{T}\int_t^{t+T}\phi dt\)
Answer: d
Explanation: Time averaging represents the average of the flow variable based on a time interval ‘T’. It uses integration to sum up the flow variables at different times and then divides by the time interval \(lim_{T→∞}\frac{1}{T}\int_t^{t+T}\phi dt\).
5. The governing equations which are averaged using these methods of averaging are used in _____________
a) DNS model
b) SST model
c) RANS model
d) k-ε model
Answer: c
Explanation: The governing Navier-Stokes equation is averaged using the Reynolds averaging techniques and these averaged equations are used in the RANS method. This is the reason why the technique is named the Reynolds-Averaged Navier-Stokes equations method.
6. Time averaging method is useful for ____________
a) unsteady turbulent flows
b) steady turbulent flows
c) turbulent boundary layer flows
d) mixing flows
Answer: b
Explanation: Time averaging method is useful when we have to decompose the turbulent flow variables into mean and fluctuating components based on time. They are particularly applicable for steady turbulent flows.
7. Which of these represent spatial averaging?
a) \
\
\
\(\frac{1}{N}\int_N \phi dN\)
Answer: b
Explanation: Spatial averaging represents the mean based on a particular space interval or volume. So, equation \(lim_{V→∞}\frac{1}{V}\int_V \phi dV\) represents spatial averaging.
8. Ensemble averaging represents the average of ____________
a) unsteady quantities
b) steady quantities
c) identical quantities
d) mean quantities
Answer: c
Explanation: This is useful for identical quantities. Identical in the sense, that they have similar properties in some concern. A number of quantities which have the simultaneous variations can be averaged using this method.
9. Which of these represent ensemble averaging if ‘N’ represents the number of identical quantities?
a) \
\
\
\(lim_{N→∞}\frac{1}{N} \sum_{i=1}^N\phi_i \)
Answer: d
Explanation: Ensemble averaging is based on identical flow variables. It sums up the identical variables and then takes the average. The equation is \(lim_{N→∞}\frac{1}{N} \sum_{i=1}^N\phi_i \).
10. Spatial averaging is suitable for ____________
a) homogeneous turbulent flows
b) unsteady turbulent flows
c) turbulent boundary layer flows
d) mixing flows
Answer: a
Explanation: Spatial averaging finds the average of a quantity based on a spatial interval or volume. It is suitable for homogeneous turbulent flows. In homogeneous flows, the properties are invariant under the arbitrary translation of the coordinate axes.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Turbulence Modelling – Averaging Rules”.
1. These rules for averaging are used to average ___________
a) fluctuations in the turbulent flow
b) variation in results of turbulent flow
c) the coefficients in FVM
d) the coefficients in FDM
Answer: a
Explanation: The flow variables in a turbulent flow are divided into mean and fluctuating components. These fluctuating components in the turbulent flow are averaged for the further solution of the system. These rules are used for averaging.
2. According to the rules for averaging, which of these will sum up to zero?
a) The mean component of the flow variable
b) The fluctuating component of the flow variable
c) The flow variable
d) Integration of the flow variable
Answer: a
Explanation: The mean component of a flow variable is the overall average of the flow variable. So, when the average of the flow variable is its mean component, the average of the fluctuating component and hence its summation will be zero.
3. The average of the mean component will be ____________
a) equal to zero
b) equal to the mean component itself
c) equal to 1
d) equal to the fluctuating component
Answer: b
Explanation: The mean component is already found by taking the arithmetic mean of the flow variables. So, if the average of only the mean component is taken, it will again be the same mean component itself.
4. The mean of the spatial partial derivative of a flow variable will be equal to ____________
a) 0
b) 1
c) the spatial partial derivative of the mean component
d) the mean component
Answer: c
Explanation: The mean of the flow variable will be equal to the mean variable. The mean of the flow variable’s spatial partial derivative will be equal to the spatial partial derivative of the mean component of that variable.
5. The mean of the summation of two flow variables will be equal to ____________
a) the summation of their mean components – the summation of the mean of their fluctuating components
b) the summation of their mean components + the summation of the mean of their fluctuating components
c) the summation of their fluctuating components
d) the summation of their mean components
Answer: d
Explanation: Consider two flow variables which can be decomposed as a=A+a’ and b=B+b’. The mean of their summation means
a+b = A+a’+B+b’
But, a’ = 0 and b’ = 0. Therefore,
a+b = A+B
Also, A = A and B = B. Hence,
a+b = A+B.
6. The mean of the space-based integral of a flow variable is equal to ____________
a) the summation of its mean component
b) the space-based integral of its fluctuating component
c) the space-based integral of its mean component
d) the summation of its fluctuating components
Answer: c
Explanation: As the mean of the fluctuating component is zero and the mean of the mean component is the mean component itself, the mean of the space-based integral of a flow variable is equal to the space-based integral of its mean component alone.
7. The mean of the product of the mean component of one variable and the fluctuating component of another variable is ____________
a) 1
b) 0
c) the product of their mean components
d) the product of their fluctuating components
Answer: b
Explanation: The mean of a fluctuating component is zero. The mean of a mean component is a variable. So, the mean of the product of the mean component of one variable and the fluctuating component of another variable will become zero.
8. The mean of the product of a flow variable and the mean component of another flow variable is ____________
a) the product of their mean components
b) the product of their fluctuating components
c) the mean of the product of their mean components
d) the mean of the product of their fluctuating components
Answer: a
Explanation: Consider two flow variables which can be decomposed as a=A+a’ and b=B+b’. The mean of the product of one flow variable and the mean component of another flow variable is represented as
aB =
B
aB = AB+a’B
As a’B =0 and AB =AB,
aB =AB.
9. Consider a vector flow variable which can be decomposed as \Missing open brace for subscript div \
\
\
\(\overline{div \vec{A}}\)
Answer: a
Explanation: From the given problem,
\(\overline{div \,\vec{a}}=div\overline{\vec{a}}=div\overline{\vec{A}+\vec{a’}} = div\overline{\vec{A}}=div\vec{A}\).
10. Consider two flow variables which can be decomposed as a=A+a’ and b=B+b’. What is ab ?
a) 0
b) 1
c) AB
d) a’b’
Answer: d
Explanation: For fluctuating variables,
ab =
ab =
But, a’B =0, Ab’ = 0 and AB = AB. So,
ab = AB+ a’b’ .
This set of Computational Fluid Dynamics MCQs focuses on “Direct Numerical Solution for Turbulent Models”.
1. Which of these equations is the starting point of the DNS method?
a) Continuity and momentum equations of homogeneous turbulent flow
b) Continuity and momentum equations of incompressible turbulent flow
c) Momentum and energy equations of incompressible turbulent flow
d) Momentum and energy equations of homogeneous turbulent flow
Answer: b
Explanation: The instantaneous continuity and Navier-Stokes equations for an incompressible turbulent flow form the initial point of the Direct Numerical Solution method. These equations form a closed set of four equations with the four unknowns .
2. DNS can solve _____________
a) transient 3-D equations
b) steady-state 3-D equations
c) transient 2-D equations
d) steady-state 2-D equations
Answer: a
Explanation: The starting set of equations is taken by the DNS system and the transient 3-D solution is done for a sufficiently fine spatial mesh and sufficiently small time-step sizes to resolve even the smallest turbulent eddies.
3. The grid size and time-step size of the DNS method depends upon the _____________
a) Schmidt number
b) Peclet number
c) Nusselt number
d) Reynolds number
Answer: d
Explanation: The grid size and time-step size of the DNS method is based on the largest and the smallest length and time scales of eddies in a turbulent flow. This, in turn, depends on the Reynolds number of the flow. So, Reynolds number decides the grid size and time-step size here.
4. When is the DNS method apt?
a) For complex flows
b) For design purposes
c) For precise details
d) For economic simulation
Answer: c
Explanation: The DNS method is used when we want the precise simulation with all the details in it. They are used for the development and validation of other turbulent models. The transport of any flow variable at any point can be precisely obtained using this model.
5. For which of these purposes, the DNS method is not suitable?
a) Designing
b) Simulation of the production of aerodynamic noise
c) Effects of compressibility on turbulence
d) To understand the mechanism of turbulence
Answer: a
Explanation: DNS method cannot be used to make the aerodynamic design of a model. They are computationally very expensive. To employ a model in designing, the same simulation should be done many times to improve the design. As DNS models are expensive, they cannot be used repeatedly.
6. DNS method is applicable for __________
a) Complex geometry and low Reynolds number
b) Simple geometry and low Reynolds number
c) Simple geometry and high Reynolds number
d) Complex geometry and high Reynolds number
Answer: b
Explanation: The DNS method needs highly refined grids to capture the detailed flow. So, it is not suitable for complex geometries as it will make the system more complicated. As the Reynolds number increases, the number of grids and time-steps also will increase. This will lead to higher complexity again.
7. Which of these is the simplest type of turbulent flows?
a) Homogeneous anisotropic turbulence
b) Incompressible turbulent flows
c) Homogeneous isotropic turbulence
d) Compressible turbulent flows
Answer: c
Explanation: The homogeneous isotropic turbulence is the simplest model of turbulence problems. It just needs a uniform grid to simulate the flow. However, the number of grids depends upon the Reynolds number of the flow.
8. The Reynolds number of a 3-D turbulent flow is 10 4 . What is the number of grid points needed?
a) 10 3
b) 10 4
c) 10 6
d) 10 9
Answer: d
Explanation: The ratio of the largest to the smallest length scale gives the number of grid points needed. The number of grid points needed in each direction is 10 3 . The total number of grid points needed in all three directions is 10 9 .
9. The Reynolds number of a 3-D turbulent flow is 10 4 . What is the number of time-steps needed?
a) 100
b) 1000
c) 10 4
d) 10 5
Answer: a
Explanation: The number of time steps depends on the ratio of the largest to the smallest time scales. This ratio depends on the Reynolds number. The times steps needed is Re 1/2 . Here, it is (10 4 ) 1/2 , which is equal to 100. 100-time steps are needed to solve this problem.
10. Which of these methods which are used for time advance discretization in DNS needs more computation per time step?
a) Adams-Bashforth
b) Runge-Kutta method
c) Leapfrog
d) Newton Raphson method
Answer: b
Explanation: Runge-Kutta method is generally used to advance in time in the DNS method. They need more computation time per unit step. In spite of this disadvantage, they are preferred because of their accuracy.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Large Eddy Simulation for Turbulent Models”.
1. LES is preferred when _____________
a) the flow is compressible
b) the flow has a high Reynolds number
c) the flow is turbulent
d) the flow has heat transfers
Answer: b
Explanation: Large Eddy Simulation method is useful to capture the large turbulent eddies. It is not as accurate as the DNS method and not computationally demanding. As high Reynolds number will have large eddies, it is preferred when the flow has a high Reynolds number.
2. LES uses _____________
a) spatial filtering
b) time averaging
c) ensemble averaging
d) reynolds averaging
Answer: a
Explanation: Les captures large eddies only. So, spatial filtering is used to separate the large and small eddies. A filtering function and the cut-off width above which the flow will be solved are selected before solving the flow.
3. What does SGS stress stand for?
a) Sub-grid-scale stress
b) Suitable-grid-scale stress
c) Suitable-grey-scale stress
d) Sub-grey-scale stress
Answer: a
Explanation: During spatial filtering, the information about the small eddies will be lost. The interaction between large eddies and small eddies leads to this SGS stress. It stands for sub-grid-scale stress.
4. Which of these will not come under the types of filter functions used by the LES model?
a) Top-hat filter
b) Leonard filter
c) Gaussian filter
d) Spectral cut-off
Answer: b
Explanation: The LES method uses a filter function to spatially filter and get the larger eddies of interest. The most common three-dimensional filters are Top-hat filter, Gaussian filter and Spectral cut-off filters.
5. What is the other name of the top-hat filter?
a) Cross filter
b) Cube filter
c) Box filter
d) Square filter
Answer: c
Explanation: Top-hat filter is otherwise called as the box filter. It is the simplest filter function given by
\
= \left\{
\right\}\)
Where,
\(\vec{x}\) → Spatial vector
\(\vec{x’}\) → Derivative of spatial vector
Δ → Cut-off width
6. Which of these represent the Gaussian filter function?
a) \
^{\frac{3}{2}} exp
\)
b) \
^{\frac{3}{2}} exp
\)
c) \
^{\frac{3}{2}} exp
\)
d) \
^{\frac{3}{2}} exp
\)
Answer: d
Explanation: The filter function, in general, is a function of the spatial vector
, its derivative
and the cut-off width . The Gaussian filter has an additional parameter . The typical value of γ is 6. The function is given as
\
=
^{\frac{3}{2}} exp
\)
7. Which of these is the spectral cut-off filter function?
a) \
\
/\Delta] \)
c) \
\
/\Delta]\)
Answer: a
Explanation: The spectral cut-off filter is the product of \(\frac{sin[
]}{
}\) for all three directions of the x-vector. This gives a sharp cut-off in the energy spectrum at a wavelength of Δ/π.
8. Which of these filters is commonly used in FVM models?
a) Gaussian filter
b) Top-hat filter
c) Spectral cut-off filter
d) Gaussian and spectral cut-off filter
Answer: b
Explanation: The top-hat or box filter is the one which is preferred for the finite volume methods in CFD packages. This is because of their simple elimination of small eddies. The Gaussian and spectral cut-off filter are used for research purposes.
9. In FVM methods, cut-off width depends on ____________
a) the PDE
b) the algebraic equation
c) the grid-size
d) the discretization method
Answer: c
Explanation: When we use the finite volume method, it is pointless to use a cut-off width which is smaller than the grid size. If such cut-off width is chosen, the accuracy of the method in capturing the eddies will be affected.
10. The cut-off width dependent on the finite volume grid size is equal to _____________
a) the square of the grid cell volume
b) the cube of the grid cell volume
c) the square root of the grid cell volume
d) the cube root of the grid cell volume
Answer: d
Explanation: If Δx, Δy and Δz are the grid sizes in the x, y and z-directions, the cut-off width should be the cube root of the cell volume . This is because the cut-off width is in a single direction.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Turbulence Modelling – Reynolds Averaged Navier-Stokes Model”.
1. Which of these properties of turbulence is ruled out in Reynolds averaged equations?
a) Fluctuations
b) Turbulence
c) Non-linearity
d) Randomness
Answer: a
Explanation: The flow properties of turbulent flow can be decomposed into mean and fluctuating components. These fluctuating components result in an unsteadiness in the flow. This unsteadiness is ruled out by means of Reynolds averaging.
2. The averaging interval in RANS equation is based on ____________
a) the grid size
b) the eddy size
c) the fluctuations
d) the time interval of the problem
Answer: c
Explanation: Time averaging is used only when the flow is steady and time-independent. The time interval taken to average the fluctuations depend upon the time scale of the fluctuations itself. If this interval is large enough, the lower limit of the integral does not even matter.
3. For unsteady turbulent flows, which of these averaging method is used?
a) Time averaging
b) Ensemble averaging
c) Spatial averaging
d) Volume averaging
Answer: b
Explanation: Time averaging is generally used to remove the fluctuations in RANS model. But it cannot be used when the problem is unsteady. In these cases, ensemble averaging is used to eliminate the fluctuations.
4. Which of these terms arise in the conservation equations when using the RANS model?
a) Reynolds stresses and turbulent scalar flux
b) Cross stresses and turbulent scalar flux
c) Leonard stresses and turbulent scalar flux
d) Leonard stresses and cross stresses
Answer: a
Explanation: While the conservation equations are Reynolds averaged, they get additional terms due to the decomposition of the flow variables. These additional terms include Reynolds stresses and turbulent scalar fluxes. This occurs because of the mean of the product of the fluctuating components.
5. Reynolds averaging makes the conservation equations ____________
a) non-conservative
b) non-linear
c) unstable
d) inconsistent
Answer: b
Explanation: Reynolds averaging add extra terms to the conservation equations. So, the number of unknowns becomes more than the number of equations. This leads to a linearity problem and makes the conservative equations non-linear.
6. How many additional terms are present in the x-momentum equation Reynolds-Averaged Navier-Stokes equations?
a) No additional terms
b) Six additional terms
c) Three additional terms
d) Two additional terms
Answer: c
Explanation: The non-reduced x-momentum equation is
\
=-\frac{\partial p}{\partial x}+div)+S\)
The Reynolds-Averaged x-momentum equation is
\
=
-\frac{\partial\tilde{p}}{\partial x} + div
+
+S\)
Here, the terms ( ρ u’ 2 ) , ( ρ u’v’ ) and ( ρ u’v’ ) are the three extra terms.
7. From which of these terms does the turbulent viscosity arise from?
a) \
\
\
\)
d) \(\frac{\partial\overline{
}}{\partial y}\)
Answer: d
Explanation: The term \
in the momentum equation. This leads to the turbulent or eddy viscosity in the turbulent models.
8. The Reynolds stress term arises in the turbulent equation only when ____________
a) two quantities are correlated
b) two quantities are uncorrelated
c) the flow is steady
d) the flow is unsteady
Answer: a
Explanation: Consider two flow properties u and v. While decomposing them using the Reynolds decomposition method, we get u= u +u’ and v= v +v’. The mean of their product is ( uv = u v + u’v’ ). The Reynolds stress \(\overline{
}\) arises when this term ( u’v’ ) is not zero. This term is not zero only when the two quantities are correlated.
9. To close the RANS equations, we need _____________
a) Incompressible flow model
b) DNS method
c) Turbulence models
d) SGS model
Answer: c
Explanation: It is impossible to derive a closed set of RANS equations. So, some approximations in the flow model are done. These approximations are called the turbulence model. This usually means prescribing the Reynolds stresses and turbulent scalar flux in terms of the mean flow quantities.
10. What is the difference between the RANS model and the Reynolds stress model?
a) The RANS model needs 5 extra transport equations
b) The Reynolds stress model needs 5 extra transport equations
c) The RANS model needs 7 extra transport equations
d) The Reynolds stress model needs 7 extra transport equations
Answer: d
Explanation: The most common RANS turbulence models are classified on the basis of the number of extra transport equations they need. These models form the basis of the current procedures for turbulence problems in CFD packages. The Reynolds stress model needs seven more equations.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Mixing Length Turbulence Model”.
1. The mixing length model defines the turbulence dynamic viscosity as a function of ____________
a) position
b) mean flow properties
c) fluctuating components
d) velocities
Answer: a
Explanation: The mixing length model is a variation of the RANS model. It does not need any additional transport equations. It describes the stresses in terms of a simple algebraic formula for the turbulent dynamic viscosity as a function of position.
2. The mixing length model links _____________ with _____________
a) length scale with mean flow properties
b) velocity scale with mean flow properties
c) length scale with position coordinates
d) velocity scale with position coordinates
Answer: b
Explanation: The large eddies directly interact with the mean flow properties and extract energy from them. So, there is a strong connection between the mean flow properties and the behaviour of the large eddies. So, the velocity scale is linked with the mean flow properties.
3. If ν t is the turbulent kinematic viscosity, l m is the mixing length and U is the mean flow velocity in the x-direction, which of these gives the Prandtl mixing length model equation?
a) \
\
\
\(ν_t =l_m^2 \Big|\frac{∂U}{∂x}\Big|\)
Answer: b
Explanation: Prandtl mixing length model is an attempt to give the transport of momentum in terms of Reynolds stresses. \(ν_t =l_m^2 \Big|\frac{∂U}{∂y}\Big|\) gives the Prandtl mixing length model.
4. The value of mixing length depends on ____________
a) small eddies
b) large eddies
c) turbulence
d) time scales
Answer: c
Explanation: The mixing length model defines the Reynolds stresses in terms of velocity gradients, mixing length and density of the fluid. Turbulence is a function of the flow. So, if the turbulence changes, the Reynolds stresses should change. This change is accounted by changing the mixing length.
5. For a 2-D flow, what is the mixing length of the mixing layer turbulence model?
a) 0.1 of layer width
b) 0.09 of layer width
c) 0.08 of layer width
d) 0.07 of layer width
Answer: d
Explanation: Mixing length varies for different turbulent flows. For free turbulent flow of the mixing layer type, the mixing length is 0.07 times of the layer width. Mixing layer turbulent flow occurs due to the interaction of two flows with various velocities.
6. What is the mixing length for the outer layer of a 2-D turbulent boundary layer?
a) 0.09 times the boundary layer thickness
b) 0.08 times the boundary layer thickness
c) 0.07 times the boundary layer thickness
d) 0.06 times the boundary layer thickness
Answer: a
Explanation: For a turbulent boundary layer problem, the mixing length varies for different layers of the flow. For the 2-D case, the mixing length of the outer boundary layer is 0.09 times the boundary layer thickness.
7. The mixing length model can be used to get the turbulent scalar fluxes also using _____________
a) turbulent Prandtl/Reynolds number
b) turbulent Reynolds/ Schmidt number
c) turbulent Prandtl/Schmidt number
d) turbulent Reynolds/Nusselt number
Answer: c
Explanation: Mixing length model uses the turbulent viscosity coefficients. If a relationship can be established between the turbulent viscosity and turbulent diffusivity, the model can be used for turbulent scalar fluxes. This relationship is established by the Turbulent Prandtl/Schmidt number.
8. Mixing length model cannot be used for _____________
a) turbulent jets
b) turbulent mixing layers
c) turbulent wakes
d) turbulent flows with separation
Answer: d
Explanation: Mixing length model is ideal for predictions of thin turbulent shear flows such as turbulent jets, wakes, boundary layers and mixing lengths. But, they will not be able to predict flows with separation or even recirculation.
9. Consider a turbulent flow of viscosity μ t , diffusivity Γ t and Prandtl/Schmidt number σ t . Let Φ be a flow property which can be decomposed into Φ=Φ+Φ’. What is the turbulent scalar flux given by?
a) – ρu’Φ’ =Γ t \
– ρu’Φ’ =Γ t \
– ρu’Φ’ =μ t \
– ρu’Φ’ =μ t \(\frac{\partial\Phi}{\partial x}\)
Answer: a
Explanation: By using the mixing length model, the scalar flux can be given by the mean flow property Φ. Since the equation is in general for a flow property, the general term diffusivity only can be given and not the viscosity. So, the relationship is given by – ρu’Φ’ =Γ t \(\frac{\partial\Phi}{\partial x}\).
10. The mixing length for a 2-D turbulent boundary layer depends on ____________
a) the distance from the wall and the boundary layer thickness
b) the distance from the wall and von Karman’s constant and dimensionless distance
c) von Karman’s constant
d) the boundary layer thickness
Answer: b
Explanation: The mixing length for a 2-D turbulent boundary layer is given by κy[1-exp(y + /26)]. Where, κ is the von Karman’s constant which is equal to 0.41. y is the distance from the wall. y + is the dimensionless distance.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Turbulence Modelling – K-epsilon Model”.
1. What does the name k-ε model signify?
a) The seven extra transport equations used in the model
b) The variation of k and ε with the flow variables
c) The variation of k with ε
d) The two extra transport equations used in the model
Answer: d
Explanation: k-ε is a turbulence model used to supplement the RANS equations in overcoming its non-linearity. This model uses two additional transport equations which govern the transport of k and ε.
2. What does k and ε stand for?
a) Turbulent kinetic energy and its dissipation rate per unit mass
b) Turbulent kinetic energy and turbulent diffusivity
c) Turbulent diffusivity and its dissipation rate per unit mass
d) Turbulent kinetic energy and mass transfer
Answer: a
Explanation: In the k-ε model, the two additional equations govern the transport of turbulent kinetic energy and the rate of dissipation of the turbulent kinetic energy . The behaviour of turbulent flow is given in terms of these two properties in this model.
3. The k-ε model focuses on the mechanism which affects ____________
a) the Reynolds stresses
b) the cross stresses
c) the transport of scalar fluxes
d) the turbulent kinetic energy
Answer: d
Explanation: The basic mixing length model cannot define a flow which involves flow separation or recirculation. So, a better turbulence model is developed in terms of k and ε. This model focuses on the dynamics of the flow and hence its turbulent kinetic energy.
4. ____________ and _____________ are used in the k-ε model in addition to k and ε to formulate the transport equations.
a) Internal thermal energy and turbulent stresses
b) Internal thermal energy and kinetic energy
c) Rate of deformation and turbulent stresses
d) Rate of deformation and kinetic energy
Answer: c
Explanation: While forming the transport equations for k and ε, the rate of deformation term and the turbulent stresses are also used. These are used in their tensor form. Both of them can be expressed in terms of the mean velocity gradients.
5. If S ij represents the rate of deformation, μ represents the dynamic viscosity and \(\vec{V}\), the velocity of the flow, What does the terms div(2μ\(\vec{V}\)S ij ) and 2μS ij account for?
a) The effect of turbulent stresses
b) The effect of viscous stresses
c) The effect of Reynolds stresses
d) The effect of kinetic energy
Answer: b
Explanation: The term div(2μ\(\vec{V}\)S ij ) represents the transport of kinetic energy due to viscous stresses. The term 2μS ij represents the viscous dissipation of kinetic energy. Together, these two terms represent the effect of viscous stresses on kinetic energy.
6. The terms accounting for turbulence effects contain ____________
a) Reynolds stresses
b) Turbulent kinetic energy
c) Dissipation of turbulent kinetic energy
d) Length scale terms
Answer: a
Explanation: The terms accounting for turbulence stresses are \
\) and \
\) represents the turbulent transport of kinetic energy by means of Reynolds stresses. \(\rho \overline{u_{i}^{‘} u_{j}^{‘}}\) represents the net decrease of kinetic energy due to deformation work by Reynolds stresses. Both of these terms contain the Reynolds stress term \(\rho \overline{u_{i}^{‘} u_{j}^{‘}}\).
7. In high Reynolds number turbulent flows _______________ terms dominate.
a) diffusion terms
b) convection terms
c) viscous stress terms
d) turbulent effect terms
Answer: d
Explanation: In high Reynolds number flows, the difference between the length scales will be very high. The large eddies are more energetic. So, the turbulent effect terms are much larger than the viscous stress terms in high Reynolds number flows.
8. Express the large scale velocity in terms of k and ε.
a) ε 1/2
b) 1/2
c) k 1/2
d) 1/2
Answer: c
Explanation: In the k-ε model, the properties of turbulence can be expressed in terms of the variables k and ε. The velocity scale of the large eddies are given by k 1/2 . k is the turbulent kinetic energy term.
9. Express the large scale length in terms of k and ε.
a) 3/2
b) k 3/2 /ε
c) ε/k 3/2
d) 3/2
Answer: b
Explanation: The length scale of the large eddies can be given as k 3/2 /ε. The small scale dissipation rate of the turbulent kinetic energy can be used to represent the large scale length as the rate at which the large eddies extract energy from the mean flow is matched with the rate of energy transfer to small eddies.
10. Let C μ be a dimensionless constant and ρ be the density of the flow. Express the eddy dynamic viscosity in terms of k and ε.
a) ρC μ k 2 /ε
b) ρC μ k/ε
c) ρC μ ε/k
d) ρC μ ε 2 /k
Answer: a
Explanation: Using dimensional analysis, the turbulent dynamic viscosity can be given as
μ t = C μ ρϑl
Where,
ϑ → Velocity scale of large eddies
l → Length scale of large eddies
Substituting these two in k and ε terms, we get
μ t = C μ ρ\(\frac{k^2}{\varepsilon}\) Where, C μ is a dimensionless constant which is adjustable. The standard k-ε model uses C μ =0.09.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Turbulence Modelling – K-omega Model”.
1. The k-ω model adds ___________ to the RANS equations.
a) three variables
b) three equations
c) two variables
d) two equations
Answer: d
Explanation: The k-ω model is a variation of the k-ε model. This also adds two equations which are the transport equations of k and ω to the RANS equations to overcome the linearity problem of the RANS equation.
2. What does the variable ω in the k-ω model stand for?
a) Turbulence eddy size
b) Turbulence eddy wavelength
c) Turbulence frequency
d) Turbulence large length scale
Answer: c
Explanation: The variable ω means the turbulence frequency. It gives the rate at which the turbulent kinetic energy is converted into turbulent internal thermal energy per unit volume and per unit time.
3. Which of these could not be modelled using the k-ε model, but can be modelled using the k-ω model?
a) Turbulent jet flows
b) Adverse pressure gradients in turbulent flows
c) Boundary layer on turbulent flows
d) Turbulent free flows
Answer: b
Explanation: For free-shear flows, k-ε models are well suited. But, it cannot model adverse pressure gradients in the turbulent flows. This problem can be overcome by the k-ω model.
4. If k is the turbulent kinetic energy, what is the relationship between the turbulence frequency and dissipation rate of the turbulent kinetic energy ?
a) ω=ε/k
b) ω=k/ε
c) ω=ε 2 /k
d) ω=k 2 /ε
Answer: a
Explanation: Turbulence frequency is the ratio of the rate of dissipation of the turbulent kinetic energy to the turbulent kinetic energy. It is given by ω=ε/k. The values of ω are easier to assume than the ε values.
5. What is the unit of turbulence frequency?
a) Turbulence frequency is dimensionless
b) 1/s 2
c) s
d) 1/s
Answer: d
Explanation: Turbulence frequency has the same unit as the frequency . The dimension of ε is m 2 /s 3 . The dimension of kinetic energy is m 2 /s 2 . Dividing both, we get 1⁄s.
6. Represent the length scale in terms of k and ω.
a) ω/k
b) k/ω
c) √k/ω
d) ω/√k
Answer: c
Explanation: The turbulence large-scale length is given in terms of
l=\(\frac{k^{3/2}}{\epsilon}\)
l=\(\frac{k.k^{1/2}}{\epsilon}\)
l=\(\frac{k^{1/2}}{\omega}\).
