Computational Fluid Dynamics Questions and Answers – Convection-Diffusion Problems – Second Order Upwind Scheme

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This set of Computational Fluid Dynamics Multiple Choice Questions & Answers (MCQs) focuses on “Convection-Diffusion Problems – Second Order Upwind Scheme”.

1. 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

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.
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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
View Answer

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
View Answer

Answer: a
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. Consider the stencil.
computational-fluid-dynamics-questions-answers-second-order-upwind-scheme-q5
What is φe according to the second-order upwind scheme?
(Note: φ is the flow variable).
a) \(\phi_e=\phi_P-\frac{\phi_P-\phi_W}{x_P-x_W}(x_e-x_w)\)
b) \(\phi_e=\phi_P+\frac{\phi_P-\phi_W}{x_P-x_W}(x_e-x_w)\)
c) \(\phi_e=\phi_P-\frac{\phi_P-\phi_W}{x_P-x_W}(x_e-x_c)\)
d) \(\phi_e=\phi_P+\frac{\phi_P-\phi_W}{x_P-x_W}(x_e-x_c)\)
View Answer

Answer: d
Explanation: The second-order upwind scheme approximates the variation to be linear and uses extrapolation for finding the values.
\(\phi_e=\phi_P+\frac{\phi_P-\phi_W}{x_P-x_W}(x_e-x_c)\).

5. Consider the stencil.
computational-fluid-dynamics-questions-answers-second-order-upwind-scheme-q5
Assume a uniform grid. What is φe according to the second-order upwind scheme?
(Note: φ is the flow variable).
a) \(\phi_e=\frac{\phi_P-\phi_W}{2}\)
b) \(\phi_e=\frac{\phi_P+\phi_W}{2}\)
c) \(\phi_e=\frac{3}{2}\phi_P-\frac{1}{2}\phi_W\)
d) \(\phi_e=\frac{3}{2}\phi_P+\frac{1}{2}\phi_W\)
View Answer

Answer: c
Explanation: In general, from the second-order upwind scheme,
\(\phi_e=\phi_P+\frac{\phi_P-\phi_W}{x_P-x_W}(x_e-x_c)\)
For a uniform grid,
\(\frac{(x_e-x_c)}{(x_P-x_W)}=\frac{1}{2}\)
Therefore,
\(\phi_e=\phi_P+\frac{\phi_P-\phi_W}{2}=\phi_e=\frac{3}{2} \phi_P-\frac{1}{2}\phi_W\).
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6. Consider the stencil.
computational-fluid-dynamics-questions-answers-second-order-upwind-scheme-q5
Assume a uniform grid. What is \(\dot{m_w} \phi_{wv}\) according to the second-order upwind scheme?
(Note: \(\dot{m}\) and φ are the mass flow rate and flow variable).
a) \(\dot{m_w}\phi_w=(\frac{3}{2}\phi_P-\frac{1}{2}\phi_W)max⁡(\dot{m_w},0)+(\frac{3}{2}\phi_W-\frac{1}{2}\phi_{WW}) max⁡(-\dot{m_w},0)\)
b) \(\dot{m_w}\phi_w=(\frac{3}{2}\phi_P-\frac{1}{2}\phi_W)max⁡(\dot{m_w},0)-(\frac{3}{2}\phi_W-\frac{1}{2}\phi_{WW}) max⁡(-\dot{m_w},0)\)
c) \(\dot{m_w}\phi_w=(\frac{3}{2}\phi_P-\frac{1}{2}\phi_W)max⁡(-\dot{m_w},0)+(\frac{3}{2}\phi_W-\frac{1}{2}\phi_{WW}) max⁡(-\dot{m_w},0)\)
d) \(\dot{m_w}\phi_w=(\frac{3}{2}\phi_P-\frac{1}{2}\phi_W)max⁡(-\dot{m_w},0)-(\frac{3}{2}\phi_W-\frac{1}{2}\phi_{WW}) max⁡(-\dot{m_w},0)\)
View Answer

Answer: b
Explanation: According to the second-order upwind scheme,
\(\phi_w=\begin{cases}
(\frac{3}{2}\phi_P-\frac{1}{2}\phi_W) & \dot{m_w}>0 \\
(\frac{3}{2}\phi_W-\frac{1}{2}\phi_{WW}) & \dot{m_w}<0
\end{cases}\) Therefore,
\(\dot{m_w}\phi_w=(\frac{3}{2}\phi_P-\frac{1}{2}\phi_W)max⁡(\dot{m_w},0)-(\frac{3}{2}\phi_W-\frac{1}{2}\phi_{WW}) max⁡(-\dot{m_w},0)\).

7. 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) \(-\frac{3}{8}(\Delta x)^2 \phi_P”’\)
b) \(-\frac{3}{8}(\Delta x) \phi_P”’\)
c) \(-\frac{3}{8}(\Delta x)^2 \phi_P”\)
d) \(-\frac{3}{8}(\Delta x) \phi_P”\)
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 \(-\frac{3}{8}(\Delta x) \phi_P”’\).

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
View Answer

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 φf and φC for a uniform grid while using the second-order upwind scheme?
a) \(\tilde{\phi_f}=\frac{1}{2}\tilde{\phi_C}\)
b) \(\tilde{\phi_f}=-\frac{1}{2}\tilde{\phi_C}\)
c) \(\tilde{\phi_f}=\frac{3}{2}\tilde{\phi_C}\)
d) \(\tilde{\phi_f}=-\frac{3}{2}\tilde{\phi_C}\)
View Answer

Answer: c
Explanation: The relationship between φf and φC in the second-order upwind scheme is
\(\phi_f=\frac{3}{2}\phi_C-\frac{1}{2}\phi_U\)
After normalizing, φf, φC and φU becomes \(\tilde{\phi_f}, \tilde{\phi_C}\) and 0 respectively. Therefore,
\(\tilde{\phi_f}=\frac{3}{2}\tilde{\phi_C}\).
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10. The flux limiter Ψ(r) of the second-order upwind scheme is __________
a) r2
b) \(\frac{1}{2}r\)
c) 2r
d) r
View Answer

Answer: d
Explanation: To find the flux limiter,
\(\phi_f=\phi_C+\frac{1}{2}\psi(r)(\phi_D-\phi_C)\)
For the second order upwind scheme,
\(\phi_f=\frac{3}{2}\phi_C-1\frac{1}{2}\phi_U\)
Equating both,
\(\frac{1}{2}\psi(r)(\phi_D-\phi_C)=\frac{3}{2}\phi_C-\phi_C-\frac{1}{2}\phi_U\)
\(\frac{1}{2}\psi(r)(\phi_D-\phi_C)=\frac{1}{2}(\phi_C-\phi_U)\)
\(\psi(r)=\frac{(\phi_C-\phi_U)}{(\phi_D-\phi_C)}=r\).

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn