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# Computational Fluid Dynamics – Turbulence Modelling – K-omega Model

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

a) three variables

b) three equations

c) two variables

d) two equations

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Answer: d

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.

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

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Answer: cExplanation: 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.

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

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Answer: bExplanation: 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.

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=ε2/k

d) ω=k2/ε

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Answer: aExplanation: 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.

What is the unit of turbulence frequency?

a) Turbulence frequency is dimensionless

b) 1/s2

c) s

d) 1/s

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Answer: dExplanation: Turbulence frequency has the same unit as the frequency (1/s). The dimension of ε is m2/s3. The dimension of kinetic energy is m2/s2. Dividing both, we get 1⁄s.

Represent the length scale in terms of k and ω.

a) ω/k

b) k/ω

c) √k/ω

d) ω/√k

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Answer: c

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

l=k3/2ϵ

l=k.k1/2ϵ

l=k1/2ω.

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}

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Answer: b

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,

μt=ρ×k−−√×k√ω

μt=ρ×kω.

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

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Answer: aExplanation: 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.

Using k-ω model is difficult for ____________

a) free stream

b) boundary layer flows

c) jet flows

d) mixing layer flows

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Answer: aExplanation: 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.

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

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Answer: bExplanation: 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.