7. Represent the turbulent dynamic viscosity in terms of k and ω.
a) ρ ω/k
b) ρ k/ω
c) ρ k 2 /ω
d) ρ ω/k 2
Answer: b
Explanation: The turbulent dynamic viscosity is
μ t = ρϑl
Where,
ρ → Density of the flow
ϑ → Length scale of the large eddies
l → Length scale of the small eddies
Replacing with the k and ω equivalents,
\(μ_t=\rho×\sqrt{k}×\frac{\sqrt{k}}{\omega}\)
\(μ_t=\rho×\frac{k}{\omega}\).
8. The values of k and ω must be specified in ___________
a) the inlet boundary conditions
b) the outlet boundary conditions
c) the wall boundary conditions
d) the symmetry boundary conditions
Answer: a
Explanation: At the inlet boundaries, the values of k and ω are specified. Zero gradient conditions are used at the outlet boundary conditions. At the wall boundaries with low Reynolds number, k is set to zero.
9. Using k-ω model is difficult for ____________
a) free stream
b) boundary layer flows
c) jet flows
d) mixing layer flows
Answer: a
Explanation: The k-ω model is sensitive to the free stream specified values. The value of ω in the free stream is zero. But, if this is set to zero, the eddy viscosity becomes infinity or indeterminate. So, a small non-zero value is specified and the whole problem becomes dependent on this non-zero value. The k-ε model does not have this problem.
10. Which of these is an advantage of the k-ω model over the k-ε model?
a) Does not depend on the ε value
b) Easier to integrate
c) Can be applied for turbulent boundary layers
d) Has only two extra equations
Answer: b
Explanation: The greatest advantage of replacing the ε value with the ω value is that this ω value is easier to integrate. It does not need additional damping functions to integrate. The k-ω model also depends on the ε value for the ω value. Both can be applied for turbulent boundary layers. Both has two extra equations.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Turbulence Modelling – Spalart Allmaras Model”.
1. The Spalart-Allmaras model differs from the RANS equations by ___________
a) four extra transport equations
b) one extra transport equation
c) two extra transport equations
d) three extra transport equations
Answer: b
Explanation: Spalart-Allmaras is a turbulence model to the RANS equations. This has an extra transport equation. This extra transport equation is used to overcome the non-linear problem of the RANS equations.
2. The transport equation in the Spalart-Allmaras model is for the transport of ___________
a) kinematic eddy viscosity parameter
b) kinematic eddy viscosity
c) dynamic eddy viscosity parameter
d) dynamic eddy viscosity
Answer: a
Explanation: The extra transport equation of the Spalart-Allmaras method describes the transport of the kinematic eddy viscosity parameter through convection, diffusion, dissipation and source terms. This way, it is different from the other turbulence models.
3. In the Spalart-Allmaras model, the dynamic eddy viscosity in terms of the kinematic eddy viscosity parameter ( v ) is given by __________ (Note: f ν1 is the wall damping function and ρ is the density of flow).
a) ρ v f ν1
b) (ρ v ) ⁄ f ν1
c) (ρf ν1 ) ⁄ v
d) v ⁄ (ρf ν1 )
Answer: a
Explanation: Dynamic viscosity is the product of the kinematic viscosity and the density of the flow. As the kinematic eddy viscosity parameter is used here, the wall function comes into the picture. So, the dynamic eddy viscosity μ t =ρ v f ν1
4. The first wall damping function in the Spalart-Allmaras model is a function of ___________
a) the product of the dynamic eddy viscosity parameter and the dynamic eddy viscosity
b) the ratio of the dynamic eddy viscosity parameter and the dynamic eddy viscosity
c) the product of the kinematic eddy viscosity parameter and the kinematic eddy viscosity
d) the ratio of the kinematic eddy viscosity parameter and the kinematic eddy viscosity
Answer: d
Explanation: The first wall damping function is introduced in the dynamic eddy viscosity. The dynamic eddy viscosity divided by the density of the flow is the kinematic viscosity. So, the function is a function of the ratio of the kinematic eddy viscosity parameter and the kinematic eddy viscosity.
5. At high Reynolds numbers, the first wall damping function becomes ___________
a) -1
b) 1
c) 0
d) ∞
Answer: b
Explanation: The first wall damping function becomes one when we consider turbulent flows at high Reynolds numbers. This is because, at these Reynolds numbers, the kinematic eddy viscosity parameter value is close to the kinematic eddy viscosity.
6. Near the wall, the first wall damping function tends to ___________
a) -1
b) 1
c) 0
d) ∞
Answer: c
Explanation: The value of the kinematic eddy viscosity parameter decreases with the Reynolds number. Near the wall, the Reynolds number is very small. So, the kinematic eddy viscosity parameter and the first wall function also tends to zero.
7. Expand the Reynolds stress term \
\
\)
b) \
\)
c) \
\)
d) \
\)
Answer: b
Explanation: The Reynolds stress term is given as
\
\)
Converting to Spalart-Allmaras terms,
\
\)
.
8. The rate of production of the kinematic eddy viscosity parameter is related to ___________
a) rate of dissipation of kinetic energy
b) turbulence frequency
c) vorticity
d) kinetic energy
Answer: c
Explanation: The rate of production of the kinematic eddy viscosity parameter is a term in the transport equation of the Spalart-Allmaras model. This is related to the local mean vorticity by means of the vorticity parameter.
9. The rate of dissipation of kinematic eddy viscosity parameter is C w1 ρ\
^2 f_w\). What is the length scale used here?
a) κy
b) 2
c) \
\(\frac{y}{C_{w1}} \)
Answer: a
Explanation: The length scale cannot be computed in the Spalart-Allmaras model. It must be specified separately. The length scale used here is κy. Where, κ is the von Karman’s constant which is equal to 0.4187 and y is the distance from the wall.
10. The Spalart-Allmaras model is best suited for ___________
a) turbulent jet flows
b) turbulent mixing layers
c) turbulent boundary layers with slight pressure gradients
d) turbulent boundary layers with adverse pressure gradients
Answer: d
Explanation: The Spalart-Allmaras model is suitable for flow near walls. So, it is suitable for turbulent boundary layers. The other models are also suitable for this case. But, they cannot model adverse pressure gradients for which Spalart-Allmaras is the best model.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Turbulence Modelling – RNG K Epsilon”.
1. The RNG k-ε model comes under which of these types of the turbulence models?
a) Two-equation model
b) One-equation model
c) Algebraic model
d) Second-order closure models
Answer: a
Explanation: The RNG k-ε model is a variation of the k-ε model. So, the RNG k-ε model also has two extra transport equations like the k-ε model. It comes under the two-equation turbulence models.
2. What does the abbreviation RNG in the RNG k-ε model stand for?
a) Random number generator
b) Renormalization group
c) Renormalization generator
d) Random number group
Answer: b
Explanation: The RNG in the RNG k-ε model means renormalization group. This method was invented by Yakhot and Orszag of Princeton University. This is invented to overcome the disadvantages of the k-ε model.
3. The RNG k-ε model makes assumptions for ___________
a) dynamic eddy viscosity
b) kinematic eddy viscosity
c) small-scale turbulence
d) large-scale turbulence
Answer: c
Explanation: The statistical mechanics approach has led to new mathematical formalizations. A limited number of assumptions regarding the statistics of small scale turbulence is made in this RNG k-ε model.
4. The effects of the small-scale turbulence are ____________
a) normalized
b) assumed
c) neglected
d) represented by random forcing functions
Answer: d
Explanation: In the Navier-Stokes equations, the effects of small-scale turbulence are replaced by random forcing functions in the RNG k-ε model. In the Large Eddy Simulation method, these are neglected.
5. The effective dynamic eddy viscosity is ___________
a) the addition of molecular dynamic viscosity and dynamic eddy viscosity
b) the ratio of molecular dynamic viscosity and dynamic eddy viscosity
c) the product of molecular dynamic viscosity and dynamic eddy viscosity
d) molecular dynamic viscosity to the power of dynamic eddy viscosity
Answer: a
Explanation: The small scales of motion in the governing equations are replaced by the larger scale motions and modified viscosity. This modified effective viscosity is the addition of molecular dynamic viscosity and dynamic eddy viscosity.
6. The dynamic eddy viscosity used in the RNG k-ε model is ____________
a) half of the that of the k-ε model
b) the same as that of the k-ε model
c) twice that of the k-ε model
d) thrice that of the k-ε model
Answer: b
Explanation: The dynamic eddy viscosity used in the RNG k-ε model is the same as that of the k-ε model. The equation used is
μ t =ρC μ \(\frac{k^2}{\varepsilon}\).
7. What is the value of C μ in the RNG k-ε model?
a) 0.0545
b) 0.0645
c) 0.0845
d) 0.0745
Answer: c
Explanation: The value of C μ is 0.0845. This is useful in finding the turbulent dynamic viscosity and hence the effective dynamic viscosity of the RNG k-ε model. This value is obtained by comparing to the existing turbulence model data.
8. div[α k μ eff grad k] is the diffusion term in the transport equation of the turbulent kinetic energy. α k represents ____________
a) effects of internal thermal energy
b) effects of turbulent kinetic energy
c) effects of large-scale turbulence
d) effects of small-scale turbulence
Answer: d
Explanation: The effect of small-scale turbulence is used while representing the diffusion of turbulent kinetic energy in its transport equation as α k . In a similar way, the transport equation for the rate of dissipation of the turbulent kinetic energy contains α ε .
9. The Reynolds stress term used in the RNG k-ε model is given by 2μ t S ij -2⁄3 ρkδ ij . What does the term δ ij represent?
a) Kronecker delta
b) Euler delta
c) Hermann delta
d) Cartesian delta
Answer: a
Explanation: The rate of dissipation of the turbulent kinetic energy in different directions are given by the equation ε ij = 2 ⁄ 3 ρδ ij . Here, the term δ ij is the Kronecker delta. This value becomes zero if i≠j. If i=j, it becomes 1.
10. The term representing the effect of small-scale eddies in diffusion terms of the k transport equation and ε transport equations take the values ___________ and ____________
a) 1.49, 1.49
b) 1.39, 1.39
c) 1.39, 1.49
d) 1.49, 1.39
Answer: b
Explanation: The terms mentioned in the question are α k and α ε . These terms take the same value for the RNG k-ε model. The values are 1.39 both. These are obtained from the experimental data.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Turbulence Modelling – Realizable K Epsilon”.
1. Which of these conditions define realizability?
a) Negative k and ε values
b) Linear k and ε transport equations
c) Non-negative k and ε values
d) Non-linear k and ε transport equations
Answer: c
Explanation: The values of turbulence quantities such as the turbulent kinetic energy and the rate of dissipation of the turbulent kinetic energy cannot be negative and must always be constrained to have values above zero.
2. Which of these statements is true?
a) The standard k-ε model and the RNG k-ε model are realizable
b) Neither the standard k-ε model nor the RNG k-ε model is realizable
c) The standard k-ε model is realizable but not the RNG k-ε model
d) The RNG k-ε model is realizable but not the standard k-ε model
Answer: b
Explanation: Both the standard k-ε model and the RNG k-ε model are not realizable. They do not satisfy the mathematical condition of realizability. Only the realizable k-ε model is realizable.
3. The realizable k-ε model falls into which of these categories?
a) Non-linear two-equation turbulence models
b) Linear two-equation turbulence models
c) Non-linear three-equation turbulence models
d) Linear two-equation turbulence models
Answer: a
Explanation: There is a division of the k-ε models which are non-linear. The transport equations are non-linear in these models. The realizable k-ε model also comes under this non-linear k-ε models.
4. The realizable k-ε model is based on ________
a) the turbulence model replacing the realizability constraint
b) the viscoelastic analogy replacing the realizability constraint
c) the realizability constraint with viscoelastic analogy
d) the realizability constraint without viscoelastic analogy
Answer: d
Explanation: The non-linear k-ε models were initially developed based on the analogy between the viscoelastic fluids and the turbulent flows. This analogy is not used by the realizable k-ε model. The realizability constraint rules out this analogy.
5. The non-linear k-ε models relate the Reynolds stresses to _________
a) the cubic vector products of strain rate and vorticity
b) the quadratic vector products of strain rate and vorticity
c) the cubic tensor products of strain rate and vorticity
d) the quadratic tensor products of strain rate and vorticity
Answer: d
Explanation: The Reynolds stresses in the non-linear k-ε models relate the Reynolds stresses to the quadratic product of local vorticity and strain rates. These two quantities are tensors. The method sensitizes the Reynolds stresses.
6. Which of these equations gives the turbulent dynamic viscosity used in the realizable k-ε model?
a) μ t ∝ ρk/ε
b) μ t ∝ k / ε
c) μ t ∝ ρk 2 /ε
d) μ t ∝ k 2 ε
Answer: c
Explanation: The standard k-ε model, RNG k-ε model and the realizable k-ε model use the same equation for the turbulent dynamic viscosity. The turbulent dynamic viscosity is the product of the turbulent kinematic viscosity and the density of the flow given by μ t ∝(ρk 2 )/ε.
7. Which of these conditions satisfy realizability?
a) \
\
\
\(-\rho \overline{u_{i}^{‘} u_{j}^{‘}}\)<0
Answer: d
Explanation: According to the realizability conditions, the properties which are physically non-negative must be numerically non-negative too. So,\(-\rho \overline{u_{i}^{‘} u_{j}^{‘}}\)>0. Therefore, \(-\rho \overline{u_{i}^{‘} u_{j}^{‘}}\) should be less than zero.
8. The realizable k-ε model is best for predicting __________
a) mixing layers
b) wake formation
c) spreading of jets
d) smooth boundary layer flows
Answer: c
Explanation: The realizable k-ε model more accurately predicts the spreading of jets, be it planar or round jets. It performs better than the other models for complex problems involving recirculation and flow separation.
9. Though the applicability of the realizable k-ε model and the RNG k-ε model are almost the same, the realizable k-ε model is ___________ when compared to the RNG k-ε model.
a) more accurate and converges easily
b) more stable
c) linear
d) more consistent
Answer: a
Explanation: The realizable k-ε model and the RNG k-ε model have the same benefits and applications. But, the realizable model gives more accurate results and it is easy to converge when compared to the RNG k-ε model.
10. Which of these conditions should be satisfied for a model to be realizable?
a) Bessel’s inequality
b) Cauchy-Schwarz inequality
c) Holder’s inequality
d) Jensen’s inequality
Answer: b
Explanation: Other than the non-negativity condition, a realizable model should also satisfy the Cauchy-Schwarz inequality. According to this, the term \
^2 ≤ \overline{u_{i}^{‘2} u_{j}^{‘2}}\). This inequality becomes important as the realizable k-ε model is non-linear.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Turbulence Modelling – Shear Stress Transport Model”.
1. The Shear Stress Transport model is a hybrid of _________
a) the standard k-ε model and the k-ω model
b) the standard k-ε model and the RNG k-ε model
c) the realizable k-ε model and the RNG k-ε model
d) the realizable k-ε model and the k-ω model
Answer: a
Explanation: For boundary layers with adverse pressure gradients, the standard k-ε model does not perform well. To overcome this problem, the other models were devised. But, those are sensitive to the arbitrary constants used. So, the Shear Stress Transport model combines the standard k-ε model and the k-ω model to get the advantages of both.
2. Which of these is unmodified for the Shear Stress Transport model and the k-ω model?
a) Reynolds stress calculation and the k-equation
b) Reynolds stress calculation and the ε-equation
c) The k-equation and the ε-equation
d) Reynolds stress calculation, the k-equation and the ε-equation
Answer: a
Explanation: For a Shear Stress Transport model, the k-equation and the calculation of the Reynolds stresses are the same as used in the standard k-ω model. The equation for the transport of ε is transformed into the ω-equation by using the relationship ε=kω.
3. Which of these statements holds true regarding the Shear Stress Transport model?
a) In the near-wall region, the k-ε model is transformed into k-ω model
b) In the near-wall region, the standard k-ε model is used
c) In regions far from the wall, the k-ε model is transformed into k-ω model
d) In regions far from the wall, the k-ω model is used
Answer: a
Explanation: The Shear Stress Transport model uses a transformation of the k-ε model into a k-ω model in the near-wall region. In the region far from the wall, it uses the standard k-ε model as it gives satisfactory results there.
4. When compared to the standard ε-equation, the transformed ω-equation has _________
a) the same number of terms
b) an extra dissipation term
c) two extra source terms
d) an extra source term
Answer: d
Explanation: While transforming the ε-equation into the ω-equation, an extra source term arises. It is called the cross-diffusion term. This cross diffusion term is modified using external blending functions.
5. The σ k value used in the Shear Stress Transport model is ________
a) -1
b) 1
c) 2
d) -2
Answer: b
Explanation: The constants used in the Shear Stress Transport model are revised to optimize the performance. The σ k value used for the k-equation is 1. The σ ω1 value used for the ω-equation of the near-wall region is 2. The σ ω2 value used for the ω-equation of the far region is 1.17.
6. Which of these problems may occur because of the hybrid nature of the Shear Stress Transport model?
a) Non-linearity
b) Inconsistency
c) Numerical instability
d) Inaccuracy
Answer: c
Explanation: As the modelling equations vary from the converted k-ω model in the near-wall region to the standard k-ω model in the far away region, numerical instabilities may arise in the computed eddy viscosities.
7. The blending function used in the Shear Stress Transport model is a function of _____________
a) Turbulent kinematic viscosity, the ratio of turbulence and distance from the wall
b) Turbulence Reynolds number, turbulent kinematic viscosity and distance from the wall
c) Turbulence Reynolds number, the ratio of turbulence and turbulent kinematic viscosity
d) Turbulence Reynolds number, the ratio of turbulence and distance from the wall
Answer: d
Explanation: The blending function is a function of l t /y and Re y . Where,
l t → Ratio of turbulence.
y→ Distance from the wall.
Re y → Reynolds number based on the y-distance.
8. The blending function is __________ at the wall and __________ in the far field.
a) 0,→1
b) 0,→∞
c) 1,0
d) 1,∞
Answer: a
Explanation: The blending function is chosen in a way that it becomes zero at the wall and tends to unity in the far-field region. Also, it should produce a smooth transition around a distance halfway between the boundary layer’s edge and the wall.
9. A limiter is imposed on _________ to improve the performance in adverse pressure gradients and wake regions.
a) the Reynolds number
b) the eddy viscosity
c) the k-value
d) the ε-value
Answer: b
Explanation: There are two limiters used in the Shear Stress Transport model. One of these is on the eddy viscosity. This is done to improve the performance of the model when there are adverse pressure gradients or with wakes. These are the places where the k-ε model fails.
10. The turbulent kinetic energy production is limited to prevent the build-up of turbulence in __________ regions.
a) far-field
b) near-wall
c) stagnation
d) trailing
Answer: c
Explanation: In the stagnation region, the velocities will be zero. So, there is no possibility of turbulence there. The model should be limited to prevent unrealistic turbulence in the stagnation region.
This set of Tricky Computational Fluid Dynamics Questions and Answers focuses on “Turbulence Modelling – Concept of Y +”.
1. The value y + is important only when we deal with ___________
a) turbulent boundary layers
b) turbulent jets
c) free turbulent mixing layers
d) turbulent wakes
Answer: a
Explanation: The value of y + depends on the distance from the wall. So, the concept of y + is also valid only when we solve turbulent boundary layer problems. The turbulent boundary layer has different sub-layers based on this y + value.
2. The value of y + is used while finding ___________
a) eddy kinematic viscosity for the turbulent boundary layers
b) mixing length for the turbulent boundary layers
c) eddy dynamic viscosity for the turbulent boundary layers
d) kinetic energy for the turbulent boundary layers
Answer: b
Explanation: The formula for finding the mixing length for turbulent boundary layers is
l m = \
] \)
While the other parameters like eddy kinematic viscosity, eddy dynamic viscosity and the kinetic energy do not depend on the y + value.
3. The concept of y + is not used in which of these laws?
a) Law of the wall
b) Law of the wake
c) Log law
d) Linear law
Answer: b
Explanation: All these laws are used in modelling turbulent boundary layers. The law of the wake depends on the function of the distance from the wall and the boundary layer thickness. It does not depend on the y + value.
4. Let y be the distance from the wall, u t be the shear velocity and ν be the kinematic viscosity. Which of these equations define y + ?
a) y + =y/u t ν
b) y + =u t ν/y
c) y + =(y u t )/ν
d) y + =u t /yν
Answer: c
Explanation: y + is the ratio of the product of the distance from the wall boundary and the shear velocity to the kinematic viscosity. This is given by the equation y + =y u t /ν. This is why the value of y + increases with the distance from the wall.
5. What is shear velocity?
a) Square of the ratio of density to wall shear stress
b) Square root of the ratio of density to wall shear stress
c) Square of the ratio of wall shear stress to density
d) Square root of the ratio of wall shear stress to density
Answer: d
Explanation: Shear velocity is used while defining the y + . It otherwise called the friction velocity. It is the square root of the ratio of wall shear stress to density. It has the same unit as that of normal velocity.
6. What is the unit of y + ?
a) y + is dimensionless
b) m
c) m 2
d) 1⁄m
Answer: a
Explanation: y + is dimensionless length. It does not have any dimensions. This can be again proved from the equation
y + = \u_t
⁄ν
\).
7. Wall function cannot be used when ___________
a) y + <30
b) y + >30
c) y + <20
d) y + >20
Answer: a
Explanation: For low Reynolds number turbulent models, wall function can be used to integrate the function. This wall function cannot be used if the value of y + is not more than 30. If the value of y + is less than 30, the wall function is invalid.
8. What is the range of y + in the buffer layer?
a) 0 < y + < 5
b) 5 < y + < 30
c) 30 < y + < 500
d) 10 < y + < 20
Answer: b
Explanation: Buffer layer has the turbulent forces and the viscous forces in equal magnitude. In this layer, the range of y + is 5 < y + <30. 0 < y + < 5 is for the linear or viscous sub-layer. The buffer layer lies just above the viscous sub-layer.
9. The value of y + is 50. Which layer does it belong to?
a) Inertia dominated layer
b) Velocity defect layer
c) Log-law layer
d) Law of the wake layer
Answer: c
Explanation: 30 < y + < 500 is the range of y + where the value of u + varies logarithmically with the y + value. It is called the log-law layer. As the value 50 falls in this range, it belongs to the log-law layer.
10. The value of y + at the intersection between the linear profile and log-law is ___________
a) 20
b) 5
c) 30
d) 11.63
Answer: d
Explanation: Though the layer varies from viscous sub-layer to buffer layer when y + crosses the value 5, the variation is still linear. This linear variation becomes logarithmic variation when y + crosses 11.63 to be exact.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Turbulence Modelling – Boundary Conditions”.
1. If n is the spatial coordinate, in the outlet or symmetry boundaries, which of these following is correct for a k-ε model?
a) \
\
\
\(\frac{\partial ^2 k}{\partial n^2}=0; \frac{\partial^2 \varepsilon}{\partial n^2}=0\)
Answer: a
Explanation: In general, for any flow property, the gradients will be zero for the outlet or symmetry boundaries. In the same way, for a k-ε model, the gradients of the variables k and ε are zero. This is given by the equations \(\frac{\partial k}{\partial n}=0; \frac{\partial\varepsilon}{\partial n}=0\).
2. Boundary conditions near the solid-walls for a k-ε model depends on ___________
a) Eddy viscosity
b) Reynolds number
c) ε-value
d) k-value
Answer: b
Explanation: The behaviour of the boundary conditions depends on the Reynolds number. Near the wall, the k-ε model does not perform well. So, wall functions are used. This again depends on the Reynolds number of the flow only.
3. Which of these values vanish near the wall boundary?
a) Velocity and turbulent viscosity
b) Velocity and Reynolds number
c) Velocity and k-value
d) k-value and Reynolds number
Answer: c
Explanation: Near the wall boundary, the velocity of the flow is reduced by the friction of the wall. So, velocity vanishes. The variable k stands for turbulent kinetic energy. As the kinetic energy depends on the velocity, when velocity vanishes, k also will vanish.
4. In the low Reynolds number turbulence models, the first internal grid point is placed in the ___________
a) log-law layer
b) buffer layer
c) inertial sub-layer
d) viscous sub-layer
Answer: d
Explanation: It is difficult to model the flow in the buffer layer. So, the low Reynolds number models place the first internal grid point in the viscous sub-layer and the high Reynolds number models place it in the inertial sub-layer skipping the buffer layer.
5. When k and ε values are not available, for inlet boundary conditions, they are ____________
a) obtained from turbulence intensity
b) assumed to be zero
c) assumed to be unity
d) obtained from Reynolds number
Answer: a
Explanation: At the inlet boundary conditions, the k and ε values must be specified for the k-ε model. In most of the industrial CFD applications, these values will not be known. So, they are obtained from the turbulence intensity and the characteristic length of the model.
6. Which of these is correct about the first internal node of a k-ε model?
a) k-equation is not solved
b) ε-equation is not solved
c) Both k and ε-equations are not solved
d) Both k and ε-equations are solved simultaneously
Answer: b
Explanation: In the k-ε model, the ε-equation is not solved at the first interior point near the wall. Instead, this value is obtained by the condition that the turbulent kinetic dissipation rate will be equal to its production rate.
7. Which of these equations give the turbulence intensity?
a) \
\
\
\(\frac{\sqrt{\vec{V}}}{\sqrt{\overline{\vec{V}^{‘}}}}\)
Answer: a
Explanation: The turbulence intensity which is used to get the k and ε-values for the inlet boundary conditions is
\(T_i=\frac{\sqrt{\overline{\vec{V}^{‘}.\vec{V}^{‘}}}}{\sqrt{\vec{V}.\vec{V}}}\) Where,
\(\vec{V}\) → Velocity vector.
8. The relationship between the turbulence intensity T i and the turbulence kinetic energy k is given by ___________
a) k=\
\)
b) k=\
\)
c) k=\
\)
d) k=\
\)
Answer: b
Explanation: To find the turbulent kinetic energy k using the turbulence intensity value, the following formula is used.
k=\
\)
Where,
\(\vec{v}\) → Velocity vector.
9. The range of values of the turbulent kinetic energy is ___________
a) 50 to 75%
b) 11 to 20%
c) 1 to 10%
d) 0 to 1%
Answer: c
Explanation: The value of turbulence intensity lies between 1% and 10%. The values below 1% are considered to be very less and the values above 10% are considered to be a high one.
10. The formula to find ω from the k-value obtained using the turbulence intensity is ____________
a) ω=\
ω=\
ω=\
ω=\(\frac{k^{1/2}}{l}\)
Answer: d
Explanation: The ε and the ω-values should be obtained from k-value in the k-ε and k-ω models. The formulae used to get these values are
ε=\(C_μ\frac{k^{\frac{3}{2}}}{l}\)
ω=\(\frac{k^{\frac{1}{2}}}{l}\)
Where,
l → Turbulent length scale.
C μ → A dimensionless constant.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Turbulence Modelling – Filtering”.
1. When is the LES filter commutative?
a) When the filter function is unity
b) When the filter function is quadratic
c) When the filter function is isotropic
d) When the filter function is uniform
Answer: d
Explanation: The LES filter function is always linear. If a uniform filter function is used, the order of filtering and differentiation can be swapped with respect to time and space coordinates. Thus, the function will be commutative.
2. How many extra stress terms occur due to the LES filtering operation?
a) No extra terms
b) Four terms
c) Three terms
d) Two terms
Answer: c
Explanation: The LES filtering operation resolves the flow variables into two – the filtered one and the unresolved spatial variations. This resolution leads to three extra stress terms. They are collectively called as Sub-Grid-Scale stresses.
3. Identify Leonard stresses from the following.
a) \
\
\
\(\rho\overline{\overline{u_i}\overline{u_j}}\, – \rho\overline{u_i}\overline{u_j}\)
Answer: a
Explanation: The term \(\rho\overline{\overline{u_i}\overline{u_j}}\, – \rho\overline{u_i}\overline{u_j}\) is named Leonard stresses. It is named after an American Scientist A. Leonard who first identified an approximate method to compute them from the filtered flow.
4. The stress term \Missing open brace for subscript Filter Reynolds stress
b) LES Reynolds stress
c) Reynolds stress
d) Sub-Reynolds stress
Answer: b
Explanation: As the term resembles the Reynolds stress term, it is called the LES Reynolds stress term. This is modelled using Sub-Grid-Scale models, a special type of turbulence model used for these stresses.
5. Which of these terms corresponds to the cross stresses?
a) \
\
\
\(\rho\overline{\overline{u_i}u_j^{‘}}\, – \rho\overline{u_i^{‘}\overline{u_j}}\)
Answer: c
Explanation: The cross-stresses among the other SGS stresses occur due to the interaction between two variables. It can be given by the equation \(\rho\overline{\overline{u_i}u_j^{‘}} + \rho\overline{u_i^{‘}\overline{u_j}}\). The filtered and the unfiltered variables come together in each of these terms.
6. The SGS stresses are obtained from ___________
a) Reynolds stress term
b) Convection term
c) Source term
d) Diffusion term
Answer: a
Explanation: Reynolds stress term comes from the Reynolds-Averaged Navier-Stokes equations. When this term is again averaged, but based on spatial coordinate this time, the SGS stresses occur.
