Engineering Questions with Answers - Multiple Choice Questions

Home » MCQs » Aeronautical Engineering » Computational Fluid Dynamics – Approaches for Non-uniform Time Steps

# Computational Fluid Dynamics – Approaches for Non-uniform Time Steps

1 - Question

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

**
View Answer**

Answer: cExplanation: 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 - Question

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

**
View Answer**

Answer: cExplanation: 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 - Question

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

**
View Answer**

Answer: dExplanation: 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 - Question

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

**
View Answer**

Answer: aExplanation: 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 - Question

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?

(Note: ρ is the density and V the volume).

a) Δt−ΔtoΔt+ΔtoρCVC

b) Δt−ΔtoΔt+ΔtoρoCVC

c) Δt−ΔtoΔt+ΔtoρoC

d) Δt+ΔtoΔt−ΔtoρoCVC

**
View Answer**

Answer: bExplanation: 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. aoC=Δt−ΔtoΔt+ΔtoρoCVC This becomes zero when a uniform transient grid is used.

6 - Question

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?

(Note: ρ is the density and V the volume).

a) ΔtoΔt(Δt+Δto)ρooCVC

b) Δt(Δt+Δto)Δt×ΔtoρooCVC

c) Δt+2Δto(Δt+Δto)ρooCVC

d) ΔtΔto(Δt+Δto)ρooCVC

**
View Answer**

Answer: dExplanation: 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 aooC=ΔtΔto(Δt+Δto)ρooCVC.

7 - Question

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?

(Note: ρ is the density and V the volume).

a) (1Δto+1Δt+Δto)ρCVC

b) (1Δt+1Δtoo+Δto)ρCVC

c) (1Δt+1Δt+Δtoo)ρCVC

d) (1Δt+1Δt+Δto)ρCVC

**
View Answer**

Answer: dExplanation: 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 (1Δt+1Δt+Δto)ρCVC.

8 - Question

Which of these is correct for the non-uniform time-steps?

(Note: δt is the distance between the centroids of two temporal elements. Δt is the size of a temporal element. The superscript o indicates the older time step).

a) δt=(Δt−Δto)2

b) δt=(Δt+Δto)2

c) δt=Δt

d) δt=Δt°

**
View Answer**

Answer: bExplanation: For the variable time-steps, δt=(Δt+Δto)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 - Question

The general discretised form of the transient term using the finite volume approach is

FluxT=FluxC ΦC+FluxC°ΦC°+FluxV.

The superscript o indicates the older time step. If the Crank-Nicolson method is used with non-uniform grids, what is FluxV? a) −ΔtooΔtoρoocVCϕoocΔt

b) ΔtooΔto+ΔtooρoocVCϕoocΔt

c) −ΔtooΔto+ΔtooρoocVCϕoocΔt

d) ΔtooΔtoρoocVCϕoocΔt

**
View Answer**

Answer: cExplanation: 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=-−ΔtooΔto+ΔtooρoocVCϕoocΔt.

10 - Question

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

**
View Answer**

Answer: aExplanation: For the finite difference approach, ac=(1Δt+1Δt+Δto)ρCVC For the finite volume approach, the central coefficient is FluxC=(1Δt+1Δt+Δto)ρCVC These two are the same irrespective of the approach.