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Old   May 9, 2006, 19:28
Default static enthalpy computation in cfx
Ben Akih
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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.
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Old   May 9, 2006, 21:59
Default Re: static enthalpy computation in cfx
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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,

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Old   May 10, 2006, 05:31
Default Re: static enthalpy computation in cfx *NM*
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