7. If a flow variable ϕ can be resolved by spatial LES filtering as Φ and Φ’. Expand the term \
\
\
\
\(\rho\overline{\overline{u_i}\overline{u_j}} + \rho\overline{\overline{u_i}u_j^{‘}} + \rho\overline{u_i^{‘}\overline{u_j}}+\rho\overline{u_i^{‘}u_j^{‘}} \)
Answer: d
Explanation: The given term \(\rho\overline{u_iu_j}\) represent the Reynolds stresses.
Resolving u i and u j , we get u i = \(\overline{u_{i}} +u_{i}^{‘}\) and \(u_j=\overline{u_{j}}+u_{j}^{‘}\). Therefore,
\(\rho\overline{u_iu_j} = \rho\overline{
}\)
\(\rho\overline{u_iu_j} = \rho\overline{\overline{u_i}\overline{u_j}} + \rho\overline{\overline{u_i}u_j^{‘}} + \rho\overline{u_i^{‘}\overline{u_j}} + \rho\overline{u_i^{‘}u_j^{‘}}\)
These terms are separated into the three components of SGS stresses.
8. Leonard stresses are caused by ____________
a) The SGS eddies
b) The effects at the resolved scale
c) The effects of the unresolved variables
d) The combined effect of the SGS eddies and the resolved scale
Answer: b
Explanation: The Leonard stress terms occur because of the term \(\rho\overline{\overline{u_i}\overline{u_j}}\). These are both the resolved velocities. So, we can conclude that the cause for Leonard stresses is the effects at the resolved scale.
9. Cross-stresses are caused by ___________
a) Convection term
b) Resolved flow
c) SGS eddies
d) Interaction of SGS eddies and resolved flow
Answer: d
Explanation: The cross-stress terms are caused by the combination of the filtered function ( Φ ) and the eddy function
. They are given by \(\rho\overline{\overline{u_i}u_j^{‘}} + \rho\overline{u_i^{‘}\overline{u_j}}\). So, we can say that they are caused by the interaction of SGS eddies and resolved flow.
10. LES Reynolds stresses are caused by ____________
a) Source term
b) Resolved flow
c) SGS eddies
d) Diffusion term
Answer: c
Explanation: The LES Reynolds stresses are caused by the convective momentum transfer due to the interaction of the SGS stresses among themselves. These SGS stresses are modelled separately as the Reynolds stresses in the RANS equations.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Turbulence Modelling – Sub Grid Models”.
1. The Smagorinsky-Lilly SGS model uses ___________
a) Boussinesq hypothesis and Prandtl mixing length model
b) Prandtl mixing length model and k-ε model
c) k-ε model and k-ω model
d) k-ω model and Boussinesq hypothesis
Answer: a
Explanation: The Smagorinsky-Lilly SGS model is built on the Prandtl mixing length model and models the SGS eddy viscosities. It uses the Boussinesq hypothesis to assume the effects of the SGS eddies.
2. According to the Smagorinsky-Lilly SGS model, the SGS stresses depend on the ___________
a) Rate of strain of the SGS eddies
b) Rate of strain of the resolved flow
c) Strain of the resolved flow
d) Strain of the SGS eddies
Answer: b
Explanation: To define the effects of the unresolved SGS eddies on the resolved flow, the Smagorinsky-Lilly SGS model uses the Boussinesq hypothesis. So, the SGS stresses depend on the local rate of strain of the resolved flow.
3. The characteristic length of the SGS eddies is __________
a) half of the filter cut-off width
b) the filter cut-off width
c) twice the filter cut-off width
d) thrice the filter cut-off width
Answer: b
Explanation: The LES filter accepts and rejects eddies based on the filter cut-off width. So, the size of the SGS eddies is determined by the filter cut-off width and the same is used as the characteristic length.
4. Which of these assumptions is made in the Smagorinsky-Lilly SGS model?
a) The changes in the flow direction are slow in the resolved flow
b) The changes in the cross-stream direction are slow in the resolved flow
c) The changes in the flow direction are slow in the SGS eddies
d) The changes in the cross-stream direction are slow in the SGS eddies
Answer: a
Explanation: The Smagorinsky-Lilly SGS model is valid only if
The changes in the flow direction of the resolved flow are very small that the production and dissipation of turbulence are more or less in balance
The turbulent structure is isotropic.
5. What is the relationship between SGS viscosity (μ SGS ), density , characteristic length and the average strain rate of the resolved flow
in the Smagorinsky-Lilly SGS model?
a) μ SGS =ρ 2 Δ\
μ SGS =ρC 2 \
μ SGS =ρ 2 \
μ SGS =ρ 2 \(\mid\overline{S}\mid \)
Answer: d
Explanation: The equation for SGS viscosity is obtained by the dimensional analysis. It is given by the equation μ SGS =ρ 2 \(\mid\overline{S}\mid \) . Where, C is the constant of SGS viscosity.
6. What is the velocity scale taken in the Smagorinsky-Lilly SGS model?
a) The ratio of the length scale and the time scale
b) The square of the average strain rate of the resolved flow
c) The product of the length scale and the average strain rate of the resolved flow
d) The square of the length scale
Answer: c
Explanation: The Smagorinsky-Lilly SGS model assumes a velocity scale equal to the Product of the length scale and the average strain rate of the resolved flow. It is given by the equation Δ×\(\mid\overline{S}\mid \) .
7. In the higher-order SGS model, what is the velocity scale used?
a) The ratio of the SGS turbulent kinetic energy to the SGS eddy viscosity
b) The product of the SGS turbulent kinetic energy and the SGS eddy viscosity
c) The square root of the SGS eddy viscosity
d) The square root of the SGS turbulent kinetic energy
Answer: d
Explanation: The major difference between the Smagorinsky-Lilly SGS model and the higher-order SGS models is the velocity scale used. The higher order models use velocity scale which is equal to the square root of the SGS turbulent kinetic energy.
8. The Smagorinsky-Lilly SGS model is ___________
a) Dissipative
b) Convective
c) Diffusive
d) Convective and diffusive
Answer: a
Explanation: The Smagorinsky-Lilly SGS model is completely dissipative. The direction of energy flow is from eddies at the resolved scale towards the sub-grid scales . This is changed in the later models.
9. The SGS model uses _________ to reduce the sub-grid-scale eddy viscosity near the wall.
a) van Karman’s constant
b) van Driest damping
c) wall function
d) Leonard stresses
Answer: b
Explanation: The purpose of the van Driest damping is to reduce the sub-grid-scale eddy viscosity near the wall in the SGS models. An alternative method is to reduce the eddy viscosity when the Reynolds number becomes small.
10. ___________ creates a problem in the SGS models.
a) Low Reynolds number flows
b) High Reynolds number flows
c) Anisotropic flow near the wall
d) Viscous flow near the wall
Answer: c
Explanation: Near the wall, the flow structure is very anisotropic. Here, regions of low and high-speed fluids are created. This needs a highly anisotropic grid. But, the choice of length scale is restricted by the cut-off width. This poses a problem in the SGS models.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “High-Resolution Schemes”.
1. A high-resolution scheme is __________
a) a higher order bounded scheme
b) a first-order accurate scheme
c) a first-order bounded scheme
d) a higher-order unbounded scheme
Answer: a
Explanation: The high-resolution scheme is a combination of a higher-order profile and the Convection Boundedness Criterion to get the most out of it. So, it can be called a higher order bounded scheme.
2. Which of these techniques cannot be used to implement a high-resolution scheme?
a) DC technique
b) DWF method
c) NWF method
d) TVD method
Answer: d
Explanation: The deferred correction, Downwind Weighing Factor and Upwind Weighing Factor are all methods to implement the higher-order and high-resolution schemes. TVD scheme is not used for this purpose.
3. A high-resolution scheme is the best suited for ___________
a) Turbulent flows
b) Problems involving shocks
c) Incompressible flows
d) Unsteady flows
Answer: b
Explanation: A normal first-order scheme produces accurate results for flows involving high discontinuities like shocks at the cost of increased grid points. This problem is overcome by high-resolution schemes.
4. Which of these diagrams are used to visualize the high-resolution scheme?
a) Leonard diagram and NVD
b) Cumulative flow diagram and NVD
c) Sweby’s diagram and NVD
d) Cumulative flow diagram and Sweby’s diagram
Answer: c
Explanation: The NVD and Sweby’s diagram are used for visualizing the high-resolution schemes in the NVF and TVD formulation respectively. They can be used to visualize the flux limiters.
5. Which of these is not a high-resolution scheme?
a) MINMOD
b) SMART
c) SIMPLE
d) MUSCL
Answer: c
Explanation: The MINMOD scheme, SMART scheme and MUSCL scheme are all high-resolution schemes. SIMPLE is an algorithm which is used to solve the Navier-Stokes equations used for pressure linked equations. It is not a high-resolution scheme.
6. To construct a high-resolution scheme, the monotonic profile in the range 0\Missing open brace for subscript points and
b) points and
c) points and
d) points and
Answer: d
Explanation: In the NVD diagram, the monotonic profile in the range 0\
and for a high-resolution scheme to be constructed. For the other ranges, the profile should follow the upwind scheme.
7. High-resolution schemes give ___________
a) solution without oscillations
b) highly converging
c) inconsistent solutions
d) unstable solution
Answer: a
Explanation: The high-resolution schemes are bounded in the sense that they result in solutions which are free from wiggles or oscillations. This the main advantage of the high-resolution schemes over the other schemes.
8. For a high-resolution scheme to have good convergence, the profile in the NVD frame should __________
a) involve sharp angles
b) avoid sharp angles
c) be smooth
d) be differentiable
Answer: b
Explanation: A high-resolution scheme can have composite nature of profiles in the NVD frame which involves cut. But, at these cuts, the angle should not be sharp for the schemes to have better convergence.
9. The SMART scheme is constructed using the ____________
a) SUPERBEE scheme
b) STOIC scheme
c) QUICK scheme
d) MINMOD scheme
Answer: c
Explanation: The SMART scheme is constructed using the QUICK scheme. The SMART scheme can be easily modified to get a higher convergence. This modification is done by modifying the vertical portion of the profile of the SMART scheme.
10. The diagram represents the NVD of a high-resolution scheme.
computational-fluid-dynamics-questions-answers-high-resolution-schemes-q10
In the diagram, φ c = \(\widetilde{\phi_c}\) and φ f =\
MINMOD
b) SMART
c) STOIC
d) OSHER
Answer: d
Explanation: The OSHER scheme in NVD terms is given by
\(\widetilde{\phi_f} = \left\{
\right\}\)
This is represented by the given diagram.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “High Resolution Schemes – Normalized Variable Formulation”.
1. The Normalized Variable Formulation is used to ___________
a) describe and analyse temporal schemes
b) describe and analyse high-resolution schemes
c) visualize high-resolution schemes
d) visualize temporal scheme
Answer: b
Explanation: The Normalized Variable Formulation is used for the description and analysis of high-resolution schemes. The Normalized Variable Diagram is used for visualizing the high-resolution schemes.
2. The NVF approach does not rely on _____________
a) far downwind node
b) far upwind node
c) upwind node
d) downwind node
Answer: a
Explanation: The NVF is a face formulation procedure based on normalizing the dependent variable for which the flow variable at the face has to be constructed. The NVF approach relies upon the upwind, downwind and far upwind nodes.
3. Consider the following diagram.
computational-fluid-dynamics-questions-answers-normalized-variable-formulation-q3
In the diagram,
phi_u → Φ u
phi_d → Φ d
phi_c → Φ c
Find the normalized flow variable \
\) at the face f as in the NVF approach.
a) \
\
\
\(\tilde{\phi_f}=\frac{
}{
}\)
Answer: d
Explanation: Normalization is a procedure of non-dimensionalizing a variable. In the NVF approach, the flow variable (Φ f ) is normalized as
\(\tilde{\phi_f}=\frac{
}{
}\) Where,
Φ u → Flow variable at the far upwind node.
Φ d → Flow variable at the downwind node.
Φ c → Flow variable at the upwind node.
4. What are the normalized values of the variables Φ d and Φ u ?
a) 1 and 0
b) 0 and ∞
c) 1 and ∞
d) 0 and 1
Answer: a
Explanation: From the normalization formula,
\(\tilde{\phi_d}=\frac{\phi_d-\phi_u}{\phi_d-\phi_u}=1\).
\(\tilde{\phi_u}=\frac{\phi_u-\phi_u}{\phi_d-\phi_u}=0\).
5. Which of these conditions represent a monotonic profile of variable Φ between the far upwind node and downwind node?
.
a) \
\
\
\(0.5\leq\tilde{\phi_c}\leq 1\)
Answer: b
Explanation: For a variable Φ to have a monotonic profile,
\(\tilde{\phi_u}\leq\tilde{\phi_c}\leq\tilde{\phi_d}\)
But, \(\tilde{\phi_u}=0 \,and\, \tilde{\phi_d}=1.\) So,
\(0\leq\tilde{\phi_c}\leq 1.\)
6. If \
.
a) Maximum at c
b) Minimum at c
c) Extremum at c
d) Global minimum at c
Answer: c
Explanation: According to the NVF approach, the value of \(\tilde{\phi_{c}}\) should be bounded between 0 and 1. If this is not the case, either local maximum or local minimum exists at the upwind node. In general, we can say that the upwind node gets an extremum.
7. What is the normalized flow variable at the face
for the upwind and downwind schemes respectively?
.
a) 1 and 0
b) 1 and \
0 and 1
d) \(\tilde{\phi_c}\) and 1
Answer: d
Explanation: For the upwind scheme,
\(\phi_f=\phi_c⟹\tilde{\phi_f}=\tilde{\phi_c}.\) For the downwind scheme,
\(\phi_f=\phi_d \Rightarrow \tilde{\phi_f}=1.\) Where,
Φ f → Flow variable at the face.
Φ c → Flow variable at the upwind node.
Φ d → Flow variable at the downwind node.
\(\tilde{\phi_f}, \tilde{\phi_c}\) → Normalized flow variables at respective nodes.
8. Normalize the following equation.
\(\phi_f=\phi_c+\frac{\phi_d-\phi_u}{4}\)
Where,
Φ f → Flow variable at the face.
Φ c → Flow variable at the upwind node.
Φ d → Flow variable at the downwind node.
Φ u → Flow variable at the far upwind node.
a) \
\
\
\(\tilde{\phi_f}=\tilde{\phi_c}+\frac{3}{4}\)
Answer: a
Explanation: Normalizing the equation,
\(\tilde{\phi_f} = \frac{\phi_c-\phi_u}{\phi_d-\phi_u}+\frac{\frac{\phi_d-\phi_u}{\phi_d-\phi_u}}{4}-\frac{\frac{\phi_u-\phi_u}{\phi_d-\phi_u}}{4}\)
\(\tilde{\phi_f}=\tilde{\phi_c}+\frac{1}{4}-\frac{0}{4}\)
\(\tilde{\phi_f}=\tilde{\phi_c}+\frac{1}{4}.\)
9. Plotting the QUICK scheme in the
plane, the profile will be ____
.
a) quadratic line
b) straight line
c) curved line
d) a parabola
Answer: b
Explanation: The normalized form of the QUICK scheme is
\(\tilde{\phi_f}=\frac{3}{8}+\frac{3}{4}\tilde{\phi_c}.\)
This is of the form \(\tilde{\phi_f}=a+b\tilde{\phi_c}.\) It is a straight line equation. So, the resulting profile will be a straight line.
10. In the Normalized Variable Diagram , all the second-order and third-order schemes pass through the point ________________
a)
b)
c)
d)
Answer: c
Explanation: Except for the first-order accurate schemes, the profiles of all the schemes pass through in the NVD. For a scheme to be second-order accurate, it must pass through this point. For a scheme to be third-order accurate, its slope at the point should be 0.75.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “High Resolution Schemes – Convection Boundedness Criterion”.
1. The numerical convection schemes should be _________
a) upwind biased
b) downwind biased
c) upwind or downwind depending on the problem
d) considering both upwind and downwind equally
Answer: a
Explanation: Physically, convection transports fluid from upstream to downstream. So, the numerical convection schemes should also be upwind biased. This is why, the NVF approach involved far upwind node also but not the far downwind node.
2. According to the Godunov’s theorem, any linear monotonic scheme will be maximum ____________
a) first derivative
b) first-order accurate
c) second-order accurate
d) second derivative
Answer: b
Explanation: Godunov theorem proved that any monotonic linear scheme cannot be higher-order. It will always be first order. Therefore, all the higher-order linear schemes will be non-monotonic. This is why limiters are used to construct high resolution schemes.
3. A higher-order scheme which is not of high-resolution will have ____________
a) overshoots
b) undershoots
c) undershoots and overshoots
d) damping ratio
Answer: c
Explanation: The major shortcoming of the higher order schemes is the unboundedness. This can be seen in the QUICK scheme. They tend to produce undershoots or overshoots and even oscillations near the sudden jumps or steep gradients in the variable.
4. For a scheme to be bounded, its functional relationship should be ____________
a) quadratic
b) linear
c) smooth
d) continuous
Answer: d
Explanation: For a scheme to be bounded, the relationship between the known and the unknown variable should be continuous. Representing in the NVF terms, the function \
\) of a scheme should be continuous.
5. Consider the following diagram.
computational-fluid-dynamics-questions-answers-convection-boundedness-criterion-q5
Line a in the diagram is ____________
a) compressive
b) anti-diffusive
c) diffusive
d) convective
Answer: c
Explanation: Line a in the diagram represents the upwind scheme. The upwind scheme is completely diffusive. Lone b in the diagram represents downwind scheme. Downwind scheme is completely anti-diffusive or compressive.
6. Consider the following NVD.
computational-fluid-dynamics-questions-answers-convection-boundedness-criterion-q6
For a bounded scheme plotted in this NVD, the profile should lie inside which of these regions?
a) D, A
b) A, B
c) B, C
d) C, D
Answer: b
Explanation: For a bounded scheme, the profile should lie in the region between the upwind scheme and ) to the downwind scheme and ) lines. This region comes under A and B.
7. In the region where \Missing open brace for subscript.
a) SMART scheme
b) QUICK scheme
c) Downwind scheme
d) Upwind scheme
Answer: d
Explanation: One of the conditions for a scheme to be bounded is given as
\
\) resembles the function of the upwind scheme. So, we can say that the scheme should follow the upwind scheme in this region.
8. For a bounded higher-order scheme, \Missing open brace for subscript.
a) ∞
b) 0
c) 1
d) 0.5
Answer: c
Explanation: A bounded higher-order scheme should pass through the points and in the NVD. So, at \(\tilde{\phi_c}=1,\) the value of \(\tilde{\phi_f}\) should also be 1. Similarly, when \(\tilde{\phi_c}=0, \, \tilde{\phi_f}=0.\)
9. In the region where \Missing open brace for subscript.
a) conduction
b) convection
c) radiation
d) anti-diffusion
Answer: b
Explanation: In this region, convection is high. This is why, the solution follows the upwind scheme. When convection is dominant, the scheme should be upwind biased as the process is from the upstream to downstream.
10. Consider the diagrams. They give the NVD. The thick line represents the scheme. Which of these schemes is unbounded?
a) computational-fluid-dynamics-questions-answers-convection-boundedness-criterion-q10a
b) computational-fluid-dynamics-questions-answers-convection-boundedness-criterion-q10b
c) computational-fluid-dynamics-questions-answers-convection-boundedness-criterion-q10c
d) computational-fluid-dynamics-questions-answers-convection-boundedness-criterion-q10d
Answer: a
Explanation: For a scheme to be bounded, it should pass through the point . This means that at \
, the profile of the scheme does not pass through the point . So, the scheme is unbounded.
computational-fluid-dynamics-questions-answers-convection-boundedness-criterion-q10a
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “High Resolution Schemes – TVD Framework”.
1. What is the total variation of a flow variable at a particular time step t?
a) TV t =∏ i Φ i+1 -Φ i
b) TV t =∫ n Φ n dn
c) TV t =∑ i Φ Φ i
d) TV t =∑ i Φ Φ i
Answer: d
Explanation: Total variation is the summation of variations of the flow variable between two consecutive nodes. This is mathematically given as
TV t =∑ i Φ i+1 -Φ i .
2. A numerical method is total variation diminishing if __________
a) the total variation remains constant with increasing time
b) the total variation increases with increasing time
c) the total variation does not increase with increasing time
d) the total variation decreases with increasing time
Answer: c
Explanation: Any numerical method is said to be total variation diminishing if the total variation diminishes with time. This means that the value of total variation should not increase with time. It can either decrease or remain the same.
3. A Total Variation Diminishing scheme is always __________
a) continuous
b) monotonic
c) stable
d) bounded
Answer: b
Explanation: A TVD scheme is monotonic and any monotonically preserving scheme is TVD. This means that the value of a local minimum is non-decreasing and the value of a local maximum is non-increasing.
4. Consider the discretized form of an equation given by \
+b
.\) For this numerical scheme to be TVD, what is the condition?
(Note: Φ u , Φ c and Φ d are the flow variables at the far upwind, upwind and downwind schemes).
a) a≥0;b≥0;0≤a+b≤1
b) a≥0;b≤0;0≤a+b≤1
c) a≥0;b≥0;0≤a-b≤1
d) a≥0;≤0;0≤a-b≤1
Answer: a
Explanation: This condition is given by Sweby and Harten. The sufficient condition for a system having the discretized equation -a(Φ c -Φ u )+b(Φ d -Φ c ) to be TVD is given by
a≥0;b≥0;0≤a+b≤1.
5. Developing a TVD scheme relies upon _________
a) the flux limiter
b) the coefficients
c) the PDE
d) the convection terms
Answer: a
Explanation: A TVD scheme should not be completely upwind or downwind. So, to develop a TVD scheme, an approach is used in which a portion of the anti-diffusive flux is added to the upwind scheme. This flux is limited by a flux limiter function. To find the best flux limiter is the work in developing a TVD scheme.
6. The flux limiter is a function of __________
a) the gradient at that central node
b) the ratio of two consecutive gradients
c) the product of two consecutive gradients
d) the difference between two consecutive gradients
Answer: b
Explanation: Flux limiter prevents the excessive use of flux in regions where oscillations might occur and maximizes the contribution in smooth areas. The flux limiter is denoted by Ψ, where r is usually taken as the ratio of two consecutive gradients.
7. The Sweby’s diagram is drawn in __________ plane.
a)
b)
c)
d)
Answer: a
Explanation: A Sweby’s diagram is used to represent the TVD. This diagram is drawn with the flux limiter (Ψ r ) in the y-direction and the variable r in the x-direction. A high-resolution scheme should lie in a particular region of this diagram to be TVD and monotonic.
8. The condition that the flux limiter of a scheme should satisfy to be TVD is __________
a) Ψ r =min & if r>0; Ψ r =0 & if r<0
b) Ψ r =min & if r>0; Ψ r =0 & if r≤0
c) Ψ r =min & if r>0; Ψ r =0 & if r≤0
d) Ψ r =min & if r>0; Ψ r =0 & if r<0
Answer: c
Explanation: Similar to the Convection Boundedness Criterion, a flux limiter should satisfy the following criterion to be a TVD. There is a list of conditions which has to be satisfied. Simplifying and combining all of them, we get
\Missing \end{matrix} & r>0 \
0 & r\leq 0
\end{matrix}\right\}.\)
9. What are the flux limiters for upwind and downwind schemes respectively?
a) 0 and 2
b) 0 and 1
c) 0 and ∞
d) 1 and ∞
Answer: a
Explanation: The TVD schemes are developed starting from the upwind scheme. If flux limiters are used, the upwind scheme will change its nature. So, no flux limiter is required in this case and hence flux limiter for an upwind scheme is 0. For downwind scheme, the whole profile in the Sweby’s diagram should be in the line Ψ=2. So, the flux limiter here is 2.
10. Give the relationship between NVF and TVD.
\(\tilde{\phi_c}\) → Normalized flow variable at the upwind node
r f → Variable of flux limiter
a) \
\
\
\(\tilde{\phi_c}=\frac{r_f}{1+r_f}\)
Answer: d
Explanation: The variable of flux limiter is given by
\(r_f=\frac{\phi_c-\phi_u}{\phi_d-\phi_c}\)
\(r_f=\frac{\phi_c-\phi_u}{\phi_d-\phi_u+\phi_u-\phi_c}\)
\(r_f=\frac{\frac{\phi_c-\phi_u}{\phi_d-\phi_u}}{\frac{\phi_d-\phi_u+\phi_u-\phi_c}{\phi_d-\phi_u}}\)
\(r_f=\frac{\frac{\phi_c-\phi_u}{\phi_d-\phi_u}}{\frac{\phi_d-\phi_u}{\phi_d-\phi_u}-\frac{\phi_c-\phi_u}{\phi_d-\phi_u}}\)
\(r_f=\frac{\tilde{\phi_c}}{1-\tilde{\phi_c}}\)
\(\tilde{\phi_c}=\frac{r_f}{1+r_f}\)
This is the relationship between TVD and NVF.
This set of Computational Fluid Dynamics online quiz focuses on “Deferred Correction for High-Resolution Schemes”.
1. The coefficients in the deferred correction procedure are based on ___________
a) the downwind scheme
b) the quick scheme
c) the upwind scheme
d) the high-resolution scheme
Answer: c
Explanation: Deferred correction procedure is used to overcome problems related to the convergence of a scheme. The coefficients used here are based on the upwind scheme which is suitable for convection.
2. What is added as a source term in the deferred correction method?
a) Difference between the high-resolution and the upwind schemes
b) Constants in the high-resolution schemes
c) Constants in the upwind schemes
d) Difference between the downwind and the upwind schemes
Answer: a
Explanation: The source term of the given flow problem is considered. In addition to this, a source term is added in the deferred correction method which is equal to the difference between the high-resolution scheme and the upwind scheme.
3. Difference between the upwind line and the high-resolution scheme line gives __________
a) the difference between the upwind and the downwind nodes
b) the normalized difference between the cell face values
c) the difference between the far upwind and the upwind nodes
d) the difference between the far upwind and the downwind nodes
Answer: b
Explanation: The effects of the deferred correction method can be seen in an NVD frame. The normalized difference between the cell face values is given by the difference between the upwind and the high-resolution scheme lines in the NVD.
4. If the normalized difference between the cell face values is high, the convergence rate ___________
a) tends to infinity
b) is undisturbed
c) is low
d) is high
Answer: c
Explanation: If the difference between the high-resolution scheme line and the upwind line increases, convergence is affected. The rate of convergence diminishes. This is when the normalized difference between the cell face values is high.
5. Deferred correction procedure for high-resolution schemes leads to __________
a) NVF and TVD
b) NWF and DWF
c) NVF and DWF
d) TVD and NWF
Answer: b
Explanation: The convergence problem of the high-resolution schemes paved the way for techniques of implementing high-resolution schemes that are more implicit such as NWF and DWF.
6. What does NWF stand for?
a) Normalized Weighing Factor
b) Northern Weighing Factor
c) Normalized West Factor
d) Normalized Weighted Formulation
Answer: a
Explanation: The abbreviation NWF stands for Normalized Weighing Factor. The abbreviation DWF stands for Downwind Weighing Factor. These two are implicit methods of formulating high-resolution schemes.
7. Which of these statements is correct?
a) The DWF method converges faster than the NWF method
b) Both the methods are of the same properties
c) The NWF method is more robust than the DWF method
d) The DWF method is more robust than the NWF method
Answer: c
Explanation: The NWF method is created to overcome the shortcomings of the DWF method. In the NWF method, instabilities do not arise. This allows the NWF method to be much more robust than the DWF method.
8. The Downwind Weighing Factor is a function of the flow variables at ___________
a) the face and upwind nodes
b) the far upwind, upwind and downwind nodes
c) the upwind and downwind nodes
d) the face, upwind and downwind nodes
Answer: d
Explanation: The Downwind Weighing Factor is given by
\(DWF_f=\frac{\phi_f-\phi_c}{\phi_d-\phi_c}\)
Where,
Φ f → Flow variable at the face.
Φ c → Flow variable at the upwind node.
Φ f → Flow variable at the downwind node.
9. What is the difficulty in the direct use of nodal values without the deferred correction method?
a) Convergence and stability
b) Stability and accuracy
c) Accuracy and Boundedness
d) Boundedness and convergence
Answer: a
Explanation: The difficulty arising when explicitly expressing a flow variable in terms of neighbouring values is the convergence rate. They violate one of the basic rules for stability and convergence which is the coefficients should be of the same sign.
10. The NWF method uses __________
a) TVD
b) NVF
c) Both NVF and TVD
d) Neither NVF nor TVD
Answer: b
Explanation: The NWF method operates by linearizing the normalized interpolation profile. This uses the simplified form of the TVD framework given by \(\tilde{\phi}_f=a\tilde{\phi}_c+b\). The correction term here is smaller than the one obtained by standard deferred correction methods.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “High Resolution Schemes – Downwind and Normalized Weighing Factor”.
1. The Downwind Weighing Factor in the normalized form is given by __________
a) \
\
\
\(\frac{\tilde{\phi_c}-\tilde{\phi_f}}{1-\tilde{\phi_f}}\)
Answer: a
Explanation: The Downwind Weighing Factor is given by
DWF f =\(\frac{\phi_f-\phi_c}{\phi_d-\phi_c}\)
Normalizing this, we get
DWF f =\(\frac{\tilde{\phi_f}-\tilde{\phi_c}}{\tilde{\phi_d}-\tilde{\phi_c}}\)
But, the value of \(\tilde{\phi_d}\) is 1. So,
DWF f =\(\frac{\tilde{\phi_f}-\tilde{\phi_c}}{1-\tilde{\phi_c}}\)
2. The value of the Downwind Weighing Factor lies between ___________
a) 0≤DWF≤∞
b) DWF≥0
c) 0≤DWF≤1
d) DWF≤1
Answer: c
Explanation: By using DWF, the high-resolution estimate of \(\tilde{\phi_f}\, or\, \phi_f\) is redistributed between the upwind and the downwind nodes. As the value of Φ f computed using Φ c and Φ c . The value of DWF always lies between 0 and 1.
