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Which pressure drop has to be chosen? (Pipe laminar flow) |
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September 1, 2018, 19:13 |
Which pressure drop has to be chosen? (Pipe laminar flow)
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#1 |
Member
Liliana de Luca Xavier Augusto
Join Date: Feb 2013
Posts: 68
Rep Power: 13 |
Hello all. I have a question about a pressure drive flow in a pipe.
I found a thesis where the author said this: "At the inlet, a static pressure equal to the total pressure drop along the pipe was specified; the boundary condition at the outlet was then a zero static pressure. These pressure values were relative to the reference pressure, which was set at 100 kPa. For gravity driven flow, the pressure drop along the pipe was chosen such that dp/dL = ρg, which gave a relative pressure at the inlet of 58.86 Pa." The geometry is a pipe with 0.006 m length (x) and 0.002 m radius (r). The rho of the fluid is 1000 kg/m³. When he use a viscosity of 1 Pa.s, the maximum velocity he gets was around 0.0098 m/s. By Hagen-Poiseuille equation vmax = -R²/4mu * dp/dx, it is correct if we use R=0.002m, mu=1Pa.s and dp/dx=9.81kPa. If the length of the pipe is 0.006m, so the difference between the inlet and outlet should be 58.86 Pa (9.81kPa = 0.006*9810). Is that right? So, if I want to reproduce his results, should I use 58.86 Pa as a "pressure-inlet" boundary condition and 0 Pa as "pressure-outlet" boundary condition in Fluent? I did this, and I got a completely different results. I got a parabolic profile for the velocity, but the maximum velocity was totally different. Does anyone now what values for boundary conditions should I use to get 9.81 kPa/m as dP/dx? Why not 58.86 Pa for a pipe of 0.006 m length? Thanks in advance! |
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September 2, 2018, 00:11 |
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#2 |
Senior Member
Behrooz Jamshidi
Join Date: Apr 2013
Posts: 110
Rep Power: 13 |
Hi
Be careful that in Fluent software, pressure inlet is total pressure boundary condition not static one. So surely your maximum velocity is less than the thesis because you have set 58.86 Pa as total pressure for inlet. |
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September 2, 2018, 00:18 |
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#3 |
Senior Member
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If is much easier if you simply use velocity inlet and specify a parabolic velocity profile instead. Strictly speaking, the pressure specified in pressure inlet is total pressure while that in pressure outlet if static pressure, so you should specify (58.86+dynamic pressure) and (0) at inlet and outlet, respectively. However, in this case the dynamic pressure is negligible. So you are doing it right. If the result differs significantly, make sure that the size of the geometry (it happens if you forget to convert the unit), the density and viscosity of the fluid is set right.
The Reynolds number based on diameter is 0.04, which is quite small. The dimension resembles to the fluid in a syringe needle driven by gravity, in which surface tension might play an important role at the inlet and outlet. |
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September 2, 2018, 00:38 |
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#4 |
Member
Liliana de Luca Xavier Augusto
Join Date: Feb 2013
Posts: 68
Rep Power: 13 |
Hi!
Thank you both for the reply. So at inlet I should set a value that represents (pressure drop)+(atmospheric pressure)? At outlet I could use 0, since it is gauge pressure. I did this and my maximum velocity was almost 6 m/s while the author and the analytical value is around 0.01 m/s. I really dont know what is going on. I double checked the viscosity and the size of the geometry and it is correct. I am uploading the files of fluent case if someone have some time do see what it could be the problem. https://drive.google.com/open?id=1V8...Dtjdbn5NBxEOf5 |
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September 2, 2018, 01:19 |
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#5 |
Senior Member
Behrooz Jamshidi
Join Date: Apr 2013
Posts: 110
Rep Power: 13 |
at inlet you must set (pressure drop)+(0.5*density*V^2)
But as Blackmask said your dynamic pressure is negligible. |
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September 2, 2018, 01:36 |
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#6 |
Member
Liliana de Luca Xavier Augusto
Join Date: Feb 2013
Posts: 68
Rep Power: 13 |
That is right, it is negligible...
I use 58.86 Pa as Inlet pressure, and 0 at outlet. I got around 1 m/s as maximum velocity, for a radius of 0.002 m and 0.006 m length, and viscosity 1 Pa.s (see picture attached). If we use: vmax = -R²/4mu * (dp/dx) dp/dx = 58.86/0.006 = 9810 Pa. vmax = -(0.002^2)/(4*1) * (-9810) = 0.00981 m/2 (which matches with the results from the author too...) So I am not getting... Why I got 1 m/s (around 100 times more) than the analytical solution Poiseuille? |
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September 2, 2018, 01:55 |
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#7 |
Senior Member
Behrooz Jamshidi
Join Date: Apr 2013
Posts: 110
Rep Power: 13 |
I will check it
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September 2, 2018, 01:57 |
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#8 |
Member
Liliana de Luca Xavier Augusto
Join Date: Feb 2013
Posts: 68
Rep Power: 13 |
Because it is not water, it is only a benchmark case. Although, I faced the same problem with higher or lower viscosity.
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September 2, 2018, 02:13 |
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#9 |
Senior Member
Behrooz Jamshidi
Join Date: Apr 2013
Posts: 110
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I have done your simulation and max velocity is 0.00976 m/s. If you want I can send you the file to compare with your simulation and easily find the problem.
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September 2, 2018, 02:18 |
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#10 |
Member
Liliana de Luca Xavier Augusto
Join Date: Feb 2013
Posts: 68
Rep Power: 13 |
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September 2, 2018, 02:22 |
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#11 |
Senior Member
Behrooz Jamshidi
Join Date: Apr 2013
Posts: 110
Rep Power: 13 |
check your inbox
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September 2, 2018, 02:28 |
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#12 |
Member
Liliana de Luca Xavier Augusto
Join Date: Feb 2013
Posts: 68
Rep Power: 13 |
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September 2, 2018, 06:20 |
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#13 | |
Senior Member
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Quote:
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June 19, 2022, 12:20 |
Cane you send me the ressult to comparit with mine please ;;:::/....????????????
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#14 |
New Member
Casablanca et région
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June 19, 2022, 12:22 |
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#15 |
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Casablanca et région
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