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static enthalpy computation in cfx

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Old   May 9, 2006, 19:28
Default static enthalpy computation in cfx
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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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Opaque
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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,

Opaque
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Old   May 10, 2006, 05:31
Default Re: static enthalpy computation in cfx *NM*
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Old   April 16, 2018, 18:53
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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!
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Old   April 17, 2018, 14:51
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What about reference temperature?

H - H_ref = Cp * (T - T_ref)

for constant Cp.
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Old   April 17, 2018, 19:14
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Quote:
Originally Posted by Opaque View Post
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?
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Old   April 18, 2018, 07:49
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Difficult to tell w/o seeing the setup.

The reference values should be listed under the material being used.
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Old   April 19, 2018, 06:23
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Thanks! Now problem solved.
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