This set of Computational Fluid Dynamics Questions and Answers for Entrance exams focuses on “FVM for Multi-dimensional Steady State Diffusion”.

1. Which of these equations represent the semi-discretized equation of a 2-D steady-state diffusion problem?

a) \(\int_A(\Gamma A\frac{\partial \phi}{\partial x})dA+\int_A(\Gamma A \frac{\partial\phi}{\partial y}) dA+\int_{\Delta V} S\,dV=0\)

b) \(\int_A\frac{\partial}{\partial x}(\Gamma A \frac{\partial\phi}{\partial x})dA+\int_A\frac{\partial}{\partial y}(\Gamma A\frac{\partial\phi}{\partial y})dA+\int_{\Delta V}S\, dV=0\)

c) \(\int_A(\Gamma A\frac{d\phi}{dx})dA+\int_A(\Gamma A \frac{d\phi}{dy})dA+\int_{\Delta V}S\, dV=0\)

d) \(\frac{\partial \phi}{\partial t}+\int_A\frac{\partial}{\partial x}(\Gamma A \frac{\partial \phi}{\partial x}) dA+\int_A\frac{\partial}{\partial y}(\Gamma A \frac{\partial \phi}{\partial y})dA+\int_{\Delta V}S\, dV=0\)

View Answer

Explanation: The general governing equation for a 2-D steady-state diffusion problem is given by

\(\frac{\partial}{\partial x}(\Gamma\frac{\partial \phi}{\partial x})+\frac{\partial}{\partial y}(\Gamma\frac{\partial \phi}{\partial y})+S=0\)

Here, partial differentiation is used as the variable φ varies in both x and y directions, but the differentiation is only in the required direction.

Integrating the equation with respect to the control volume,

\(\int_{\delta V}\frac{\partial}{\partial x}(\Gamma\frac{\partial\phi}{\partial x})dV+\int_{\delta V}\frac{\partial}{\partial y}(\Gamma\frac{\partial \phi}{\partial y})dV+\int_{\Delta V} S \,dV=0\)

Applying Gauss Divergence theorem,

\(\int_A(\Gamma A\frac{\partial\phi}{\partial x})dA+\int_A(\Gamma A\frac{\partial\phi}{\partial y})dA+\int_{\Delta V}S \,dV=0\)

This is the semi-discretized form of the equation.

2. The area in the western face of a 2-D steady-state diffusion stencil (uniform) is _______________

a) grid size in the x-direction

b) grid size in the y-direction

c) product of the grid sizes in the x and y-directions

d) ratio of the grid sizes in the x and y-directions

View Answer

Explanation: In the one-dimensional case, the area is taken to be unity. In the two-dimensional case, the area is the grid size in the perpendicular direction multiplied by unity. So, for area A

_{e}=A

_{w}=Δy and A

_{n}=A

_{s}=Δx.

3. Consider the following stencil.

What is the flux across the northern face?

a) \(\Gamma_Na_N\frac{(\phi _N-\phi _P)}{\delta x_{PN}}\)

b) \(\Gamma_Na_N\frac{(\phi _N-\phi _P)}{\delta y_{PN}}\)

c) \(\Gamma_Na_N\frac{(\phi _E-\phi _P)}{\delta y_{PN}}\)

d) \(\Gamma_Na_N\frac{(\phi _E-\phi _P)}{\delta x_{PN}}\)

View Answer

Explanation: Flux across the northern face is \(\Gamma_N a_N\frac{\partial\phi}{\partial y}\Big|_n\). Expanding this using the central difference scheme, we get

\(\Gamma_N a_N\frac{\partial\phi}{\partial y}\Big|_n = \Gamma_Na_N \frac{(\phi _N-\phi _P)}{\delta y_PN}\).

4. Consider the following stencil.

For a source-less 2-D steady-state diffusion problem, the coefficient of the flow variable Φ_{P} is ____

a) \(\frac{\Gamma_W A_W}{\delta x_{WP}}+\frac{\Gamma_E A_E}{\delta x_{PE}}+\frac{\Gamma_S A_S}{\delta y_SP}+\frac{\Gamma_N A_N}{\delta y_{PN}}\)

b) \(\frac{\Gamma_W A_W}{\delta y_{WP}}+\frac{\Gamma_E A_E}{\delta y_{PE}}+\frac{\Gamma_S A_S}{\delta x_SP}+\frac{\Gamma_N A_N}{\delta x_{PN}}\)

c) \(\frac{\Gamma_W A_W}{\delta y_{WP}}+\frac{\Gamma_S A_S}{\delta y_{SP}}+\frac{\Gamma_E A_E}{\delta x_{PE}}+\frac{\Gamma_N A_N}{\delta x_{PN}}\)

d) \(\frac{\Gamma_W A_W}{\delta x_{WP}}+\frac{\Gamma_S A_S}{\delta x_{SP}}+\frac{\Gamma_E A_E}{\delta y_{PE}}+\frac{\Gamma_N A_N}{\delta y_{PN}}\)

View Answer

Explanation: The general form is given by a

_{P}Φ

_{P}=a

_{E}Φ

_{P}+a

_{W}Φ

_{W}+a

_{N}Φ

_{N}+a

_{S}Φ

_{S}

Here, for source-less problem, a

_{P}is the addition of all fluxes given by

\(\frac{\Gamma_W A_W}{\delta x_{WP}}+\frac{\Gamma_E A_E}{\delta x_{PE}}+\frac{\Gamma_S A_S}{\delta y_SP}+\frac{\Gamma_N A_N}{\delta y_{PN}}\).

