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-   -   Pressure Outlet Guage pressure (https://www.cfd-online.com/Forums/fluent/92099-pressure-outlet-guage-pressure.html)

 Mohsin September 2, 2011 04:54

Pressure Outlet Guage pressure

Absolute Pressure=Operating pressure+Guage pressure

I have 2 inlets and 1 outlet. The domain includes 2 phases, air and liquid solution. Vapors are generated due to high temperature in the domain. The guage pressures are known at the 2 inlets and 1 outlet (all boundaries). Velocities at the inlets are also known. hence, "velocity Inlet" boundary condition" at the inlets is used. Incompressible ideal gas law is used, which calculates pressure by operating Pressure, so i used the Pressures of the inlets (which is 2 Bar) as an operating pressure.

For the outlet (for air and vapor flow) I dont know the rate of mass flow of air and vapor at the outlet so I cannot use "Outflow" boundary condition. Now, i am left with "Pressure outlet boundary condition" at the outlet. For pressure outlet, I need guage pressure which is -6666Pa in my case.

My question is: what guage pressure should i use for Pressure outlet boundary condition used at the outlet. Would it be (-6666pa). Would it be ok to use negative value and would the change in operating presssure from 101325 Pa (default) to 2 bar effect the guage pressure here?

Thanks

 Amir September 2, 2011 07:50

Hi,

FLUENT accept negative signs; as you said, they are relative pressures.
Because of your specific state law which you have used, changing operating pressure would change results; i.e, in this state law, you have to set operating pressure wisely because it's used directly in state law. So if you want to use this state law, it's better to set a value which you'd have little pressure variation regard to that in whole domain; maybe an average of inlet and outlet will be good as first guess and you can improve that after results will be obtained.

Bests,

 Mohsin September 2, 2011 10:26

Amir, Thank you very much for your time.

The 2 inlets for N2 gas has a pressure of Pguage=200Kpa each.
The outlet is a suction which creates a vacuum pressure Pguage=-6666Pa. The operating Pressure used by me is Po=200Kpa (as the 2 inlets had this pressure)

For outlet boundary condition Pguage @the outlet is required.
Pguage=-6666Pa but It might be relative to 101,325 Pa. As operating pressure was changed by me from 101,325 Pa to 200 Kpa so Pguage may also be changed like this:

Pguage@outlet=-6666-200K =-206.66KPa

This negative value was input by me into the pressure outlet boundary condition for suction at the outlet.

Now, you mean to say that, I should not use 200Kpa value as operating pressure, rather, I should use the average of 200Kpa and -6666pa as the operating pressure. and you said, I can improve it after results. Could u elaborate on that? how can I improve it? Could you tell me the procedure.

thank you very much. So nice of you.

 Amir September 2, 2011 15:01

Quote:
 Originally Posted by Mohsin (Post 322668) For outlet boundary condition Pguage @the outlet is required. Pguage=-6666Pa but It might be relative to 101,325 Pa. As operating pressure was changed by me from 101,325 Pa to 200 Kpa so Pguage may also be changed like this: Pguage@outlet=-6666-200K =-206.66KPa
P_static@outlet=101.325-6.666=94.659K; P_gauge@outlet=94.659-200=-105.341K; right?

As I said before; FLUENT accept negative signs.
Quote:
 Originally Posted by Mohsin (Post 322668) Now, you mean to say that, I should not use 200Kpa value as operating pressure, rather, I should use the average of 200Kpa and -6666pa as the operating pressure. and you said, I can improve it after results. Could u elaborate on that? how can I improve it? Could you tell me the procedure.
with op pressure=200K :
P_gauge@inlet=101.325K
P_gauge@outlet=-105.341K
I meant that; if you insist on using incompressible ideal gas law, in order to achieve more accurate density, set an operating pressure with minimum pressure deviation; I wasn't sure about static or gauge kind of pressure you've reported, but now it seems that 200kPa is a good choice as your first guess because deviation of pressure are almost equal. when you have got new results; you'll have pressure for each cell and by a statistical procedure you can find new op pressure which has minimum pressure deviation about. But this procedure wouldn't need if you use ideal gas law instead.;)

Bests,

 Mohsin September 3, 2011 02:01

Quote:
 Originally Posted by Amir (Post 322693) P_static@outlet=101.325-6.666=94.659K; P_gauge@outlet=94.659-200=-105.341K; right? As I said before; it's not important to set negative values. with op pressure=200K : P_gauge@inlet=101.325K P_gauge@outlet=-105.341K Bests,
Experimental Given Max guage pressure @2 inlets=200Kpa
Experimental Given Min guage Pressure @ 1 outlet suction= -6666Pa
Operating pressure used by me in FLUENT=200Kpa (as both inlets have same pressure so it seems a better option as opposed to the default 101,325Pa)

Based on these aforementionentioned values,

For Velocity_Inlet_Boundary_condition:
P_guage@inlet1=0
P_guage@inlet2=0

For Pressure Outlet Boundary condition:
P_guage@outelt-suction=-6666-200K =-206.66KPa

You said negative sign is not important but if i dont use negative sign with 206.66Kpa, it would mean that the P_Absolute @outelt-suction=206.66K+200K=406.66Kpa. Rather than, P_Absolute=-206.66K+200K=-6.66KPa

Secondly, could you please tell me that the values calculated by me are correct?

