# 2/3div(u)

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 February 3, 2006, 19:31 2/3div(u) #1 George Guest   Posts: n/a 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 Navier-Stokes 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 Opaque Guest   Posts: n/a 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 pressure-velocity 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 non-linear 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 George Guest   Posts: n/a Thank you for your response. Also when I calculate density from the equation of state,I should use p_abs?

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