5. If a_{P}Φ_{P}=a_{E}Φ_{P}+a_{W}Φ_{W}+a_{N}Φ_{N}+a_{S}Φ_{S}+S is the general form of a 2-D steady-state diffusion problem, what is a_{E} by considering the following stencil?

a) \(\frac{\Gamma_E A_E}{\delta y_{PE}}\)

b) \(\frac{\Gamma_E A_E}{\delta y_{PE}}\)

c) \(\frac{\Gamma_E A_E}{\delta x_{PE}}\)

d) \(\frac{\Gamma_E A_E}{\delta x_{WP}}\)

View Answer

Explanation: Flux in the eastern direction is given by

\(\Gamma_E A_E\frac{\partial\phi}{\partial x}\Big|_e=\Gamma_E A_E\frac{(\phi _E-\phi _P)}{\delta x_{PE}}\)

\(\Gamma_E A_E\frac{\partial\phi}{\partial x}\Big|_e=\Gamma _e A_E\frac{\phi_E}{\delta x_{PE}}-\Gamma_E a_E\frac{\phi_P}{\delta x_{PE}}\)

Expanding this while forming the general equation, we will get

\(a_E=\frac{\Gamma_E A_E}{\delta x_{PE}}\).

6. Consider the following 2-D surface with the numbers inside as the global indices of their cells.

1 | 2 | 3 | 4 |

5 | 6 | 7 | 8 |

9 | 10 | 11 | 12 |

13 | 14 | 15 | 16 |

The general discretized equation is of the form a_{P}Φ_{P}=a_{E}Φ_{P}+a_{W}φ_{W}+a_{N}Φ_{N}+a_{S}Φ_{S}+S. Which of the following is correct regarding the cell numbered “13”?

a) a_{E}=0; a_{W}=0

b) a_{W}=0; a_{N}=0

c) a_{N}=0; a_{S}=0

d) a_{S}=0; a_{W}=0

View Answer

Explanation: For the control volumes adjacent to the boundary of the global domain, the boundary-side coefficient is set to zero. Therefore, for the cell numbered “13”, the southern and the western coefficients are zero (a

_{S}=0; a

_{W}=0).

7. I general, for all the steady-state diffusion problems, the discretized equation can be given as a_{P}Φ _{P} = ∑a_{nb}Φ_{nb}-S. For a one-dimensional problem, which of these is wrong?

a) ∑a_{nb} =a_{T}+a_{B}

b) ∑a_{nb} =a_{S}+ a_{N}

c) ∑a_{nb} =a_{W}+a_{E}

d) ∑a_{nb} =a_{P}+a_{E}

View Answer

Explanation: For a one-dimensional problem is x-direction, ∑a

_{nb}=a

_{W}+a

_{E}. For a one-dimensional problem is y-direction, ∑a

_{nb}=a

_{S}+ a

_{N}. For a one-dimensional problem is z-direction, ∑a

_{nb}=a

_{T}+a

_{B}.

8. In a control volume adjacent to the boundary, the flux crossing the boundary is _______________ in the discretized equation.

a) set to some arbitrary constant

b) set to zero

c) introduced as a source term

d) introduced as a convective flux

View Answer

Explanation: As the boundary-side coefficients are set to zero in the discretized equations of the boundary-based control volumes, the information in the boundary may be lost. To avoid this, the flux crossing the boundary is introduced as a source term in the equation.

9. Consider a source-less 3-D steady-state diffusion problem. The general discretized equation is a_{P} Φ_{P} = ∑a_{nb} Φ_{nb}. What is a_{P}?

a) a_{P}=a_{W}+a_{E}+a_{S}+a_{N}+a_{T}+a_{B}

b) a_{P}=a_{W}+a_{E}+a_{S}+a_{N}

c) a_{P}=a_{W}+a_{E}+a_{S}+a_{N}+a_{T}

d) a_{P}=0

View Answer

Explanation: For all steady-state diffusion problems, in the absence of source term, a

_{P}=∑a

_{nb}. Therefore, for the three-dimensional case, a

_{P}=a

_{W}+a

_{E}+a

_{S}+a

_{N}+a

_{T}+a

_{B}which includes the coefficients of all the neighbouring flow variables.

10. Consider the stencil.

The values of \(\vec{A_w}\, and\, \vec{A_s}\) are _____________

a) \(\vec{A_w}=\Delta x; \vec{A_s}=\Delta y\)

b) \(\vec{A_w}=-\Delta x; \vec{A_s}=-\Delta y\)

c) \(\vec{A_w}=-\Delta y; \vec{A_s}=-\Delta x\)

d) \(\vec{A_w}=\Delta y; \vec{A_s}=\Delta x\)

View Answer

Explanation: The values of A

_{w}and A

_{s}are Δ x and Δ y respectively. The signs of the area vectors depend on their directions. Therefore, \(\vec{A_w}=-\Delta x; \vec{A_s}=-\Delta y\).

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