Pressure drop - simpleFoam
1 Attachment(s)
I am currently working on matching the analytical solution for pressure drop in a pipe. I created a mesh of a pipe with 0.0254m diameter and 0.254m length with an inlet and outlet. The mesh was created using Ansys and refined in the near wall region to help with wall effects. The analytical solution indicates that the pressure drop with water flowing at 2 m/s should be about 1600 Pa/m. The simulation with simpleFoam using k-epsilon are giving an absurdly low value (usually less than 10 Pa/m). Things I have tried:
1. Ensured the viscosity value and units are correct 2. Calculated the correct k and epsilon values for the ICs 3. BCs are all okay, (no slip at boundaries, zero pressure outlet, etc.) 3. Corrected the kinematic pressure values to Pascal for final comparison 5. Ran checkMesh to ensure there are no irregularities in the mesh. Everything is OK according to checkMesh Suggestions on what other things I could try would be greatly appreciated. All run files except for the mesh (too big, ~2M cells) are attached for reference. Here is the last time in the log file with the residuals: Code:
Time = 362 Many thanks in advance, E.Lowry |
Pressure drop - simpleFoam
1 Attachment(s)
I tried your case with a simple pipe similar to what you describe, 0.0254 D x 0.254 L. For that length of pipe the drop was 0.44 m^2/s^2 by plotting p along the center line. To get the pressure drop in Pa multiply by the density of fluid (water) 983 kg/m^3 (@60 C) -> 433 kg/m * s^2 or Pa. Since the length of the pipe is 0.254 m, divide by this to determine the pressure drop for 1 m of pipe -> 1705 Pa/m. Attach is a screen shot for paraFoam.
https://www.dropbox.com/s/ggoo45qa8d...Dx254LPipe.png |
kcjarvis56,
Thank you for running my case. It seems that it is indeed working as expected. I guess the way OpenFOAM calculates and displays dP in the log file was confusing. I know see that in order to get the dP, I need to divide the value printed in the log file by the cross-sectional area prior to multiplying by the density of the fluid. Thanks again for your help |
Quote:
Hi, No, you have to divide the result by the length of your pipe (not the cross section area) in order to have a pressure drop / linear meter And of course, you have to multiply it by the density of the fluid. Fred. |
All times are GMT -4. The time now is 19:41. |