3. The value of DWF for the downwind scheme is __________
a) 0
b) 1
c) 2
d) 3
Answer: b
Explanation: The relation between the DWF formulation and the TVD formulation is given by
DWF f =\(\frac{1}{2}\) ψ(r f )
The ψ(r f ) value for downwind scheme is 2. Therefore, the DWF f value is 1.
4. DWF f for the FROMM scheme is ___________
a) \
\
\
\(\frac{1}{4}\)
Answer: b
Explanation: For FROMM scheme,
\(\tilde{\phi_f}=\tilde{\phi_c}+\frac{1}{4}\)
Therefore,
DWF f =\(\frac{
-\tilde{\phi_c}}{1-\tilde{\phi_c}}\)
DWF f =\(\frac{1}{4
}.\)
5. For a scheme modelled using the DWF method, the diagonal coefficient becomes zero when ___________
a) DWF f > 0
b) DWF f > 1
c) DWF f > 0.5
d) DWF f > 2
Answer: c
Explanation: For values of DWF f larger than 0.5, results in a system with negative diagonal coefficients. So, the system becomes unsolvable by iterative methods. This happens whenever Φ f > 0.5(Φ c +Φ d ).
6. The value of DWF f for the central difference scheme is __________
a) 1
b) \
\
\(\frac{1}{2}\)
Answer: d
Explanation: For the central difference scheme,
ψ(r f )=1
So, the value of DWF f for this scheme is ½.
7. The deferred correction source term of the NWF method using he normalized interpolation profile \Missing open brace for subscript Φ u
b) Φ u
c) Φ u
d) Φ u
Answer: a
Explanation: We have the equation
\(\tilde{\phi_f}=l\tilde{\phi_c}+k\)
This can be expanded as
\(\frac{\phi_f-\phi_u}{\phi_d-\phi_u}=l \frac{\phi_c-\phi_u}{\phi_d-\phi_u}+k\)
\(\frac{\phi_f-\phi_u}{\phi_d-\phi_u}=l \frac{\phi_c-\phi_u}{\phi_d-\phi_u}+k\frac{\phi_d-\phi_u}{\phi_d-\phi_u}\)
Φ f =l(Φ c -Φ u )+k(Φ d -Φ u )+Φ u
Φ f =l(Φ c ))+k(Φ d ))+Φ u
The term Φ u in this equation is the deferred correction source term.
8. The high-resolution schemes formulated using the NWF method with the equation \Missing open brace for subscript k>2
b) l>2
c) k>l
d) l>k
Answer: d
Explanation: The NWF formulation of the high-resolution schemes, when the value of l is greater than the value of k, the diagonal coefficients are all positive and hence the solution is highly stable. This is the case everywhere except a narrow region in NVD.
9. What is DWF f for the second-order upwind scheme?
a) \
\
\
\(\frac{1}{4
}\)
Answer: a
Explanation: For the second order upwind scheme,
\(\tilde{\phi_f}=\frac{3}{2} \tilde{\phi_c}\)
Therefore,
DWF f =\(\frac{\frac{3}{2}\tilde{\phi_c}-\tilde{\phi_c}}{1-\tilde{\phi_c}}\)
DWF f =\(\frac{\tilde{\phi_c}}{2
}\).
10. Along the downwind line of the NVD, the values of _____________ are changed to make the system stable.
a) a c
b)
c) Φ c
d) Φ f
Answer: b
Explanation: Along the downwind line of NVD, the values of =, a value of zero is obtained for the diagonal coefficient and the system becomes unstable. To overcome this problem, the values of are set equal to (L,1-LΦ f ). The value of L can be chosen which is usually set to l in the previous interval.
This set of Computational Fluid Dynamics Questions and Answers for Campus interviews focuses on “FVM for 1-D Steady State Diffusion”.
1. Which of these statements is true?
a) 1-D steady-state diffusion is the simplest of all transport equations
b) 1-D steady-state diffusion is the toughest of all transport equations
c) 1-D steady-state convection is the simplest of all transport equations
d) 1-D transient diffusion is the simplest of all transport equations
Answer: a
Explanation: The one-dimensional steady-state diffusion is the simplest preliminary problem in CFD. This cancels out most of the terms in the general transport equation. In heat flow problems it means conduction and in mass-flow problems, it means diffusion.
2. Which of these equations represent 1-D steady state diffusion?
a) div+S=0
b) \
+S=0\)
c) \
+S=0\)
d) \
+S=0\)
Answer: b
Explanation: The term div represents diffusion in all three directions. One-dimensional diffusion is given by the equation \
\) is the transient term. So, this should not be present in the steady-state equation. Considering all these, the correct equation is
\
=0\).
3. Which of these theorems is used to transform the general diffusion term into boundary based integral in the FVM?
a) Gauss divergence theorem
b) Stokes’ theorem
c) Kelvin-Stokes theorem
d) Curl theorem
Answer: a
Explanation: The general diffusion term is div. Integrating for the finite volume method, it becomes
∫ CV divdV
Applying the Gauss divergence theorem,
∫ A \
dA
This is the boundary based integration as the boundaries will be areas.
4. Which of these gives the statement of one-dimensional steady-state diffusion problem?
a) The diffusive flux of Φ leaving the exit face is the same as the diffusive flux of Φ entering the inlet face
b) The diffusive flux of Φ leaving the exit face plus the diffusive flux of Φ entering the inlet face is equal to the generation of Φ
c) The diffusive flux of Φ leaving the exit face minus the diffusive flux of Φ entering the inlet face is equal to the generation of Φ
d) The diffusive flux of Φ leaving the exit face is the same in magnitude and opposite in direction as the diffusive flux of Φ entering the inlet face
Answer: c
Explanation: The diffusive flux of Φ leaving the exit face minus the diffusive flux of Φ entering the inlet face is equal to the generation of Φ. It constitutes the balance equation over the control volume. This ensures conservation.
5. Consider the following stencil.
computational-fluid-dynamics-questions-answers-campus-interviews-q5
Discretize the diffusive term of the one-dimensional steady-state diffusion problem based on this stencil. .
a) \
_e-
_w \)
b) \
_E-
_W\)
c) \
_E-
_W\)
d) \
_e-
_w\)
Answer: d
Explanation: The diffusive term in one dimension is \
.\) Applying integration over the control volume,
\
dV=\int_w^e
dA=
_e-
_w.\)
6. Consider the following stencil.
computational-fluid-dynamics-questions-answers-campus-interviews-q5
Apply linear interpolation to the term Γ e .
a) \
\
\
\(\frac{\Gamma_P+\Gamma_E}{2}\)
Answer: d
Explanation: The value of Γ at the face e can be obtained from the values of Γ at the nodes E and P by linear interpolation.
\(\Gamma_e = \frac{\Gamma_P+\Gamma_E}{2}\)
Similarly,
\(\Gamma_w = \frac{\Gamma_W+\Gamma_P}{2}\).
7. Consider the following stencil.
computational-fluid-dynamics-questions-answers-campus-interviews-q5
Get the discretized form of \
_w\) using the central differencing scheme.
a) \
\
\
\(\frac{\phi_P+\phi_W}{2}\)
Answer: c
Explanation: To get \
_i\) using central difference scheme,
\(\frac{d\phi}{dx}_i=\frac{\phi_{i+1}-\phi_{i-1}}{2\Delta x} \)
Applying it to the stencil, i=w, i+1=P, i-1=W and 2Δx=δx Ww +δx wP =δx WP . Therefore,
\
_w=\frac{\phi_P-\phi_W}{\delta x_{WP}}\).
8. The general discretized equation is modified for ____________
a) the central control volume
b) the boundary control volumes
c) the non-boundary control volumes
d) the interior control volumes
Answer: b
Explanation: There is special attention needed at the boundary nodes. For control volumes that are adjacent to the domain boundaries, the equation is modified so that the boundary values are incorporated into the equation without any problem.
9. Which of these equations govern the problem of source-free one-dimensional steady-state heat conduction?
a) \
\)
b) \
\)
c) \
\)
d) \
\)
Answer: a
Explanation: The general one-dimensional steady-state diffusion equation is:
\
+S=0 \)
For heat conduction problem, the diffusion constant is the heat conductivity and the flow variable is temperature . As the given problem is source free, S=0. Therefore, the equation becomes
\
=0\).
10. Consider the general discretized equation a P Φ P =a W Φ W +a E Φ E +S. Which of these will become zero for the left boundary node?
a) Φ E
b) a E
c) Φ W
d) a W
Answer: d
Explanation: For the left boundary node, there is no western node present. It has only one neighbour on the eastern side. So, the western node coefficient a W is set to zero. Φ W is not zero as we do not know the flow variable at that point and cannot assume.
This set of Computational Fluid Dynamics Questions and Answers for Entrance exams focuses on “FVM for Multi-dimensional Steady State Diffusion”.
1. Which of these equations represent the semi-discretized equation of a 2-D steady-state diffusion problem?
a) \
dA+\int_A
dA+\int_{\Delta V} S\,dV=0\)
b) \
dA+\int_A\frac{\partial}{\partial y}
dA+\int_{\Delta V}S\, dV=0\)
c) \
dA+\int_A
dA+\int_{\Delta V}S\, dV=0\)
d) \
dA+\int_A\frac{\partial}{\partial y}
dA+\int_{\Delta V}S\, dV=0\)
Answer: a
Explanation: The general governing equation for a 2-D steady-state diffusion problem is given by
\
+\frac{\partial}{\partial y}
+S=0\)
Here, partial differentiation is used as the variable φ varies in both x and y directions, but the differentiation is only in the required direction.
Integrating the equation with respect to the control volume,
\
dV+\int_{\delta V}\frac{\partial}{\partial y}
dV+\int_{\Delta V} S \,dV=0\)
Applying Gauss Divergence theorem,
\
dA+\int_A
dA+\int_{\Delta V}S \,dV=0\)
This is the semi-discretized form of the equation.
2. The area in the western face of a 2-D steady-state diffusion stencil is _______________
a) grid size in the x-direction
b) grid size in the y-direction
c) product of the grid sizes in the x and y-directions
d) ratio of the grid sizes in the x and y-directions
Answer: b
Explanation: In the one-dimensional case, the area is taken to be unity. In the two-dimensional case, the area is the grid size in the perpendicular direction multiplied by unity. So, for area A e =A w =Δy and A n =A s =Δx.
3. Consider the following stencil.
computational-fluid-dynamics-questions-answers-entrance-exams-q4
What is the flux across the northern face?
a) \
\
\
\(\Gamma_Na_N\frac{
}{\delta x_{PN}}\)
Answer: b
Explanation: Flux across the northern face is \(\Gamma_N a_N\frac{\partial\phi}{\partial y}\Big|_n\). Expanding this using the central difference scheme, we get
\(\Gamma_N a_N\frac{\partial\phi}{\partial y}\Big|_n = \Gamma_Na_N \frac{
}{\delta y_PN}\).
4. Consider the following stencil.
computational-fluid-dynamics-questions-answers-entrance-exams-q4
For a source-less 2-D steady-state diffusion problem, the coefficient of the flow variable Φ P is ____
a) \
\
\
\(\frac{\Gamma_W A_W}{\delta x_{WP}}+\frac{\Gamma_S A_S}{\delta x_{SP}}+\frac{\Gamma_E A_E}{\delta y_{PE}}+\frac{\Gamma_N A_N}{\delta y_{PN}}\)
Answer: a
Explanation: The general form is given by a P Φ P =a E Φ P +a W Φ W +a N Φ N +a S Φ S
Here, for source-less problem, a P is the addition of all fluxes given by
\(\frac{\Gamma_W A_W}{\delta x_{WP}}+\frac{\Gamma_E A_E}{\delta x_{PE}}+\frac{\Gamma_S A_S}{\delta y_SP}+\frac{\Gamma_N A_N}{\delta y_{PN}}\).
5. If a P Φ P =a E Φ P +a W Φ W +a N Φ N +a S Φ S +S is the general form of a 2-D steady-state diffusion problem, what is a E by considering the following stencil?
computational-fluid-dynamics-questions-answers-entrance-exams-q4
a) \
\
\
\(\frac{\Gamma_E A_E}{\delta x_{WP}}\)
Answer: c
Explanation: Flux in the eastern direction is given by
\(\Gamma_E A_E\frac{\partial\phi}{\partial x}\Big|_e=\Gamma_E A_E\frac{
}{\delta x_{PE}}\)
\(\Gamma_E A_E\frac{\partial\phi}{\partial x}\Big|_e=\Gamma _e A_E\frac{\phi_E}{\delta x_{PE}}-\Gamma_E a_E\frac{\phi_P}{\delta x_{PE}}\)
Expanding this while forming the general equation, we will get
\(a_E=\frac{\Gamma_E A_E}{\delta x_{PE}}\).
6. Consider the following 2-D surface with the numbers inside as the global indices of their cells.
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
The general discretized equation is of the form a P Φ P =a E Φ P +a W φ W +a N Φ N +a S Φ S +S. Which of the following is correct regarding the cell numbered “13”?
a) a E =0; a W =0
b) a W =0; a N =0
c) a N =0; a S =0
d) a S =0; a W =0
Answer: d
Explanation: For the control volumes adjacent to the boundary of the global domain, the boundary-side coefficient is set to zero. Therefore, for the cell numbered “13”, the southern and the western coefficients are zero (a S =0; a W =0).
7. I general, for all the steady-state diffusion problems, the discretized equation can be given as a P Φ P = ∑a nb Φ nb -S. For a one-dimensional problem, which of these is wrong?
a) ∑a nb =a T +a B
b) ∑a nb =a S + a N
c) ∑a nb =a W +a E
d) ∑a nb =a P +a E
Answer: d
Explanation: For a one-dimensional problem is x-direction, ∑a nb =a W +a E . For a one-dimensional problem is y-direction, ∑a nb =a S + a N . For a one-dimensional problem is z-direction, ∑a nb =a T +a B .
8. In a control volume adjacent to the boundary, the flux crossing the boundary is _______________ in the discretized equation.
a) set to some arbitrary constant
b) set to zero
c) introduced as a source term
d) introduced as a convective flux
Answer: c
Explanation: As the boundary-side coefficients are set to zero in the discretized equations of the boundary-based control volumes, the information in the boundary may be lost. To avoid this, the flux crossing the boundary is introduced as a source term in the equation.
9. Consider a source-less 3-D steady-state diffusion problem. The general discretized equation is a P Φ P = ∑a nb Φ nb . What is a P ?
a) a P =a W +a E +a S +a N +a T +a B
b) a P =a W +a E +a S +a N
c) a P =a W +a E +a S +a N +a T
d) a P =0
Answer: a
Explanation: For all steady-state diffusion problems, in the absence of source term, a P =∑a nb . Therefore, for the three-dimensional case, a P =a W +a E +a S +a N +a T +a B which includes the coefficients of all the neighbouring flow variables.
10. Consider the stencil.
computational-fluid-dynamics-questions-answers-entrance-exams-q4
The values of \Missing open brace for subscript \
\
\
\(\vec{A_w}=\Delta y; \vec{A_s}=\Delta x\)
Answer: b
Explanation: The values of A w and A s are Δ x and Δ y respectively. The signs of the area vectors depend on their directions. Therefore, \(\vec{A_w}=-\Delta x; \vec{A_s}=-\Delta y\).
This set of Computational Fluid Dynamics Question Bank focuses on “Diffusion Problem – Discretization Equation Rules”.
1. The zero sum rule and the opposite signs rule are applicable to _______________
a) each discretized equation
b) the global matrix
c) the coefficients of each discretized equation
d) the coefficient matrix
Answer: c
Explanation: A proper discretization should result in a discretized algebraic equation that reflects the characteristics of the original conservation equation. The coefficients of each discretized equation should satisfy the zero sum and the opposite signs rules.
2. Which of these is a sufficient condition for a discretized equation?
a) Neither the opposite sign rule nor the zero sum rule
b) Both the opposite sign and the zero sum rules
c) The opposite signs rule
d) The zero sum rule
Answer: c
Explanation: For of the discretized equations to be bounded, the sufficient condition is the opposite signs rule. Sufficient condition means that the rule will be enough to make sure that the equation is bounded.
3. Which of these assumptions is made regarding the variation of Φ over a domain?
a) Linear profile
b) Central differencing
c) Quadratic profile
d) Downwind differencing
Answer: a
Explanation: The major approximation made while discretizing a governing equation is “variation of the flow variable is considered to be linear”. The linear interpolation method is used to get the unknown values near the known values.
4. Why a parabolic profile is not used to model the variation of Φ?
a) Accuracy comes at the cost of computation
b) Exact results
c) Unphysical results
d) Difficult to use
Answer: c
Explanation: Consider a parabolic profile is used to find the value of a flow variable at the face in between two centroids. It will lead to a value higher or lower than the values at the centroids which will be unphysical.
5. In the absence of any source or sink, the steady-state diffusion problem is governed by _______________
a) Fourier series
b) Linear interpolation
c) Taylor series
d) Second order interpolation
Answer: a
Explanation: For a source-less steady state diffusion problem, the transfer of the flow variable occurs only by diffusion. So, the transfer of Φ will be in the direction opposite to the increasing Φ. This is governed by Fourier series.
6. Consider the general discretized equation given by a P Φ P +∑ F~NB a F Φ F =0. Which of these statements is correct?
a) When the value of Φ F is increased, the value of Φ P remains the same
b) When the value of Φ F is increased, the value of Φ P increased
c) When the value of Φ F is increased, the value of Φ P decreases
d) When the value of Φ F is decreased, the value of Φ P decreases
Answer: c
Explanation: The value of Φ P varies with the variation of the values of Φ F . So, it will not remain the same. When the value of Φ F is increased, the value of Φ P decreases. This is the statement of the opposite signs rule taken physically.
7. Boundedness is ensured in the steady-state diffusion problem _______________
a) only when the source term is non-negative
b) only when the source term is negative
c) only when the source term is non-zero
d) only when the source term is zero
Answer: d
Explanation: When there is a source term, the linear profile of the flow variable may fail. The source or sink term may lead to increased or decreased values than that guesses by the linear interpolation. So, in this case, boundedness is not possible.
8. Consider the general discretized equation given by a P Φ P +∑ F~NB a F Φ F =0. Which of these statements is correct according to the opposite signs rule?
a) a P and a F are of opposite signs
b) a P and Φ P are of opposite signs
c) a F and Φ F are of opposite signs
d) Φ P and Φ F are of opposite signs
Answer: a
Explanation: The boundedness property is enforced only when the coefficients a P and a F are of opposite signs. So, the opposite sign rule says that the coefficients of the flow variables Φ P and Φ F are of opposite signs.
9. Consider the general discretized equation given by a P Φ P +∑ F~NB a F Φ F = 0. According to the zero sum rule, which of these is correct?
a) a P +∑ F~NB a F = ∞
b) a P +∑ F~NB a F = 0
c) a P +∑ F~NB a F = 1
d) a P +∑ F~NB a F = -1
Answer: b
Explanation: A consistent discretization method should yield an equation which incorporates the property of the overall domain – conservation. To ensure this, the equation should satisfy the condition a P +∑ F~NB a F = 0.
10. Consider the general discretized equation given by a P Φ P +∑ F~NB a F Φ F =0. According to the zero sum rule, which of these is correct?
a) ∑ F~NB \
∑ F~NB \
∑ F~NB \
∑ F~NB \(\frac{a_F}{a_P}\) = -∞
Answer: c
Explanation: From the zero sum rule,
\(a_P+\sum_{)}a_F = 0\)
Divided by a P , the equation becomes
\(1+\frac{\sum_{F \sim NB}a_F}{a_P} = 0 \)
\(\frac{\sum_{F \sim NB}a_F}{a_P} =-1\)
This can be written as
\(\sum_{F \sim NB}\frac{a_F}{a_P} =-1\).
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Diffusion Problem – Orthogonal and Non-Orthogonal Grids”.
1. Which of these statements is true?
a) The Cartesian and non-Cartesian orthogonal grids lead to the same discretized equation
b) A non-Cartesian orthogonal grid leads to an extra source term when compared to the Cartesian orthogonal grids
c) The equations of the Cartesian and non-Cartesian orthogonal grids differ by a trigonometric function
d) The equations obtained from a non-Cartesian grid has fewer terms when compared to that obtained from a Cartesian grid
Answer: a
Explanation: The discretized equation for the non-Cartesian grids should be exactly the same as obtained from the Cartesian grids. The solution should also be the same if the boundary conditions match.
2. Non-orthogonality creates a problem in _________ of the steady-state diffusion equation.
a) the neighbouring terms
b) the source term
c) the direction of the surface vector
d) the magnitude of the surface vector
Answer: c
Explanation: The surface vector and the vector joining the owner and the neighbouring elements are not collinear for non-orthogonal grids. Thus, non-orthogonal grids need special attention in the steady-state diffusion equation.
3. Non-orthogonality leads to ________ in diffusion problems.
a) cubic-diffusion
b) less-diffusion
c) additional-diffusion
d) cross-diffusion
Answer: d
Explanation: In general, the surface vector of the non-orthogonal grids is given as the sum of the vector connecting the owner and the neighbour elements and an additional vector. This leads to an extra term in the diffusion equation called the cross-diffusion or non-orthogonal diffusion.
4. Which of these statements is false?
a) Unstructured grids are always non-orthogonal
b) Structured grids are always orthogonal
c) Non-orthogonal grids can be structured or unstructured
d) Curvilinear structured grids are non-orthogonal
Answer: b
Explanation: Non-orthogonality can exist in structured grids also when it is curvilinear. Structured curvilinear grids and unstructured grids are non-orthogonal. So, the statement “Structured grids are always orthogonal” is wrong.
5. In the minimum correction approach of decomposing the surface vector of a non-orthogonal grid, the relationship between the vector connecting the owner and the neighbour node \
\) and the surface vector \
\) is given as _________
a) \
\
\
\vec{e}\)
d) \
\vec{e}\)
Answer: c
Explanation: Here, a right-angled triangle is formed by the vectors \
\vec{e}=
\vec{e}\).
6. Which of these is correct regarding the minimum correction approach?
a) The non-orthogonal correction is kept as small as possible
b) The non-orthogonal correction is kept as large as possible
c) The surface vector is kept as small as possible
d) The surface vector is kept as large as possible
Answer: a
Explanation: In the minimum correction approach, the decomposition of the surface vector is done in a way that the non-orthogonal correction is as small as possible, thus making the \(\vec{E_f} \,and\, the\, \vec{T_f}\) orthogonal.
7. In the orthogonal correction approach, the relationship between \Missing open brace for subscript \
\
\
\(\vec{E_f}=\vec{S_f}.\vec{e}\)
Answer: c
Explanation: In this approach, the contribution of the term involving Φ F and Φ C are kept the same as that of the orthogonal mesh. This is achieved by the relation
\(\vec{E_f}=S_f \vec{e}\).
8. In the over-relaxed approach, the importance of the term involving Φ F and Φ C _________ as the non-orthogonality _________
a) decreases, increases
b) remains the same, increases
c) increases, remains the same
d) increases, increases
Answer: d
Explanation: The importance of the term involving Φ F and Φ C decreases when the non-orthogonality decreases for the minimum correction approach. For the over-relaxed approach, the importance increases.
9. What is the relationship between \
\
.\vec{e}\)
b) \
\vec{e}\)
c) \
×\vec{e}\)
d) \
.\vec{e} ) \vec{e}\)
Answer: b
Explanation: The relationship is given by \
\vec{e}\). This is calculated in CFD packages as \
\vec{e}=
\vec{e} =\frac{\vec{S_f}.\vec{S_f}}{\vec{e}.\vec{S_f}}\vec{e}\).
10. Which of these methods is used to treat the non-orthogonal diffusion term?
a) Deferred correction
b) Predictor–corrector
c) Green-gauss
d) Trial and error method
Answer: a
Explanation: The cross-diffusion term cannot be expressed in terms of nodal values. So, deferred correction is used here. Its value is computed using the current gradient field and this is added as a source term in the right-hand side of the algebraic equation.
This set of Basic Computational Fluid Dynamics Questions and Answers focuses on “Diffusion Problem – Green-Gauss Gradient for Cartesian Grids”.
1. To get the gradient of the flow variable using the Green-Gauss Theorem, which of these theorems is used?
a) Mean value theorem
b) Stolarsky mean
c) Racetrack principle
d) Newmark-beta method
Answer: a
Explanation: The Green-Gauss theorem states that for a closed volume V with the surrounding surface ∂V and outward pointing incremental surface vector d\(\vec{S}\),
∫ V \(\nabla\Phi dV=∮_{∂V} \Phi d\vec{S}\)
Using the mean value theorem,
∫ V ∇ΦdV=\(\overline{\nabla\Phi} V\)
Where, \(\overline{\nabla\Phi} V\) is the average gradient over the volume V.
2. What is the final form of the Green-Gauss gradient method for finding the gradient of Φ over element C?
a) ∇Φ C =∑ f~nb Φ f \
∇Φ C =1/V C ∑ f~nb Φ f \
∇Φ C =1/V C ∑ f~nb Φ f
d) ∇Φ C =1/V C ∑ f~nb a f Φ f
Answer: b
Explanation: The final form of the Green-Gauss gradient method is given by
∇Φ C =1/V C ∑ f~nb Φ f \(\vec{S_f}\)
Where,
\(\vec{S_f}\) is known from the geometry of the grids.
Φ f for all the faces should be known to compute ∇Φ C .
3. The gradient at the face of an element is obtained using ________
a) Linear interpolation
b) Geometric values
c) Green-Gauss theorem
d) Weighted average
Answer: c
Explanation: Gradient at a face of an element is given by the weighted average of the gradients at the centroids of the element sharing that surface.
∇Φ f =g C ∇Φ C +g f ∇Φ f .
4. The face-based stencil used for computing Φ f in the Green-Gauss Gradient formula is ________
a) more accurate and needs a large stencil
b) less accurate and needs a large stencil
c) more accurate and needs a compact stencil
d) less accurate and needs a compact stencil
Answer: d
Explanation: The face-based computation of Φ f uses a compact stencil involving face neighbours. This is less accurate than the vertex-based computation involving a large stencil of vertex neighbours.
5. It is easy to construct _________ in the face-based computation.
a) Grids
b) Stencil
c) Global matrix
d) Jacobian matrices
Answer: d
Explanation: Compact stencil uses implicit methods. So, it is easy to construct compact Jacobian matrices. But the large stencil brings more information into the reconstruction and therefore is more accurate.
6. The value of the flow variable at face centre (Φ f ) in terms of the flow variable at the owner cell’s centre (Φ C ) and the neighbouring cell’s centre (Φ f ) as given by the face-based stencil is ________
.
a) Φ f =g C Φ C +g C Φ f
b) Φ f =g C Φ C +g f Φ f
c) Φ f =g C Φ C +(1-g C )Φ f
d) Φ f =g C Φ C +(1+g C )Φ f
Answer: c
Explanation: In compact stencil, the value of Φ f is calculated using the weighted average values of the two cells sharing the face. It is given by
Φ f =g C Φ C +(1-g C ) Φ f
Where,
g C =\(\frac{distance_{Ff}}{distance_{FC}}\).
7. The face-centred stencil is second-order _________
a) always
b) only when the centroid of the face and the line connecting the two cells meet
c) only when the centroid of the face and the line connecting the two cells do not meet
d) never
Answer: b
Explanation: The face-centred stencil is second-order accurate when the centroid of the face is the same as the point of intersection of the face and the line connecting the two cells. This happens when the centroid and the line connecting the two cells meet.
8. In the vertex-based stencil, the flow variable at the face centroid is computed as ___________
a) the mean of the values at the vertices defining the surface
b) the mean of the values at the cell centres
c) the mean of the values at the vertices of the cells
d) the mean of the values at the centroids of the neighbouring faces
Answer: a
Explanation: In the enlarged stencil, the value Φ f is calculated as the mean of the values at the vertices defining the surface. Mathematically representing it,
Φ f =\(\frac{\phi_{n1}+\phi_{n2}}{2}\).
9. The flow variable at the vertex node is calculated using the weighted average of the values at the cells sharing it. What is the weight used here?
a) Inverse of the distance of the vertex from the cell centroid
b) Distance of the vertex from the cell
c) Centroids of the cells
d) Mass of the cells
Answer: d
Explanation: The impact decreases as the centroid of the cell moves away from the vertex. So, the inverse is used as the weight here. This is why the method needs an enlarged stencil. But it leads to more accurate approximation.
10. To overcome the disadvantage caused by the information from the wrong side of cells ____________ is used in the vertex-based method.
a) upwind biased scheme
b) weighted average
c) downwind biased scheme
d) central scheme
Answer: a
Explanation: The major disadvantage of using the enlarged stencil is that information from the wrong side of the face may also contribute to the weighted average while calculating the vertex values. To avoid this, the upwind biased method is used.
This set of Advanced Computational Fluid Dynamics Questions and Answers focuses on “Diffusion Problem – Least-Square Gradient for Cartesian Grids”.
