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-   -   Total pressure in rel frame and total pressure (http://www.cfd-online.com/Forums/cfx/98482-total-pressure-rel-frame-total-pressure.html)

 Salut March 12, 2012 09:45

Total pressure in rel frame and total pressure

Hi! Please let me know what is a principal difference between Total pressure in rel frame and Total Pressure functions in a case of axial turbine stage problem. CFX Post returns different fields of theese variables on turbo surface with constant span =0.5, but i think they must be identical. Thanks.

 ghorrocks March 12, 2012 17:25

Total pressure = p + 1/2 rho v^2

The velocity term is different depending on what frame of reference you measure it in. This means total pressure is different in different frames of reference.

 Salut March 13, 2012 09:53

Thanks, ghorrocks. I was made a mistake in my question, in fact I told about only rotating domain. The velocity term on the rotating frame of reference is the same for total pressure and total pressure in rel frame, isn't it? So why do the total pressure and total pressure in rel frame fields for look different in a same rotating domain?

 ghorrocks March 13, 2012 17:31

My previous post explains why. Because the velocity in the two frames of reference is different.

 Salut March 14, 2012 04:35

Ok, probably it's my english problems.

What is the differnce between:

Total Pressure in Stn Frame=p+1/2*rho*Ustn^2 (1)

Total Pressure in Rel Frame=p+1/2*rho*Urel^2 (2)

Total Pressure = p+1/2*rho*(Urel^2-(omega*R)^2) (3)

I know, that Urel=Ustn-omega*R in rotating systems,

Physical meaning of 1 term - absolute total pressure

Physical meaning of 2 term - relative total pressure

But what's the physical meaning of term 3 ????!!!!

Best wishes!

 ghorrocks March 14, 2012 06:09

Eqn 3 is not total pressure. It is rothalpy. Look it up in the documentation (eqn 1-56 in the theory guide) or on google.

 sfallah December 1, 2014 06:11

Quote:
 Originally Posted by ghorrocks (Post 349019) Total pressure = p + 1/2 rho v^2 The velocity term is different depending on what frame of reference you measure it in. This means total pressure is different in different frames of reference.
Hi Dear ghorrocks;
Total pressure in rotating frame is less than static pressure (In the outlet ), whereas according to equation: Total Pressure in Rel Frame=p+1/2*rho*Urel^2, it should be greater than static pressure because of positive values of "1/2*rho*Urel^2" in any condition(rotating or stationary frame) .
My test case is NASA rotor37. What is the reason for this contradiction?
Another question:
Urel=Ustn-omega*R and Ptotal(rot)=pstatic+1/2*rho*Urel^2, as aresult: Ptotal(rot)=pstatic+1/2*rho*(Ustn-omega*R)^2
but in cfx theory guide(equation 1-53): Ptot(rot)=Pstatic+1/2*rho*(Urel*Urel-(omega*R.omega*R)). What is the reason for this contradiction?

 ghorrocks December 1, 2014 17:07

Are you saying you are getting total pressure in a rotating frame less than inlet static pressure in your case? Can you post an image of what you are modelling? And the boundary conditions?

 sfallah December 2, 2014 04:38

1 Attachment(s)
Yes, In CFD-Post, 3 Total pressure is exist: Total Pressure, Total pressure in Rel Frame and Total Pressure in stn Frame. Total pressure is less than static pressure but total pressure in rel frame and total pressure in stn frame is greater than static pressure. Image of Mach contour in 80% span is attached. Boundary condition:
Test case: NASA ROTOR 37
Reference pressure:0 [pa]
Inlet: Total pressure:1[atm]
Outlet: Static pressure:106.5 [kpa]
High resolution discretization is used for all equations.
High speed numeric, clip pressure and high resolution rhio chow algorithm are activated via advance parameter tab in solution tab.
I can not reach to nominal mass flow rate (20.19 kg/s) and total pressure ratio (2.1) by variation in static pressure. I reached to pressure ratio 1.88 and mass flow rate 20.38 at best condition. What is its solution?

 schorschele October 6, 2015 04:26

Hi guys,
I was reading the forum and have still a question regarding the total pressure in rotating frame.

I could reproduce the values for "TotalPressure in relative and stationary frame" with the given formulas. But for the TotalPressure itself I didnt get the value in [Pa] as CFX gives me, I get to high values.
Total Pressure in Stn Frame=p_stat+1/2*rho*Ustn^2 OK
Total Pressure in Rel Frame=p_stat+1/2*rho*Urel^2 OK
Total Pressure = p_stat+1/2*rho*(Urel^2-(omega*R)^2) Fail

For R I use the maxValue at the plane I want to investigate.
Would be nice if someone could give me some advice.

Thanks

 sodynamic February 19, 2016 21:38

Rothalpy shows the loss without output work

Quote:
 Originally Posted by sfallah (Post 522008) Yes, In CFD-Post, 3 Total pressure is exist: Total Pressure, Total pressure in Rel Frame and Total Pressure in stn Frame. Total pressure is less than static pressure but total pressure in rel frame and total pressure in stn frame is greater than static pressure. Image of Mach contour in 80% span is attached. Boundary condition: Test case: NASA ROTOR 37 Reference pressure:0 [pa] Inlet: Total pressure:1[atm] Outlet: Static pressure:106.5 [kpa] Angular velocity:-1800[rad/s] High resolution discretization is used for all equations. High speed numeric, clip pressure and high resolution rhio chow algorithm are activated via advance parameter tab in solution tab. I can not reach to nominal mass flow rate (20.19 kg/s) and total pressure ratio (2.1) by variation in static pressure. I reached to pressure ratio 1.88 and mass flow rate 20.38 at best condition. What is its solution?
I believe another post had the similar discussion and offerred more detailed resources.

Difference between total pressure, total pressure in Stn and in Rel frames

Anyhow, if you understand what rothalpy is, these three definitions of "total pressure" should not be a problem. The definiton of Total Pressure here, which is used in CFD-Post, is deduced from rothalpy. It gets rid off the work which the fluid output against the rotors.

The work will be computed differently in the stationary frame and rotational frame. If the stationary work is added to rothalpy, the result is the total enthalpy in stationary frame. And the difference between rothalpy and the total enthalpy in relative frame showes the effect of Coriolis force. The equation of rothalpy includes a part that (Omega X Radius)^2 should be taken away. So if the calculation is about rotors, and the output work is very large, it is quite possible that the rothalpy is negative, or the total pressure is less than the static pressure.

I think this Total Pressure definition based on rothalpy can be very convenient for loss assessment, especially for the centrifugal pump or radial flow turbine. For these flowfields, the Omega X Radius may be quite different on the same streamline within a rotational domain.

 turbo February 20, 2016 10:09