# basic question: Specifying Pressure for comressible solvers

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 October 13, 2010, 12:16 basic question: Specifying Pressure for comressible solvers #1 Senior Member   Nilesh Rane Join Date: Apr 2010 Posts: 122 Rep Power: 7 Hello All, This is a basic question but just to clear the doubt once and for all: The way we specify pressure for any compressible solver is with the units of [1 -1 -2 0..] i.e. p/rho. So we should specify the value of pressure as well after dividing p value by rho. Meaning if i have p = 1MPa and rho = 10 kg/m3 then, 1) for compressible solver: p = 1/10 = 0.1 2) for incompressible solver: p = 10 Is this correct?? __________________ Imagination is more important than knowledge.. Last edited by nileshjrane; October 13, 2010 at 12:51.

 October 14, 2010, 05:00 #2 New Member   Join Date: Jul 2010 Posts: 20 Rep Power: 7 I think you confused things here. For incompressible solvers, you have to specify the kinematic pressure, i.e. p/rho units [0 2 2 0 0 0 0] wheres as for compressible solvers, the actual pressure p [ 1 -1 -2 0 0 0 0 ] is used. Therefore, for your case with p = 1MPa and rho = 10 kg/m3: 1) for compressible solver: p = 1 MPa = 1e6 2) for incompressible solver: p = 1e6/10 = 1e5 However, when running incompressible simulations, only pressure differences count so you could also use 0 as a reference pressure. If you want to obtain real pressure differences, multiply the result obtained with foam (kinematic pressure) with rho. Hope that helps.

 October 14, 2010, 10:16 #3 Senior Member   Nilesh Rane Join Date: Apr 2010 Posts: 122 Rep Power: 7 Hi, You are damn right.. i was indeed confuse . I realised my mistake but didnt get time to edit my post. In fact i realised that mentioning p/rho is the only way to specify density in incomressible solvers. Thanks for the reply. __________________ Imagination is more important than knowledge..

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