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# Computational Fluid Dynamics – Turbulence Modelling – Reynolds Averaged Navier-Stokes Model

Which of these properties of turbulence is ruled out in Reynolds averaged equations?

a) Fluctuations

b) Turbulence

c) Non-linearity

d) Randomness

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Answer: aExplanation: The flow properties of turbulent flow can be decomposed into mean and fluctuating components. These fluctuating components result in an unsteadiness in the flow. This unsteadiness is ruled out by means of Reynolds averaging.

The averaging interval in RANS equation is based on ____________

a) the grid size

b) the eddy size

c) the fluctuations

d) the time interval of the problem

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Answer: cExplanation: Time averaging is used only when the flow is steady and time-independent. The time interval taken to average the fluctuations depend upon the time scale of the fluctuations itself. If this interval is large enough, the lower limit of the integral does not even matter.

For unsteady turbulent flows, which of these averaging method is used?

a) Time averaging

b) Ensemble averaging

c) Spatial averaging

d) Volume averaging

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Answer: bExplanation: Time averaging is generally used to remove the fluctuations in RANS model. But it cannot be used when the problem is unsteady. In these cases, ensemble averaging is used to eliminate the fluctuations.

Which of these terms arise in the conservation equations when using the RANS model?

a) Reynolds stresses and turbulent scalar flux

b) Cross stresses and turbulent scalar flux

c) Leonard stresses and turbulent scalar flux

d) Leonard stresses and cross stresses

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Answer: aExplanation: While the conservation equations are Reynolds averaged, they get additional terms due to the decomposition of the flow variables. These additional terms include Reynolds stresses and turbulent scalar fluxes. This occurs because of the mean of the product of the fluctuating components.

Reynolds averaging makes the conservation equations ____________

a) non-conservative

b) non-linear

c) unstable

d) inconsistent

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Answer: bExplanation: Reynolds averaging add extra terms to the conservation equations. So, the number of unknowns becomes more than the number of equations. This leads to a linearity problem and makes the conservative equations non-linear.

How many additional terms are present in the x-momentum equation Reynolds-Averaged Navier-Stokes equations?

a) No additional terms

b) Six additional terms

c) Three additional terms

d) Two additional terms

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Answer: c

Explanation: The non-reduced x-momentum equation is

∂(ρu)∂t+div(ρuV⃗ )=−∂p∂x+div(μgrad(u))+S

The Reynolds-Averaged x-momentum equation is

∂(ρ¯¯¯u~)∂t+div(ρ¯¯¯u~(V⃗ )˜)=−∂p~∂x+div(μgrad(V⃗ )˜)+(−∂(ρ¯¯¯u‘2)¯¯¯¯¯¯¯¯¯¯¯¯¯¯∂x–∂(ρ¯¯¯u‘v‘)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∂y–∂(ρ¯¯¯u‘w‘)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∂z)+S

Here, the terms (ρu’^{2}), (ρu’v’) and (ρu’v’) are the three extra terms.

From which of these terms does the turbulent viscosity arise from?

a) ∂(ρ¯¯¯u~)∂t

b) −∂p~∂x

c) div(μgrad(V⃗ )˜)

d) ∂(ρ¯¯¯u‘v‘)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∂y

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Answer: d

Explanation: The term (ρ¯¯¯u′v′)¯¯¯¯¯¯¯¯¯¯¯¯¯¯ represents the turbulent shear stress (Reynolds stress) in the momentum equation. This leads to the turbulent or eddy viscosity in the turbulent models.

The Reynolds stress term arises in the turbulent equation only when ____________

a) two quantities are correlated

b) two quantities are uncorrelated

c) the flow is steady

d) the flow is unsteady

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Answer: a

Explanation: Consider two flow properties u and v. While decomposing them using the Reynolds decomposition method, we get u=u+u’ and v=v+v’. The mean of their product is (uv=uv+u’v’). The Reynolds stress (ρ¯¯¯u′v′)¯¯¯¯¯¯¯¯¯¯¯¯¯¯ arises when this term (u’v’) is not zero. This term is not zero only when the two quantities are correlated.

To close the RANS equations, we need _____________

a) Incompressible flow model

b) DNS method

c) Turbulence models

d) SGS model

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Answer: cExplanation: It is impossible to derive a closed set of RANS equations. So, some approximations in the flow model are done. These approximations are called the turbulence model. This usually means prescribing the Reynolds stresses and turbulent scalar flux in terms of the mean flow quantities.

What is the difference between the RANS model and the Reynolds stress model?

a) The RANS model needs 5 extra transport equations

b) The Reynolds stress model needs 5 extra transport equations

c) The RANS model needs 7 extra transport equations

d) The Reynolds stress model needs 7 extra transport equations

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Answer: dExplanation: The most common RANS turbulence models are classified on the basis of the number of extra transport equations they need. These models form the basis of the current procedures for turbulence problems in CFD packages. The Reynolds stress model needs seven more equations.