|
[Sponsors] |
September 2, 2011, 04:54 |
Pressure Outlet Guage pressure
|
#1 |
Senior Member
Mohsin Mukhtar
Join Date: Mar 2010
Location: South Korea
Posts: 249
Rep Power: 17 |
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 |
|
September 2, 2011, 07:50 |
|
#2 |
Senior Member
|
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,
__________________
Amir Last edited by Amir; September 3, 2011 at 05:46. |
|
September 2, 2011, 10:26 |
|
#3 |
Senior Member
Mohsin Mukhtar
Join Date: Mar 2010
Location: South Korea
Posts: 249
Rep Power: 17 |
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. |
|
September 2, 2011, 15:01 |
|
#4 | ||
Senior Member
|
Quote:
As I said before; FLUENT accept negative signs. Quote:
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,
__________________
Amir Last edited by Amir; September 3, 2011 at 05:47. |
|||
September 3, 2011, 02:01 |
|
#5 | |
Senior Member
Mohsin Mukhtar
Join Date: Mar 2010
Location: South Korea
Posts: 249
Rep Power: 17 |
Quote:
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 Last edited by Mohsin; September 3, 2011 at 03:26. |
||
September 3, 2011, 05:45 |
|
#6 | ||
Senior Member
|
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; 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:
@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. maybe it's better to compute staic pressures with previous op pressure and then find new gauge ones; as done above. Quote:
Bests,
__________________
Amir |
|||
September 3, 2011, 06:47 |
|
#7 |
Senior Member
Mohsin Mukhtar
Join Date: Mar 2010
Location: South Korea
Posts: 249
Rep Power: 17 |
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 |
|
September 21, 2011, 02:21 |
|
#8 |
Senior Member
Mohsin Mukhtar
Join Date: Mar 2010
Location: South Korea
Posts: 249
Rep Power: 17 |
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.... |
|
September 21, 2011, 02:39 |
|
#9 |
Senior Member
|
Dear Mohsin,
First of all, the above relation is not correct; correct form is something like this: Code:
Absolute pressure=Gauge Pressure +Operating pressure Bests,
__________________
Amir |
|
September 21, 2011, 02:49 |
|
#10 |
Senior Member
Mohsin Mukhtar
Join Date: Mar 2010
Location: South Korea
Posts: 249
Rep Power: 17 |
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..... |
|
September 21, 2011, 03:02 |
|
#11 |
Senior Member
|
This equation is not correct again!
Code:
Total guage pressure=Static guage pressure+Dynamic guage Pressure Bests,
__________________
Amir |
|
September 21, 2011, 03:14 |
|
#12 |
Senior Member
Mohsin Mukhtar
Join Date: Mar 2010
Location: South Korea
Posts: 249
Rep Power: 17 |
Dear Amir,
Thank you for your reply. let me correct the equations: 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. |
|
September 21, 2011, 03:26 |
|
#13 | |
Senior Member
|
Quote:
Bests,
__________________
Amir |
||
September 21, 2011, 03:36 |
|
#14 |
Senior Member
Mohsin Mukhtar
Join Date: Mar 2010
Location: South Korea
Posts: 249
Rep Power: 17 |
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. |
|
September 21, 2011, 03:49 |
|
#15 | |
Senior Member
|
Quote:
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,
__________________
Amir |
||
September 21, 2011, 04:04 |
|
#16 |
Senior Member
Mohsin Mukhtar
Join Date: Mar 2010
Location: South Korea
Posts: 249
Rep Power: 17 |
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? |
|
September 21, 2011, 04:29 |
|
#17 | ||
Senior Member
|
Quote:
Both total pressure equations have similar trends up to mach number 0.3; you can simply examine that. Quote:
Bests,
__________________
Amir Last edited by Amir; September 21, 2011 at 08:01. |
|||
September 21, 2011, 09:46 |
|
#18 |
Senior Member
Mohsin Mukhtar
Join Date: Mar 2010
Location: South Korea
Posts: 249
Rep Power: 17 |
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? Please assist. |
|
September 21, 2011, 10:41 |
|
#19 |
Senior Member
|
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,
__________________
Amir |
|
September 21, 2011, 10:55 |
|
#20 |
Senior Member
Mohsin Mukhtar
Join Date: Mar 2010
Location: South Korea
Posts: 249
Rep Power: 17 |
Thank you for your reply.
I will use Constant density and will see the result and post here by tomorow. Just a question: what do u think, the results for "velocity inlet" case were fine? even though the Absolute pressure contours were not accurate in the contour report? Can they be used or they may be highly inaccurate?? |
|
Thread Tools | Search this Thread |
Display Modes | |
|
|
Similar Threads | ||||
Thread | Thread Starter | Forum | Replies | Last Post |
Outlet pressure for compressible flow | Michelle | CFX | 12 | September 1, 2015 18:38 |
Pressure Outlet setting | CoG | STAR-CCM+ | 4 | June 9, 2010 21:47 |
what actually is the 'zero pressure outlet b. c.' | hwe001 | CFX | 4 | June 7, 2010 15:22 |
Pressure Rise Error | emueller | CFX | 0 | May 5, 2009 11:08 |
UDF in Fluent to Match Mass Flow at Pressure Outlet | Jonas Larsson | Main CFD Forum | 1 | April 29, 1999 10:44 |