Engineering Questions with Answers - Multiple Choice Questions

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

1 - Question

The k-ω model adds ___________ to the RANS equations.
a) three variables
b) three equations
c) two variables
d) two equations

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 - Question

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
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 - Question

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
c) Boundary layer on turbulent flows
d) Turbulent free flows
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 - Question

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
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 - Question

What is the unit of turbulence frequency?
a) Turbulence frequency is dimensionless
b) 1/s2
c) s
d) 1/s
Explanation: 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.

6 - Question

Represent the length scale in terms of k and ω.
a) ω/k
b) k/ω
c) √k/ω
d) ω/√k

Explanation: The turbulence large-scale length is given in terms of
l=k3/2ϵ
l=k.k1/2ϵ
l=k1/2ω.

7 - Question

Represent the turbulent dynamic viscosity in terms of k and ω.
a) ρ ω/k
b) ρ k/ω
c) ρ k2
d) ρ ω/k2

Explanation: The turbulent dynamic viscosity is
μt = 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−−√×kω
μt=ρ×kω.

8 - Question

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
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 - Question

Using k-ω model is difficult for ____________
a) free stream
b) boundary layer flows
c) jet flows
d) mixing layer flows
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 - Question

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