Computational Fluid Dynamics Questions and Answers – Incompressible Flows – Pressure Correction Equation

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This set of Computational Fluid Dynamics Multiple Choice Questions & Answers (MCQs) focuses on “Incompressible Flows – Pressure Correction Equation”.

1. The pressure correction equation is used to ensure _________
a) energy conservation
b) velocity conservation
c) momentum conservation
d) mass conservation
View Answer

Answer: d
Explanation: The pressure correction uses a pressure equation to make sure that mass is conserved at each time step. This is to get the correct solution from the assumed initial values at that time step.
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2. Consider a one-dimensional flow with two bounding faces in the eastern (e) and the western sides (w). Applying pressure correction to the mass conservation equation, which of these equations will be obtained?
(Note: \(\dot{m}\) represents the mass flow rate and the signs * and ‘ represent the initial guess and the correction terms respectively).
a) \(\dot{m}_{e}^{‘}+\dot{m}_{w}^{‘}=+\dot{m_e}*+\dot{m}_{w}^{*}\)
b) \(\dot{m}_{w}^{‘}=-\dot{m}_{e}^{*}\)
c) \(\dot{m}_{e}^{‘}+\dot{m}_{w}^{‘}=-\dot{m_e}*-\dot{m}_{w}^{*}\)
d) \(\dot{m}_{e}^{‘}=-\dot{m}_{e}^{*}\)
View Answer

Answer: c
Explanation: The correction in any flow variable is given by
Φ=Φ*+Φ’
By the continuity equation,
\(\dot{m_e}+\dot{m_w}=0\)
\(\dot{m}_{e}^{‘}+\dot{m}_{e}^{*}+\dot{m}_{w}^{‘}+\dot{m}_{w}^{*}=0\)
Therefore,
\(\dot{m}_{e}^{‘}+\dot{m}_{w}^{‘}=-\dot{m_e}*-\dot{m}_{w}^{*}\).

3. In the incompressible flows, the correction implies a correction in _________
a) momentum
b) velocity
c) mass
d) density
View Answer

Answer: b
Explanation: In general,
\(\dot{m}=\rho uA\)
A correction in mass flow rate cannot be the correction in density as density is constant in the incompressible flows. It cannot be a correction in area also as it is geometrically defined by the grids. So, it must be a correction in velocity.

4. State the condition obtained by applying the correction to the continuity equation.
a) When the mass flow rate reaches an exact solution, the correction field becomes zero
b) When the velocity reaches an exact solution, the correction field becomes zero
c) When the mass flow rate reaches an exact solution, the correction field becomes infinity
d) When the velocity reaches an exact solution, the correction field becomes infinity
View Answer

Answer: a
Explanation: The correction equation obtained for the mass flow rate is
\(\dot{m}_{e}^{‘}+\dot{m}_{w}^{‘}=-\dot{m_e}*-\dot{m}_{w}^{*}\)
When the solution reaches the exact answers,
\(-\dot{m}_{e}^{*}-\dot{m}_{w}^{*}=\dot{m_e}*+\dot{m}_{w}^{*}=0\)
Therefore, the correction field
\(\dot{m}_{e}^{‘}+\dot{m}_{w}^{‘}=0\).

5. The continuity equation drives the correction field of __________
a) density
b) velocity
c) pressure
d) energy
View Answer

Answer: b
Explanation: The correction field becomes zero when the mass flow rate values satisfy the continuity equation. This is useful only for the velocity correction as the mass flow rate correction and velocity correction means the same.
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6. The momentum equation drives the correction field of __________
a) density
b) temperature
c) pressure
d) energy
View Answer

Answer: c
Explanation: To establish the correction field for pressure, the mass flow rate cannot be used. So, the pressure correction field is defined by the momentum equation. But, the momentum equation is used to get the initial guess of velocities.

7. In which of these terms of the momentum equation will the correction have no impact?
a) Diffusion terms
b) Source terms
c) Velocity terms
d) Surface flux terms
View Answer

Answer: d
Explanation: Corrections based on the surface fluxes become zero when the solution converges. The other terms of the momentum equation do not become zero here. Therefore, the surface fluxes have no impact on the final solution.

8. The pressure used to find the velocities from the momentum equations is of __________
a) the previous time step
b) the oldest value
c) the latest value
d) the current time step
View Answer

Answer: a
Explanation: The momentum equations are used to get the initial guesses of the velocities. To get these velocities, other terms in the equation should be known. As the pressure at the current time step is unknown, the pressure values of the previous time step are used.

9. The correction in the velocity field is used to _____________
a) to find the pressure field of the next time step
b) correct the pressure field
c) to get the velocity field of the next time step
d) to correct the velocity field in the previous iteration
View Answer

Answer: b
Explanation: Pressure correction equation resembles the predictor-corrector method. The velocities are corrected until they satisfy the continuity equation. Then the corrected velocity is used to correct the pressure field of the previous time step.
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10. The pressure correction is an __________
a) explicit time-independent method
b) implicit time-independent method
c) implicit time-dependent method
d) explicit time-dependent method
View Answer

Answer: c
Explanation: The pressure correction method is based on time. It uses time steps to move towards a steady solution for steady problems. If the problem is transient it uses time steps to move to the desired interval of time. It is an implicit method as it solves all the equations simultaneously.

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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