# Total pressure in real gas (compressible flow)

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 November 14, 2000, 05:31 Total pressure in real gas (compressible flow) #1 Bart Prast Guest   Posts: n/a I have a (small) problem: With CFD I want to simulate internal, transonic flow with real gas effects and viscous effects. An important parameter is the (loss of) total pressure. When I look at the expressions for total pressure, I can derive it from local variables (gamma, Mach, static pressure). However in real gasses gamma (Cp/Cv) changes with pressure and temperature. This means that in isentropic flow my (local) total pressure will change even without viscous effects. Does anybody have a clou how to resolve this? Does this mean I have to solve an integral relation to get the real total pressure (only changes due to viscous effects/dissipation)? Your comments please

 November 14, 2000, 06:22 Re: Total pressure in real gas (compressible flow) #2 Richard Clayton Guest   Posts: n/a You're right to expect the total pressure in a real gas to be found via an integral relation. Taking the gas isentropically to rest implies changing its state from (h,p,u), where u is the velocity, to (h0,p0,0). So you can expect a process of integration with respect to u along a line of constant entropy. I think this is the required set of coupled ODEs: dh/du = -u dp/du = -rho u which can be integrated between the limits given above. The equation set is closed by T=T(h,p) and rho=rho(p,T).

 November 14, 2000, 11:01 Re: Total pressure in real gas (compressible flow) #3 Bernard Parent Guest   Posts: n/a Starting from the momentum equation along a particle path, it can be shown that (see a standard textbook in fluid mechanics): -dp=rho*v*dv where dp and dv are ordinary derivatives. For a constant density flow the above can be integrated to yield: Pstag=P+0.5*rho*v^2 which corresponds of course to Bernoulli's equation. On the other hand, for an isentropic process assuming calorically and thermally perfect gas (hence, P*rho^(-gamma)=constant): Pstag=P*(1-(gamma-1)/(2*gamma)*rho*v^2/P)^(gamma/(gamma-1)) For a real gas however, the above would not hold since P/rho^gamma is not a constant along a constant entropy path. However, for any gas the density can be expressed as a function of two other thermodynamic properties which we can choose to be the pressure and the entropy, i.e. rho=f(P,s). Since s is constant throughout the integration, the integral of dP/f(P,s) can be evaluated easily numerically provided that a function giving rho from P and s is available. -bernard

 November 14, 2000, 11:44 Re: Total pressure in real gas (compressible flow) #4 Bart Prast Guest   Posts: n/a OK thsi is very valuable information. However, if I would like to know pressure losses in an internal pipe with transonic flow and I would have Joule-Thompson cooling/heating, than there is no other means of getting this except for these integral equations? Is there no-one in this field with hands on experience? Poeple have been calculating turbomachinery for some years now. They want to know pressure losses to a high level of accuracy. They must have been dealing with real gas effects. From my experience real gas effects are not handled very well in commercial codes. If at all, they yust let you specify an equation of state and d_p/d_rho (speed of sound) and that's it. Then they give you an Total pressure (local) based on constant gamma. There must be more available.

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