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-   -   Momentum source coefficient, cylindrical coordinates, circumferential component? (http://www.cfd-online.com/Forums/cfx/91988-momentum-source-coefficient-cylindrical-coordinates-circumferential-component.html)

polakse August 29, 2011 13:12

Momentum source coefficient, cylindrical coordinates, circumferential component?
 
Hi,

I have a fluid (air) sub-domain defined, with a general momentum source applied to it (using a cylindrical coordinate system). My goal is to simulate the air flow through an axial fan, including an estimation of the swirl (circumferential) velocity. I have the fan curve (volume flow vs system pressure delta) defined as a 1-D interpolation function. I am calculating system pressure delta as:

PressureRiseSystem = massFlowAve(Total Pressure)@Outlet - massFlowAve(Total Pressure)@Inlet

I am applying the result of the FanCurve function (volume flow) to the axial component of the momentum source, using the momentum source coefficient approach. i.e.

Axial Component = -C*(massFlowAve(w)@FanOutlet - FanCurve(PressureRiseSystem)/area()@FanOutlet)

Where -C is the momentum source coefficient (i.e. 10^5)

The radial component is set to 0 [kg m^-2 s^-2]

I have an estimation for the tangential/circumferential component of the velocity (it is a function of fan RPM and radial location with some scaling factors), but it's not clear to me how use this estimated/spec tangential velocity to apply the Theta Component of the momentum source. In other words, if the Theta Component of the momentum source is something like:

Theta Component = -C*(VelocityThetaActual - VelocityThetaSpec)

can I make both of these (VelocityThetaActual and VelocityThetaSpec) functions of the radial position, r? Also, how do I determine the VelocityThetaActual value? Convert the cartesian velocites u, v, and w to cylindrical components? Does that make sense?

Any help is greatly appreciated. Thanks in advance!

ghorrocks August 29, 2011 18:47

Firstly, pressure rise is best done on an area average basis rather than mass flow ave. Mass flow ave is meaningless for pressure.

Quote:

can I make both of these (VelocityThetaActual and VelocityThetaSpec) functions of the radial position, r?
Yes. You can either define your own CEL expression to work out r, or define a local coordinate system and get r in that.

Quote:

Also, how do I determine the VelocityThetaActual value?
Same as above.

Sounds like you need to define a local coordinate system.

polakse September 9, 2011 15:30

Hi Glenn, thanks for the reply. With the help of your comments, I think I have a model successfully defined, and converging nicely.

I ended up calculating the tangential/circumferential velocity in an expression, and using that value in the momentum source definition, along with the estimated "spec" or target tangential velocity imparted to the air from the rotating fan...

Theta Component = -C*(VelocityThetaSpec - VelocityTheta)

where the expression

VelocityTheta = -u*sin(atan2(y,x)) + v*cos(atan2(y,x))

To achieve convergence stability, I had to define a non-constant momentum source coefficient (C), that steadily increased (eventually to a value around 10e5) by a factor of the accumulated time step (atstep). Otherwise, if I set the momentum source coefficient constant at 10^5, the simulation would oscillate out of control, and if I set the value lower (i.e. 10^3), the model would converge with a significant error/differential between V and Vspec.

Thanks again!

stumpy September 9, 2011 15:43

I would say Total Pressure should be mass flow averaged, while only static pressure should be area averaged.

polakse September 9, 2011 15:45

Quote:

Originally Posted by stumpy (Post 323595)
I would say Total Pressure should be mass flow averaged, while only static pressure should be area averaged.

I agree. In this situation, I used MassFlowAve particularly because I have uneven flow with locally high velocity streams, and thus high dynamic pressure.

ghorrocks September 12, 2011 19:56

Quote:

Total Pressure should be mass flow averaged
What physical basis do you have for this statement? Taking a mass flow average of ANY type of pressure is meaningless. The area integral of pressure is force, which is physically meaningful.

Massflow averages of things like enthalpy, temperature, concentrations and the like are physically meaningful as it gives you the total flux of the quantity. But this does not apply to pressures. The flux of pressure is not physically meaningful.

brunoc September 14, 2011 18:34

Hi Glenn,

Some people argue that mass flow averages should be used for any transported variable. It's like saying "ok, enthalpy is advected so I'll use it's advection (the local mass flow) to compute the local average". I believe I even read something like this on a CFX training course.

If you follow that thought you can think of dynamic pressure as a transported variable, specially if you look at velocity as a type of transported momentum. In this case, it would be ok to use mass flow average on dynamic pressure, and if you stretch it a little, also on total pressure.

Quote:

Originally Posted by ghorrocks (Post 323846)
The area integral of pressure is force, which is physically meaningful.

Massflow averages of things like enthalpy, temperature, concentrations and the like are physically meaningful as it gives you the total flux of the quantity. But this does not apply to pressures.

