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# Computational Fluid Dynamics – Transient Flows – Euler Schemes

1 - Question

Which of these equations give the derivative of the function T at time t as given by the Crank-Nicolson scheme?

a) T(t+Δt)−T(t−Δt)2Δt

b) T(t+Δt)+T(t−Δt)2Δt

c) T(t+Δt)−T(t−Δt)Δt

d) T(t+Δt)+T(t−Δt)Δt

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Answer: aExplanation: The Crank-Nicolson scheme uses the previous and the next steps to get the derivative at the current step. Expressing it mathematically, ∂T(t)∂t=T(t+Δt)−T(t−Δt)2Δt.

2 - Question

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

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Answer: dExplanation: 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 - Question

The Crank-Nicolson scheme is ________

a) fourth-order accurate

b) third-order accurate

c) second-order accurate

d) first-order accurate

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Answer: cExplanation: 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 - Question

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

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Answer: bExplanation: 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 - Question

For the transient convection problems, the Crank-Nicolson scheme is stable when _________

a) CFLconv≤2

b) CFLconv≤1

c) CFLconv≥-2

d) CFLconv≥-1

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Answer: aExplanation: The Crank-Nicolson scheme is not stable always. It is only conditionally stable. For the transient convection problems, the scheme is stable only when CFLconv≤2.

6 - Question

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

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Answer: cExplanation: 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 - Question

According to the Adams-Moulton scheme, the derivative of a function T at time-step t is given by _________

a) 3T(t)+4T(t−Δt)−T(t−2Δt)2Δt

b) 3T(t)−4T(t−Δt)−T(t−2Δt)2Δt

c) 3T(t)+4T(t−Δt)+T(t−2Δt)2Δt

d) 3T(t)−4T(t−Δt)+T(t−2Δt)2Δt

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Answer: dExplanation: To find the derivative, the Adams-Moulton method uses the previous and the second previous steps. The mathematical expression is ∂T(t)∂t=3T(t)−4T(t−Δt)+T(t−2Δt)2Δt.

8 - Question

The Adams-Moulton scheme is __________

a) explicit

b) implicit

c) a two-level scheme

d) a three-level scheme

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Answer: bExplanation: 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 - Question

The Adams-Moulton scheme comes under ____________

a) Backward schemes

b) Forward schemes

c) Multipoint schemes

d) Runge-Kutta methods

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Answer: cExplanation: 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 - Question

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

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Answer: bExplanation: 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.