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Same pressure gradient but different velocity field 

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April 19, 2018, 19:35 
Same pressure gradient but different velocity field

#1 
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Hello guys,
I am running a simple steadystate laminar case within a cubic box in openFoam with fixed static pressure. When I assign fixed static pressure at inlet, let's say with constant pressure gradient = 1, and it will generate large velocity within the domain. When I specified total pressure at the inlet which will result same static pressure gradient = 1 in the end, yet the velocity field is way more smaller than the previous one. I am confused: since the static pressure gradients at steadystate for both cases are exactly the same, and the purpose of pressure gradient is balancing the viscous force, so I think both cases should have same velocity fields. However the computations give complete different results. Why is that?  I am using SIMPLE algorithm solving a simple laminar flow in a box, so it will be a 1D flow. I compare the results from 2 different BCs: 1. fixed static pressure at inlet; 2. fixed total pressure at inlet. And fixed static pressure at outlet for both cases. First one will give a constant pressure difference of 1. 2nd one has pressure difference of 3.8 initially and will reduce to 1 when the flow reaches steady state. Since pressure gradient is solely balancing with viscous force, these 2 cases has exactly static pressure difference(or pressure gradient), it should have same velocity field. However, my results shows that the 1st case gives a much larger velocity field. I have no idea why this is happening. Last edited by TurbJet; April 23, 2018 at 14:00. 

April 23, 2018, 03:34 

#3 
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Filippo Maria Denaro
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I don't understand your problem without more details. What type of formulation are you using?


April 23, 2018, 13:58 

#4  
Senior Member

Quote:
1. fixed static pressure at inlet; 2. fixed total pressure at inlet. And fixed static pressure at outlet for both cases. First one will give a constant pressure difference of 1. 2nd one has pressure difference of 3.8 initially and will reduce to 1 when the flow reaches steady state. Since pressure gradient is solely balancing with viscous force, these 2 cases has exactly static pressure difference(or pressure gradient), it should have same velocity field. However, my results shows that the 1st case gives a much larger velocity field. I have no idea why this is happening. 

April 23, 2018, 14:05 

#5 
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Filippo Maria Denaro
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In summary, you have an incompressible flow at a steady state but I don't knwo if you have up and down walls.
Remember that for incompressible flow you have not thermodinamic meaning of the pressure and the realtion between static and total pressure p0=p+ k =constant is valid only in absence of dissipative effects therefore assuming regular inviscid flows 

April 23, 2018, 14:18 

#6  
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Quote:
I am not sure I understand your point. My pressures are not constant. My pressure gradient is constant. 

April 23, 2018, 14:27 

#7 
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Filippo Maria Denaro
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For incompressible flows you set either velocity or pressure as BC.s. If you set at the inlet a static pressure you do not set the velocity profile that must be computed from the solution developing in the interior. Of course, if you set the total pressure (sum of static and dynamic pressure) you should have difference in the generated velocity profile. That depends on the way you set the outlet and the global pressure difference acts.


April 23, 2018, 14:28 

#8  
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Quote:


April 23, 2018, 14:34 

#9  
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Filippo Maria Denaro
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Quote:
IF you want to work setting only the pressure at the inlet and outlet you need to know that the difference gives the flow rate. 

April 23, 2018, 14:43 

#10  
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Quote:


April 23, 2018, 14:46 

#11 
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Filippo Maria Denaro
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Since the second derivative is constant, the velocity is parabolic u(y)=c0+c1*y+c2*y^2. Two conditions are determined by the no slip condition at the walls but the third depends on the flow rate, that is on the max velocity at the centerline. Therefore you need to provide a third condition by setting the difference p_outlet  p_inlet 

April 23, 2018, 14:52 

#12  
Senior Member

Quote:
But just as I stated, eventually, both cases will have same p_out  p_in. It should give the same c0. I am thinking about, with constant pressure gradient (the 1st case), would it be possible that it is solving NS incorrectly? Since pressure gradient is constant, basically it is solving NS without pressure term. 

April 23, 2018, 15:11 

#13  
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Filippo Maria Denaro
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Quote:
Yes. Setting the static pressure difference between inlet and outlet and fixing the Re number will give you the third condition. Of course, owing to the dissipative effects, also the total pressure has a loss. 

April 23, 2018, 15:17 

#14  
Senior Member

Quote:
Let's simplify my question to: with same pressure gradient at steady state, why these 2 cases have different velocity field? 1st case has much larger velocity, which means it has much larger velocity gradient along ydirection to balance the pressure gradient; 2nd case has smaller velocity gradient along ydirection, which means the pressure gradient only needs smaller viscous force to balance. So which one is correct? 

April 23, 2018, 15:21 

#15  
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Filippo Maria Denaro
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Quote:
Well, if the two velocity profiles are different, the integrals are different too and hence the flow rates are different. That corresponds to a different (p_outletp_inlet) values. Hence, you are setting different problems. Check some bug in the setting. 

April 23, 2018, 16:01 

#16 
Senior Member

I double checked, and I don't think I have bugs in my settings.


April 23, 2018, 16:14 

#17 
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Filippo Maria Denaro
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Analytically there is only a different pressure difference that produces a different flow rate.
The first thing to assess is if the two velocity profiles are both a parabola or one of them is wrong. If they are both a parabola with different max values, the setting of the pressure gives a different flow rates and you have to check what happens. Of course check also that at the steady state the vertical velocity is always zero and the continuity equation is satisfied. 

April 23, 2018, 16:19 

#18  
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Quote:
So maybe the 1st case will result in incorrect solutions? 

April 23, 2018, 16:27 

#19  
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Filippo Maria Denaro
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Quote:
You have an analytical solution, you can check all you want to assess which solution is correct 

April 23, 2018, 19:32 

#20 
Senior Member

I kind of figured this out. 1st case will be Poiseuille flow and it follows the parabola profile. But for 2nd case, it's not Poiseuill flow, the velocity profile will not be parabola. It has much larger gradient at near wall region to balance the pressure gradient, so the entire velocity field is smaller than the 1st case.


Tags 
pressure and velocity, pressure boundaries, viscous laminar 
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