Third, I would use "ideal gas" instead of "Incompressible ideal gas law" as per our discussion of presure variations.

Thanks a lot of your kindness.
Mohsin

 Amir September 3, 2011 05:45

Quote:
 Originally Posted by Mohsin (Post 322724) You said negative sign is not important
:eek: I didn't say that! I meant FLUENT accept both positive and negative values and you don't need to concern about that, so it's not important but it's necessary:D; I thought you were not confident of setting negative values. I just wanted to say these are relative pressures and it's obvious that they may be positive or negative and you have to set correct sign hope that clear now! (I edited this doubting sentence)
Quote:
 Originally Posted by Mohsin (Post 322724) Secondly, could you please tell me that the values calculated by me are correct?
I don't think so.
@inlet:P_static=301.325kPa so new gauge pressure=101.325kPa
@outlet:P_static=94.659KPa so new gauge pressure=-105.341kPa
and you have to consider signs as well.:D
maybe it's better to compute staic pressures with previous op pressure and then find new gauge ones; as done above.
Quote:
 Originally Posted by Mohsin (Post 322724) Third, I would use "ideal gas" instead of "Incompressible ideal gas law" as per our discussion of presure variations. Mohsin
yes, it's better for this case.;)

Bests,

 Mohsin September 3, 2011 06:47

Dear Amir

I agree with your values. I had made a mistake. So nice of you to help me to solve this issue.

Could you also put some light on this thread also:

http://www.cfd-online.com/Forums/flu...ry-inputs.html

Thanks

 Mohsin September 21, 2011 02:21

Dear Amir,

As I have already told u in the previous posts that my geometry contains 2 inlets and 1 outlet.

As you suggested the following values:

P operating= 200kPa
@inlet:P_static=301.325kPa so new gauge pressure@inlet=101.325kPa
@outlet:P_static=94.659KPa so new gauge pressure@outlet=-105.341kPa

Now, as

Absolute pressure=Static Pressure +Operating pressure

which in my case should be around 301.325Kpa at the inlets. However, after convergence, the contour plots for "Absolute pressure" at the inlets are not showing it to be anywhere near 301.325Kpa rather the contour plots are giving me a very lower value (close to outlet Pressure-rather it should be 301.325 Kpa). Why is it so? I am completly baffeled by this..Please help....

 Amir September 21, 2011 02:39

Quote:
 Originally Posted by Mohsin (Post 324976) Absolute pressure=Static Pressure +Operating pressure
Dear Mohsin,

First of all, the above relation is not correct; correct form is something like this:
Code:

`Absolute pressure=Gauge Pressure +Operating pressure`
Secondly, you have to note that the value you've set in FLUENT for pressure-inlet boundaries is gauge total pressure and for pressure-outlet is gauge static pressure.

Bests,

 Mohsin September 21, 2011 02:49

Yes thank you for the correction

Total guage pressure=Static pressure+Dynamic Pressure
Absolute pressure=guage Pressure + Operating Pressure

Now, at the inlets, as the absolute pressure is about 3 bar, then why the absolute pressure is being shown at about 1 bar at the inlets. rather it should show 3 bar.

It is a little confusing.....

 Amir September 21, 2011 03:02

Quote:
 Originally Posted by Mohsin (Post 324984) Total guage pressure=Static pressure+Dynamic Pressure
This equation is not correct again!
Code:

`Total guage pressure=Static guage pressure+Dynamic guage Pressure`
It's obvious, you've set gauge total pressure and you're seeing static pressure.

Bests,

 Mohsin September 21, 2011 03:14

Dear Amir,

Total guage pressure=Static guage pressure+Dynamic guage Pressure
Absolute pressure=guage Pressure + Operating Pressure

I am using "Velocity inlet BC" at the inlets and "Pressure outlet BC" at the outlets. For Velocity inlet BC "Initial Guage pressure" was required (which is the guage static pressure at the inlet).

Could you please elaborate on this:

"It's obvious, you've set gauge total pressure and you're seeing static pressure".

How can i see the correct Pabs. I will be really glad for your guidance.

 Amir September 21, 2011 03:26

Quote:
 Originally Posted by Mohsin (Post 324989) I am using "Velocity inlet BC" at the inlets and "Pressure outlet BC" at the outlets. For Velocity inlet BC "Initial Guage pressure" was required (which is the guage static pressure at the inlet).
I thought your inlet condition is pressure inlet; anyway, you're using velocity inlet and in density based solver it need out flow gauge pressure which is not a boundary condition for pressure it's just a threshold to enhance stability; here, you can see that inlet velocity will be shown correctly, right? so to achieve your pressure, maybe it's better to use pressure-inlet condition.

