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# Computational Fluid Dynamics – Numerical Methods – Variable Arrangements and Velocity Components

1 - Question

Which of these coordinate systems will best suit to model the flow in a pipe with a swirl?

a) Cylindrical coordinates

b) Cartesian coordinates

c) Spherical coordinates

d) Polar coordinates

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Answer: aExplanation: In the case of flow over a pipe, cylindrical coordinates do the best as the dependent spatial variables are just 2 which would be 3 if it is Cartesian coordinates. But, in this case, a swirl component is included which makes the flow dependent on three dimensions anyway. To keep the complexities less, Cartesian coordinates should be chosen for this problem.

2 - Question

While using grid oriented velocity components, conservation of which of these equations will be lost?

a) Energy equation

b) Euler equations

c) Momentum equation

d) Continuity equation

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Answer: cExplanation: If grid-oriented velocity components are used, non-conservative source or sink terms will appear in Momentum equations. This will affect the conservation of Momentum equation. Euler equations is a set including all three – continuity, momentum, energy equations.

3 - Question

In the polar-cylindrical momentum equation for the r-component, the term ρvθ2r represents ___________

a) Centrifugal force

b) Centripetal force

c) Tangential force

d) Coriolis force

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Answer: aExplanation: When the momentum equation is expressed in cylindrical terms, due to the transformation from Cartesian coordinates to polar coordinates, a centrifugal term arises. This is the term ρvθ2r. It describes the transfer of θ-momentum into r-momentum due to the change of direction of angular velocity.

4 - Question

Which of these terms in the momentum equation for θ – component represents the Coriolis force?

a) ρvr∂vθ∂r

b) ρvθr∂vθ∂θ

c) ρvrvθr

d) ∂vθ∂r

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Answer: cExplanation: The term ρvrvθr represents Coriolis force in the θ – momentum equation. Coriolis force arises due to the motion of the coordinate system taken. The term ρvrvθr represents the source or sink of θ -momentum.

5 - Question

Curvilinear coordinates do not suit __________

a) Unstructured grids

b) Structured grids

c) Orthogonal grids

d) Non-orthogonal grids

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Answer: aExplanation: For curvilinear coordinates, the grid should be very smooth and the change of grid direction from point to point must be very small. Since the grid direction in the unstructured grids varies much, they do not suit unstructured grids.

6 - Question

While changing the coordinates from Cartesian to non-orthogonal in the Finite Difference Method, which of the following remains the same?

a) all the terms in the equation

b) conservation properties of the equation

c) number of terms in the equation

d) source terms of the equation

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Answer: bExplanation: In the Finite Difference Method, while changing from Cartesian to non-orthogonal grids, the terms change and the number of terms increases. But, the conservation properties of the equation remain the same.

7 - Question

Staggered grid arrangements are used to establish strong coupling between ____________ and __________

a) velocities and momentum components

b) velocities and pressure gradients

c) velocities and viscosity terms

d) viscosity and source terms

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Answer: dExplanation: Coupling problem between velocities and pressure gradients is the reason why staggered grids are formed. Staggered grids ensure the coupling between these two terms. The velocity components normal to the cell face should lie between the pressure coordinates on either side of that face.

8 - Question

Which of these arrangements of the velocity and pressure gradient components is suitable for non-orthogonal grids?

a) Staggered arrangement with covariant base

b) Collocated arrangement

c) Staggered arrangement

d) Covariant arrangement

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Answer: bExplanation: The collocated arrangement is the simplest as all the variables share the same control volume. But this requires more interpolation. When the grid is non-orthogonal, the collocated arrangement is the best as the other arrangements are more difficult.

9 - Question

Which of these flow properties does not suit covariant or contra-variant bases?

a) Velocity

b) Density

c) Stress

d) Viscous forces

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Answer: bExplanation: Covariant or contra-variant bases can be used only with vectors or tensors. They cannot be used with scalars. Here, Velocities stress and forces all come under either vectors or tensors. Density is a scalar property. So, it cannot be represented using covariant or contra-variant bases.

10 - Question

Stress components cannot be expressed in terms of __________

a) Cartesian tensors

b) Contra-variant tensors

c) Covariant tensors

d) Metric tensors

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Answer: dExplanation: In general, Stress components are expressed in the Cartesian tensors form. It is also possible to define them in covariant or contra-variant terms. But they cannot be expressed in metric tensor. Metric tensors are used to transform covariant tensors into contra-variant tensors.