Engineering Questions with Answers - Multiple Choice Questions

Home » MCQs » Aeronautical Engineering » Computational Fluid Dynamics – Turbulence Modelling – Realizable K Epsilon

# Computational Fluid Dynamics – Turbulence Modelling – Realizable K Epsilon

1 - Question

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: cExplanation: 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.

2 - Question

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: bExplanation: 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 - Question

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: aExplanation: 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 - Question

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: dExplanation: 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 - Question

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: dExplanation: 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.

6 - Question

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/ε

d) μt ∝ k2ε

**
View Answer**

Answer: cExplanation: 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 - Question

Which of these conditions satisfy realizability?

a) −ρu‘iu‘j¯¯¯¯¯¯¯¯¯<1 b) −ρu‘iu‘j¯¯¯¯¯¯¯¯¯>1

c) −ρu‘iu‘j¯¯¯¯¯¯¯¯¯>0

d) −ρu‘iu‘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,−ρu‘iu‘j¯¯¯¯¯¯¯¯¯>0. Therefore, −ρu‘iu‘j¯¯¯¯¯¯¯¯¯ should be less than zero.

8 - Question

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: cExplanation: 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 - Question

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: aExplanation: 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.

10 - Question

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: bExplanation: Other than the non-negativity condition, a realizable model should also satisfy the Cauchy-Schwarz inequality. According to this, the term (u‘iu‘j¯¯¯¯¯¯¯¯¯)2≤u‘2iu‘2j¯¯¯¯¯¯¯¯¯¯¯¯. This inequality becomes important as the realizable k-ε model is non-linear.