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IV.3 The k - e model

The standard model used for turbulence is to solve additional transport equations for k and e. These equations relate transport of k (or e) to diffusion ( \~/ 2 term) and net production terms.

@k- -- nt- 2 @t + \~/ .k u = sk \~/ k + (Production - Removal)k @e n ---+ \~/ .eu = -t \~/ 2e + (Production - Removal)e @t se
Production and Removal terms actually very similar between the two equations - regions of high k are also regions of high dissipation of k, i.e. of high e. The model includes 5 adjustable constants (Cm = 0.09 is one) - values for which have been derived by fitting the model for a range of flows, and are taken to be(more or less) universal constants.

Having solved these equations to find k and e, and thus found a value of nt, we are now in a position to close the Reynolds equations. The eddy viscosity concept is usually taken to mean

2 \~/ .R = nt \~/ 2u-- - \~/ k 3
We are now solving for 3 scalars (p, k, e) and the 3 components of mean velocity u-. We have 6 equations in all, so we can find solutions to this problem.

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