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I Why is DNS more computationally expensive than k-ε?

  1. Jul 3, 2016 #1
    In the modelling of turbulent flows, DNS is often too computationally expensive.

    Therefore, we take a 'statistical' approach and use mean flow equations, which require closure on account of the Reynolds stress. One way to do that is through the Eddy Viscosity Hypothesis, and one of the ways of determining this is "Eddy Viscosity" is the k-ε model.

    Why is the former more computationally expensive than the latter? I know that for high Re, the smallest eddy sizes are very small – but don't you need to calculate the equations (Navier-Stokes or k-ε) at these small scales whether using DNS or k-ε?

    Last edited: Jul 3, 2016
  2. jcsd
  3. Jul 4, 2016 #2
    Simply put: in DNS, you need to resolve all the length and time scales. In Large Eddy Simulations, you resolve the largest scales and you have a model for the small ones (all scales smaller than the grid size are modeled). In Reynolds Stress Models, you have a lumped model for all turbulence scales and you assume that turbulence is governed by a single length and time scale. That is why for k-epsilon simulations, you only need to resolve the mean turbulence structures.
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