This set of Computational Fluid Dynamics Multiple Choice Questions & Answers (MCQs) focuses on “Finite Volume Method”.
1. 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
View Answer
Explanation: 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. 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
View Answer
Explanation: 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. 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
View Answer
Explanation: 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. 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 ∫Sfd\(\vec{S}\)?
a) \(\Sigma_{k=1}^4 \int_{S_k} f d\vec{S}\)
b) \(\Sigma_{k=1}^2 \int_{S_k} f d\vec{S}\)
c) \(\Sigma_{k=1}^6 \int_{S_k} f d\vec{S}\)
d) \(\Sigma_{k=1}^8 \int_{S_k} f d\vec{S}\)
View Answer
Explanation: 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. Approximate the surface integral ∫Snfd\(\vec{S}\) using the midpoint rule.
a) fn Sn
b) Sn (fne+fnw)
c) \(\frac{S_n}{2}\) (fne+fnw)
d) \(\frac{S_n}{2}\) fn
View Answer
Explanation: 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,
∫Snfd\(\vec{S}\)=fnSn
6. Approximate the surface integral in the eastern face ∫Sefd\(\vec{S}\) of a two-dimensional problem using the trapezoidal rule.
a) \(\frac{3}{2}\)(fne+fse)
b) 3 \(\frac{S_e}{2}\)(fne+fse)
c) \(\frac{1}{2}\)(fne+fse)
d) \(\frac{S_e}{2}\) (fne+fse)
View Answer
Explanation: 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.
∫Sefd\(\vec{S}=\frac{1}{2}\) (fne+fse).
7. 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
View Answer
Explanation: 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. 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
View Answer
Explanation: 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. 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
View Answer
Explanation: 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. Approximate the surface integral ∫Swf d\(\vec{S}\) using the Simpson’s rule.
a) \(\frac{S_w}{6}\)(2fnw+2fw+2fsw)
b) \(\frac{S_w}{4}\)(2fnw+2fsw)
c) \(\frac{S_w}{6}\)(fnw+4fw+fsw)
d) \(\frac{S_w}{4}\)(fnw+2fw+fsw)
View Answer
Explanation: 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
\(\frac{S_w}{6}\)(fnw+4fw+fsw).
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