Engineering Questions with Answers - Multiple Choice Questions
Computational Fluid Dynamics – Convection-Diffusion Problems – Second Order Upwind Scheme
1 - Question
The Second Order Upwind (SOU) scheme uses ____________
a) asymmetric linear profile
b) symmetric linear profile
c) asymmetric quadratic profile
d) symmetric quadratic profile
View Answer
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 - Question
The value at the face in the second order upwind scheme is calculated using _____________
a) interpolation
b) extrapolation
c) weighted average
d) geometric mean
View Answer
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 - Question
The second-order upwind scheme is ___________ than the general upwind scheme.
a) less diffusive
b) more diffusive
c) less accurate
d) less stable
View Answer
Explanation: The second-order upwind scheme is more accurate (second-order) accurate than the general (first-order) upwind scheme. But, it is less diffusive when compared to the general upwind scheme.
4 - Question
Consider the stencil.
What is φe according to the second-order upwind scheme?
(Note: φ is the flow variable).
a) ϕe=ϕP−ϕP−ϕWxP−xW(xe−xw)
b) ϕe=ϕP+ϕP−ϕWxP−xW(xe−xw)
c) ϕe=ϕP−ϕP−ϕWxP−xW(xe−xc)
d) ϕe=ϕP+ϕP−ϕWxP−xW(xe−xc)
View Answer
Answer: d
Explanation: The second-order upwind scheme approximates the variation to be linear and uses extrapolation for finding the values.
ϕe=ϕP+ϕP−ϕWxP−xW(xe−xc).
5 - Question
Consider the stencil.
Assume a uniform grid. What is φe according to the second-order upwind scheme?
(Note: φ is the flow variable).
a) ϕe=ϕP−ϕW2
b) ϕe=ϕP+ϕW2
c) ϕe=32ϕP−12ϕW
d) ϕe=32ϕP+12
View Answer
Answer: c
Explanation: In general, from the second-order upwind scheme,
ϕe=ϕP+ϕP−ϕWxP−xW(xe−xc)
For a uniform grid,
(xe−xc)(xP−xW)=12
Therefore,
ϕe=ϕP+ϕP−ϕW2=ϕe=32ϕP−12ϕW.
6 - Question
Consider the stencil.
Assume a uniform grid. What is mw˙ϕwv according to the second-order upwind scheme?
(Note: m˙ and φ are the mass flow rate and flow variable).
a) mw˙ϕw=(32ϕP−12ϕW)max(mw˙,0)+(32ϕW−12ϕWW)max(−mw˙,0)
b) mw˙ϕw=(32ϕP−12ϕW)max(mw˙,0)−(32ϕW−12ϕWW)max(−mw˙,0)
c) mw˙ϕw=(32ϕP−12ϕW)max(−mw˙,0)+(32ϕW−12ϕWW)max(−mw˙,0)
d) mw˙ϕw=(32ϕP−12ϕW)max(−mw˙,0)−(32ϕW−12ϕWW)max(−mw˙,0)
View Answer
Answer: b
Explanation: According to the second-order upwind scheme,
ϕw={(32ϕP−12ϕW)(32ϕW−12ϕWW)mw˙>0mw˙<0 Therefore,
mw˙ϕw=(32ϕP−12ϕW)max(mw˙,0)−(32ϕW−12ϕWW)max(−mw˙,0).
7 - Question
What is the first term in the truncation error of the second-order upwind scheme?
(Note: φP is the flow variable at the central node).
a) −38(Δx)2ϕP”′
b) −38(Δx)ϕP”′
c) −38(Δx)2ϕP”
d) −38(Δx)ϕP”
View Answer
Answer: a
Explanation: The truncation error can be obtained by using the exact solution (Taylor Series expansion) of the gradients. Since the scheme is second-order accurate, the first term of the error should have (Δx)2 in it. Associated with this we have, ΦP”’. Therefore, the term is −38(Δx)ϕP”′.
8 - Question
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
View Answer
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 - Question
Find the normalized functional relationship between φf and φC for a uniform grid while using the second-order upwind scheme?
a) ϕf~=12ϕC~
b) ϕf~=−12ϕC~
c) ϕf~=32ϕC~
d) ϕf~=−32ϕC~
View Answer
Answer: c
Explanation: The relationship between φf and φC in the second-order upwind scheme is
ϕf=32ϕC−12ϕU
After normalizing, φf, φC and φU becomes ϕf~,ϕC~ and 0 respectively. Therefore,
ϕf~=32ϕC~.
10 - Question
The flux limiter Ψ(r) of the second-order upwind scheme is __________
a) r2
b) 12r
c) 2r
d) r
View Answer
Answer: d
Explanation: To find the flux limiter,
ϕf=ϕC+12ψ(r)(ϕD−ϕC)
For the second order upwind scheme,
ϕf=32ϕC−112ϕU
Equating both,
12ψ(r)(ϕD−ϕC)=32ϕC−ϕC−12ϕU
12ψ(r)(ϕD−ϕC)=12(ϕC−ϕU)
ψ(r)=(ϕC−ϕU)(ϕD−ϕC)=r.