1. Which of these statements is true?
a) The Gauss-Gradient method is a special case of the least-square gradient method
b) The least-square gradient method is a special case of the Gauss-Gradient method
c) The least-square method is not connected with the Gauss-Gradient method
d) The least-square method is suitable only for the Cartesian grids
Answer: a
Explanation: The least-square method is a general method. This can be reduced to the Gauss-Gradient method by altering some values. It is obtained by applying the least-square method to a Cartesian grid system.
2. What is the advantage of the least-square method over the other methods?
a) Computational ease
b) Accuracy
c) Flexibility
d) Stability
Answer: c
Explanation: The least-square method offers more flexibility over the other methods to compute gradients for choosing between the order of accuracy used and the stencil on which to work. According to the user’s ease, these two can be chosen.
3. What is the disadvantage of using the least-square method?
a) Inconsistent
b) Less convergence rate
c) Instability
d) Computational cost
Answer: d
Explanation: The cost for the advantage of the least-square gradient methods is its high computational cost. This high computational cost is due to its separate calculation of the weighted average.
4. The weight used in the least-square method is a function of __________
a) twice the distance between the vertex and the centroid of the cells
b) square of the distance between the vertex and the centroid of the cells
c) inverse of the distance between the vertex and the centroid of the cells
d) the distance between the vertex and the centroid of the cells
Answer: c
Explanation: Generally, the weight used is based on the distances. We can choose it to be the inverse of the distance between the vertex and the cell centroids. Mostly, the weight is a function of the distance between the vertex and the cell centroids.
5. In the least-square method, the gradient is computed using ___________
a) trial and error method
b) optimization method
c) weighted average
d) predictor-corrector method
Answer: b
Explanation: The least-square method is based on an optimization procedure. This optimization carried over a function which is a function of the weight used, the gradients and the grid sizes in the x, y and z-directions.
6. The gradient is found in the least-square method by solving ____________
a) a system of n equations
b) a single equation
c) a system of n 2 equations
d) a system of 2n equations
Answer: a
Explanation: A set of equations with the number of equations equal to the number of dimensions is formed using some conditions. This is written in the matrix form and the matrix is solved to get the gradients.
7. The least-square method is exact when ____________
a) the system is two-dimensional
b) the system is linear
c) the system is two-dimensional
d) the system is quadratic
Answer: b
Explanation: Unless the field where we solve the problem is linear, the solution using the least square method will not be exact. This is because the number of columns in the coefficient matrix will be more than the number of rows.
8. While using the Cartesian grids, the coefficient matrix becomes ____________
a) a square matrix
b) an upper triangular matrix
c) a diagonal matrix
d) a lower triangular matrix
Answer: c
Explanation: The coefficient matrix in the least-square method gives a square matrix. While substituting the required values in this square matrix for the Cartesian grids, the matrix is reduced into a diagonal one.
9. The least-square method is ____________
a) at least first-order accurate
b) at least second-order accurate
c) first-order accurate
d) second-order accurate
Answer: a
Explanation: The accuracy of the gradient found using this method is at least first-order. This can be proved by expanding the values of the flow variable using the Taylor series. It can be of higher orders also.
10. The diagonal elements of the coefficient matrix obtained by applying the least-square to Cartesian grids are ___________
a) the ratio of the grid sizes and the weights in the x, y, z-directions
b) the product of the grid sizes and the weights in the x, y, z-directions
c) the weights in the x, y, z-directions
d) the grid sizes in the x, y, z-directions
Answer: d
Explanation: The coefficient matrix obtained by applying the least-square to Cartesian grids has non-zero elements only in the diagonal. These non-zero elements are Δx, Δy and Δz .
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Convection-Diffusion Problems – Upwind and Downwind Schemes”.
1. The upwind scheme is suitable for _____________
a) convection term
b) diffusion term
c) both convection and diffusion terms
d) either convection or diffusion term
Answer: a
Explanation: The upwind scheme is not suitable for non-directional phenomena. The diffusion scheme is a non-directional phenomenon. The convection scheme is a directional phenomenon. So, the scheme is suitable for the convection term.
2. The upwind scheme is dependent on the _____________
a) Convection term
b) Peclet number
c) Flow direction
d) Gradient
Answer: c
Explanation: The upwind scheme reflects the physics of advection. The cell-face value is dependent on the upwind nodal value. So, we can say it is dependent on the flow direction and it is suitable for directional flows.
3. Consider the stencil.
computational-fluid-dynamics-questions-answers-upwind-downwind-schemes-q7
Give the advection flux at face e \
\) using the upwind scheme.
.
a) \
\phi_C-max
\phi_E\)
b) \
\phi_C-max
\phi_E\)
c) \
\phi_C-min
\phi_E\)
d) \
\phi_C-min
\phi_E\)
Answer: a
Explanation: According to the upwind scheme,
\
\phi_C-max
\phi_E\).
4. The neighbour coefficients yielded by the upwind scheme for convection is _____________
a) zero
b) cannot predict
c) negative
d) positive
Answer: c
Explanation: The upwind scheme leads to negative neighbouring coefficients. If continuity is ensured, the diagonal coefficients are the addition of the neighbouring coefficients.
5. Consider the following stencil.
computational-fluid-dynamics-questions-answers-upwind-downwind-schemes-q7
Give the relationship between \
for the upwind scheme while used for the convection term.
.
a) \
\
\
\(\frac{\phi_C-\phi_W}{\phi_E-\phi_W} = \frac{2+max}{4+\big|Pe\big|} \)
Answer: d
Explanation: For the upwind scheme,
Φ C -Φ W = 2+max and Φ E -Φ C = 2+max
Therefore,
\(\frac{\phi_C-\phi_W}{\phi_E-\phi_C+\phi_C-\phi_W} = \frac{2+max}{2+max+2+max}\)
\(\frac{\phi_C-\phi_W}{\phi_E-\phi_W}=\frac{2+max}{4+\big|Pe\big|}\).
6. The order of accuracy of the upwind scheme is _____________
a) first-order
b) second-order
c) third-order
d) fourth-order
Answer: a
Explanation: The upwind scheme is first-order accurate. This is why, even though the fashion of the upwind scheme matches with that of the exact solution, it varies much. The downwind scheme is also first-accurate.
7. Consider the stencil.
computational-fluid-dynamics-questions-answers-upwind-downwind-schemes-q7
Give the advection flux at face w \
\) using the downwind scheme.
.
a) \
\phi_C+max
\phi_W\)
b) \
\phi_C+max
\phi_W\)
c) \
\phi_C+max
\phi_W\)
d) \
\phi_C+max
\phi_W\)
Answer: c
Explanation: According to the downwind scheme,
\
\phi_C+max
\phi_W\).
8. The advantage of the upwind scheme is over the central-difference scheme is _____________
a) accuracy
b) stability
c) high convergence rate
d) consistency
Answer: a
Explanation: The upwind scheme is less accurate than the central difference schemes. But the central difference schemes are oscillatory. They do not give answers which are physically correct. This is the advantage of the upwind scheme over the central-difference scheme.
9. The upwind scheme is _____________
a) conservative but wiggles
b) bounded and conservative
c) bounded but not conservative
d) neither conservative nor bounded
Answer: b
Explanation: The upwind scheme does not produce results which wiggle. So, it is bounded. It uses consistent expressions to calculate fluxes through cells. Therefore, it is sure that the formulation is conservative.
10. When the flow is not aligned with the grid lines, the diffusion produced by the upwind scheme is ____________
a) false advection
b) false convection
c) anti-diffusion
d) false diffusion
Answer: c
Explanation: A major drawback of the scheme is that it does not produce correct results when the flow is not aligned with the grid lines. The error has a diffusion-like appearance and is referred to as false diffusion.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Convection-Diffusion Problems – Central Difference Schemes”.
1. Consider the following stencil.
computational-fluid-dynamics-questions-answers-central-difference-schemes-q1
What is Φ e as given by the central difference scheme?
.
a) \
\)
b) \
\)
c) \
\)
d) \
\)
Answer: c
Explanation: The central difference scheme incorporates values both behind and ahead of the central node in the equation. Therefore,
\
\).
2. Consider the following stencil.
computational-fluid-dynamics-questions-answers-central-difference-schemes-q1
Assume that the grid is a uniform Cartesian grid. What is φ w as given by the central difference scheme?
.
a) Φ c
b) \
\
\(\frac{\phi_w-\phi_c}{2}\)
Answer: b
Explanation: The formula for central differencing is
\
\)
For uniform grids,
\
\)
\(\phi_E=\frac{
}{2}\).
3. What is the central differencing scheme similar to?
a) Interpolation profile
b) Linear interpolation profile
c) Weighted average method
d) Geometric mean
Answer: b
Explanation: The central difference scheme matches the linear interpolation profile. The general form of both are the same.
Φ=k 0 +k 1 (x-x C ).
4. What is the relationship between \
when the grid is uniform?
a) \
\)
b) \
\)
c) \
\)
d) \
\)
Answer: a
Explanation: For the central difference scheme applied to the convection-diffusion problems
\
\).
5. The central difference approximation goes wrong when _____________
a) Peclet number is negative
b) Peclet number is positive
c) Peclet number is low
d) Peclet number is high
Answer: d
Explanation: When the Peclet number is low in the positive or negative direction, the central differencing scheme is valid. If the Peclet number goes beyond a certain value both in the positive and negative direction, this approximation gives unphysical answers.
6. Which of these is correct about the central differencing scheme?
a) The importance of upwind and downwind nodes depends on the problem
b) It gives more importance to the downwind nodes
c) It gives equal importance to upwind and downwind nodes
d) It gives more importance to the upwind nodes
Answer: c
Explanation: The central differencing scheme gives equal importance to the upwind and the downwind nodes. The contribution of all the neighbouring nodes is considered for this approximation.
7. The central differencing scheme becomes inconsistent when the Peclet number _____________
a) is higher than 2
b) is less than 2
c) is higher than 5
d) is less than 5
Answer: a
Explanation: When the Peclet number goes beyond 2, the central difference approximation fails. The discretization process becomes inconsistent here. In this case, an increase in the neighbouring value will lead to a decrease in the value at the central node.
8. The central difference scheme gives unphysical results when the problem is _____________
a) depends on the boundary conditions
b) equally dominated by diffusion and convection
c) diffusive dominant
d) convective dominant
Answer: d
Explanation: In diffusion problems, both the upwind and the downwind neighbours will be equally important. As this is the case in the central differencing scheme, the scheme fits diffusive dominant problems.
9. The order of accuracy of the central differencing scheme is _____________
a) fourth-order
b) third-order
c) second-order
d) first-order
Answer: c
Explanation: The central differencing scheme is second-order accurate. This can be proved by using the Taylor series expansion. This is more accurate when compared to the upwind or the downwind schemes.
10. The central differencing scheme gives good results when _____________
a) the grid is coarse
b) the grid is very fine
c) the grid is Cartesian and uniform
d) the gird is on-Cartesian
Answer: b
Explanation: The central differencing scheme is good when the cell Peclet number is less than 2. For this, the grid should be very fine. So, the central differencing scheme is good to use when the grid is fine.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Convection-Diffusion Problems – Hybrid Differencing Scheme”.
1. The hybrid differencing scheme is a combination of ____________ and ___________
a) upwind and downwind schemes
b) downwind and central difference schemes
c) central difference and upwind schemes
d) two types of central difference schemes
Answer: c
Explanation: Hybrid differencing scheme was introduced by Spalding in 1970s. It is the hybrid between upwind and central differencing schemes so that the advantages of both of these schemes is utilized.
2. The difference scheme to be used in the hybrid system is chosen by evaluating the ____________
a) Local Peclet number
b) Global Peclet number
c) Reynolds number
d) Nusselt number
Answer: a
Explanation: The central differencing scheme works well with low Peclet numbers and the upwind scheme works well for high Peclet numbers. So, the differencing scheme to be used in this method is chosen using the Peclet number.
3. The hybrid differencing scheme is ____________
a) never bounded
b) bounded unconditionally
c) bounded in the low Peclet number
d) bounded in the high Peclet number
Answer: b
Explanation: The coefficients of the hybrid differencing scheme is always positive. So, it is unconditionally bounded. There is no particular region for the boundedness of the hybrid differencing scheme.
4. What is the order of accuracy of the hybrid differencing scheme?
a) Fourth-order
b) Third-order
c) Second-order
d) First-order
Answer: d
Explanation: The major disadvantage of the hybrid difference scheme is its low order of accuracy based on the Taylor series truncation term. It is first-order accurate. Yet, it is useful for solving practical flow problems.
5. Which of these is correct about the hybrid differencing scheme?
a) It is conservative but not transportive
b) It is conservative and transportive
c) It is transportive but not conservative
d) It is neither transportive nor conservative
Answer: b
Explanation: The hybrid differencing scheme is fully conservative. It satisfies the transportiveness condition by using upwind scheme for high Peclet numbers. So, the scheme is conservative and transportive as well.
6. What is the advantage of the hybrid differencing scheme compared to the QUICK scheme?
a) Transportiveness
b) Accuracy
c) Stability
d) Conservativeness
Answer: c
Explanation: The QUICK scheme also possesses conservativeness and transportiveness. QUICK scheme has a higher order of accuracy. But, it is not stable. Stability is the advantage of the hybrid scheme over this scheme.
7. The Peclet number is calculated at the ____________
a) control volume
b) cell centres
c) vertices
d) faces
Answer: d
Explanation: The hybrid differencing scheme uses piecewise formulae based on the Peclet number evaluated at the faces of each control volume. Based on this Peclet number, a scheme is chosen.
8. In which of these ranges is the central differencing scheme used?
a) -2≤Pe≤2
b) -1≤Pe≤1
c) -0.5≤Pe≤0.5
d) -5≤Pe≤5
Answer: a
Explanation: The central differencing scheme is valid until the Peclet number reaches a value of two. So, in the hybrid difference scheme, in the range -2≤Pe≤2, the central differencing scheme is used.
9. Consider the following stencil.
Which is correct about the hybrid differencing scheme?
.
a) q w =F w φ C if Pe≥2; q w =F w φ C if Pe≤-2
b) q w =F w φ W if Pe≥2; q w =F w φ C if Pe≤-2
c) q w =F w φ W if Pe≥2; q w =F w φ W if Pe≤-2
d) q w =F w φ C if Pe≥2; q w =F w φ W if Pe≤-2
Answer: b
Explanation: When the Peclet number goes more than positive two or less than negative two, the upwind scheme is employed. Therefore,
\(q_w =
\)
10. Consider the following stencil.
For steady two-dimensional convection-diffusion problem, if the general discretized equation is a P φ P =a W φ W +a E φ E +a S φ S +a N φ N , what is a N using the hybrid differencing scheme?
.
a) max(F n ,(D n –\
,0)
b) max(-F n ,(D n +\
,0)
c) max(-F n ,(D n –\
,0)
d) max(F n ,(D n +\
,0)
Answer: c
Explanation: The coefficients are decided by the Peclet numbers. Put into a simple form, they are given as the maximum among the three as
a N =max(-F n ,(D n –\
,0).
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Convection-Diffusion Problems – Second Order Upwind Scheme”.
1. The Second Order Upwind scheme uses ____________
a) asymmetric linear profile
b) symmetric linear profile
c) asymmetric quadratic profile
d) symmetric quadratic profile
Answer: a
Explanation: The second order upwind scheme, like the central difference scheme, uses a linear profile. But, unlike the central differencing scheme, it uses an asymmetric linear profile. This is why it got the name upwind scheme.
2. The value at the face in the second order upwind scheme is calculated using _____________
a) interpolation
b) extrapolation
c) weighted average
d) geometric mean
Answer: b
Explanation: Second order upwind scheme uses an upwind biased stencil. Therefore, it needs linear extrapolation to guess the values at the faces instead of interpolation. This is where it is different from the central differencing scheme.
3. The second-order upwind scheme is ___________ than the general upwind scheme.
a) less diffusive
b) more diffusive
c) less accurate
d) less stable
Answer: a
Explanation: The second-order upwind scheme is more accurate accurate than the general upwind scheme. But, it is less diffusive when compared to the general upwind scheme.
4. Consider the stencil.
computational-fluid-dynamics-questions-answers-second-order-upwind-scheme-q5
What is φ e according to the second-order upwind scheme?
.
a) \
\)
b) \
\)
c) \
\)
d) \
\)
Answer: d
Explanation: The second-order upwind scheme approximates the variation to be linear and uses extrapolation for finding the values.
\
\).
5. Consider the stencil.
computational-fluid-dynamics-questions-answers-second-order-upwind-scheme-q5
Assume a uniform grid. What is φ e according to the second-order upwind scheme?
.
a) \
\
\
\(\phi_e=\frac{3}{2}\phi_P+\frac{1}{2}\phi_W\)
Answer: c
Explanation: In general, from the second-order upwind scheme,
\
\)
For a uniform grid,
\(\frac{
}{
}=\frac{1}{2}\)
Therefore,
\(\phi_e=\phi_P+\frac{\phi_P-\phi_W}{2}=\phi_e=\frac{3}{2} \phi_P-\frac{1}{2}\phi_W\).
6. Consider the stencil.
computational-fluid-dynamics-questions-answers-second-order-upwind-scheme-q5
Assume a uniform grid. What is \
.
a) \
max
+
max
\)
b) \
max
-
max
\)
c) \
max
+
max
\)
d) \
max
-
max
\)
Answer: b
Explanation: According to the second-order upwind scheme,
\Missing \end{cases} & \dot{m_w}>0 \
& \dot{m_w}<0
\end{cases}\) Therefore,
\
max
-
max
\).
7. What is the first term in the truncation error of the second-order upwind scheme?
.
a) \
^2 \phi_P”’\)
b) \
\phi_P”’\)
c) \
^2 \phi_P”\)
d) \
\phi_P”\)
Answer: a
Explanation: The truncation error can be obtained by using the exact solution of the gradients. Since the scheme is second-order accurate, the first term of the error should have 2 in it. Associated with this we have, ΦP”’. Therefore, the term is \
\phi_P”’\).
8. Which statement is correct?
a) The second-order upwind scheme is never stable
b) The second-order upwind scheme is always stable
c) The second-order upwind scheme is conditionally stable
d) The second-order upwind scheme is always unstable
Answer: b
Explanation: The numerical stability of a scheme can be analysed by using the rate of change of influx. If the derivative of the influx with respect to the flow variable is negative, the scheme is stable. For the second-order upwind scheme, this is always negative.
9. Find the normalized functional relationship between φ f and φ C for a uniform grid while using the second-order upwind scheme?
a) \
\
\
\(\tilde{\phi_f}=-\frac{3}{2}\tilde{\phi_C}\)
Answer: c
Explanation: The relationship between φ f and φ C in the second-order upwind scheme is
\(\phi_f=\frac{3}{2}\phi_C-\frac{1}{2}\phi_U\)
After normalizing, φ f , φ C and φ U becomes \(\tilde{\phi_f}, \tilde{\phi_C}\) and 0 respectively. Therefore,
\(\tilde{\phi_f}=\frac{3}{2}\tilde{\phi_C}\).
10. The flux limiter Ψ of the second-order upwind scheme is __________
a) r 2
b) \
2r
d) r
Answer: d
Explanation: To find the flux limiter,
\
\)
For the second order upwind scheme,
\
=\frac{3}{2}\phi_C-\phi_C-\frac{1}{2}\phi_U\)
\
=\frac{1}{2}
\)
\=\frac{
}{
}=r\).
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Convection-Diffusion Problems – QUICK Scheme”.
1. What does QUICK stand for?
a) Quadratic Upstream Interpolation for Convective Kinetics
b) Quadratic Upstream Interval for Convective Kinetics
c) Quadratic Upwind Interval for Convective Kinetics
d) Quadratic Upwind Interpolation for Convective Kinetics
Answer: a
Explanation: QUICK is a higher-order differencing scheme introduced by Brian P. Leonard in his paper in the year 1979. It is the abbreviation of Quadratic Upstream Interpolation for Convective Kinetics.
2. Which is correct about the QUICK scheme?
a) A two-point upwind biased interpolation
b) A three-point upwind biased interpolation
c) A three-point downwind biased interpolation
d) A two-point downwind biased interpolation
Answer: b
Explanation: QUICK scheme uses a three-point upstream-weighted quadratic interpolation to approximate the cell face values. It uses two immediate neighbours of the face and an extra upstream node .
3. According to the QUICK scheme, the flow variable is given by ____
.
a) \
+\frac{
}{
}
\)
b) \
+\frac{
}{
}
\)
c) \
+\frac{
}{
}
\)
d) \
+\frac{
}{
}
\)
Answer: d
Explanation: The scheme should reduce to
\
+\frac{
}{
}
\) .
4. Consider the stencil.
computational-fluid-dynamics-questions-answers-quick-scheme-q4
Assume a uniform grid. What is φ e according to the QUICK scheme?
a) \
\
\
\(\phi_e=\frac{\phi_P-\phi_E}{2}-\frac{\phi_E-2\phi_P+\phi_W}{8}\)
Answer: c
Explanation: According to the QUICK scheme,
\
+ \frac{
}{
}
\)
For a uniform grid,
x e -x W =3(x e -x P ); x E -x W = 4(x e -x P ); x E -x P =2(x e -x P );
x E -x E = -(x e -x P );x P -x W =2(x e -x P );
Applying all these,
\(\phi_e=\frac{\phi_P+\phi_E}{2}-\frac{\phi_E-2\phi_P+\phi_W}{8}\)
5. What is the order of accuracy of the QUICK scheme?
a) second-order
b) first-order
c) fourth-order
d) third-order
Answer: a
Explanation: As the QUICK scheme is based on a quadratic function, its accuracy in terms of Taylor Series truncation error is third-order. This has a higher order of accuracy than the upwind and second-order upwind schemes.
6. How many terms does the discretized form of source-free 1-D convection problem modelled using the QUICK scheme has?
a) 3
b) 5
c) 2
d) 4
Answer: b
Explanation: The discretized form of a source-free 1-D convection problem modelled using the QUICK scheme involves the far upstream and the far downstream nodes too. Therefore, it contains extra terms than the upwind and the second-order upwind schemes. The stencil is
computational-fluid-dynamics-questions-answers-quick-scheme-q4
The discretized equation is
a P Φ P +a E Φ E +a W Φ W +a EE Φ EE +a WW Φ WW =0
It contains 5 terms.
7. What is the first term of the truncation error of the QUICK scheme?
a) \
^2 \phi_C”’\)
b) \
^3 \phi_C”’\)
c) \
^3 \phi_C^{iv}\)
d) \
^2 \phi_C^{iv}\)
Answer: c
Explanation: The order of accuracy is 3. Therefore, 3 should be there in the first term of the truncation error. The truncation error is obtained using the Taylor series. Therefore, this 3 comes along with Φ C iv .
8. Which of these is correct about the QUICK scheme?
a) Stable and bounded
b) Stable and unbounded
c) Unstable and bounded
d) Unstable and unbounded
Answer: d
Explanation: The QUICK scheme is not bounded. It involves undershoots and overshoots. The main coefficients are not guaranteed to be positive. The coefficients a EE and a WW are negative. Therefore, the solution is not stable.
9. Consider the stencil.
computational-fluid-dynamics-questions-answers-quick-scheme-q4
Assume a uniform grid. Using the QUICK scheme, what is the convective flux at the western face \
\
×max
-
×max
\)
b) \
×max
-
×max
\)
c) \
×max
-
×max
\)
d) \
×max
-
×max
\)
Answer: a
Explanation: Using QUICK scheme, for a flow in the positive x-direction,
\
×max
-\)
\
× max
\)
10. Which of these is correct for a QUICK scheme?
a) False diffusion is zero
b) False diffusion is small
c) False diffusion is big
d) False diffusion is infinity
Answer: b
Explanation: The QUICK scheme involves one downwind node also. So, there will be a false-diffusion in this method. But this false diffusion value is not big as it is an upwind biased scheme .
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Convection-Diffusion Problems – FROMM Scheme”.
1. Which of these profiles is used by the FROMM scheme?
a) Φ=k 0 +k 1 (x-x c )+k 2 (x-x c ) 2
b) Φ=k 1 (x-x c )+k 2 (x-x c ) 2
c) Φ=k 0 +k 1 (x-x c )
d) Φ=k 1 (x-x c )
Answer: c
Explanation: The FROMM scheme uses a linear interpolation method to approximate the cell face values. So, Φ=k 0 +k 1 (x-x c ) is the profile used by the FROMM scheme. But, the approach is different from the other profiles using a linear profile.
2. What is the order of accuracy of the FROMM scheme?
a) First-order
b) Second-order
c) Third-order
d) Fourth-order
Answer: b
Explanation: The first term of the truncation error while implementing the Taylor series in the FROMM scheme is of order two. Therefore, the FROMM scheme is second-order accurate using a linear profile.
3. FROMM scheme ____________
a) gives weighted importance to the upwind and downwind schemes
b) gives equal importance to upwind and downwind scheme
c) is downwind biased
d) is upwind biased
Answer: d
Explanation: The FROMM scheme is upwind biased. It gives more importance to the upwind nodes than the downwind nodes. IT uses two upwind nodes and one downwind node .
4. Which of these is correct about the FROMM scheme?
a) A linear profile is obtained between the immediate upwind and the far downwind nodes
b) A linear profile is obtained between the far upwind and the immediate downwind nodes
c) A linear profile is obtained between the far upwind and the immediate upwind nodes
d) A linear profile is obtained between the far upwind and the far downwind nodes
Answer: b
Explanation: A linear profile is obtained by connecting the values of the far upwind node and the immediate downwind node. A profile with the same slope obtained here is created between the immediate upwind and the current node to get the required value.
5. Consider the following stencil.
computational-fluid-dynamics-questions-answers-fromm-scheme-q5
What is Φ e according to the QUICK scheme?
a) Φ e =\(\phi_P+\frac{x_e-x_P}{x_E-x_W}\)(Φ E -Φ W )
b) Φ e =\(\phi_P+\frac{x_e-x_P}{x_E-x_W}\)(Φ E +Φ W )
c) Φ e =\(\phi_P-\frac{x_e-x_P}{x_E-x_W}\)(Φ E -Φ W )
d) Φ e =\(\phi_P-\frac{x_e-x_P}{x_E-x_W}\)(Φ E +Φ W )
Answer: a
Explanation: To find Φ e , the FROMM scheme first finds Φ P using the profile between Φ W and Φ E given by
Φ P =Φ W +\(\frac{x_P-x_W}{x_E-x_W}\)(Φ E -Φ W )
Φ W =Φ P –\(\frac{x_P-x_W}{x_E-x_W}\) (Φ E -Φ W )
Now, Φ e is given by,
Φ e =Φ W +\(\frac{x_e-x_W}{x_E-x_W}\)(Φ E -Φ W )
Which becomes
Φ e =Φ P –\(\frac{x_P-x_W}{x_E-x_W}\)(Φ E -Φ W )+\(\frac{x_e-x_W}{x_E-x_W}\)(Φ E -Φ W )
Therefore,
Φ e =Φ P +\(\frac{x_e-x_P}{x_E-x_W}\)(Φ E -Φ W ).
6. Consider the following stencil.
computational-fluid-dynamics-questions-answers-fromm-scheme-q5
Assume a uniform grid. What is Φ e according to the QUICK scheme?
a) Φ P +\(\frac{2}{3}\)(Φ E -Φ W )
b) Φ P +\(\frac{1}{2}\)(Φ E -Φ W )
c) Φ P +\(\frac{1}{4}\)(Φ E -Φ W )
d) Φ P +\(\frac{3}{4}\)(Φ E -Φ W )
Answer: c
Explanation: In general,
Φ e =Φ P +\(\frac{x_e-x_P}{x_E-x_W}\) (Φ E -Φ W )
For a uniform grid,
\(\frac{x_e-x_P}{x_E-x_W}\)=1/4
So,
Φ e =Φ P +\(\frac{1}{4}\)(Φ E -Φ W ).
7. Which of these is correct about the FROMM scheme?
a) Stable and bounded for a variable velocity system
b) Stable but not bounded for a variable velocity system
c) Stable but not bounded for a constant velocity system
d) Stable and bounded for a constant velocity system
Answer: c
Explanation: The FROMM scheme is numerically stable when the velocity field is constant. But, when the velocity is varying, the scheme is unstable. It includes undershoots and overshoots and hence not bounded.
8. What is the normalized relationship between Φ f and Φ c for the FROMM scheme?
a) \
\
\
\(\tilde{\phi_f}=\frac{1}{4} \tilde{\phi_c}\)
Answer: a
Explanation: The relationship between Φ f and Φ c is
\
\)
The normalized forms of Φ f , Φ c , Φ D and Φ U are \(\tilde{\phi_f}, \tilde{\phi_c},\) 1 and 0 respectively. Therefore,
\(\tilde{\phi_f}=\tilde{\phi_c}+\frac{1}{4}.\)
9. For the FROMM scheme, what is the flux limiter ψ equal to?
a) 1-\
1+\
\
\(\frac{1+r}{2}\)
Answer: d
Explanation: To find the flux limiter,
Φ f =Φ c +\
(Φ D -Φ c )
For the FROMM scheme,
Φ f =Φ c +\(\frac{1}{4}\)(Φ D -Φ U )
Comparing both,
Ψ(Φ D -Φ c )=\(\frac{1}{2}\)(Φ D -Φ U )
Ψ=\
=\
=\frac{1}{2}\).
10. Consider the following stencil.
computational-fluid-dynamics-questions-answers-fromm-scheme-q5
Assume a uniform grid. What is the convective flux at the western face
using the FROMM scheme?
a) \
max
-
max
\)
b) \
max
-
max
\)
c) \
max
-
max
\)
d) \
max
-
max
\)
Answer: a
Explanation: When the flow direction is positive,
\
max
-
max
\).
