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If n is the spatial coordinate, in the outlet or symmetry boundaries, which of these following is correct for a k-ε model?

(a) \(\frac{\partial k}{\partial n}=0; \frac{\partial\varepsilon}{\partial n}=0\)

(b) \(\frac{\partial^2 k}{\partial n^2}=0; \frac{\partial\varepsilon}{\partial n}=0\)

(c) \(\frac{\partial k}{\partial n}=0; \frac{\partial^2 \varepsilon}{\partial n^2}=0\)

(d) \(\frac{\partial ^2 k}{\partial n^2}=0; \frac{\partial^2 \varepsilon}{\partial n^2}=0\)

This question was addressed to me in an interview for internship.

Asked question is from Turbulence Modelling in section Turbulence Modelling of Computational Fluid Dynamics

1 Answer

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by (65.4k points)

The correct answer is:

(a) ∂k∂n = 0; ∂ε∂n = 0

Explanation:

In the context of turbulence modeling using the k-ε model, the boundary conditions at the outlet or symmetry boundaries for the turbulence variables kk (turbulent kinetic energy) and ϵ\epsilon (turbulent dissipation rate) are typically applied as follows:

  • For kk (turbulent kinetic energy): The gradient of kk in the normal direction (nn) is set to zero at the outlet or symmetry boundaries. This is because the turbulence kinetic energy kk does not have a natural boundary condition at the outlet, and it's assumed that there is no flux of turbulent energy across the boundary. Thus, ∂k∂n=0\frac{\partial k}{\partial n} = 0.

  • For ϵ\epsilon (turbulent dissipation rate): Similarly, the gradient of ϵ\epsilon in the normal direction (nn) is also set to zero at these boundaries, meaning ∂ϵ∂n=0\frac{\partial \epsilon}{\partial n} = 0. This is because the dissipation rate ϵ\epsilon is typically assumed to decay towards zero away from the boundary in the case of an outlet or symmetry condition.

Thus, both kk and ϵ\epsilon have a zero gradient in the normal direction at the outlet or symmetry boundaries.

Correct Answer: (a) ∂k∂n = 0; ∂ε∂n = 0

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