# Theorical pressure drop Doesn't match numerical.

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 April 15, 2002, 06:27 Theorical pressure drop Doesn't match numerical. #1 july Guest   Posts: n/a Sponsored Links I'm in a big trouble uhhmmm...! I want to know the pressure drop in a 2D channel (1mm height, 30mm length) Air is the gas, Vinlet is .335m/s isothermal, laminar, steady state flow... I calculate the pressure drop by the formula: delta p= (32 V L mu)/d**2 and it gives me 5.82 Pa Numerical gives me 2.267. Anybody know the origin of mismatching? Can anyone try to do this simple calculation to compare my results? thank's a lot

 April 15, 2002, 07:37 Re: Theorical pressure drop Doesn't match numerica #2 Ossi Guest   Posts: n/a I do not know about the equation you are using to compare with the computed result but remember to compare only the fully developed part of your channel flow. I think there is no data available for the pressure drop during the developing phase of the flow. Additionally, you should have dense enough grid (y+ ~1) near the wall. You can also make the developing phase shorter if you make the mesh initially more dense. Also, remember to make to grid long enough so that the section you are comparing with is not interfered by the outflow (or the inflow).

 April 15, 2002, 08:48 Re: Theorical pressure drop Doesn't match numerica #3 july Guest   Posts: n/a thank's but everything you suggest to me i made... the flow is fully developed cause i use L/h =30 I think the grid is ok (24x100) structured... Any advise? can you send me some rough calculation for this problem? thanks

 April 15, 2002, 10:30 Re: Theorical pressure drop Doesn't match numerica #4 bosko Guest   Posts: n/a July, I get 2.34 Pa, along 30mm fully developed. The problem is that the equation you are using is wrong, from Massey pp 160ish "Steady Laminar Flow Between Parallel Planes" gives dp/dx = -12*MU/C**2 which gives delta P = 2.18286 for this problem, which is about right.... All to easy to blame the CFD don't you think. Cheers B.

 April 15, 2002, 10:41 Re: Theorical pressure drop Doesn't match numerica #5 bosko Guest   Posts: n/a Sorry should be dp/dx = -12*U*MU/C**2 You can do fully developed flow calcs with just two cells in the downstream direction, pressure boundary for inflow and a fixed flow mass outlet for outflow (chosen to give correct mean velocity). The profile you get at inflow is automatically fully developed. B.

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