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February 3, 2006, 19:31 
2/3div(u)

#1 
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Hi everybody! I have a question.I am writing a code for 3D,turbulent,unsteady,compressible flow.In the source term of the momentum equations of the NavierStokes equations there is the term 2/3*(viscosity)*div(u) and also the term 2/3*(density)*k ,where k is the turbulent kinetic energy.I have read in some articles that we can incorporate these two terms in the pressure term of the momentum equations, so that: pressure=pressure+2/3*(viscosity)*div(u)+2/3*(density)*k . Is that correct?The source term is then more simple, but is there any wrong with the above expression? I also correct the pressure using the PISO algorithm.No changes will occur there,wright? Thank you for reading my problem and waiting for your reply


February 3, 2006, 21:36 
Re: 2/3div(u)

#2 
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Since the term can be written as
div ( p delta_ij  2/3 * mu * div U  2/3 * rho * K) there is nothing wrong with calling p_star and using it in the momemtum equation. Your pressurevelocity algorithm does not change either. You may also include the buoyancy term (incompressible) as div ( p delta_ij  2/3 * mu * div U  2/3 * rho * K + rho * grav dot R) where R is the position vector.. However, there is the small problem that if you need p_abs for thermodynamic calculation, you have to back it out of p_star.. You cannot use p_star within any thermodynamic relation that is nonlinear in p.. For example, stagnation conditions, entropy calculation and so on.. If you are solving compressible flows the pressure term in the total enthalpy equation is for p, not p_star. It goes away from momemtum but it comes back at you in the energy transport equation Good luck, Opaque.. 

February 4, 2006, 08:19 
Re: 2/3div(u)

#3 
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Thank you for your response. Also when I calculate density from the equation of state,I should use p_abs?


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