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Convection-Diffusion Problems – Error Sources”.
1. Numerical diffusion causes __________
a) smearing of sharp gradients
b) oscillations
c) undershoots and overshoots
d) inaccuracy
Answer: a
Explanation: The sources of numerical errors caused by the convection flux can be divided into two – numerical diffusion and numerical dispersion. The numerical diffusion leads to smearing of sharp gradients.
2. Which of these ways can be used to overcome stream-wise numerical diffusion?
a) Decreasing the order of interpolation
b) Increasing the order of interpolation
c) Increasing the number of neighbours considered
d) Decreasing the number of neighbours considered
Answer: b
Explanation: The numerical diffusion error can be further divided into two – stream-wise and cross-stream numerical diffusions. The stream-wise numerical diffusion can be reduced by using a higher-order interpolation profile.
3. What is the cause of cross-stream numerical diffusion?
a) One-dimensional nature of the assumed profiles
b) Multi-dimensional nature of the assumed profiles
c) Higher order of accuracy
d) Higher order of interpolation
Answer: a
Explanation: One of the types of numerical diffusion errors is the cross-stream numerical diffusion. Cross-stream numerical diffusion is caused by the one-dimensional nature of interpolation when the grid is multi-dimensional. This is caused by cross-flow diffusion or false diffusion.
4. Which of these methods cannot be used to reduce the errors due to cross-stream numerical diffusion?
a) Interpolation in the direction of flow
b) Higher order interpolation profile
c) Multi-dimensional interpolation profiles
d) Changing the direction of interpolation
Answer: d
Explanation: The cross-stream numerical diffusion can be decreased by either interpolating in the direction of the flow which means multi-dimensional interpolation profiles or using a one-dimensional higher-order interpolation profile.
5. Numerical dispersion error causes __________
a) convergence problems
b) accuracy problems
c) boundedness problems
d) stability problems
Answer: c
Explanation: One of the types of sources of the errors is numerical dispersion. This is shown out through oscillations in the resulting profiles in the presence of large gradients in the profile resulting in an unbounded solution.
6. Which of these methods can be used to evaluate the errors in convection-diffusion schemes?
a) Using the exact solution of the source-free problem
b) Using the exact solution of the source and diffusion-free problem
c) Using the exact solution of the diffusion-free problem
d) Using the first-order schemes
Answer: b
Explanation: To evaluate the errors, a simplified version of the problem where there is no source and diffusion is taken and further velocity and density fields are assumed to be constants. The exact solution of this problem is used for the analysis.
7. Which is correct regarding the upwind scheme?
a) Neither numerical dispersion nor numerical diffusion error arises
b) Only numerical diffusion error arises
c) Both numerical dispersion and numerical diffusion errors arise
d) Only numerical dispersion error arises
Answer: c
Explanation: In the analysis for errors, while comparing the exact and numerical solutions, the first-order upwind scheme gives a complex k-value. Therefore, it will have both numerical dispersion and numerical diffusion problems.
8. Which of these is correct for the central difference scheme?
a) Neither numerical dispersion nor numerical diffusion error arises
b) Only numerical diffusion error arises
c) Both numerical dispersion and numerical diffusion errors arise
d) Only numerical dispersion error arises
Answer: d
Explanation: For the central differencing scheme, there is no problem of numerical diffusion. Only numerical dispersion error arises here. This is because, the k-value is completely imaginary and free from a real part.
9. What is the problem of numerical diffusivity?
a) The simulated model has a higher diffusivity than the actual flow
b) The simulated model has a lower diffusivity than the actual flow
c) The simulated model has a different diffusivity than the actual flow
d) The simulated model has a zero diffusivity
Answer: a
Explanation: Numerical diffusivity is a problem which results in a simulated solution with a higher diffusivity than the actual flow problem. This becomes an important criteria to be considered when the flow is free from diffusion.
10. Numerical dispersion is a result of __________
a) higher-order interpolation profile
b) unphysical behaviour of assumed interpolation profile
c) first-order interpolation profile
d) quadratic interpolation profile
Answer: b
Explanation: Numerical dispersion error arises with all interpolation profiles except the upwind scheme. It is the result of the unphysical behaviour of the assumed interpolation profile. It leads to unphysical dispersion.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Incompressible Flows – Staggered Grid”.
1. The staggered grid can be used to overcome __________
a) decoupling of pressure and velocities
b) coupling of pressure and velocities
c) interpolation problems
d) boundedness problems
Answer: a
Explanation: The uncoupling between pressure and velocities cause a number of problems. Coupling can be enforced if the different flow variables are stored in the staggered grid form. Here, pressure and velocity are stored in different places.
2. Which of these statements is correct when using the staggered grids?
a) Interpolation is enough to find the pressure gradient in the continuity equation
b) No interpolation is needed to find the pressure gradient in the continuity equation
c) No interpolation is needed to find the pressure gradient in the momentum equation
d) Interpolation is enough to find the pressure gradient in the momentum equation
Answer: c
Explanation: The momentum equation has this pressure gradient term given by \(\frac{\partial p}{\partial x}, \frac{\partial p}{\partial y}\, and \frac{\partial p}{\partial z}\). To find this pressure gradient, there is no need for any interpolation. The pressure gradients can be easily found out.
3. The staggered grid needs no interpolation to get the __________
a) velocity field in the momentum equation
b) velocity field in the continuity equation
c) kinetic energy field in the momentum equation
d) kinetic energy in the continuity equation
Answer: b
Explanation: In the staggered grid formation, as the pressure is already stored in a different place from the pressure, there is no need of any interpolation to find the velocity field required for the continuity equation.
4. Which of these is not stored at the cell centres in the staggered grids?
a) Density
b) Pressure
c) Temperature
d) Velocity
Answer: d
Explanation: In the staggered grid, all the scalar flow quantities are stored in the cell centres. Therefore, temperature, pressure and density are all stored in the cell centres. As velocity is a vector quantity, it is not stored here.
5. While using a staggered grid, the velocities are stored in the ___________
a) face centres
b) cell centres
c) edge centres
d) vertices
Answer: a
Explanation: In the staggered grid arrangement, the velocities are stores in the face centres. This is why there will not be any need for interpolation to get the velocity field for the continuity equation.
6. What is the advantage of the staggered grid arrangement?
a) Helps to avoid stability problems
b) Helps to avoid convergence and consistency problems
c) Helps to avoid convergence problems and oscillations
d) Helps to avoid boundedness problems and oscillations
Answer: c
Explanation: The biggest advantage of the staggered grid arrangement is the coupling of pressure and velocities. It also helps to avoid convergence problems and oscillations in pressure and velocity fields.
7. Arbitrary Lagrangian-Eulerian method comes under __________
a) Staggered grid arrangements
b) Partially staggered grid arrangements
c) Collocated arrangements
d) Orthogonal arrangement
Answer: b
Explanation: The Arbitrary Lagrangian-Eulerian method is a partially staggered grid arrangement. It uses different locations to store the velocity and pressure fields but the arrangement varies from the completely staggered arrangement.
8. Which of these statements is correct about the Arbitrary Lagrangian-Eulerian method?
a) It produces non-converging pressure or velocity fields
b) It produces inconsistent pressure or velocity fields
c) It produces erroneous pressure or velocity fields
d) It produces oscillatory pressure or velocity fields
Answer: d
Explanation: One of the main disadvantages of the Arbitrary Lagrangian-Eulerian method is the possibility of producing oscillatory pressure or velocity fields. It is advantageous only in some special cases.
9. Which of these statements is incorrect about the Arbitrary Lagrangian-Eulerian method?
a) Velocities are stored in the face centres
b) All the components of velocities are stored at the same point
c) Velocities are stored in the vertices
d) Pressures are stored in the cell centres
Answer: a
Explanation: In the Arbitrary Lagrangian-Eulerian method, the scalar quantities are all stored at the cell centres like the completely staggered arrangement. All the velocity components are stored at the same point which are the vertices.
10. The Arbitrary Lagrangian-Eulerian method is advantageous for ___________
a) Curvilinear grids
b) Non-orthogonal grids
c) Cartesian grids
d) Orthogonal grids
Answer: b
Explanation: The Arbitrary Lagrangian-Eulerian method has some advantages for the non-orthogonal grids. One of these advantages is that the pressure at the boundaries need not be specified here.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Incompressible Flows – Special Features of Navier Stokes Equation”.
1. Which of these statements is correct?
a) Body force term in the momentum equation is non-linear
b) Rate of change term in the momentum equation is non-linear
c) Convective term in the momentum equation is linear
d) Convective term in the momentum equation is non-linear
Answer: d
Explanation: The convective term in the momentum equation is given by \(\frac{\partial
}{\partial x_j}\). Where, ρ and u are density and velocity respectively. This term is a non-linear term. The rate of change and the body force terms are linear.
2. For the incompressible flows, which of these terms will be zero?
a) Pressure force
b) Body force
c) Bulk viscosity
d) Shear force
Answer: c
Explanation: A part of the viscous terms in the momentum equation is present in the diffusive term. Among these comes the bulk viscosity terms. These terms will be non-zero only for the compressible flows.
3. In which of these approaches is the pressure force treated as a body force?
a) Finite volume method – non-conservative approach
b) Finite volume method – conservative approach
c) Finite difference method – conservative approach
d) Finite difference method – non-conservative approach
Answer: b
Explanation: In the general integral form of the conservation equations, the pressure force is taken as a volume integral. While using the finite volume method, this is changed into a surface force by using the Gauss-Divergence theorem.
4. If the pressure force is not treated as a surface force in the finite volume method, what will happen?
a) Non-conservative error
b) Stability issues
c) Boundedness problems
d) Convergence issues
Answer: a
Explanation: Without using the Gauss-Divergence theorem, the pressure force can be treated as a volume integral itself. But, this will lead to a non-conservative form of equations which, in turn, will result in non-conservative errors.
5. The difference between the conservative and the non-conservative approaches occurs in the ___________
a) finite difference method
b) finite volume method
c) finite element method
d) spectral element method
Answer: b
Explanation: The conservative and non-conservative approaches gives rise to a considerable difference only in the finite volume method. In the finite difference approach, they do not result in any variation.
6. Which of these is correct for extra viscous terms in cylindrical coordinates?
a) The implicit and explicit treatments do not depend on the sign of the coefficients
b) It is treated implicitly when its contribution to the coefficient of the central node is negative
c) It is treated implicitly when its contribution to the coefficient of the central node is positive
d) It is treated explicitly when its contribution to the coefficient of the central node is positive
Answer: c
Explanation: While using the non-Cartesian coordinate systems, extra terms occur such as that of the extra viscous term in the momentum equation. This term is treated implicitly when it leads to a positive coefficient for the central node. Otherwise, it is treated explicitly.
7. Which of these changes do not contribute to a change in momentum in the momentum equation?
a) Surface fluxes
b) Surface forces
c) Body forces
d) Rate of change term
Answer: d
Explanation: The rate of change term represents the change in momentum in the momentum equations. This change is affected by any changes in the surface fluxes, surface forces and body forces acting on the element.
8. In which of these flows is the kinetic energy important?
a) Compressible flows
b) Compressible isothermal flows
c) Incompressible isothermal flows
d) Incompressible flows
Answer: c
Explanation: When the flow is not isothermal, thermal energies play an important role in energy conservation. When the flow is compressible, internal energies play an important role. Only when the flow is incompressible and isothermal, kinetic energies become significant.
9. An equation for the conservation of kinetic energy can be obtained by ___________
a) the product of momentum equations and mass
b) the product of momentum equations and velocities
c) the product of continuity equations and velocities
d) the product of continuity equations and mass
Answer: b
Explanation: Momentum is the product of mass and velocity. Kinetic energy is half of the product of mass and the square of velocities. Therefore, by multiplying the momentum equation with the velocity and further simplifying the resultant equations, we can get the conservation equations of kinetic energy.
10. For incompressible flows with no body forces, the volume integral term is ____________
a) viscous terms
b) pressure forces
c) body forces
d) flux terms
Answer: a
Explanation: Pressure forces and body forces are not present as the flow is incompressible and has no body forces. The flux terms are usually treated with the Gauss-Divergence theorem and converted into surface integrals. Therefore, the viscous terms remain as volume integrals here.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Incompressible Flows – Pressure Correction Equation”.
1. The pressure correction equation is used to ensure _________
a) energy conservation
b) velocity conservation
c) momentum conservation
d) mass conservation
Answer: d
Explanation: The pressure correction uses a pressure equation to make sure that mass is conserved at each time step. This is to get the correct solution from the assumed initial values at that time step.
2. Consider a one-dimensional flow with two bounding faces in the eastern and the western sides . Applying pressure correction to the mass conservation equation, which of these equations will be obtained?
.
a) \
\
\
\(\dot{m}_{e}^{‘}=-\dot{m}_{e}^{*}\)
Answer: c
Explanation: The correction in any flow variable is given by
Φ=Φ*+Φ’
By the continuity equation,
\(\dot{m_e}+\dot{m_w}=0\)
\(\dot{m}_{e}^{‘}+\dot{m}_{e}^{*}+\dot{m}_{w}^{‘}+\dot{m}_{w}^{*}=0\)
Therefore,
\(\dot{m}_{e}^{‘}+\dot{m}_{w}^{‘}=-\dot{m_e}*-\dot{m}_{w}^{*}\).
3. In the incompressible flows, the correction implies a correction in _________
a) momentum
b) velocity
c) mass
d) density
Answer: b
Explanation: In general,
\(\dot{m}=\rho uA\)
A correction in mass flow rate cannot be the correction in density as density is constant in the incompressible flows. It cannot be a correction in area also as it is geometrically defined by the grids. So, it must be a correction in velocity.
4. State the condition obtained by applying the correction to the continuity equation.
a) When the mass flow rate reaches an exact solution, the correction field becomes zero
b) When the velocity reaches an exact solution, the correction field becomes zero
c) When the mass flow rate reaches an exact solution, the correction field becomes infinity
d) When the velocity reaches an exact solution, the correction field becomes infinity
Answer: a
Explanation: The correction equation obtained for the mass flow rate is
\(\dot{m}_{e}^{‘}+\dot{m}_{w}^{‘}=-\dot{m_e}*-\dot{m}_{w}^{*}\)
When the solution reaches the exact answers,
\(-\dot{m}_{e}^{*}-\dot{m}_{w}^{*}=\dot{m_e}*+\dot{m}_{w}^{*}=0\)
Therefore, the correction field
\(\dot{m}_{e}^{‘}+\dot{m}_{w}^{‘}=0\).
5. The continuity equation drives the correction field of __________
a) density
b) velocity
c) pressure
d) energy
Answer: b
Explanation: The correction field becomes zero when the mass flow rate values satisfy the continuity equation. This is useful only for the velocity correction as the mass flow rate correction and velocity correction means the same.
6. The momentum equation drives the correction field of __________
a) density
b) temperature
c) pressure
d) energy
Answer: c
Explanation: To establish the correction field for pressure, the mass flow rate cannot be used. So, the pressure correction field is defined by the momentum equation. But, the momentum equation is used to get the initial guess of velocities.
7. In which of these terms of the momentum equation will the correction have no impact?
a) Diffusion terms
b) Source terms
c) Velocity terms
d) Surface flux terms
Answer: d
Explanation: Corrections based on the surface fluxes become zero when the solution converges. The other terms of the momentum equation do not become zero here. Therefore, the surface fluxes have no impact on the final solution.
8. The pressure used to find the velocities from the momentum equations is of __________
a) the previous time step
b) the oldest value
c) the latest value
d) the current time step
Answer: a
Explanation: The momentum equations are used to get the initial guesses of the velocities. To get these velocities, other terms in the equation should be known. As the pressure at the current time step is unknown, the pressure values of the previous time step are used.
9. The correction in the velocity field is used to _____________
a) to find the pressure field of the next time step
b) correct the pressure field
c) to get the velocity field of the next time step
d) to correct the velocity field in the previous iteration
Answer: b
Explanation: Pressure correction equation resembles the predictor-corrector method. The velocities are corrected until they satisfy the continuity equation. Then the corrected velocity is used to correct the pressure field of the previous time step.
10. The pressure correction is an __________
a) explicit time-independent method
b) implicit time-independent method
c) implicit time-dependent method
d) explicit time-dependent method
Answer: c
Explanation: The pressure correction method is based on time. It uses time steps to move towards a steady solution for steady problems. If the problem is transient it uses time steps to move to the desired interval of time. It is an implicit method as it solves all the equations simultaneously.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Incompressible Flows – Pressure Calculation”.
1. Which of these is a disadvantage of the Navier-Stokes equations?
a) No independent-equation for pressure
b) No independent-equation for temperature
c) No equation to find the density
d) No equation to find the velocity
Answer: a
Explanation: The Navier-stokes equations do not have a separate equation to find the pressure values at different points. Pressure gradients are present in different momentum equations making it a dependent form.
2. Which of these statements is true for an incompressible flow?
a) Absolute mass flux is not important
b) Absolute density is not important
c) Absolute temperature is not important
d) Absolute pressure is not important
Answer: d
Explanation: There is no need for the absolute value of pressure in an incompressible flow. This is because the flow is not affected by this value. But, their gradients in all the directions are important as it will affect the flow.
3. How is pressure calculated in a compressible flow?
a) Pressure correction equation
b) Equation of state
c) Momentum equation
d) Energy equation
Answer: b
Explanation: In the compressible flows, the density terms of the continuity equation do not cancel out. So, density can be determined using the continuity equation. Using this density value and the equation of state, we can get the pressure values.
4. Which of these equations are used to get the pressure values in the incompressible flows?
a) Continuity and momentum equations
b) Momentum and energy equations
c) Energy equation and equation of state
d) Equation of state and continuity equations
Answer: a
Explanation: The divergence of the momentum equation is taken. The equation obtained from this is then simplified using the continuity equation to get the pressure values. So, a combination of continuity and momentum equations is needed here.
5. The pressure equation for the incompressible equation is _________
a) Eulerian equation
b) Divergence equation
c) Lagrangian equation
d) Poisson equation
Answer: d
Explanation: The resultant equation obtained by modifying the existing governing equations to get the pressure values is a Poisson equation for pressure. A Poisson equation is a partial differential equation of elliptic type.
6. Which of these terms in the pressure equation become zero?
a) Viscous and pressure terms
b) Viscous and transient terms
c) Pressure and source terms
d) Source and transient terms
Answer: b
Explanation: The viscous terms in the pressure equation are cancelled because of constant viscosity. The transient terms are cancelled out as the density is constant. These cancellations are made by implementing the continuity equation.
7. The pressure equation in the incompressible flows contain _________
a) Taylor series terms
b) Hermitian operator
c) Laplacian operator
d) Fourier series terms
Answer: c
Explanation: The Poisson’s equation which is formulated for finding the pressure values has a Laplace operator. This is the product of the divergence operator in the continuity equation and the gradient operator in the momentum equation.
8. According to the explicit time-advanced method for getting pressure, which of these is correct?
a) Continuity is enforced at the previous step after starting to solve the current step
b) Momentum conservation is enforced at the previous step after starting to solve the current step
c) Momentum equation is enforced at each step before moving to the next step
d) Continuity is enforced before at each step moving to the next step
Answer: d
Explanation: The divergence of the velocity field at the current step should be made zero. For this, the divergence of the velocity field at the previous step must be zero. So, continuity is zero at each step before moving to the next step.
9. The explicit method is preferred when __________
a) an accurate velocity field is needed
b) the pressure value is needed
c) an accurate time history of the flow is needed
d) only less memory is available
Answer: c
Explanation: Explicit methods are used to solve the Navier-Stokes equations when accurate time history of the flow variables is needed. The explicit time advancement method is more accurate than the first-order Euler method.
10. Which of these is correct for the implicit method of solving pressure in the compressible flows?
a) The Poisson equation alone is solved
b) The Poisson equation and the momentum equation are solved simultaneously
c) The momentum equation is solved first
d) The Poisson equation is solved first
Answer: b
Explanation: In the implicit method, the values in the previous time steps are altered after moving to the next step. The Poisson equation and the momentum equation are solved simultaneously here. No time-advancement is done here.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Incompressible Flows – Rhie-Chow Interpolation”.
1. The checkerboard problem arises due to __________
a) linear interpolation in a staggered grid
b) quadratic interpolation in a staggered grid
c) linear interpolation in a collocated grid
d) quadratic interpolation in a collocated grid
Answer: c
Explanation: The decoupling between the pressure and velocity values at the cell level happens due to linear interpolation used in the collocated grid. This gives rise to the checkerboard problem in the collocated grids.
2. Which of these methods use Rhie-Chow interpolation?
a) QUICK scheme
b) NVF method
c) TVD method
d) SIMPLE algorithm
Answer: d
Explanation: Rhie-Chow interpolation is used in the collocated grid without leading to the uncoupled pressure and velocity fields. Rhie-Chow interpolation leads to the formulation of the SIMPLE algorithm.
3. Which of these terms is extra while using the Rhie-Chow interpolation?
a) Diffusion term
b) Dissipative term
c) Advection term
d) Source term
Answer: b
Explanation: Two different grid stencils are used in the Rhie-Chow interpolation method. The difference between the values obtained from these two grid stencils is added as a dissipative term in the result.
4. The Rhie-Chow interpolation is equivalent to __________
a) pseudo-momentum equation with the coefficients linearly interpolated
b) pseudo-momentum equation with the coefficients quadratically interpolated
c) pseudo-continuity equation with the coefficients linearly interpolated
d) pseudo-continuity equation with the coefficients quadratically interpolated
Answer: a
Explanation: The Rhie-Chow interpolation is used on collocated grids. It is equivalent to constructing a pseudo-momentum equation at the face. It has coefficients which are obtained using linear interpolation of the momentum equations at the centroids.
5. The Rhie-Chow interpolation mimics __________
a) pressure-velocity coupling of the collocated grid arrangement
b) pressure-velocity coupling of the staggered grid arrangement
c) pressure-velocity decoupling of the staggered grid arrangement
d) pressure-velocity decoupling of the collocated grid arrangement
Answer: b
Explanation: The pressure gradients in the Rhie-Chow interpolation are calculated using small grid stencil with two different stencils. This resembles the pressure-velocity coupling of the staggered grid arrangement.
6. The coefficients of the equation formed in the Rhie-Chow interpolation is obtained using ___________
a) Weighted average
b) Quadratic interpolation
c) Rhie-Chow interpolation
d) Linear interpolation
Answer: d
Explanation: The Rhie-Chow interpolation forms momentum equations. The coefficients of this new momentum equation cannot be directly computed. They are assumed using interpolation of the values in the neighbouring nodes.
7. The coefficient approximation in the Rhie-Chow interpolation is ___________
a) fourth-order accurate
b) third-order accurate
c) second-order accurate
d) first-order accurate
Answer: c
Explanation: The coefficient approximation in the Rhie-Chow interpolation results in an equation which is the same as the one obtained using the staggered grids. This coefficient approximation is of second-order accuracy.
8. The collocated arrangement with Rhie-Chow interpolation will result in an equation which is dependent on the ___________
a) staggered grid
b) linear interpolation
c) average
d) under-relaxation factor
Answer: d
Explanation: When a collocated grid arrangement is used with the Rhie-Chow interpolation to get the momentum equation at the faces, the resulting solution becomes dependent on the under-relaxation factor present in the momentum equation.
9. The Rhie-Chow interpolation needs a redistribution of the _________
a) source term
b) pressure gradient term
c) viscous force term
d) body force term
Answer: b
Explanation: The body force in the staggered grid term has the exact stencil of the pressure gradients. Therefore, a redistribution of the body force term in the Rhie-Chow interpolation is needed to match the staggered grids.
10. When the transient term in the Rhie-Chow interpolation is not accounted for separately, the result will ___________
a) be oscillatory
b) be stable
c) have overshoots
d) have undershoots
Answer: a
Explanation: When the transient term in the Rhie-Chow interpolation is not properly accounted for the result will be time step-dependent and will lead to an oscillatory solution for small time-steps.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Incompressible Flows – SIMPLE Algorithm”.
1. What does SIMPLE stand for?
a) Semi-Implicit Method for Pressure-Linked Equations
b) Simple Implicit Method for Pressure-Linked Equations
c) Solution in Iterative Method for Pressure-Links Explicitly
d) Simple Iterative Method for Pressure-Links Explicitly
Answer: a
Explanation: The SIMPLE in SIMPLE algorithm expands as “Semi-Implicit Method for Pressure-Linked Equations”. This method was introduced by Suhas Patankar and Brian Spalding in the year 1972 at Imperial College.
2. The SIMPLE algorithm is a ____________
a) Weighted average method
b) Predictor-corrector method
c) Euler method
d) Heun’s method
Answer: b
Explanation: The algorithm is a guess and correct method used to calculate the pressure values in a staggered grid arrangement. It involves continuous iterations of predicting the values and correcting them.
3. In the momentum equation for the correction field, which of these terms are neglected?
a) Neighbouring correction terms
b) All the coefficients
c) The source terms
d) The velocity terms
Answer: a
Explanation: The discretized momentum equation is rewritten for the initial guesses. Now, the correction is introduced into this equation and the correction equations are formed. In these correction equations, the corrections for the neighbouring terms are omitted. This is applicable as it does not make any changes globally.
4. Which of these equations are used in the SIMPLE algorithm?
a) Momentum and energy equations
b) Energy equation and equation of state
c) Equation of state and continuity equation
d) Continuity and momentum equations
Answer: d
Explanation: The SIMPLE algorithm involves the calculations for pressure and velocity. It uses both the continuity and momentum equations. The momentum equation is used for the initial guesses. While applying the constraints, the continuity equation is also needed.
5. Which of these equations is/are obtained from the continuity equation?
a) Both the pressure and velocity-correction equations
b) Velocity-correction equation
c) Pressure-correction equation
d) Density-correction equation
Answer: c
Explanation: The equation for the pressure correction arises from the discretized continuity equation. It contains a source term which is the result of the continuity imbalance from the incorrect velocity field.
6. Which of these statements is correct about the pressure correction equation?
a) An over-relaxation factor is used to avoid divergence
b) An under-relaxation factor is used to avoid divergence
c) An under-relaxation factor is used to avoid undershoots
d) An over-relaxation factor is used to avoid undershoots
Answer: b
Explanation: The pressure correction equation may diverge if an under-relaxation factor is not used. So, an under-relaxation factor is used with the correction term which is added to the guessed term to get the correct value.
7. The range of the relaxation factor in the pressure correction equation is ____________
a) 0 ≤ factor < 1
b) 0 < factor < 1
c) 0 < factor ≤ 1
d) 0 ≤ factor ≤ 1
Answer: b
Explanation: An under-relaxation factor equal to 1 leads to no relaxation. The whole correction factor is taken into account. An under-relaxation factor equal to 0 leads to no correction. The guessed value is taken as the correct value. So, the value should lie in between these two.
8. Which of these statements is correct about a small relaxation factor?
a) Computation is unstable but convergence is fast
b) Computation is stable and convergence is fast
c) Computation is stable but convergence is slow
d) Computation is unstable and convergence is slow
Answer: c
Explanation: When the under-relaxation factor is small, the speed of computation is very slow. Because, to get the solutions converged, a large number of iterations are needed. But, this small under-relaxation factor will lead to a stable computation.
9. A large relaxation factor will lead to ____________
a) large and oscillatory iteration steps
b) large and stable iteration steps
c) small and oscillatory iteration steps
d) small and stable iteration steps
Answer: a
Explanation: The under-relaxation factor should be large enough to move the iteration steps forward at a fast rate. But, if it is too large, the iteration steps will be unstable and oscillatory or even divergent at times.
10. Which of these is correct about the SIMPLE algorithm?
a) It is a looped algorithm
b) It is iterative
c) It is sequential
d) It is simultaneous
Answer: d
Explanation: The SIMPLE algorithm is sequential. The final solution of each iteration satisfies both the continuity and the momentum equations. Only when these two are satisfied, the algorithm moves forward for the next iteration.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Incompressible Flows – SIMPLER Algorithm”.
1. What does the R in the SIMPLER algorithm stand for?
a) Rewritten
b) Revised
c) Republished
d) Refresh
Answer: b
Explanation: The SIMPLER algorithm stands for SIMPLE – Revised algorithm. This was introduced by Patankar who had worked for building the SIMPLE algorithm also. This was introduced in the year 1980.
2. What is the difference between the SIMPLE and the SIMPLER algorithms?
a) No velocity-correction equation
b) No relaxation factor
c) Pressure is directly calculated
d) No pressure-correction equation
Answer: c
Explanation: The SIMPLE algorithm uses a correction factor for correcting the initial guessed pressure. But, this SIMPLER algorithm does not involve any guess and correction of pressure. It has an equation which can be solved to directly get the pressure values.
3. What is the disadvantage of using the SIMPLE algorithm over the SIMPLER one?
a) The pressure field does not satisfy the momentum equation
b) The pressure field does not satisfy the continuity equation
c) The velocity field does not satisfy the continuity equation
d) The velocity field does not satisfy the momentum equation
Answer: d
Explanation: In the SIMPLE algorithm, the velocity field is initially guessed using the momentum equation. Then, corrections are made using the continuity equation. But, once corrected, the velocity field does not satisfy the momentum equations.
