Engineering Questions with Answers - Multiple Choice Questions
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
View Answer
Explanation: 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
View Answer
Explanation: 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
View Answer
Explanation: 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
View Answer
Explanation: 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
View Answer
Explanation: 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
View Answer
Explanation: 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
View Answer
Explanation: 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
View Answer
Explanation: 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 ¯¯¯¯¯¯¯¯¯¯¯
View Answer
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’
View Answer
Explanation: 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’.