Turbulence models k-epsilon k-omega sst

cfdnoob · January 8, 2016, 15:22

Hi, I am using ansys cfx. Now I want to know in general how the two equation models work, especially the k-e model, because this is used often.

davidwilcox · January 14, 2016, 05:18

firstly, k=1/2(u_i'u_i') and not k= 1/2 u_i'u_j'.
Pertaining to your first question, yes. The transport equation solves for the scalars they transport which in your case is k, epsilon or omega.

If you look at the raw RANS equations, you realize that the Reynolds stresses are on the right hand side of the equations together with pressure and viscous diffusion. The way these Reynolds stresses are modelled in 2-equation eddy viscosity models like the k-epsilon, k-omega is via the eddy viscosity ( Boussinesq assumption). In short, it says that Reynolds stress=some constant times Strain rate. that constant is the eddy viscosity.

Let's look at modelled Reynolds stresses in 2D case:

u'u'=-2 * eddyviscosity * du/dx
v'v'=-2 * eddyviscosity * dv/dy
u'v'=-2 * eddyviscosity * 0.5(du/dy+dv/dx)
NOTE that k = 0.5(u'u'+v'v'). no u'v' involved!

The two equation models "do not transport" the Reynolds stresses. You need to compute them via the relations above. If you do not want a modelled form of the Reynolds stresses, you use the RSM models.

Notice also in the raw RANS (momentum), you do not have have k. where is k? K is actually coupled with the pressure term so it becomes a modified pressure term. Once you solve the transport equations, it spits k out.

January 8, 2016, 15:22	Turbulence models k-epsilon k-omega sst	#1
cfdnoob New Member Rich Join Date: Jul 2014 Posts: 23 Rep Power: 11	Hi, I am using ansys cfx. Now I want to know in general how the two equation models work, especially the k-e model, because this is used often. Last edited by cfdnoob; February 2, 2016 at 06:35.

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January 14, 2016, 05:18		#2
davidwilcox Senior Member david Join Date: Oct 2012 Posts: 142 Rep Power: 13	firstly, k=1/2(u_i'u_i') and not k= 1/2 u_i'u_j'. Pertaining to your first question, yes. The transport equation solves for the scalars they transport which in your case is k, epsilon or omega. If you look at the raw RANS equations, you realize that the Reynolds stresses are on the right hand side of the equations together with pressure and viscous diffusion. The way these Reynolds stresses are modelled in 2-equation eddy viscosity models like the k-epsilon, k-omega is via the eddy viscosity ( Boussinesq assumption). In short, it says that Reynolds stress=some constant times Strain rate. that constant is the eddy viscosity. Let's look at modelled Reynolds stresses in 2D case: u'u'=-2 * eddyviscosity * du/dx v'v'=-2 * eddyviscosity * dv/dy u'v'=-2 * eddyviscosity * 0.5(du/dy+dv/dx) NOTE that k = 0.5(u'u'+v'v'). no u'v' involved! The two equation models "do not transport" the Reynolds stresses. You need to compute them via the relations above. If you do not want a modelled form of the Reynolds stresses, you use the RSM models. Notice also in the raw RANS (momentum), you do not have have k. where is k? K is actually coupled with the pressure term so it becomes a modified pressure term. Once you solve the transport equations, it spits k out.