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# Computational Fluid Dynamics – Finite Volume Method

1 - Question

Which of these models will directly give the conservative equations suitable for the finite volume method?

a) Finite control volume moving along with the flow

b) Finite control volume fixed in space

c) Infinitesimally small fluid element moving along with the flow

d) Infinitesimally small fluid element fixed in space

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Answer: bExplanation: Finite volume method uses the conservation equation in the integral form without any substantial derivative. This can be given by a finite control volume fixed in space. This directly results in the conservative integral form of the governing equations.

2 - Question

Which of these terms need a surface integral?

a) Diffusion and rate of change terms

b) Convection and source terms

c) Convection and diffusion terms

d) Diffusion and source terms

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Answer: cExplanation: The convection and diffusion terms have fluxes flowing along the surfaces of the control volume. So, they have to be integrated over the faces of the control volume. So, the convection and diffusion terms need a surface integral.

3 - Question

Which of these terms need a volume integral while modelling steady flows?

a) Convection term

b) Diffusion term

c) Source term

d) Rate of change term

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Answer: cExplanation: The convection and diffusion terms need surface integrals and not volume integrals. The source and rate of change terms need volume integral. Since the flow taken is a steady flow, the rate of change term will be equal to zero. So, in this case, only the source term needs a volume integral.

4 - Question

Consider a two-dimensional flow. If f is the component of the flux vector normal to the control volume faces, which of these terms represent ∫SfdS⃗ ?

a) Σ4k=1∫SkfdS⃗

b) Σ2k=1∫SkfdS⃗

c) Σ6k=1∫SkfdS⃗

d) Σ8k=1∫SkfdS⃗

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Answer: aExplanation: In a two-dimensional flow, the number of faces bounding a control volume is four. So, the summation of the integrals along these four faces will be equal to the total flux of the control volume. Here, f may be convective or diffusive flux.

5 - Question

Approximate the surface integral ∫_{Sn}fdS⃗ using the midpoint rule.

a) f_{n} S_{n}

b) S_{n} (f_{ne}+f_{nw})

c) Sn2 (f_{ne}+f_{nw})

d) Sn2 f_{n}

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Answer: aExplanation: For surface integral, the value of the integrand at the face centre is used in the midpoint approximation. The neighbouring values are not used. So, ∫SnfdS⃗ =fnSn

6 - Question

Approximate the surface integral in the eastern face ∫_{Se}fdS⃗ of a two-dimensional problem using the trapezoidal rule.

a) 32(f_{ne}+f_{se})

b) 3 Se2(f_{ne}+f_{se})

c) 12(f_{ne}+f_{se})

d) Se2 (f_{ne}+f_{se})

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Answer: cExplanation: The trapezoidal rule is a second-order accurate approximation. It needs the values of the integrand at two points. Here, as we need the surface integral in the eastern face, the value is approximated using the northern and the southern nodes of the eastern face. ∫SefdS⃗ =12 (fne+fse).

7 - Question

In a two dimensional flow, how many terms does Simpson’s rule need to approximate a surface integral?

a) four terms

b) one term

c) two terms

d) three terms

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Answer: dExplanation: For a two-dimensional flow and the surface integral, the midpoint rule needs only one term (the value of the integrand at the face centre). The trapezoidal rule needs two terms (the value of the integrand at the nodes above and below the face centre). The Simpson’s rule needs three terms (the value of the integrand at the face centre and at the nodes above and below the face centre).

8 - Question

For three-dimensional flows, what is the approximation of the volume integral using the midpoint rule?

a) Product of the integrand at the face centre and the volume of the control volume

b) Product of the integrand at the control volume centre and the volume of the control volume

c) Product of the integrand at the control volume centre and the surface area of the control volume

d) Product of the integrand at the face centre and the surface area of the control volume

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Answer: bExplanation: Using the midpoint rule, the volume integral is approximated as the product of the integrand at the centre of the cell (the control volume) and the volume of the cell. Representing it mathematically, ∫VqdV = qP×ΔV.

9 - Question

In a one-dimensional flow, the volume integral becomes __________

a) a line integral

b) an area integral

c) a surface integral

d) a surface integral and the Gauss divergence theorem

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Answer: aExplanation: For a one-dimensional flow, the volume of the control volume (cell) is in a single dimension which is the length of that cell. So, there will be no need for a volume, area or surface integrals. It is enough to integrate over the length of that cell.

10 - Question

Approximate the surface integral ∫_{Sw}f dS⃗ using the Simpson’s rule.

a) Sw6(2f_{nw}+2f_{w}+2f_{sw})

b) Sw4(2f_{nw}+2f_{sw})

c) Sw6(f_{nw}+4f_{w}+f_{sw})

d) Sw4(f_{nw}+2f_{w}+f_{sw})

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Answer: cExplanation: The Simpson’s rule uses values of the integrand at three points – centre of the face and the two vertices in the same face. It is given by Sw6(fnw+4fw+fsw).