4. The equation for pressure of the SIMPLER algorithm is obtained using _________
a) continuity equation in the PDE form
b) discretized continuity equation
c) momentum equation in the PDE form
d) discretized momentum equation
Answer: b
Explanation: Obtaining the pressure values in the SIMPLER algorithm is simpler than in the SIMPLE algorithm. In the SIMPLER case, a discretized equation for pressure is obtained using the continuity equation. Solving this, pressure values can be obtained.
5. Which of these is correct about the SIMPLER algorithm?
a) Velocity is obtained without any correction
b) Velocity is obtained using the pressure values and the continuity equation
c) Prediction-correction is used for the velocity field
d) Velocity is obtained using the pressure values and the momentum equation
Answer: c
Explanation: Though the pressure field in the SIMPLER algorithm does not need any guess and correction, the velocity field still needs a prediction-correction method. It is found in a similar way as in the SIMPLE algorithm.
6. Which of these equations is used in the SIMPLER algorithm for updating the pressure values?
a) Pressure-correction equation
b) Discretized pressure equation
c) Momentum equation
d) Continuity equation
Answer: a
Explanation: In the SIMPLER algorithm, the velocity field should be corrected. For this correction and updating purpose, the pressure correction equation is used as in the SIMPLE algorithm. Therefore, the momentum equations are also satisfied.
7. The source term in the pressure equation of the SIMPLER algorithm is obtained using ___________
a) velocities in the previous iteration
b) pressure in the previous iteration
c) pseudo-neighbours
d) pseudo-velocities
Answer: d
Explanation: The discretized pressure equation of the SIMPLER algorithm is the same as the pressure correction equation obtained for the SIMPLE algorithm. The only difference is that the source term here is obtained from the pseudo velocities.
8. The pseudo-velocities of the SIMPLER algorithm are obtained from _________
a) continuity equation in the PDE form
b) discretized momentum equation
c) discretized pressure equation
d) discretized pressure-correction equation
Answer: b
Explanation: The discretized momentum equation is taken. From this equation, the pseudo-velocities of the SIMPLER algorithm are obtained. The momentum equation is then written in terms of these pseudo-velocities for further use.
9. Which of these terms is not included in the pseudo-velocities?
a) Pressures in the current and previous iterations
b) Momentum sources
c) Neighbouring velocities
d) Coefficients of the current velocities
Answer: a
Explanation: The pseudo-velocity is given as
\(\hat{u}_{i,j}=\frac{\sum a_{nb} u_{nb}+b_{i,j}}{a_{i,j}} \)
Where,
u nb → Neighbouring velocities.
b i,j → Source of momentum.
a i,j → Coefficient of the current velocity.
10. The velocity-corrections in the SIMPLER algorithm depends on ____________
a) pressure equation
b) pressure-correction equation
c) pressure corrections
d) pressure values
Answer: c
Explanation: There is pressure correction involved in the SIMPLER algorithm. But, this pressure correction is not used to correct the pressure values. Instead, it is used to correct the velocity values.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Incompressible Flows – SIMPLEC Algorithm”.
1. What does the ‘C’ in SIMPLEC algorithm?
a) Continuation
b) Converging
c) Corrected
d) Consistent
Answer: d
Explanation: SIMPLEC expands as SIMPLE-Consistent algorithm. It was built by Von Doormal and Raithby in the year 1984. It is advantageous over both the SIMPLE and SIMPLER algorithms.
2. Which of these equations of the SIMPLE algorithm is manipulated in the SIMPLEC algorithm?
a) Continuity equation
b) Momentum equation
c) Pressure equation
d) Pressure-correction equation
Answer: b
Explanation: The SIMPLEC algorithm uses the same procedure as the SIMPLE algorithm. The major difference is that the momentum equation which is used by the SIMPLE algorithm is changed in the SIMPLEC algorithm.
3. Which of these is affected by the change in the SIMPLEC algorithm?
a) Velocity correction
b) Pressure equation
c) Pressure-correction equation
d) Velocity-correction equation
Answer: d
Explanation: The momentum equation used by the SIMPLEC algorithm is modified. This modification directly affects the velocity-correction equation. This gives an advantage to the SIMPLEC algorithm.
4. The SIMPLEC algorithm _________ in the SIMPLE algorithm.
a) omits over-relaxed terms
b) adds terms
c) omits less significant terms
d) omits under-relaxed terms
Answer: c
Explanation: The SIMPLEC algorithm omits some less significant terms in the SIMPLE algorithm. So, the velocity correction equations of the SIMPLEC model in all the three directions has fewer terms than the SIMPLE algorithm.
5. The velocity correction in the SIMPLEC algorithm is __________
a) the weighted average of the neighbouring corrections
b) the weighted average of the current pressure values
c) the weighted average of the older pressure values
d) the weighted average of the older velocity values
Answer: a
Explanation: The SIMPLEC algorithm assumes that the velocity correction at the central node as the weighted average of the corrections at the neighbouring nodes. This changes the momentum and the velocity-correction equations.
6. Which of these is an advantage of the SIMPLEC algorithm over the SIMPLE algorithm?
a) No over-relaxation
b) No under-relaxation
c) No undershoots
d) No extra terms
Answer: b
Explanation: In the SIMPLE algorithm, the whole momentum equation is used. So, for the correction of pressure and velocity, an under-relaxation factor is used. But, as the SIMPLEC algorithm omits all the insignificant terms in the momentum equation, this under-relaxation becomes unnecessary.
7. Which of these performance-based advantages does the SIMPLEC algorithm have?
a) More stable
b) Faster convergence
c) Bounded solutions
d) Conservativeness
Answer: b
Explanation: As some of the terms in the momentum equation are omitted in the SIMPLEC algorithm, the velocity correction suits the momentum equations more. Therefore, the rate of convergence increases in this algorithm.
8. What is the advantage of the SIMPLEC algorithm over the SIMPLER algorithm?
a) No extra equations
b) No extra terms
c) No momentum equation
d) More stable
Answer: a
Explanation: The SIMPLER algorithm has an extra equation when compared to the SIMPLEC and SIMPLE algorithms. But, the SIMPLEC algorithm increases the performance of the SIMPLE algorithm without any extra terms.
9. Which of these facts about the SIMPLEC algorithm is correct?
a) The discretized pressure-correction equation is difficult to solve than that in the SIMPLE algorithm
b) It does not need a pressure correction equation
c) It is computationally expensive
d) Wrong pressure field will influence the velocity field
Answer: d
Explanation: The effect of the pressure field on the velocity field in the SIMPLEC algorithm is the same as the SIMPLE algorithm. A wrong pressure field will result in a bad velocity field too.
10. Which of these performance characters of the SIMPLEC algorithm match with the SIMPLE algorithm?
a) Convergence per iteration
b) Number of iterations
c) Cost per iteration
d) Total computational cost
Answer: c
Explanation: As the SIMPLEC algorithm is more converging than the SIMPLE algorithm, the SIMPLEC has more convergence per iteration, less number of iterations and hence less total computational cost than the SIMPLE algorithm. But, the computational cost per iteration is the same for both.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Incompressible Flows – PISO Algorithm”.
1. What does PISO stand for?
a) Pressure Implicit with SIMPLE Operators
b) Pressure Implicit with Settling of Operators
c) Pressure Implicit with Splitting of Operators
d) Pressure Indication with Splitting of Operators
Answer: c
Explanation: The PISO algorithm is expanded as Pressure Implicit with Splitting of Operators. It was introduced by Issa in the year 1986. This is also an extension of the SIMPLE algorithm with some changes in it.
2. The PISO algorithm is first built for ___________
a) steady compressible flows
b) unsteady compressible flows
c) steady incompressible flows
d) unsteady incompressible flows
Answer: b
Explanation: This algorithm was actually built for the unsteady compressible flows. It included a non-iterative computation process. This method was then adapted to solve the incompressible flows with an iterative process.
3. How many predictor and corrector steps does the PISO algorithm involve?
a) One predictor and one corrector steps
b) Two predictor and one corrector steps
c) Two predictor and two corrector steps
d) One predictor and two corrector steps
Answer: d
Explanation: The SIMPLE algorithm has one predictor and one corrector steps. But, the PISO algorithm has an extra corrector step. Therefore, it can be called an extension of the SIMPLE algorithm.
4. Which of these procedures is done in the predictor step of the discretized momentum equation?
a) Initial guess for velocity and the first corrected pressure are obtained
b) First corrected values of pressure and velocity are obtained
c) Initial guesses for pressure and velocity are obtained
d) Initial guess for pressure and the first corrected velocity are obtained
Answer: c
Explanation: In the first step of the PISO algorithm, an initial guess is made on the pressure and the velocity values. This is corrected later in the following corrector steps. This is exactly the same for the SIMPLE algorithm.
5. The guessed velocities will satisfy the continuity equation when ___________
a) the pressure field is correct
b) the pressure correction is correct
c) the velocity correction is correct
d) the pressure-correction equation satisfies the continuity equation
Answer: a
Explanation: The momentum equation is solved to get the guessed pressure values. Then the velocities are assumed from this pressure field. This assumed velocity field will not satisfy the continuity equations until the pressure field is correct.
6. The initial corrector step is used to __________
a) get a velocity field that satisfies the momentum equation
b) get a velocity field that satisfies the continuity equation
c) get a pressure field that satisfies the continuity equation
d) get a pressure field that satisfies the momentum equation
Answer: b
Explanation: The first corrector step of the PISO algorithm results in a velocity field which will satisfy the continuity equation. The equations resulting from this step is the same as the equations used in the SIMPLE algorithm.
7. In which of these steps are the pressure-correction equations obtained in the PISO algorithm?
a) The first predictor step
b) The first corrector step
c) The second predictor step
d) The second corrector step
Answer: b
Explanation: The pressure-correction equation is obtained by applying the corrected velocities to the continuity equation. This pressure-correction equation is then used to get the first pressure correction value.
8. Which of these equations is solved by the PISO algorithm in the extra corrector step?
a) Pressure-correction equation
b) Pressure equation
c) Continuity equation
d) Momentum equation
Answer: d
Explanation: The initial pressure guesses are obtained from the momentum equation. This pressure field is used to get the velocity field approximations. This is corrected in the first corrector step using the pressure-correction equation. It is again corrected using the momentum equation in the second corrector step.
9. Which of these is a disadvantage of the PISO algorithm?
a) Additional storage
b) Slow convergence
c) Less consistent
d) Not accurate
Answer: a
Explanation: The PISO algorithm corrects the pressure and velocity fields twice in the two corrector steps. So, to store the intermediate values and the intermediate equations, additional memory requirements is needed. This is a disadvantage.
10. Which of these characteristics of the PISO algorithm match with the SIMPLE algorithm?
a) No extra equation
b) Extra equation
c) Under-relaxation factor
d) High convergence
Answer: c
Explanation: The PISO algorithm uses the under-relaxation procedure followed in the SIMPLE algorithm. This is the only algorithm that is modified from the SIMPLE algorithm which uses this under-relaxation factor.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Compressible Flows – Conservation Equation”.
1. Which of these equations are needed for compressible flows?
a) Mass, momentum and energy conservations and equation of state
b) Mass and momentum conservations and equation of state
c) Momentum and energy conservations and equation of state
d) Mass, momentum and energy conservations
Answer: a
Explanation: For the compressible flows, the basic governing equations – the mass, momentum and energy conservations are solved first. From these known values, to find the unknowns, an extra governing equation – the equation of state is also used.
2. Which of these differences occur in the momentum equation of the compressible flows when compared to that of the incompressible flows?
a) Temperature term
b) Source term
c) Bulk viscosity
d) Bulk modulus
Answer: c
Explanation: In the incompressible flows, the bulk viscosity term is omitted as the density is constant. But, in the compressible flows, this cannot be omitted. This is an additional term in the momentum equation of the compressible flows.
3. Which of these terms is newly interpolated for the compressible flows?
a) Temperature
b) Density
c) Pressure
d) Mass
Answer: b
Explanation: In the incompressible flows, the density was constant everywhere. But pressure values were calculated. But, for the compressible flows, the density values vary at different points. So, at the interfaces, they should be interpolated.
4. Which of these methods is used in the compressible flows to find the densities at the faces?
a) Central difference
b) Upwind difference
c) Rhie-chow interpolation
d) Weighted average
Answer: b
Explanation: The central difference scheme produces oscillations at high speeds. So, a bounded upwind first or higher-order equation is used to get the density values at the interfaces.
5. The bulk viscosity term in the compressible flows are discretized as ___________
a) summation of bulk viscosity and area at the faces
b) product of bulk viscosity and volume at the cell centre
c) product of bulk viscosity and area at the faces
d) summation of fluxes over the faces
Answer: d
Explanation: The bulk viscosity term in the momentum equation is a volume integral. This is converted into a surface integral. This surface integral is then discretized as the summation of fluxes over the faces of the control volume.
6. The extra term in the pressure-correction equation of the compressible flows ___________
a) density-correction field
b) pressure-correction field
c) velocity-correction field
d) viscosity-correction field
Answer: a
Explanation: The variable nature of the density in the compressible flows adds an extra term to the pressure-correction equation. Density-correction is that extra term. It is obtained using the pressure-density relation.
7. The pressure-correction equation of the compressible flows is ___________
a) a hyperbolic equation
b) an elliptic equation
c) a combined elliptic and hyperbolic equation
d) a parabolic equation
Answer: a
Explanation: The pressure-correction equation in the incompressible flows is an elliptic equation. This is transformed into a hyperbolic equation in the compressible flows which is capable of resolving shock waves.
8. In the mass flow rate correction term at low Mach numbers, which of these terms dominate?
a) The density correction
b) The gradient of density correction
c) The pressure correction
d) The gradient of pressure correction
Answer: d
Explanation: There are two terms involved in the mass flow rate correction. When the Mach number is low, the gradient of pressure correction term dominates the flow and makes the equation elliptic.
9. An extra term in the energy equation of the compressible flows is ___________
a) convection term
b) viscous dissipation term
c) diffusion term
d) unsteady term
Answer: b
Explanation: The viscous dissipation term is an additional term in the energy conservation equation of the compressible flows like the bulk viscosity term of the momentum equation. The discretization procedure is the same as that of the bulk viscosity term.
10. Which of these properties of the SIMPLE scheme matches with the energy equation of the compressible flows?
a) Instability
b) Overshoots
c) Extra equation
d) Under-relaxation term
Answer: c
Explanation: Like the SIMPLE scheme, the energy equation of the compressible flows also needs an under-relaxation term. The overall solution of the compressible flows follows the SIMPLE family of algorithms.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Transient Flows – Two-level Methods”.
1. For which kind of problems are the two-level methods used?
a) Spatial integrations
b) Spatial problems in ODEs
c) Temporal initial value problems in ODEs
d) Temporal initial value problems in integration
Answer: c
Explanation: The two-level methods are used to discretize the ordinary differential equations which of the initial value types. They proceed using two steps in time starting from the available initial value.
2. Which of these methods will not come under a two-level method?
a) Forward Euler method
b) Adams method
c) Trapezoidal method
d) Midpoint rule
Answer: b
Explanation: The Adams method is a multipoint method. Some of the two-level methods are the forward and backward Euler methods, the midpoint rule and the trapezoidal rule. They do not use more than two points to solve the system.
3. Which of these methods is the basis of the leapfrog method?
a) Midpoint rule
b) Trapezoidal rule
c) Implicit Euler method
d) Explicit Euler method
Answer: a
Explanation: The midpoint rule uses the midpoint of the interval to approximate the results. This forms the basis of the leapfrog method which is a very important method for solving the partial differential equations.
4. Which of these methods is derived from the trapezoidal rule?
a) Euler method
b) Adams method
c) Runge-Kutta method
d) Crank-Nicolson method
Answer: d
Explanation: The Crank-Nicolson method is another method for solving the partial differential equations. They are derived from the trapezoidal rule of numerical approximation. The trapezoidal rule is a straight point interpolation.
5. Which of these is an explicit method of solving initial value problems?
a) Forward Euler method
b) Adams method
c) Trapezoidal method
d) Midpoint rule
Answer: a
Explanation: The forward Euler method needs the value of the flow variable at the endpoint. Therefore, it cannot be calculated without any interpolation or approximation. This is an explicit method.
6. What is the condition of stability for the forward Euler method when the function is real?
a) \
\
Always stable
d) Never statble
Answer: b
Explanation: The forward Euler method is conditionally stable. For this method to be stable, it needs the following condition to be satisfied.
\(\big|1+\Delta t\frac{\partial f}{\partial\phi} \big|<1\)
When the function f is real, this becomes
\(\big|\Delta t\frac{\partial f}{\partial\phi} \big|<2\).
7. The trapezoidal rule is ___________
a) stable when Δ t>1
b) stable when Δ t<1
c) always stable
d) never stable
Answer: c
Explanation: The trapezoidal rule, the midpoint rule and even the backward implicit Euler method are all unconditionally stable. They do not need any condition for the solution to be bounded when the input is bounded.
8. Which of these methods is stable for non-linear systems?
a) Forward Euler method
b) Backward Euler method
c) Trapezoidal method
d) Midpoint rule
Answer: b
Explanation: Though the trapezoidal rule is unconditionally stable, it is not the case for non-linear problems. But, the backward Euler method behaves well and smooth for non-linear systems also. They produce smooth results for large time steps too.
9. What is the order of accuracy of the forward Euler method?
a) First-order
b) Second-order
c) Third-order
d) Fourth-order
Answer: a
Explanation: The forward Euler method has the least accuracy in the two-level methods of approximating ODEs. They are first-order accurate. But, the Taylor series expansion of the forward Euler method says it to be a second-order accurate scheme.
10. What is the maximum possible accuracy for the two-level methods?
a) Fifth-order
b) Fourth-order
c) Third-order
d) Second-order
Answer: d
Explanation: The two-level schemes can at most give an accuracy of order two. The trapezoidal and midpoint rules and also the backward Euler method are second-order accurate. But, this does not determine the accuracy of the method solely.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Transient Flows – Predictor-Corrector and Multipoint Methods”.
1. The predictor-corrector method is a combination of ______________
a) midpoint and trapezoidal rules
b) backward Euler method and Trapezoidal rule
c) implicit and explicit methods
d) forward and backward Euler methods
Answer: c
Explanation: Explicit methods are very easy to program and they need less computational cost. But they are not stable. The implicit methods are unconditionally stable but more expensive and iterative. So, the positives of both of these methods are combined by the predictor-corrector method.
2. In the two-level predictor-corrector method, the prediction is done using _____________
a) trapezoidal rule
b) explicit Euler method
c) midpoint rule
d) implicit Euler method
Answer: b
Explanation: The prediction step uses the forward Euler method. The formula for this method is given as
Φ n+1* =Φ n +f(t n ,Φ n )Δt.
Here, ^*indicates that this answer is not the final one. This predicted value is corrected later.
3. Which of these formulae is used for the corrector step of the two-level predictor-corrector method?
a) Φ n+1 =Φ n +\(\frac{1}{3}\) [2f(t n ,Φ n )+f(t n+1 ,Φ n+1* )]Δt
b) Φ n+1 =Φ n +\(\frac{1}{2}\) [2f(t n ,Φ n )+f(t n+1 ,Φ n+1* )]Δt
c) Φ n+1 =Φ n +\(\frac{1}{3}\) [f(t n ,Φ n )+2f(t n+1 ,Φ n+1* )]Δt
d) Φ n+1 =Φ n +\(\frac{1}{2}\) [f(t n ,Φ n )+f(t n+1 ,Φ n+1* )]Δt
Answer: d
Explanation: The predicted result of the two-level prediction-correction method is corrected using the trapezoidal rule. The trapezoidal rule uses the linear interpolation between two time points. The formula is
Φ n+1 =Φ n +\(\frac{1}{2}\) [f(t n ,Φ n )+f(t n+1 ,Φ n+1* )]Δt.
4. The two-level predictor-corrector method is __________
a) second-order accurate
b) first-order accurate
c) fourth-order accurate
d) third-order accurate
Answer: a
Explanation: The order of accuracy of the two-level predictor-corrector method is the same as the trapezoidal rule. This is because they employ the trapezoidal rule. The trapezoidal rule is second-order accurate.
5. The stability of the two-level predictor-corrector method matches with that of the __________
a) midpoint rule
b) trapezoidal rule
c) backward Euler method
d) forward Euler method
Answer: d
Explanation: The predictor-corrector method takes the stability of the explicit Euler method. Though this is not advantageous, at least, the accuracy is better for the two-level predictor-corrector method.
6. The predictor-corrector method is maximum ___________
a) second-order accurate
b) cannot be defined
c) third-order accurate
d) fourth-order accurate
Answer: b
Explanation: The two-level predictor-corrector method has the highest possible order of accuracy as two. But, in general, there are many predictor-corrector methods are of a higher-order. So, the order of accuracy cannot be decided before.
7. To increase the order of accuracy, the multipoint method uses ___________
a) highly stable two-level methods for prediction and correction
b) higher-order two-level methods for prediction and correction
c) additional points where data is already available
d) additional points where data is interpolated
Answer: c
Explanation: To increase the order of accuracy of the predictor-corrector method, multiple points are used instead of two points. The extra points are those obtained from the previous calculations.
8. The Adam-Bashforth method is ____________
a) an explicit method
b) an implicit method
c) a first-order accurate method
d) a second-order accurate method
Answer: a
Explanation: The Adam-Bashforth method is a multipoint method. Therefore, its order of accuracy will be more than two. It is an explicit method of approximation. But it is a lot more advantageous than the explicit Euler method.
9. Which of these is used by the Adam-Bashforth method?
a) Newton’s method
b) Frobenious covariant
c) Frobenious norm
d) Lagrange polynomial
Answer: d
Explanation: A Lagrange polynomial is used to get the polynomials at the required number of points in the Adam-Bashforth method. The order of accuracy of this method depends on the number of polynomials used.
10. Which of these is correct for the multipoint method?
a) multiple derivatives at each time step
b) only one evaluation of derivative per time step
c) order of accuracy is restricted to four
d) extremely unstable
Answer: b
Explanation: The multipoint approach has a lot of advantages. The main advantage is that any order of accuracy can be obtained by using a different number of points. Also, at one time-step, only one new derivative has to be evaluated. Others are those stored from the older time-steps.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Transient Flows – Runge Kutta Method”.
1. The second-order Runge-Kutta method uses __________ as a predictor.
a) backward order method
b) forward Euler method
c) midpoint rule
d) multipoint method
Answer: b
Explanation: The second-order Runge-Kutta method includes two steps. The first step can be called a half-step predictor. This is based on the forward Euler method which is an explicit method of first-order accuracy.
2. Which of these correctors does the second-order Runge-Kutta method use?
a) Backward Euler corrector
b) Forward Euler corrector
c) Trapezoidal corrector
d) Midpoint rule corrector
Answer: d
Explanation: The second step of the second-order Runge-Kutta method is the corrector step. For this correction, midpoint rule is used. This step makes this Runge-Kutta method a second-order method.
3. How many steps does the fourth-order Runge-Kutta method use?
a) Two steps
b) Five steps
c) Four steps
d) Three steps
Answer: c
Explanation: All the Runge-Kutta methods are of high orders. The fourth-order Runge-Kutta method is a method which uses four steps. These four steps include the predictor and the corrector steps.
4. The first two steps of the fourth-order Runge-Kutta method finds the value at which point?
a) At the th point
b) At the th point
c) At the th point
d) At the n th point
Answer: a
Explanation: The first two steps of the fourth-order Runge-Kutta method find the values at the th point. It does not directly move to the next step. It finds the value at an intermediate point between the current and the next points.
5. How many predictor and corrector steps does the fourth-order Runge-Kutta method use?
a) Three predictor and one corrector steps
b) One predictor and three corrector steps
c) Two predictor and two corrector steps
d) One predictor and two corrector steps
Answer: c
Explanation: The fourth-order Runge-Kutta method totally has four steps. Among these four steps, the first two are the predictor steps and the last two are the corrector steps. All these steps use various lower order methods for approximations.
6. The first two steps of the fourth-order Runge-Kutta method use __________
a) Euler methods
b) Forward Euler method
c) Backward Euler method
d) Explicit Euler method
Answer: a
Explanation: All the steps of the Runge-Kutta method use the two-level formulae for initial value problems. The first step uses the forward Euler method and the second step uses the backward Euler method. Collectively, we can say that these two steps use Euler methods.
7. The final corrector of the fourth-order Runge-Kutta method uses ___________
a) Midpoint rule
b) Backward Euler method
c) Simpson’s rule
d) Trapezoidal rule
Answer: c
Explanation: The third step of the fourth-order Runge-Kutta method uses midpoint rule to correct the values and the last step uses Simpson’s rule. This renders a fourth-order accuracy to the Runge-Kutta method.
8. Consider an n th order accurate Runge-Kutta method. How many times is the derivative evaluated at the fourth time-step?
a) one time
b) two times
c) four times
d) n times
Answer: d
Explanation: At each step of the Runge-Kutta method, the derivate has to be evaluated n times. Here, ‘n’ is the order of accuracy of the Runge-Kutta method. This is a major disadvantage of Runge-Kutta methods.
9. Which of these statements is correct?
a) When the order of accuracy is the same for two methods, the accuracy is also the same
b) Runge-Kutta method interpolates at more than one point in a time interval
c) Runge-Kutta method is not a multipoint method
d) An n th order Runge-Kutta method is more accurate than the n th order multipoint method
Answer: d
Explanation: When comparing the Runge-Kutta method and the multipoint method, even if the order of accuracy is the same, the Runge-Kutta method is more accurate. This is because the coefficient of the Runge-Kutta method is small.
10. Which of these is a disadvantage of the Runge-Kutta method over the multipoint method?
a) Computational stability
b) Computational cost
c) Accuracy
d) Convergence
Answer: b
Explanation: Though the Runge-Kutta methods are advantageous in the accuracy and stability perspectives, this comes at the cost of computational cost. Computationally, the Runge-Kutta methods are very expensive when compared to the other methods.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Transient SIMPLE and PISO Algorithms”.
1. The SIMPLE algorithm used for transient problems is __________
a) implicit and iterative
b) implicit and direct
c) explicit and iterative
d) explicit and direct
Answer: a
Explanation: For the implicit solution of transient flow problems, the SIMPLE, SIMPLER and SIMPLEC algorithms can be used to get the solution at each time-step. The solution is iterated until we get them converged at one time-step.
2. Which of these statements is correct about the transient SIMPLE algorithm?
a) No additional terms are needed in the continuity and pressure-correction equations
b) No additional terms are needed in the continuity equation
c) An additional term is needed in the continuity and pressure-correction equations
d) No additional terms are needed in the pressure-correction equation
Answer: c
Explanation: The transient terms are added to the continuity equation. As the transient SIMPLE algorithm derives its pressure-correction equation from the continuity equation, the pressure-correction equation also gets some additional terms.
3. The PISO algorithm for the transient problem is _________
a) iterative
b) non-iterative
c) never converging
d) fast converging
Answer: b
Explanation: The PISO algorithm was actually developed for transient problems. When taken for the steady-state problems, it was made iterative. In the original form, the PISO algorithm does not involve iterations.
4. Which of these techniques is used by the PISO algorithm?
a) Neighbour splitting
b) Source splitting
c) Coefficient splitting
d) Operator splitting
Answer: d
Explanation: The temporal accuracy of the PISO algorithm depends on the differencing scheme used. This includes the operator splitting technique used. Operator splitting uses different methods to calculate different variables.
5. Which of these terms of the momentum equations are altered by the transient term in the PISO algorithm?
a) neighbouring coefficients
b) central coefficients
c) weighted average of the neighbouring coefficients
d) velocity correction term
Answer: b
Explanation: A term with the initial density, change in volume and change in time is added to the central coefficients of the momentum equations in all the directions. The neighbouring coefficients are not altered.
6. Which of these terms is added to the source term of the momentum equations of the transient PISO algorithm?
a) product of the latest central coefficient and latest velocity
b) product of the current central coefficient and previous velocity
c) product of the previous central coefficient and previous velocity
d) product of the initial central coefficient and current velocity
Answer: c
Explanation: To the source term of the u-momentum equation, the term \(a_P^0 \, u_P^0\) is added. Where, \(a_P^0\) is the initial central coefficient and \(u_P^0\) is the initial velocity in u-direction. A similar term is added to the sources of the other momentum equations also.
7. Which of these equations are altered by the transient term in the PISO algorithm?
a) both the first and second pressure-correction equations
b) only the first pressure-correction equation
c) only the second pressure-correction equation
d) neither the first nor the second pressure-correction equations
Answer: a
Explanation: The transient PISO algorithm has two corrector steps and hence two pressure-correction equations. Both of these equations are affected by the transient term. The source terms of these equations are altered.
8. The order of temporal accuracy achieved by the PISO algorithm for pressure and momentum are ___________ and ___________ respectively.
a) four, three
b) two, three
c) three, four
d) four, two
Answer: c
Explanation: The PISO procedure is repeated at each time-step to calculate the velocity and pressure values. This algorithm results in third-order temporally accurate pressure values and fourth-order temporally accurate momentum values.
9. Which of these is a disadvantage of the PISO algorithm?
a) memory required
b) small time-steps
c) computational cost
d) time required
Answer: b
Explanation: The PISO algorithm is advantageous compared to the SIMPLE algorithm that they do not need an iterative process at each time step. But the disadvantage is that they need the time steps to be small enough to produce accurate results.
10. To overcome the performance issues of the PISO algorithm, which of these methods is used?
a) Time-steps based on temporal schemes are used
b) Large time-steps are used
c) First-order temporal differencing is used
d) Higher-order temporal differencing is used
Answer: d
Explanation: The need for a small time-step in the PISO algorithm is overcome by using higher-order temporal differencing schemes. For example, a second-order implicit scheme with three time levels can be used.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Transient Flows – Euler Schemes”.
