CFD Online Logo CFD Online URL
www.cfd-online.com
[Sponsors]
Home > Forums > Software User Forums > ANSYS > CFX

static enthalpy computation in cfx

Register Blogs Members List Search Today's Posts Mark Forums Read

Like Tree2Likes
  • 1 Post By Opaque
  • 1 Post By Opaque

Reply
 
LinkBack Thread Tools Search this Thread Display Modes
Old   May 9, 2006, 20:28
Default 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.
  Reply With Quote

Old   May 9, 2006, 22:59
Default 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.
  Reply With Quote

Old   May 10, 2006, 06:31
Default Re: static enthalpy computation in cfx *NM*
  #3
CFD-student
Guest
 
Posts: n/a
  Reply With Quote

Old   April 16, 2018, 19:53
Default
  #4
Member
 
phd
Join Date: Oct 2013
Posts: 76
Rep Power: 12
lostking18 is on a distinguished road
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!
lostking18 is offline   Reply With Quote

Old   April 17, 2018, 15:51
Default
  #5
Senior Member
 
Join Date: Jun 2009
Posts: 1,785
Rep Power: 31
Opaque will become famous soon enough
What about reference temperature?

H - H_ref = Cp * (T - T_ref)

for constant Cp.
Opaque is offline   Reply With Quote

Old   April 17, 2018, 20:14
Default
  #6
Member
 
phd
Join Date: Oct 2013
Posts: 76
Rep Power: 12
lostking18 is on a distinguished road
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?
lostking18 is offline   Reply With Quote

Old   April 18, 2018, 08:49
Default
  #7
Senior Member
 
Join Date: Jun 2009
Posts: 1,785
Rep Power: 31
Opaque will become famous soon enough
Difficult to tell w/o seeing the setup.

The reference values should be listed under the material being used.
lostking18 likes this.
Opaque is offline   Reply With Quote

Old   April 19, 2018, 07:23
Default
  #8
Member
 
phd
Join Date: Oct 2013
Posts: 76
Rep Power: 12
lostking18 is on a distinguished road
Thanks! Now problem solved.
lostking18 is offline   Reply With Quote

Reply

Thread Tools Search this Thread
Search this Thread:

Advanced Search
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are On
Refbacks are On


Similar Threads
Thread Thread Starter Forum Replies Last Post
Temperature gradient in CFX Chander CFX 5 January 7, 2015 04:00
fprint problem MASOUD Fluent UDF and Scheme Programming 5 October 30, 2011 17:08
Static entropy in CFX BalanceChen CFX 2 October 5, 2011 08:01
ATTENTION! Reliability problems in CFX 5.7 Joseph CFX 14 April 20, 2010 16:45
potential energy& static enthalpy in buoyant flow Atit CFX 0 May 3, 2006 11:05


All times are GMT -4. The time now is 04:48.