This set of Computational Fluid Dynamics Multiple Choice Questions & Answers (MCQs) focuses on “Turbulence Modelling – K-omega Model”.

1. The k-ω model adds ___________ to the RANS equations.

a) three variables

b) three equations

c) two variables

d) two equations

View Answer

Explanation: The k-ω model is a variation of the k-ε model. This also adds two equations which are the transport equations of k and ω to the RANS equations to overcome the linearity problem of the RANS equation.

2. What does the variable ω in the k-ω model stand for?

a) Turbulence eddy size

b) Turbulence eddy wavelength

c) Turbulence frequency

d) Turbulence large length scale

View Answer

Explanation: The variable ω means the turbulence frequency. It gives the rate at which the turbulent kinetic energy is converted into turbulent internal thermal energy per unit volume and per unit time.

3. Which of these could not be modelled using the k-ε model, but can be modelled using the k-ω model?

a) Turbulent jet flows

b) Adverse pressure gradients in turbulent flows

c) Boundary layer on turbulent flows

d) Turbulent free flows

View Answer

Explanation: For free-shear flows, k-ε models are well suited. But, it cannot model adverse pressure gradients in the turbulent flows. This problem can be overcome by the k-ω model.

4. If k is the turbulent kinetic energy, what is the relationship between the turbulence frequency (ω) and dissipation rate of the turbulent kinetic energy (ε)?

a) ω=ε/k

b) ω=k/ε

c) ω=ε^{2}/k

d) ω=k^{2}/ε

View Answer

Explanation: Turbulence frequency is the ratio of the rate of dissipation of the turbulent kinetic energy to the turbulent kinetic energy. It is given by ω=ε/k. The values of ω are easier to assume than the ε values.

5. What is the unit of turbulence frequency?

a) Turbulence frequency is dimensionless

b) 1/s^{2}

c) s

d) 1/s

View Answer

Explanation: Turbulence frequency has the same unit as the frequency (1/s). The dimension of ε is m

^{2}/s

^{3}. The dimension of kinetic energy is m

^{2}/s

^{2}. Dividing both, we get 1⁄s.

6. Represent the length scale in terms of k and ω.

a) ω/k

b) k/ω

c) √k/ω

d) ω/√k

View Answer

Explanation: The turbulence large-scale length is given in terms of

l=\(\frac{k^{3/2}}{\epsilon}\)

l=\(\frac{k.k^{1/2}}{\epsilon}\)

l=\(\frac{k^{1/2}}{\omega}\).

7. Represent the turbulent dynamic viscosity in terms of k and ω.

a) ρ ω/k

b) ρ k/ω

c) ρ k^{2}/ω

d) ρ ω/k^{2}

View Answer

Explanation: The turbulent dynamic viscosity is

μ

_{t}= ρϑl

Where,

ρ → Density of the flow

ϑ → Length scale of the large eddies

l → Length scale of the small eddies

Replacing with the k and ω equivalents,

\(μ_t=\rho×\sqrt{k}×\frac{\sqrt{k}}{\omega}\)

\(μ_t=\rho×\frac{k}{\omega}\).

8. The values of k and ω must be specified in ___________

a) the inlet boundary conditions

b) the outlet boundary conditions

c) the wall boundary conditions

d) the symmetry boundary conditions

View Answer

Explanation: At the inlet boundaries, the values of k and ω are specified. Zero gradient conditions are used at the outlet boundary conditions. At the wall boundaries with low Reynolds number, k is set to zero.

9. Using k-ω model is difficult for ____________

a) free stream

b) boundary layer flows

c) jet flows

d) mixing layer flows

View Answer

Explanation: The k-ω model is sensitive to the free stream specified values. The value of ω in the free stream is zero. But, if this is set to zero, the eddy viscosity becomes infinity or indeterminate. So, a small non-zero value is specified and the whole problem becomes dependent on this non-zero value. The k-ε model does not have this problem.

10. Which of these is an advantage of the k-ω model over the k-ε model?

a) Does not depend on the ε value

b) Easier to integrate

c) Can be applied for turbulent boundary layers

d) Has only two extra equations

View Answer

Explanation: The greatest advantage of replacing the ε value with the ω value is that this ω value is easier to integrate. It does not need additional damping functions to integrate. The k-ω model also depends on the ε value for the ω value. Both can be applied for turbulent boundary layers. Both has two extra equations.

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