Engineering Questions with Answers - Multiple Choice Questions
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
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 - 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
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 - 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
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 - 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
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 - 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
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. 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
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 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
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 (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
Explanation: 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
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=-−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
Explanation: 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.