This set of Computational Fluid Dynamics MCQs focuses on “Direct Numerical Solution for Turbulent Models”.

1. Which of these equations is the starting point of the DNS method?

a) Continuity and momentum equations of homogeneous turbulent flow

b) Continuity and momentum equations of incompressible turbulent flow

c) Momentum and energy equations of incompressible turbulent flow

d) Momentum and energy equations of homogeneous turbulent flow

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Explanation: The instantaneous continuity and Navier-Stokes equations (momentum equations) for an incompressible turbulent flow form the initial point of the Direct Numerical Solution method. These equations form a closed set of four equations with the four unknowns (pressure and three components of velocity).

2. DNS can solve _____________

a) transient 3-D equations

b) steady-state 3-D equations

c) transient 2-D equations

d) steady-state 2-D equations

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Explanation: The starting set of equations is taken by the DNS system and the transient 3-D solution is done for a sufficiently fine spatial mesh and sufficiently small time-step sizes to resolve even the smallest turbulent eddies.

3. The grid size and time-step size of the DNS method depends upon the _____________

a) Schmidt number

b) Peclet number

c) Nusselt number

d) Reynolds number

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Explanation: The grid size and time-step size of the DNS method is based on the largest and the smallest length and time scales of eddies in a turbulent flow. This, in turn, depends on the Reynolds number of the flow. So, Reynolds number decides the grid size and time-step size here.

4. When is the DNS method apt?

a) For complex flows

b) For design purposes

c) For precise details

d) For economic simulation

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Explanation: The DNS method is used when we want the precise simulation with all the details in it. They are used for the development and validation of other turbulent models. The transport of any flow variable at any point can be precisely obtained using this model.

5. For which of these purposes, the DNS method is not suitable?

a) Designing

b) Simulation of the production of aerodynamic noise

c) Effects of compressibility on turbulence

d) To understand the mechanism of turbulence

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Explanation: DNS method cannot be used to make the aerodynamic design of a model. They are computationally very expensive. To employ a model in designing, the same simulation should be done many times to improve the design. As DNS models are expensive, they cannot be used repeatedly.

6. DNS method is applicable for __________

a) Complex geometry and low Reynolds number

b) Simple geometry and low Reynolds number

c) Simple geometry and high Reynolds number

d) Complex geometry and high Reynolds number

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Explanation: The DNS method needs highly refined grids to capture the detailed flow. So, it is not suitable for complex geometries as it will make the system more complicated. As the Reynolds number increases, the number of grids and time-steps also will increase. This will lead to higher complexity again.

7. Which of these is the simplest type of turbulent flows?

a) Homogeneous anisotropic turbulence

b) Incompressible turbulent flows

c) Homogeneous isotropic turbulence

d) Compressible turbulent flows

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Explanation: The homogeneous isotropic turbulence is the simplest model of turbulence problems. It just needs a uniform grid to simulate the flow. However, the number of grids depends upon the Reynolds number of the flow.

8. The Reynolds number of a 3-D turbulent flow is 10^{4}. What is the number of grid points needed?

a) 10^{3}

b) 10^{4}

c) 10^{6}

d) 10^{9}

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Explanation: The ratio of the largest to the smallest length scale gives the number of grid points needed. The number of grid points needed in each direction is 10

^{3}. The total number of grid points needed in all three directions (as the problem is 3-D) is 10

^{9}.

9. The Reynolds number of a 3-D turbulent flow is 10^{4}. What is the number of time-steps needed?

a) 100

b) 1000

c) 10^{4}

d) 10^{5}

View Answer

Explanation: The number of time steps depends on the ratio of the largest to the smallest time scales. This ratio depends on the Reynolds number. The times steps needed is Re

^{1/2}. Here, it is (10

^{4})

^{1/2}, which is equal to 100. 100-time steps are needed to solve this problem.

10. Which of these methods which are used for time advance discretization in DNS needs more computation per time step?

a) Adams-Bashforth

b) Runge-Kutta method

c) Leapfrog

d) Newton Raphson method

View Answer

Explanation: Runge-Kutta method is generally used to advance in time in the DNS method. They need more computation time per unit step. In spite of this disadvantage, they are preferred because of their accuracy.

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