1. Which of these equations give the derivative of the function T at time t as given by the Crank-Nicolson scheme?
a) \
\
\
\(\frac{T+T}{\Delta t}\)
Answer: a
Explanation: The Crank-Nicolson scheme uses the previous and the next steps to get the derivative at the current step. Expressing it mathematically,
\(\frac{\partial T}{\partial t} = \frac{T-T}{2 \Delta t}\).
2. The approximation of the derivative taken by the Crank-Nicolson scheme is the same as the __________ of spatial derivative.
a) second order forward difference approximation
b) backward difference approximation
c) forward difference approximation
d) central difference approximation
Answer: d
Explanation: The central difference scheme of the spatial derivative uses the previous and the next neighbours for the approximation. The Crank-Nicolson scheme also uses such an approximation for its time derivative.
3. The Crank-Nicolson scheme is ________
a) fourth-order accurate
b) third-order accurate
c) second-order accurate
d) first-order accurate
Answer: c
Explanation: The order of accuracy of the Crank-Nicolson scheme is two. It has better accuracy than the forward Euler scheme. This can be obtained using the Taylor series expansion of the temporal derivative.
4. To find the values at the current time-step, the Crank-Nicolson scheme uses ___________
a) t-Δt and t+Δ t steps
b) t-Δ t and t-2Δ t steps
c) t+Δ t and t+2Δ t steps
d) t and t+Δt steps
Answer: b
Explanation: Like the forward Euler scheme, the Crank-Nicolson scheme also uses the older steps only to get the values at the current node. It uses the values at the previous step and at the step previous to it.
5. For the transient convection problems, the Crank-Nicolson scheme is stable when _________
a) CFL conv ≤2
b) CFL conv ≤1
c) CFL conv ≥-2
d) CFL conv ≥-1
Answer: a
Explanation: The Crank-Nicolson scheme is not stable always. It is only conditionally stable. For the transient convection problems, the scheme is stable only when CFL conv ≤2.
6. For which of these problems is the Crank-Nicolson scheme unconditionally stable?
a) Compressible flows
b) Advection problems
c) Diffusion problems
d) Convection-Diffusion problems
Answer: c
Explanation: When the Crank-Nicolson scheme is applied to the diffusion problems, there is no restriction to the time-step from stability side. It is unconditionally stable for this case. This is why the scheme is often used for diffusion problems.
7. According to the Adams-Moulton scheme, the derivative of a function T at time-step t is given by _________
a) \
\
\
\(\frac{3 T-4T+T}{2\Delta t}\)
Answer: d
Explanation: To find the derivative, the Adams-Moulton method uses the previous and the second previous steps. The mathematical expression is
\(\frac{\partial T}{\partial t}=\frac{3 T-4T+T}{2\Delta t}\).
8. The Adams-Moulton scheme is __________
a) explicit
b) implicit
c) a two-level scheme
d) a three-level scheme
Answer: b
Explanation: The Adams-Moulton scheme wants all of its equations to be solved simultaneously. It is not a time marching scheme. It is an implicit scheme. So, it is computationally more expensive.
9. The Adams-Moulton scheme comes under ____________
a) Backward schemes
b) Forward schemes
c) Multipoint schemes
d) Runge-Kutta methods
Answer: c
Explanation: The Adams-Moulton scheme is a multipoint predictor-corrector method. These methods use more than two time-steps for their prediction. Many methods use this scheme as their corrector step too.
10. Which of these statements about the Adams-Moulton method is correct?
a) It involves the terms at the older time-step only
b) It involves the terms at the next time-step
c) It does not involve iterations
d) It does not involve older steps
Answer: b
Explanation: The Adams-Moulton method involves the information at the upcoming steps also. The advantage of the Adams-Moulton method is that it needs only n steps to get an order of accuracy equal to n+1.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Transient Flows – First Order Finite Volume Schemes”.
1. The discretization of the transient term using the finite volume approach is more like the spatial discretization of __________
a) the convection term
b) the diffusion term
c) the source term
d) the anti-diffusion term
Answer: a
Explanation: The finite volume approach for the discretization of the rate of change per unit time is more similar to the discretization of convection term while doing a spatial discretization. The only difference is that the first case is based on time and the second is based on spatial coordinates.
2. Consider the following equation representing the temporal integration over the time interval t-\
dt=0\)
If the first term is discretized using the difference of fluxes and the second term is evaluated using the midpoint rule, what is the discretized form?
a) \
^{t-\frac{\Delta t}{2}}+L
\Delta t\)
b) \
^{t+\frac{\Delta t}{2}}-L
\Delta t\)
c) \
^t+L
\Delta t\)
d) \
^{t+\frac{\Delta t}{2}}-V_C
^{t-\frac{\Delta t}{2}}+L
\Delta t\)
Answer: d
Explanation: The given equation is
\
dt=0\)
Discretizing the first term using the difference of fluxes,
\
^{t+ \frac{\Delta t}{2}}-V_C
^{t-\Delta\frac{\Delta t}{2}}\)
Discretizing the second term using the midpoint rule,
\
dt=L
\Delta t\)
Therefore, the final term is
\
^{t+\frac{\Delta t}{2}}-V_C
^{t-\frac{\Delta t}{2}}+L
\Delta t\).
3. Which of these changes should be made in the semi-discretized equation to get the fully discretized equation?
a) Express the face values in terms of the neighbouring face values
b) Express the face values in terms of the cell values
c) Express the cell values in terms of the face values
d) Express the cell values in terms of the neighbouring cell values
Answer: b
Explanation: While discretizing the transient term, the semi-discretized equation contains the values at the cell faces. If these face values are expressed in terms of the cell values, the complete discretized form of the equation can be obtained.
4. If the first-order implicit Euler scheme is used, the value at t+Δt/2 is replaced by the value at _________
a) t
b) t-\
t+Δt
d) t-Δt
Answer: c
Explanation: In the first-order implicit Euler scheme, the values at the cell faces are approximated by the values at cell centres of the backward direction. Therefore, the value at t+\(\frac{\Delta t}{2}\) is replaced by the value at t.
5. Which of these equations is the discretized form of the transient term using the first-order implicit Euler scheme?
a) \
\)
b) \
\)
c) \
\)
d) \
\)
Answer: b
Explanation: The first-order implicit Euler scheme gives its terms using the older terms. Before using this scheme, the terms are
\
\)
When the scheme is applied to these equations,
\
\).
6. The first-order implicit Euler schemes to discretize the transient term creates ________
a) cross-flow diffusion
b) cross-diffusion
c) numerical anti-diffusion
d) numerical diffusion
Answer: d
Explanation: As the transient term behaves like the convection term while discretizing, numerical diffusion is produced by the first-order implicit Euler schemes. The value of the numerical diffusion can be obtained using the Taylor series expansion.
7. When the first-order implicit Euler scheme is unconditionally stable, the solution is ________
a) stationary for large time-steps
b) oscillatory for large time-steps
c) stationary for small time-steps
d) oscillatory for small time-steps
Answer: a
Explanation: A numerical diffusion term scales with the time-step in a similar fashion to the upwind scheme for the advection term. Therefore, when this scheme is unconditionally stable, the solution using this scheme is stationary for large steps.
8. The extra term added while discretizing the transient term of a flow with density ρ and flow variable φ using the first-order explicit Euler scheme is _________
a) \
\
\
\(-\frac{\Delta t}{2}\frac{\partial^2}{\partial t^2}\)
Answer: d
Explanation: While using the first-order explicit Euler scheme, an extra term called the numerical anti-diffusion occurs. This term can be obtained by using the Taylor series expansion. The term is
\(-\frac{\Delta t}{2}\frac{\partial^2}{\partial t^2}\).
9. According to the first-order explicit Euler scheme, the value at time-step t-\Missing open brace for subscript t+\
t
c) t-Δt
d) t+Δt
Answer: b
Explanation: The value at t-\(\frac{\Delta t}{2}\) is at the interface of two cells. One has the cell centre t and the other has the cell centre t-Δ t. The first-order explicit Euler scheme is downstream biased. Therefore, the value at t is taken to approximate the value at t-\(\frac{\Delta t}{2}\).
10. The numerical diffusion and numerical anti-diffusion terms are equal for the first-order Euler scheme are equal in magnitude when __________
a) the courant number of diffusion is equal to one
b) the courant number of diffusion is equal to two
c) the courant number of convection is equal to one
d) the courant number of convection is equal to two
Answer: c
Explanation: The numerical diffusion term of the first-order implicit Euler scheme and numerical anti-diffusion term of the first-order explicit Euler scheme are equal in magnitude and opposite in sign when the courant number of convection is equal to one.
This set of Computational Fluid Dynamics Questions & Answers for Exams focuses on “Transient Flows – Second Order Finite Volume Schemes”.
1. What is the equivalent of (ρ C Φ C ) t+Δt/2 using the Crank-Nicolson scheme for finite volume approach?
a) \(\frac{1}{2}\)(ρ C Φ C ) t +\(\frac{1}{2}\)(ρ C Φ C ) t+Δ t
b) (ρ C Φ C ) t +(ρ C Φ C ) t+Δt
c) (ρ C Φ C ) t -(ρ C Φ C ) t+Δt
d) \(\frac{1}{2}\)(ρ C Φ C ) t –\(\frac{1}{2}\)(ρ C Φ C ) t+Δt
Answer: a
Explanation: The Crank-Nicolson scheme gives equal weight to both the cells which share the face. The formula is given by,
(ρ C Φ C ) t+Δt/2 =\(\frac{1}{2}\)(ρ C Φ C ) t +\(\frac{1}{2}\)(ρ C Φ C ) t +Δt.
2. Which of these time-steps are used to approximate the value at time-step t-\
t and t+Δ t
b) t and t-Δ t
c) t and t-\
t and t+\(\frac{\Delta t}{2}\)
Answer: b
Explanation: The Crank-Nicolson scheme uses the cell centres of both the cells which share the face which is considered. The face considered here is t-\(\frac{\Delta t}{2}\)
. It is shared by the faces t and t-Δ t. So, the scheme uses both of these values for the approximation.
3. The stability of the Crank-Nicolson scheme for finite volume approach is constrained by ________
a) CFL number
b) Peclet number
c) Time-step size
d) Spatial grid size
Answer: a
Explanation: The Crank-Nicolson scheme is an explicit scheme which uses the values before the face and after the face to find the value at the face. Therefore, its stability is constrained by the CFL or Courant number.
4. The results using the Crank-Nicolson scheme for finite volume approach can be reformulated using the ________
a) implicit first-order Euler scheme
b) implicit and explicit first-order Euler schemes
c) explicit first-order Euler scheme
d) central difference scheme
Answer: b
Explanation: The finite volume approach using the Crank-Nicolson scheme can be reformulated using the first-order implicit Euler scheme and then the explicit Euler scheme which is modified and used in the form of extrapolation.
5. Which of these terms cause instability in the Crank-Nicolson scheme when used for finite volume approach?
a) Anti-diffusion term
b) Anti-dispersive term
c) Diffusion term
d) Dispersive term
Answer: d
Explanation: By expanding the results of the Crank-Nicolson scheme using the Taylor series expansion, the scheme is proved to be a second-ordered scheme. The third-ordered term omitted here is a dispersive term causing instabilities.
6. What is the equivalent of (ρ C Φ C ) t+Δt/2 using the second-order upwind Euler scheme for finite volume approach?
a) \(\frac{3}{2}\) (ρ C Φ C ) t +(ρ C Φ C ) t-Δt
b) (ρ C Φ C ) t +\(\frac{1}{2}\) (ρ C Φ C ) t-Δt
c) \(\frac{3}{2}\) (ρ C Φ C ) t +\(\frac{1}{2}\) (ρ C Φ C ) t-Δ t
d) \(\frac{1}{2}\)(ρ C Φ C ) t +\(\frac{1}{2}\) (ρ C Φ C ) t-Δ t
Answer: c
Explanation: The second-order upwind Euler scheme gives more importance to the immediate upwind than the far upwind. The formula is
(ρ C Φ C ) t+Δ t/2 =\(\frac{3}{2}\)(ρ C Φ C ) t +\(\frac{1}{2}\)(ρ C Φ C ) t-Δt .
7. Which of these time-steps are needed to approximate the value at time-step \
t-\
t and t-Δ t
c) t-Δ t and t-2Δ t
d) t and t-2Δ t
Answer: c
Explanation: The second-order upwind Euler scheme uses two upwind nodes to approximate the values. The immediate upwind and the far upwind of t-\(\frac{\Delta t}{2}\) are t-Δ t and t-2Δ t. The vales at these two nodes are used for the approximation.
8. How many numerical diffusion terms does the second-order upwind Euler scheme have?
a) Infinity
b) No diffusion term
c) One term
d) Two terms
Answer: b
Explanation: The transient term discretization also has similar properties like the convection term discretization. The second-order upwind Euler scheme does not have any numerical diffusion or anti-diffusion terms. These terms are present for the first-order schemes only.
9. The numerical dispersion term of the second-order upwind Euler scheme is of ____________
a) third-order
b) second-order
c) first-order
d) no dispersion
Answer: a
Explanation: When the terms in the discretized form of the transient term using the second-order upwind Euler scheme is further expanded with the Taylor series expansion, the dispersion term of the third order arises.
10. When the finite volume approach is used, if the general form is given as
FluxT=FluxC Φ C +FluxC° Φ C °+FluxV
The superscript o indicates the older time step, the value of FluxC° while using the second-order upwind Euler scheme is ________
a) \
\
\
\(-\frac{2\rho_C^o V_C}{\Delta t}\)
Answer: d
Explanation: When the values for the (ρ C Φ C ) t+Δt/2 and the (ρ C Φ C ) t-Δt/2 are substituted in the semi-discretized equation, the final form obtained is
FluxT=FluxC Φ C +FluxC° Φ C °+FluxV
Where FluxC°=\(-\frac{2\rho_C^o V_C}{\Delta t}\).
This set of Computational Fluid Dynamics Problems focuses on “Approaches for Non-uniform Time Steps”.
1. Discretization of the transient term is not affected by uniform or non-uniform grids when _________
a) the scheme is downwind
b) the scheme is upwind
c) the scheme is first-order
d) the scheme is second-order
Answer: c
Explanation: Since the second-order schemes use a stencil with two time-steps in the same direction, only the second-order schemes are affected by the non-uniformity of the grids. The first-order schemes are not affected.
2. For which of these schemes is the interpolation profile need not be modified for the non-uniform transient grid?
a) Downwind scheme
b) Upwind scheme
c) Crank-Nicolson scheme
d) second-order schemes
Answer: c
Explanation: For all the second-order schemes, the interpolation profile has to be modified when the transient grid is non-uniform. But this is not the case when the Crank-Nicolson scheme is used. There is no change in the interpolation profile needed here.
3. Which of these characteristics of the Crank-Nicolson scheme is affected by the non-uniform transient grids?
a) Consistency
b) Convergence
c) Stability
d) Accuracy
Answer: d
Explanation: When the Crank-Nicolson scheme is used on the non-uniform transient grids, for each of the two steps, a different time-step is used. So, the spatial derivative is not at the centre of the temporal elements. Therefore, the accuracy is affected.
4. Which of these statements is correct about the variable time-steps?
a) The finite volume and finite difference schemes do not yield equivalent algebraic equations
b) The finite volume scheme yields equivalent algebraic equations irrespective of the non-uniformity
c) The finite difference scheme yields equivalent algebraic equations irrespective of the non-uniformity
d) All the second-order schemes result in the same algebraic equation when the grid is non-uniform
Answer: a
Explanation: The finite volume and finite difference schemes yield equivalent algebraic equations when the transient grid is uniform. If the grid is non-uniform, the algebraic equations are not the same.
5. If the Crank-Nicolson scheme is used with the finite difference approach to get the discretized equation of the transient term with the non-uniform grid, what is the central coefficient of the previous time-step?
.
a) \
\
\
\(\frac{\Delta t+\Delta t^o}{\Delta t-\Delta t^o}\rho_C^o V_C\)
Answer: b
Explanation: When the transient grid system is not uniform,
Δt≠Δ t°
Therefore, when the Crank-Nicolson scheme is used, density varies but the volume does not vary.
\(a_C^o=\frac{\Delta t-\Delta t^o}{\Delta t+\Delta t^o}\rho_C^o V_C\) This becomes zero when a uniform transient grid is used.
6. If the Crank-Nicolson scheme is used with the finite difference approach to get the discretized equation of the transient term with the non-uniform grid, what is the central coefficient of the second previous time-step?
.
a) \
\
\
\(\frac{\Delta t}{\Delta t^o
}\rho_C^{oo}V_C\)
Answer: d
Explanation: While using the Crank-Nicolson scheme, the previous and the next time steps are t+Δt and t-Δ t°. Using these in the semi-discretized equation, the coefficient of the second previous time-step is
\(a_C^{oo}=\frac{\Delta t}{\Delta t^o
}\rho_C^{oo}V_C\).
7. If the Adams-Moulton scheme is used with the finite difference approach to get the discretized equation of the transient term with the non-uniform grid, what is the central coefficient of the current time-step?
.
a) \
\rho_C V_C\)
b) \
\rho_C V_C\)
c) \
\rho_C V_C\)
d) \
\rho_C V_C\)
Answer: d
Explanation: The Adams-Moulton scheme uses the time-steps Δt-Δt° and Δt-Δt°. Applying these time-steps, the central coefficient of the current time-step is given by
\
\rho_C V_C\).
8. Which of these is correct for the non-uniform time-steps?
.
a) \
\
δt=Δt
d) δt=Δt°
Answer: b
Explanation: For the variable time-steps, \(\delta t=\frac{
}{2}\) as the size of the temporal element is not the same. For the uniform transient grids, this reduces to δt=Δt as Δ t=Δt°.
9. The general discretised form of the transient term using the finite volume approach is
FluxT=FluxC Φ C +FluxC°Φ C °+FluxV.
The superscript o indicates the older time step. If the Crank-Nicolson method is used with non-uniform grids, what is FluxV?
a) \
\
\
\(\frac{\Delta t^{oo}}{\Delta t^o}\frac{\rho_c^{oo} V_C \phi_c^{oo}}{\Delta t}\)
Answer: c
Explanation: The Crank-Nicolson scheme uses the average of the two central values to get the values at the interface. Using this with the non-uniform time-steps,
FluxV=-\(-\frac{\Delta t^{oo}}{\Delta t^o+\Delta t^{oo}}\frac{\rho_c^{oo} V_C \phi_c^{oo}}{\Delta t}\).
10. Which of these statements is correct regarding the Adams-Moulton scheme used on the non-uniform grids?
a) The current central coefficient for the finite difference and the finite volume schemes are the same
b) The current central coefficient for the uniform and the non-uniform grid is the same
c) The variation in time-steps does not result in any change
d) There is no variation in the values when the grid is uniform
Answer: a
Explanation: For the finite difference approach,
\
\rho_C V_C\)
For the finite volume approach, the central coefficient is
FluxC=\
\rho_C V_C\)
These two are the same irrespective of the approach.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Source Term Discretization”.
1. Which of these transport phenomena do not involve any source term?
a) Heat conduction
b) Heat radiation
c) Turbulence models
d) Chemical reactions
Answer: a
Explanation: Many of the physical phenomena include the source and sinks. The equations of turbulence models, chemical reactions, radiation heat transfer and multiphase flows contain source terms. Pure heat conduction problem does not include any source term.
2. Which of these physical properties of the flow problems is affected by the source term?
a) Order of accuracy
b) Reduction of error
c) Stability
d) Accuracy
Answer: c
Explanation: The physics of the problem is greatly affected by the source term. Moreover, talking from the analysis side, the stability condition of the problem differs for the problems with sources and the problems without sources.
3. Which of these methods is recommended for the source term treatment in general?
a) Treat sinks explicitly and sources implicitly
b) Treat sinks implicitly and sources explicitly
c) Treat the source term explicitly
d) Treat the source term implicitly
Answer: b
Explanation: If the source term is properly treated, the robustness of the problem will increase. In general, it is better to treat the sinks implicitly and the sources explicitly.
4. The source term is ____________
a) equal to the square of the source term
b) equal to the source term
c) a function of the flow variable
d) a function of the product of flow variable and the sink term
Answer: c
Explanation: The source term Q C is a function of the dependent variable which is the flow variable. Giving this mathematically,
Q C =Q(Φ C ).
5. When the source term does not vary much, the source term is ___________
a) implicitly calculated
b) calculated from the previous iteration’s values
c) solved simultaneously
d) calculated form the current values
Answer: b
Explanation: As the source term is a function of the flow variable, it can be calculated using the available Φ values. In an iterative process, the available values are the previous iteration’s flow variable values.
6. When the source term varies highly, which of these properties is affected?
a) Consistency
b) Boundedness
c) Stability
d) Rate of convergence
Answer: d
Explanation: When the source term is a constant or small, the analysis is not affected by it. But, when the variation of the source term is large, the rate of convergence of the numerical process is greatly affected.
7. For linearizing the source term, which of these methods is used?
a) Weighted average
b) Series expansion
c) Normalization
d) Central differencing
Answer: b
Explanation: To reduce the effects of the highly varying source term on the process, the term should be linearized. For this linearization, a series expansion like the Taylor series expansion is used.
8. The general form of the equation for the source term is
[a C -FluxC C ] Φ C +∑ F~NB a F Φ F =FluxV C .
What constraint does the term FluxC C have?
a) FluxC C must be greater than one
b) FluxC C must be less than one
c) FluxC C must be less than zero
d) FluxC C must be greater than zero
Answer: c
Explanation: The diagonal terms are those which are the coefficients of Φ C . In the above equation, the diagonal coefficients are given by a C -FluxC C . For this value to be high, FluxC C should be less than one.
9. The general form of the equation for the source term is
[a C -FluxC C ] Φ C +∑ F~NB a F Φ F =FluxV C .
The constraint on the term FluxC C is imposed by ______________
a) Scarborough criterion
b) Richardson extrapolation
c) Stability criterion
d) Gauss-Siedel criterion
Answer: a
Explanation: The constraint on the term FluxC C for diagonal dominance. Diagonal dominance is needed by the Scarborough criterion. The Scarborough criterion is used to ensure the convergence of the linear system of equations.
10. The general form of the equation for the source term is
[a C -FluxC C ] Φ C +∑ F~NB a F Φ F =FluxV C .
When the variable Φ is positive-definite, the FluxV C should be ______________
a) FluxV C must be greater than one
b) FluxV C must be less than one
c) FluxV C must be less than zero
d) FluxV C must be greater than zero
Answer: d
Explanation: In some of the cases, the value of the flow variable Φ C will always be a positive non-zero value. In these cases, the value of FluxV C must be greater than zero to ensure that the value of Φ will also be greater than zero.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Under-Relaxation”.
1. Relaxation techniques are used to ensure __________
a) convergence
b) stability
c) accuracy
d) boundedness
Answer: a
Explanation: Relaxation techniques are used to increase the convergence of the solution by changing the values of the variables during the iterative process. It actually relaxes the conditions which should be satisfied to get the solutions.
2. The relaxation techniques slow down the __________
a) whole iterative process
b) effect of the sources
c) effect of the neighbouring elements and the source terms
d) effect of the neighbouring elements
Answer: c
Explanation: It slows down the effects of the neighbouring elements and the source term on the central element. It is because these two impose the constraint on the value of the central element.
3. The explicit under-relaxation method means that __________
a) the changes are made before each iteration
b) the changes are made before the iteration process
c) the changes are made after the iteration process
d) the changes are made after each iteration
Answer: c
Explanation: Relaxation can be performed either after obtaining the solution of each iteration or before the process of iteration at each step . Implicit methods are most commonly used.
4. When the relaxation factor is less than one, which of these will occur?
a) Increase the speed of convergence
b) Slow down the speed of convergence
c) Decrease the stability
d) Increase the divergence
Answer: b
Explanation: The process is called under-relaxation when the relaxation factor is less than one. This decreases the rate of convergence of the results. But, the advantage is that the stability of the solution is increased.
5. There is no relaxation when the relaxation factor is __________
a) positive infinity
b) negative infinity
c) one
d) zero
Answer: d
Explanation: If the relaxation factor is zero, it means that the effect of the external factors is completely removed. There is no relaxation if the relaxation factor is one. The same effect as the original one is taken for consideration.
6. What is the disadvantage of over-relaxation?
a) Stability is decreased
b) Divergence is increased
c) Oscillations are increased
d) Convergence is increased
Answer: a
Explanation: When the relaxation factor is more than one, the process is called over-relaxation. This will lead to a higher rate of convergence and leads to faster convergence. But, the disadvantage is that stability will be decreased.
7. Which of these relaxation techniques is used in the SIMPLE algorithm?
a) Explicit over-relaxation
b) Implicit over-relaxation
c) Explicit under-relaxation
d) Implicit under-relaxation
Answer: c
Explanation: The very common algorithm which uses the relaxation technique is the SIMPLE algorithm. It uses the explicit under-relaxation technique to decrease the effect of the correction factors in the pressure and velocity correction equations.
8. Which of these statements about the Patankar’s under-relaxation is true?
a) The value form the previous iteration is modified
b) The diagonal coefficients are modified
c) The whole equation is modified
d) The neighbouring coefficient is modified
Answer: b
Explanation: Patankar’s under-relaxation modifies the value from the previous iteration, the diagonal coefficients and the neighbouring coefficients. The whole equation remains unmodified.
9. The time-step advancement of the E-factor relaxation is dependent on ___________
a) the convergence
b) the stability
c) the accuracy
d) the cell volume
Answer: d
Explanation: The time-step advancement of the E-factor relaxation depends on the grid size. The solution in the smaller element advances more slowly than the solution in the larger element.
10. What is the relation between the E of the E-factor relaxation and the relaxation factor λ?
a) E=\
E=\
E=\
E=\(\frac{1}{1+\lambda}\)
Answer: a
Explanation: The relaxation factor is expressed in terms of E factors in the E-factor relaxation. In general, the E factors range from 4 to 10. This range corresponds to the under-relaxation factors of range 0.75 and 0.9.
This set of Computational Fluid Dynamics Multiple Choice Questions & Answers focuses on “Relaxation – Residuals”.
1. Which of these equations can be written in the residual form?
a) The discretized equations
b) The governing equations
c) The PDEs
d) The ODEs
Answer: a
Explanation: The residual form can be taken by the discretized algebraic equation. While the other form of the equation is called the standard or direct form, the residual form can also be called correction form.
2. The standard form of the equations is solved for __________ and the residual form is solved for _________
a) flow variables, residuals
b) flow variables, corrections
c) corrections, flow variables
d) residuals, flow variables
Answer: b
Explanation: The standard form of the equations is solved for the flow variables so that the algebraic equations are satisfied. The residual form of the discretized equations is solved for the corrections needed to satisfy those equations.
3. When the field of the flow variable is exact, the residual will be __________
a) negative
b) positive
c) zero
d) one
Answer: c
Explanation: When the solution field is exact, it means that the solutions converged. Therefore, the residual will be zero. The correction will also be zero in this case. But, in real, the solution will not reach the exact solution. Therefore, the residual will never be zero.
4. Why is the residual form of the discretized equations used?
a) It increases the rate of convergence
b) It increases the stability
c) It increases the accuracy
d) Numerical errors are less
Answer: d
Explanation: The residual form of the discretized equations are mathematically equivalent to the standard form. In this form, the numerical errors are less when the variations are small for large flow variable values.
5. The absolute residual is calculated __________
a) to overcome high values of residuals
b) to decelerate the convergence process
c) to overcome the sign problems in the actual residual value
d) to increase the rate of convergence
Answer: c
Explanation: The residual may be a positive or a negative quantity. As the sign is not needed, the absolute value of the residual is found. If the absolute residual decreases with iterations, the solution converges.
6. The maximum value of residual is used to __________
a) decrease the importance of sign
b) compare with the vanishing value
c) include the sign of the residual
d) increase the importance of sign
Answer: b
Explanation: The maximum value of residual is found by comparing the values of the absolute residuals. If this maximum value becomes less than the vanishing quantity , then the result is converged.
7. Which of these statements states the need for the normalization of the residual?
a) Different variables result in different level of residuals
b) Different variables result in the same level of residuals
c) Different variables result in different residuals
d) Different variables result in same residuals
Answer: a
Explanation: When the influence of the sign of the residual can be ruled out by the absolute residual, the level of the absolute residual varies with the value of the flow variables. To overcome this problem, the residuals are normalized.
8. Which of these values is used to find the normalized residual?
a) The minimum of all the residuals
b) The maximum of all the residuals
c) The sum of all the residuals
d) The product of all the residuals
Answer: b
Explanation: The normalized residual is obtained by using the absolute residual divided by the maximum of all the residuals. When the maximum scaled value is less than the vanishing value, the solution is said to be converged.
9. Which of these is the denominator while finding the root mean square residual?
a) Sum of the absolute residuals
b) Sum of the actual residuals
c) Number of elements
d) Sum of the scaled residuals
Answer: c
Explanation: The root mean square is the ratio of the square root of the sum of squares of the absolute residuals to the number of residuals. The number of residuals is the same as the number of elements.
10. When the residuals are scaled, the vanishing value is ___________
a) less than 10
b) less than 10
c) more than 10 -5
d) less than 10 -5
Answer: d
Explanation: The common range of the vanishing value is from 10 -3 to 10 -5 for the scaled residuals. I can also range below this level. The integrated quantities are also used to ensure the convergence of the solution.
