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-   -   CAN'T CONVERGE WITH REVERSE FLOW AT OUTLET! (http://www.cfd-online.com/Forums/fluent/30619-cant-converge-reverse-flow-outlet.html)

XIAOYI LI December 4, 2002 15:46

CAN'T CONVERGE WITH REVERSE FLOW AT OUTLET!
 
I am doing simulation that there maybe a reverse flow occurs at outlet. I am using pressure outlet boundary conditon. The results can't get converged.. Any one have experience with the reverse flow at outlet? Thanks!

Anders December 4, 2002 16:00

Re: CAN'T CONVERGE WITH REVERSE FLOW AT OUTLET!
 
Sure, that can be a problem. The classic answer to that is to try and extend your domain so that any potential recirculation resides within your domain and not across a boundary, i.e. to extrude some layers downstream if it is a pipe-like geometry, or to build some box onto the end to simulate the rest of the world... Is that possible?

XIAOYI LI December 4, 2002 21:15

Re: CAN'T CONVERGE WITH REVERSE FLOW AT OUTLET!
 
THANKS A LOT! I WILL TRY IT TOMORROW.

Tom December 5, 2002 10:00

Re: CAN'T CONVERGE WITH REVERSE FLOW AT OUTLET!
 
I have done a lot of work with reverse flow at a pressure outlet. What type of discretization are you using? I find in most of my simulations that I need to use second order upwind rather than first order to achieve the desired level of reverse flow. Maybe you could try this.

Bharath December 5, 2002 13:23

Re: CAN'T CONVERGE WITH REVERSE FLOW AT OUTLET!
 
Hello ..

I have been working on Spray modelling and It is required to obtain a Uniformly Distributed Fine Spray at the Exit of the nozzle.To achieve this..I tried to extend my nozzle exit by constructing an Visualisation area of order 100mm*60 mm ( and gave pressure outlet conditions at the edes )while my nozzle dimensions are of the order 1-5 mm....do u think such a big Visualization area is required to see the flow at the exit ?..would it have any impact on the Convergence of the program and hence the modelling ??

Thanks

Bharath

lixiaoyi December 10, 2002 13:17

SECOND ORDER UPWIND DOES WORK! THANKS
 
This does work! Thanks

rk December 15, 2002 09:43

Re: SECOND ORDER UPWIND DOES WORK! THANKS
 
When there is a reversal flow at the pressure outlet, the static pressure mentioned at the outlet will be taken as total pressure. So essentially, the pressure outlet acts as pressure inlet. But the velocity component will be taken as normal to the boundary. This creates problem if there is a significant tangential component at the outlet. The above mentioned assumption destroys the other 2 components, tangential & radial. Moreover, you will observe different flow field at the outlet if there is physically reverse flow at the outlet.

So only solution is to extend the domain little bit faraway and define that as pressure outlet. When you look at the results, create a surface at the actual outlet!

Thanks,

rk

Chris December 18, 2002 12:27

Re: SECOND ORDER UPWIND DOES WORK! THANKS
 
Hi, this is a question...

I modeled a square box and put in boundary conditions at 2 opposites ends. One was a total pressure, and the other was a static pressure (lower than the total pressure). -- I had already set my operating pressure to zero earlier, so all these are gauge pressures.

I keep getting 'reversed flow' during my iterations. I'm not sure if lixiaoyi was purposely trying to force reversed flow at her exit, but in MY case, I don't expect reversed flow...

Did I make any careless mistakes somewhere? Or is it because I'm using a square box, and that I should try lengthening my box like what rk suggested? Or are my boundary conditions dodgy in the first place?

Please advise :) Gracias.


siddharameshwara January 31, 2011 07:57

Reverse flow
 
Quote:

Originally Posted by rk
;103797
When there is a reversal flow at the pressure outlet, the static pressure mentioned at the outlet will be taken as total pressure. So essentially, the pressure outlet acts as pressure inlet. But the velocity component will be taken as normal to the boundary. This creates problem if there is a significant tangential component at the outlet. The above mentioned assumption destroys the other 2 components, tangential & radial. Moreover, you will observe different flow field at the outlet if there is physically reverse flow at the outlet.

So only solution is to extend the domain little bit faraway and define that as pressure outlet. When you look at the results, create a surface at the actual outlet!

Thanks,

rk

Hello

could you please explain in detail what do you mean by that? Right now i am facing the same problem

siddharameshwara February 9, 2011 05:53

[QUOTE=rk
;103797]When there is a reversal flow at the pressure outlet, the static pressure mentioned at the outlet will be taken as total pressure. So essentially, the pressure outlet acts as pressure inlet. But the velocity component will be taken as normal to the boundary. This creates problem if there is a significant tangential component at the outlet. The above mentioned assumption destroys the other 2 components, tangential & radial. Moreover, you will observe different flow field at the outlet if there is physically reverse flow at the outlet.

So only solution is to extend the domain little bit faraway and define that as pressure outlet. When you look at the results, create a surface at the actual outlet!

Thanks,

Hi rk,

Even i am facing the same problem. I tried to extrude the outlet but invain. Could you please tell me the reason i didnt understand "But the velocity component will be taken as normal to the boundary. This creates problem if there is a significant tangential component at the outlet." this statement

Dinesh_Dhande September 14, 2012 16:39

Reverse flow problem
 
Quote:

Originally Posted by Tom
;103645
I have done a lot of work with reverse flow at a pressure outlet. What type of discretization are you using? I find in most of my simulations that I need to use second order upwind rather than first order to achieve the desired level of reverse flow. Maybe you could try this.

Dear Tom,
I am simulating water flow through journal bearing. The shaft radius is 50mm and the bearing radius is 50.43 mm. The flow domain is clearance between shaft (rotary member with velocity 108.91 rad/sec) and bearing (stationary member.) The length is 200mm. I am using segregated solver and the flow is laminar. The inlet BC is 50 KPa and outlet BC is 42kPa. When i simulate the system i am getting reverse flow first on outlet and then inlet. Kindly help.

Dinesh_Dhande September 14, 2012 16:43

Dear Tom,
I am simulating water flow through journal bearing. The shaft radius is 50mm and the bearing radius is 50.43 mm. The flow domain is clearance between shaft (rotary member with velocity 108.91 rad/sec) and bearing (stationary member.) The length is 200mm. I am using segregated solver and the flow is laminar. The inlet BC is 50 KPa and outlet BC is 42kPa. When i simulate the system i am getting reverse flow first on outlet and then inlet. Kindly help.


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