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

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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

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.

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

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

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

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

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.

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

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

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
d) smooth boundary layer flows

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

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.

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

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.