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# Computational Fluid Dynamics – Turbulence Modelling – Averaging Rules

1 - Question

These rules for averaging are used to average ___________

a) fluctuations in the turbulent flow

b) variation in results of turbulent flow

c) the coefficients in FVM

d) the coefficients in FDM

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Answer: aExplanation: The flow variables in a turbulent flow are divided into mean and fluctuating components. These fluctuating components in the turbulent flow are averaged for the further solution of the system. These rules are used for averaging.

2 - Question

According to the rules for averaging, which of these will sum up to zero?

a) The mean component of the flow variable

b) The fluctuating component of the flow variable

c) The flow variable

d) Integration of the flow variable

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Answer: aExplanation: The mean component of a flow variable is the overall average of the flow variable. So, when the average of the flow variable is its mean component, the average of the fluctuating component and hence its summation will be zero.

3 - Question

The average of the mean component will be ____________

a) equal to zero

b) equal to the mean component itself

c) equal to 1

d) equal to the fluctuating component

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Answer: bExplanation: The mean component is already found by taking the arithmetic mean (average) of the flow variables. So, if the average of only the mean component is taken, it will again be the same mean component itself.

4 - Question

The mean of the spatial partial derivative of a flow variable will be equal to ____________

a) 0

b) 1

c) the spatial partial derivative of the mean component

d) the mean component

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Answer: cExplanation: The mean of the flow variable will be equal to the mean variable. The mean of the flow variable’s spatial partial derivative will be equal to the spatial partial derivative of the mean component of that variable.

5 - Question

The mean of the summation of two flow variables will be equal to ____________

a) the summation of their mean components – the summation of the mean of their fluctuating components

b) the summation of their mean components + the summation of the mean of their fluctuating components

c) the summation of their fluctuating components

d) the summation of their mean components

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Answer: dExplanation: Consider two flow variables which can be decomposed as a=A+a’ and b=B+b’. The mean of their summation means a+b = A+a’+B+b’ But, a’ = 0 and b’ = 0. Therefore, a+b = A+B Also, A = A and B = B. Hence, a+b = A+B.

6 - Question

The mean of the space-based integral of a flow variable is equal to ____________

a) the summation of its mean component

b) the space-based integral of its fluctuating component

c) the space-based integral of its mean component

d) the summation of its fluctuating components

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Answer: cExplanation: As the mean of the fluctuating component is zero and the mean of the mean component is the mean component itself, the mean of the space-based integral of a flow variable is equal to the space-based integral of its mean component alone.

7 - Question

The mean of the product of the mean component of one variable and the fluctuating component of another variable is ____________

a) 1

b) 0

c) the product of their mean components

d) the product of their fluctuating components

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Answer: bExplanation: The mean of a fluctuating component is zero. The mean of a mean component is a variable. So, the mean of the product of the mean component of one variable and the fluctuating component of another variable will become zero.

8 - Question

The mean of the product of a flow variable and the mean component of another flow variable is ____________

a) the product of their mean components

b) the product of their fluctuating components

c) the mean of the product of their mean components

d) the mean of the product of their fluctuating components

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Answer: aExplanation: Consider two flow variables which can be decomposed as a=A+a’ and b=B+b’. The mean of the product of one flow variable and the mean component of another flow variable is represented as aB=(A+a’)B aB=AB+a’B As a’B=0 and AB=AB, aB=AB.

9 - Question

Consider a vector flow variable which can be decomposed as a⃗ =A⃗ +a′→.diva⃗ ¯¯¯¯¯¯¯¯¯¯ will be equal to ____________

a) div A⃗

b) diva′→¯¯¯¯¯¯¯¯¯¯¯¯

c) diva⃗ ¯¯¯¯¯¯¯¯¯

d) divA⃗ ¯¯¯¯¯¯¯¯¯¯¯

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Answer: a

Explanation: From the given problem,

diva⃗ ¯¯¯¯¯¯¯¯¯¯=diva⃗ ¯¯¯=divA⃗ +a′→¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=divA⃗ ¯¯¯¯=divA⃗ .

10 - Question

Consider two flow variables which can be decomposed as a=A+a’ and b=B+b’. What is ab?

a) 0

b) 1

c) AB

d) a’b’

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Answer: dExplanation: For fluctuating variables, ab = (A+a’)(B+b’) ab = (AB+a’ B+Ab’+a’b’) But, a’B=0, Ab’ = 0 and AB = AB. So, ab = AB+a’b’.