porous jump simple case

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September 14, 2013, 12:55
porous jump simple case
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I set up a simple case with inlet/outlet and porous jump face in between. It is a simple 5x10x100ft rectangular duct and medium is air. According to Fluent manual, dP across porous jump is:

dp=( mu*v/K+0.5*c*ro*v^2 ) * l

my inputs are:

mu=0.000012 lb/ft.s
ro=0.0765 lb/ft3
vel=5 ft/s
K=1 ft2
C=100 1/ft
l=0.03 ft

using the formula above gives me 2.86 lb/ft2 of dp. Now if I run this Fluent with realizable k-e model I get only 0.079 lb/ft2. I can not figure out why the difference is. Obviously I am doing something wrong.

Thank you for your help...

semo
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 September 15, 2013, 10:56 #2 New Member   Join Date: Jun 2010 Posts: 26 Rep Power: 8 can anyone help me with this matter? please...

 September 16, 2013, 05:34 #3 Senior Member   OJ Join Date: Apr 2012 Location: United Kindom Posts: 475 Rep Power: 12 Your premise is wrong. The formula in FLUENT is for local dp at the element level. During the iterations, this drop is taken into account at every element of the porous surface and thus, the "v" will be different at different places. It also depends on what angle the flow is on the porous surface. So to generalise this formula to use the average velocity would be unreasonable. OJ

 September 16, 2013, 08:17 #4 New Member   Join Date: Jun 2010 Posts: 26 Rep Power: 8 oj.bulmer, thank you for your response. You are absolutely right, the velocity will be different at different locations, however, for the test case I generated, it is not going to be so much different than the average velocity. In any case, the difference does not justify or explain the huge difference in dP. thanks again. semo

 September 16, 2013, 15:21 #5 Senior Member   OJ Join Date: Apr 2012 Location: United Kindom Posts: 475 Rep Power: 12 How have you specified the BCs for the walls? Try specifying all 4 walls as free slip boundaries or symmetries. The walls will have effect on velocity distribution and the free slip boundaries will ensure your velocities remain same everywhere. OJ

 September 16, 2013, 18:28 #6 Senior Member   Andrew Kokemoor Join Date: Aug 2013 Posts: 121 Rep Power: 6 Looks like a unit problem. One of the dangers of working in British/Imperial is that it's far too easy to mix up pounds force and pounds mass. Your math is off by a factor of g, 32.2 ft/s^2. Following that equation with accurate units will give you 0.089 lbf/ft^2.

 September 18, 2013, 10:12 #7 New Member   Join Date: Jun 2010 Posts: 26 Rep Power: 8 of course! I cant believe how I missed that. Thanks a million for your response Kokemoor. Now everything is in line. semo

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