I've never given it much thought. What you say does makes sense, that the integral of static pressure is force. But dynamic pressure doesn't translate to force, that is, by looking at the dynamic pressure alone one cannot calculate any type of force (as far as I know - maybe I don't know much? :o).
About total pressure, you can also look at it as an energy on your flow. Higher total pressure, higher energy. Integrating it with the mass flow would give you the total flux of this energy, which is phisically meaningful.

I'm not sure I'm right on this, so I'd like to hear your thoughts.

Cheers.

ghorrocks September 14, 2011 18:52

Quote:

you can think of dynamic pressure as a transported variable
Density * velocity ^2, ie pressure? I cannot see how this can be regarded as a transported variable.

Quote:

But dynamic pressure doesn't translate to force
My argument is purely dimensional. How the pressure is derived is not important.

Quote:

About total pressure, you can also look at it as an energy on your flow.
No you cannot! Pressure and energy are not equivalent, the units are different. Just as length and time are different, so is pressure and energy. If you want the energy in a flow (which you can then mass flow average) then integrate velocity^2.

brunoc September 15, 2011 15:48

Quote:

Originally Posted by ghorrocks (Post 324168)
Density * velocity ^2, ie pressure? I cannot see how this can be regarded as a transported variable.

Density is transported. Momentum is transported. Velocity could be seen as momentum per density. So isn't \rho U^2 transported too?

Quote:

Originally Posted by ghorrocks (Post 324168)
Pressure and energy are not equivalent, the units are different. Just as length and time are different, so is pressure and energy. If you want the energy in a flow (which you can then mass flow average) then integrate velocity^2.

Yes, units are different. But what I meant was "energy contained in the flow". A vessel filed with fluid under high pressure has lots of stored "energy" (maybe I'm using the wrong word, english is not my maiden language). Take Bernoulli on inviscid flow. It relates dynamic pressure, static pressure and potential energy. Decreasing one type of "energy" adds to another type.

ghorrocks September 15, 2011 18:55

My argument is purely dimensional, and your points are entirely valid also. So I guess whether you think the dimensional argument is convincing will determine which way you think on this issue.

AliLemprex April 16, 2015 09:04

Hello,

I stumbled across this thread looking for a solution to a similar problem.

I would like to implement Momentum Sources in a cylindrical geometry. The solution to my problem is not converging, though, so I would like to steadily increase my Momentum Source Terms up to the desired value to achieve convergence.

How do I apply this in CFX? Any help would be appreciated!

Thank you very much

brunoc April 16, 2015 10:18

Implement the source terms as described in this thread. You can then use a 1D table to ramp the value of your source coefficient according to the iteration number. The tutorials show how to use a 1D table.

If you're not used to 1D tables (or CFX in general) you can skip that and manually change the source coefficient value as the simulation runs by going to 'Tools > Edit Run in Progress' in the Solver Manager.

AliLemprex April 17, 2015 05:35

Thank you very much brunoc for the quick help!

Would you recommend any slope for increasing? I was wondering whether the speed with which the values were increased had any influence on my convergence, i.e. do I have to achieve a convergent state with my first value to continue on to the next one?

Cheers,

AliLemprex

ghorrocks April 17, 2015 05:41

If you have introduced a source term and it is not converging properly then you should use a source term coefficient. Correctly implemented that can make the difference between poor convergence and quick easy convergence.

Have you implemented a source term coefficient?

AliLemprex April 17, 2015 05:56

Hello Glenn,

yes, I have implemented a Source Coefficient. I think my problem may be that the original Permeability for my porous media (calculating pressure losses with Darcy Permeability model) may be too little for an initial value so I am aiming to approach the desired permeability value. Calculations have already converged for higher permeabilities on the same geometry and mesh.

I was just wondering whether there were certain strategies to safely achieve convergence when decreasing the permeability during the simulation run. Do I have to achieve a converged solution for a given permeability to safely continue to a smaller value?

Thanks a lot for the help

Antech January 18, 2016 03:40

Hello. My "5 cents" about pressure averaging (sorry for offtopic).
For integral pressure drop estimations we need the total pressure averaging on inlet and outlet BCs of our models. At first, I tried a simple mass flow averaging for total pressure but it seems to be absolutely incorrect if we look at the underlying physics.

In my opinion, the static pressure component should be area-averaged and the dynamic pressure component should be mass-flow-averaged, adding these two parts we get correct integral total pressure. I think, its clear about static pressure, but how about dynamic one? It can be treated as a unit kinetic energy (per 1 m^3) of fluid (E=0.5*V*Rho*w^2, Pdyn=0.5*Rho*w^2). Unit kinetic energy is definitely a conserved value and can be transported so, IMHO, the dynamic pressure should be mass-flow-averaged.

I tried an area-averaging of total pressure on one of our models (HRSG gas ducts, irregular velocity profile at inlet) and it gives non-physical results: total pressure increases in flow direction although there are only aerodynamic resistances, no any sources like fans e.t.c. Using areaAve for static pressure + massFlowAve for dynamic pressure (0.5*Rho*w^2) helps to eliminate this non-physical behavior.


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