# static enthalpy computation in cfx

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 May 9, 2006, 19:28 static enthalpy computation in cfx #1 Ben Akih Guest   Posts: n/a hi folks, has somebody insight on the way cfx computes the static enthalpy? the usual engineering approach defining static enthalpy as the integral from Tref to T of cp times dt seems to lead to differnt values. CFX assumes that cp is a function of temperature and pressure, thus a second term is added to the that described above. would be glad for some thoughts shared on this. regards, ben akih.

 May 9, 2006, 21:59 Re: static enthalpy computation in cfx #2 Opaque Guest   Posts: n/a Dear Ben Akih, What approach are you referring to? As I understand, from thermodynamic relations, static enthalpy change for a single component pure substance is only a function of p, and T. Therefore, a differential change is expressed as h = h(T,P) dh = (dh/dT)|p * dT + (dh/dp)|T * dp where Cp= (dh/dT)|p dh = Cp(T) * dT + (dh/dp)|T * dp From thermodynamic relationships, dh = T * ds + v * dp Since above is an exact differential (dh/dp)|T = T * (ds/dp)|T + v From one of Maxwell relations (ds/dp)|T = - (dv/dT)|P Substituting as required, you get for dh dh = Cp * dT + (v - T * (ds/dT)|p) * dp For an ideal gas, the second term on the RHS is zero..For other materials (like real gas, or liquids) the second term cannot be neglected..Otherwise, how can you explain the work input on a pump? Hope this helps, Opaque lostking18 likes this.

 May 10, 2006, 05:31 Re: static enthalpy computation in cfx *NM* #3 CFD-student Guest   Posts: n/a

April 16, 2018, 18:53
#4
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phd
Join Date: Oct 2013
Posts: 76
Rep Power: 10
Quote:
 Originally Posted by Opaque ;76309 Dear Ben Akih, What approach are you referring to? As I understand, from thermodynamic relations, static enthalpy change for a single component pure substance is only a function of p, and T. Therefore, a differential change is expressed as h = h(T,P) dh = (dh/dT)|p * dT + (dh/dp)|T * dp where Cp= (dh/dT)|p dh = Cp(T) * dT + (dh/dp)|T * dp From thermodynamic relationships, dh = T * ds + v * dp Since above is an exact differential (dh/dp)|T = T * (ds/dp)|T + v From one of Maxwell relations (ds/dp)|T = - (dv/dT)|P Substituting as required, you get for dh dh = Cp * dT + (v - T * (ds/dT)|p) * dp For an ideal gas, the second term on the RHS is zero..For other materials (like real gas, or liquids) the second term cannot be neglected..Otherwise, how can you explain the work input on a pump? Hope this helps, Opaque

Hi, Opaque:

Thanks for the derivation. Do you know why in CFX static enthalpy is not equal to 'specific heat capacity at constant pressure * Temperature'? I use Air ideal gas, total Energy model and reference pressure is 1atm. It seems in ideal gas, the second term (v - T * (dv/dT)|p) * dp should be zero.

Before I thought it is because my reference pressure is not 0 so that in CFX with total energy model, the solver will first obtain total enthalpy ht, then use
hs = ht - 1/2* velocity * velocity
to obtain static enthalpy. While tempareture is obtained by absolute pressure divided by density and R:
T = P/(rho * R)
But after trying reference pressure equals to 0, there are still discrepancies.

Thanks!

 April 17, 2018, 14:51 #5 Senior Member   Join Date: Jun 2009 Posts: 1,558 Rep Power: 28 What about reference temperature? H - H_ref = Cp * (T - T_ref) for constant Cp.

April 17, 2018, 19:14
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phd
Join Date: Oct 2013
Posts: 76
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Quote:
 Originally Posted by Opaque What about reference temperature? H - H_ref = Cp * (T - T_ref) for constant Cp.
Hi, Opaque, thanks for your reply! it seems for the model without Buoyancy, the reference temperature and reference enthalpy are default to be 0 [K] and 0[J/kg], right?

 April 18, 2018, 07:49 #7 Senior Member   Join Date: Jun 2009 Posts: 1,558 Rep Power: 28 Difficult to tell w/o seeing the setup. The reference values should be listed under the material being used. lostking18 likes this.

 April 19, 2018, 06:23 #8 Member   phd Join Date: Oct 2013 Posts: 76 Rep Power: 10 Thanks! Now problem solved.

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