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June 11, 2019, 04:55 |
negetive pressure loss
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#1 |
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hi every one
I encountered a problem with pressure loss my geometry is vertical channel. In this case, the phase change occurs,my boundary condition is attached. my problem: pressure at the outlet of channel flow be higher than the pressure at the inlet so the pressure loss is negetive thanks in advance Last edited by regard; June 15, 2019 at 02:42. |
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June 15, 2019, 02:47 |
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#2 |
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any help, hint and comment is appreciated
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June 17, 2019, 05:32 |
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#3 |
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Does the fluid slow down towards the outlet? I dont't know you inlet velocity. But if the fluid is slowing down in the pipe due to gravity, you will gain pressure.
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June 17, 2019, 15:07 |
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#4 |
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thanks for your reply
The flow direction is vertical upward,and my Reynolds number in inlet is 100 (velocity=.03) and pressure loss is negetive in Some time steps. |
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June 20, 2019, 03:31 |
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#5 |
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Ingo Riess
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It is difficult to reply to your question. I gather it is not a steady state simulation, as the pressure drop varies between time steps. So, maybe, the expression "pressure drop" is wrong in this instance. I would apply the term "pressure drop" only to steady state or time averaged flow problems.
As I understand, the static pressure (without hydrostatic pressure) is higher at the outlet than on the inlet. For a dynamic flow (e.g. LES), this may be due to acceleration terms in your flow equation. But then you mention a phase change, so it is at least a two phase flow problem. What is the phase change, liquid -> gas? bubbly flow? |
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June 20, 2019, 12:32 |
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#6 |
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very thanks for your kind reply
Exactly, it is not a steady state simulation yes liquid->gas (evaporation) & gas->liquid(Condensation) this picture show the phase change (Reference of this picture:Pressure drop and void fraction during flow boiling in rectangular minichannels in weightlessness,D. Brutin a,, V.S. Ajaev b, L. Tadrist) thanks |
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June 20, 2019, 13:16 |
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#7 |
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Ingo Riess
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Now I see what you're trying to model. And contrary to the photo, you include gravity. When you analyse the flow, you should clearly distinguish between total pressure, static pressure, hydrostatic pressure, static pressure without (rho g h) etc. In your flow regime, you have the buoyancy of the bubbles, gravity and friction. I can imagine that you may have a higher static pressure on the outlet for some time steps.
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June 20, 2019, 13:49 |
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#8 |
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special thanks to you... In your opinion, the boundary conditions that I set it ,correct?
and I have obtained the pressure drop from the following code: deltap { type patchExpression; variables ( "pout{patch'outlet}=sum(p*area())/sum(area());" ); patches ( inlet ); expression "p-pout"; accumulations ( average ); verbose true; outputControlMode timeStep;//outputTime;// outputInterval 100; } |
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June 21, 2019, 03:19 |
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#9 |
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Ingo Riess
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While in the past, I did some experimental research (PhD) on air-bubble/water flow, I am fairly new to OpenFoam. So, I cannot advise you on the boundary conditions.
For your flow problem, I would use the pressure condition static pressure p=0 (or atmospheric pressure) on the outlet (assuming this is a free stream outlet). But this really depends on the physics you are trying to model. |
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June 21, 2019, 05:35 |
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#10 |
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Thank you so much.
best. |
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Tags |
negetive pressure loss, pressure loss |
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