Bests,

 Mohsin September 21, 2011 03:36

oh so it means I have to simulate all over again by "Pressure inlet" boundary condition in order to simulate the correct pressure.

But

I have been given velocities at inlet and my goal is to change the velocities for various simulations to get the optimum condition for the geometry. If, I will use Pressure inlet boundary condition, how can it calculate the velocity? or which bounadry condition would be feasible in this case?

Secondly, which value of "Initial Guage pressure" should be specified in my case? or should it be 0?

thank you very much.

 Amir September 21, 2011 03:49

Quote:
 Originally Posted by Mohsin (Post 324994) oh so it means I have to simulate all over again by "Pressure inlet" boundary condition in order to simulate the correct pressure. But I have been given velocities at inlet and my goal is to change the velocities for various simulations to get the optimum condition for the geometry. If, I will use Pressure inlet boundary condition, how can it calculate the velocity? or which bounadry condition would be feasible in this case? Secondly, which value of "Initial Guage pressure" should be specified in my case? or should it be 0? thank you very much.
Dear Mohsin,
I think inlet pressure is more feasible for your purpose, you can set desired velocity by adjusting static and total pressure @ inlet; but an important note, these definition provided so far for total pressure are valid for incompressible flow, for compressible flow total pressure is evaluated via isentropic relation as a function of mach No.
So it depend on your case that which relation to use for computing total pressure.

Bests,

 Mohsin September 21, 2011 04:04

Dear Amir,

Thank you so so much.

before concluding i have 2 short querries:

1. I am using Ideal gas law for compressible fluids to calculate density. As we discussed in the following thread that my domain has pressure gardients so Ideal gas law for compressible fluids would be effective in my case.

http://www.cfd-online.com/Forums/flu...essiblity.html

So, these relations for Total pressure are not valid in compressible case instead Mach number and isentropic flow relations are used.

Hence, Do u think, for calculating density I may change to "incompressible ideal gas law" (as Mach number is less than 0.1, but pressure variations are there) so that the equations for total pressure are valid? without much loss of accuracy?

2. Secondly, which value of "Initial Guage pressure" should be specified in my case? for pressure inlet boundary conditions?

 Amir September 21, 2011 04:29

Quote:
 Originally Posted by Mohsin (Post 325002) 1. I am using Ideal gas law for compressible fluids to calculate density. As we discussed in the following thread that my domain has pressure gardients so Ideal gas law for compressible fluids would be effective in my case. http://www.cfd-online.com/Forums/flu...essiblity.html So, these relations for Total pressure are not valid in compressible case instead Mach number and isentropic flow relations are used. Hence, Do u think, for calculating density I may change to "incompressible ideal gas law" (as Mach number is less than 0.1, but pressure variations are there) so that the equations for total pressure are valid? without much loss of accuracy?
As you know, ideal gas law is more general in comparison with incompressible type; I think you don't have better choice because incompressible type is proper just for cases in which pressure gradient is negligible.
Both total pressure equations have similar trends up to mach number 0.3; you can simply examine that.
Quote:
 Originally Posted by Mohsin (Post 325002) 2. Secondly, which value of "Initial Guage pressure" should be specified in my case? for pressure inlet boundary conditions?
There are some inconsistency in your boundary condition. Supersonic gauge pressure will be ignored in subsonic because it's not necessary for subsonic regime! If you have subsonic regime, you cannot set velocity inlet and also pressures on your boundaries simultaneously! In other words, in subsonic regime, it's impossible to change velocity with fixed pressure boundaries unless you would change pressure, this is result of elliptic nature of equations.

Bests,

 Mohsin September 21, 2011 09:46

Thank you Amir,

I followed what you said, I used "pressure inlet Bounadry condition" instead of "velocity Inlet" boundary condition. The continuity residual was difficult to reduce however it reduced to 10-3. and the other residulas for velocity and energy were well below 10^-5, along with mass flux showing convergence.

The absolute pressure at the inlet was shown properly. However, this time the velocity at inlet is being shown 60m/s on the average (In real situation it is 0.5m/s).

Pressure inlet boundary condition is giving velocity error now. Can you tell me what might be the reason of this higher velocity? how is the velocity calculated by FLUENT and/or do i need to more iterations or do i need to change the discretization scheme?

 Amir September 21, 2011 10:41

As I said before, because you're modelling subsonic regime, you cannot set both pressure and velocity. Here, you've set pressure and achieve velocity... . The difference between achieved velocity magnitude and expected one is high, so I think there should be fundamental issues. I have some suggestions which may help:
1) ensure parameters value which you have provided and their dimension are correct ....
2) use pressure based technique for coupling ...
3) use constant density and see what will happen!
4) estimate approximate average velocity if you have a pipe with one inlet and one outlet with outlet diameter and with such pressure difference via analytic solution.
5) use resolved grid near boundaries ...

Bests,

 Mohsin September 21, 2011 10:55