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August 18, 2016, 10:40 
simpleFoam  pressure

#1 
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Igor Leonardo
Join Date: Jul 2016
Location: Brazil  SP  SJC
Posts: 20
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Hello everyone,
So i been reading around the forum, regarding pressure using simpleFoam. Imagine a simple model, like a straight tube, with one inlet, one outlet and some walls. If we use such boundary conditions, for U and P, as the follwing: U: Code:
dimensions [0 1 1 0 0 0 0]; internalField uniform (0 0 0); boundaryField { inlet { type fixedValue; value uniform (54.87 0 0); } outlet { type zeroGradient; } walls { type noSlip; } } Code:
dimensions [0 2 2 0 0 0 0]; internalField uniform 0; boundaryField { inlet { type zeroGradient; } outlet { type fixedValue; value uniform 0; } walls { type zeroGradient; } } My questions are the follwing: 1 With this kind of setup, using simpleFoam, the resulting pressure i would get when i looked at it on paraview for example, would also be p/rho right? So if wanted the actual pressure difference, or head loss, i would have to multiply everything by the density correct? 2 Now suppose that for a simple tube like that, i know the inlet velocity, and also the static pressure on the inlet. What would be the optimal boundary conditions for that situation? Or perhaps, i could keep the same setup, and then just subtract the value of the pressure that i know on the inlet with the values from pressure i obtained on the simulation? I hope i could make my self clear, if i couldn't, please let me know and i will ty to elaborate further. Thanks everyone in advance, Best regards, Igor Carvalho 

August 19, 2016, 02:31 

#2  
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Mohammed Gowhar
Join Date: Feb 2014
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Quote:
The pressure difference also same (Pressure_out Pressure_inlet). It is not required to multiply by density. Quote:


August 19, 2016, 09:34 
s

#3  
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Igor Leonardo
Join Date: Jul 2016
Location: Brazil  SP  SJC
Posts: 20
Rep Power: 6 
Quote:
I didn't understand your second statement. pout/rho  pin/rho = Delta p/rho So the actual Delta p would have to be multiplied by rho, to be just Delta p, right? Or am i confusing something? If pressure and velocity are known at the inlet, then i could use a fixed value for p and u on the inlet, and let the solver calculate both parameters on the outlet for example, what do you think? Something like this: U: Code:
dimensions [0 1 1 0 0 0 0]; internalField uniform (0 0 0); boundaryField { inlet { type fixedValue; value uniform (54.87 0 0); } outlet { type zeroGradient; // or calculated perhaps? } walls { type noSlip; } } Code:
dimensions [0 2 2 0 0 0 0]; internalField uniform 0; boundaryField { inlet { type fixedValue; value uniform 100; //value of actual p/rho } outlet { type zeroGradient; //or calculated perhaps? } walls { type zeroGradient; } } This kinda of setup is just something that i though of for a situation of which both velocity and pressure are known on the inlet and not known for the outlet, but i never tried this i don't know if it would work. My question was if there are appropriate boundary conditions for such case, or if we can assume what i said on my first post, that is, if by using pressure boundary conditions of zeroGradient on the inlet, and fixed value zero, on the outlet, then we could just subtract the value of the pressure that i know on the inlet with the values from pressure i obtained on the simulation, to obtain the actual pressure distribution. Thanks for looking at this thread. Best regards, Igor 

August 20, 2016, 07:57 

#4  
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Mohammed Gowhar
Join Date: Feb 2014
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I am sorry, we have to multiply by density to get actual Pressure.
Quote:
You can have boundary conditions as follow: 1. Velocity Inlet  Pressure Outlet 2. Mass flow rate Inlet  Pressure Outlet 3. Stagnation Inlet  Pressure Outlet In conclusion, you have to use zeroGradient for Inlet for P 

August 20, 2016, 10:19 
Velocity driven flow x Pressure driven flow

#5  
New Member
Igor Leonardo
Join Date: Jul 2016
Location: Brazil  SP  SJC
Posts: 20
Rep Power: 6 
Quote:
Sorry to bother you further but i have just 2 more questions: Consider the case i mentioned above, a simple elbow tube, 1 inlet, 1 outlet, fixed walls. The known parameters are inlet velocity and pressure. The value we would be looking for would be the force acting on the elbow, for example. 1 If i use boundary conditions such as the ones i mentioned: U: Inlet: fixed velocity. Outlet: zeroGradient. Walls: noSlip. p Inlet: zeroGradiente; Outlet: fixedValue zero; Walls: zeroGradient; This setup would be a velocity driven flow right? With this setup i can assume the velocity field would be correct. The pressure gradient i would obtain would be the head loss to that specific portion of the tube right? Here let me elaborate further about my question to try make my question very clear: Suppose that with this setup, i obtained the following data: Pressure on the outlet: zero (Because it was fixed 0) Pressure on the inlet: 500Pa. So i could say that the head loss of this part of the tube is 500Pa. But i know for a fact, that the pressure on the inlet is, for example, 10000Pa, so the real pressure i would have on my tube, considering this real pressure that i know would be: Pressure on the inlet: 10000Pa Pressure on the outlet 9500Pa. and the velocity field for this pressure would be correct too right? This concludes question number 1. If i need to elaborate further just tell me. Question 2 Now assume i would want to make a pressure driven flow, in your opinion, the following boundary conditions would be correct? p: Inlet: fixedValue 10000Pa. Outlet: fixedValue 9500Pa. Walls: zeroGradient. U: Inlet: pressureInletVelocity; Outlet: zeroGradient; Walls: zeroGradient; And the results i should be expect is now a correct pressure gradient. And for the velocity i'm not sure what i should expect? It should get to a inlet velocity same as the velocity i know that is there for a fact? This concludes question two. Sorry for asking so many questions... but i'm sure this will clarify things not only for me but for many other. Thanks for the awesome help in advantage. Best regards, Igor Carvalho. 

August 24, 2016, 07:12 

#6 
Member
Mohammed Gowhar
Join Date: Feb 2014
Posts: 47
Rep Power: 8 
Hi,
Since I was very busy with my work , I could not able to reply on time. Let me put it in simple words, Total Pressure = Static pressure + Dynamic Pressure Now consider the Bernoulli Equation, It always along the streamline. Do you agree? Rearrange the Bernoulli equation as follows Code:
(P1P2)/rho = 0.5*(V2^2  V1^2) Next question I will post the answer soon. Since I am in the middle of the work, I could not elaborate it right now. 

Tags 
boundary conditions, incompressible, p/rho, pressure, simplefoam 
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