Computational Fluid Dynamics Questions and Answers – Turbulence Modelling – Realizable K Epsilon

«
»

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

1. Which of these conditions define realizability?
a) Negative k and ε values
b) Linear k and ε transport equations
c) Non-negative k and ε values
d) Non-linear k and ε transport equations
View Answer

Answer: c
Explanation: The values of turbulence quantities such as the turbulent kinetic energy (k) and the rate of dissipation of the turbulent kinetic energy (ε) cannot be negative and must always be constrained to have values above zero.
advertisement

2. Which of these statements is true?
a) The standard k-ε model and the RNG k-ε model are realizable
b) Neither the standard k-ε model nor the RNG k-ε model is realizable
c) The standard k-ε model is realizable but not the RNG k-ε model
d) The RNG k-ε model is realizable but not the standard k-ε model
View Answer

Answer: b
Explanation: Both the standard k-ε model and the RNG k-ε model are not realizable. They do not satisfy the mathematical condition of realizability. Only the realizable k-ε model is realizable.

3. The realizable k-ε model falls into which of these categories?
a) Non-linear two-equation turbulence models
b) Linear two-equation turbulence models
c) Non-linear three-equation turbulence models
d) Linear two-equation turbulence models
View Answer

Answer: a
Explanation: There is a division of the k-ε models which are non-linear. The transport equations are non-linear in these models. The realizable k-ε model also comes under this non-linear k-ε models.

4. The realizable k-ε model is based on ________
a) the turbulence model replacing the realizability constraint
b) the viscoelastic analogy replacing the realizability constraint
c) the realizability constraint with viscoelastic analogy
d) the realizability constraint without viscoelastic analogy
View Answer

Answer: d
Explanation: The non-linear k-ε models were initially developed based on the analogy between the viscoelastic fluids and the turbulent flows. This analogy is not used by the realizable k-ε model. The realizability constraint rules out this analogy.

5. The non-linear k-ε models relate the Reynolds stresses to _________
a) the cubic vector products of strain rate and vorticity
b) the quadratic vector products of strain rate and vorticity
c) the cubic tensor products of strain rate and vorticity
d) the quadratic tensor products of strain rate and vorticity
View Answer

Answer: d
Explanation: The Reynolds stresses in the non-linear k-ε models relate the Reynolds stresses to the quadratic product of local vorticity and strain rates. These two quantities are tensors. The method sensitizes the Reynolds stresses.
advertisement

6. Which of these equations gives the turbulent dynamic viscosity used in the realizable k-ε model?
a) μt ∝ ρk/ε
b) μt ∝ k / ε
c) μt ∝ ρk2
d) μt ∝ k2ε
View Answer

Answer: c
Explanation: The standard k-ε model, RNG k-ε model and the realizable k-ε model use the same equation for the turbulent dynamic viscosity. The turbulent dynamic viscosity is the product of the turbulent kinematic viscosity and the density of the flow given by μt∝(ρk2)/ε.

7. Which of these conditions satisfy realizability?
a) \(-\rho \overline{u_{i}^{‘} u_{j}^{‘}}\)<1
b) \(-\rho \overline{u_{i}^{‘} u_{j}^{‘}}\)>1
c) \(-\rho \overline{u_{i}^{‘} u_{j}^{‘}}\)>0
d) \(-\rho \overline{u_{i}^{‘} u_{j}^{‘}}\)<0
View Answer

Answer: d
Explanation: According to the realizability conditions, the properties which are physically non-negative must be numerically non-negative too. So,\(-\rho \overline{u_{i}^{‘} u_{j}^{‘}}\)>0. Therefore, \(-\rho \overline{u_{i}^{‘} u_{j}^{‘}}\) should be less than zero.

8. The realizable k-ε model is best for predicting __________
a) mixing layers
b) wake formation
c) spreading of jets
d) smooth boundary layer flows
View Answer

Answer: c
Explanation: The realizable k-ε model more accurately predicts the spreading of jets, be it planar or round jets. It performs better than the other models for complex problems involving recirculation and flow separation.

9. Though the applicability of the realizable k-ε model and the RNG k-ε model are almost the same, the realizable k-ε model is ___________ when compared to the RNG k-ε model.
a) more accurate and converges easily
b) more stable
c) linear
d) more consistent
View Answer

Answer: a
Explanation: The realizable k-ε model and the RNG k-ε model have the same benefits and applications. But, the realizable model gives more accurate results and it is easy to converge when compared to the RNG k-ε model.
advertisement

10. Which of these conditions should be satisfied for a model to be realizable?
a) Bessel’s inequality
b) Cauchy-Schwarz inequality
c) Holder’s inequality
d) Jensen’s inequality
View Answer

Answer: b
Explanation: Other than the non-negativity condition, a realizable model should also satisfy the Cauchy-Schwarz inequality. According to this, the term \((\overline{u_{i}^{‘} u_{j}^{‘}})^2 ≤ \overline{u_{i}^{‘2} u_{j}^{‘2}}\). This inequality becomes important as the realizable k-ε model is non-linear.

Sanfoundry Global Education & Learning Series – Computational Fluid Dynamics.

To practice all areas of Computational Fluid Dynamics, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement
advertisement
advertisement
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn