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August 3, 2011, 09:25 |
negative static pressure
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#1 |
New Member
Juan
Join Date: Dec 2010
Posts: 18
Rep Power: 15 |
Hello
I am trying to compute head losses in a pipe 3D. When i have made the simulation I have got good results but i do not understand very well because i have got negative static pressure. Could anybody explain to me the meaning of a negative static pressure? |
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August 3, 2011, 12:50 |
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#2 | |
Senior Member
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Quote:
In incomperssible flow solvers, gauge pressure is solved, so negative pressure means that the pressure is lower that your operating pressure. Amir |
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August 5, 2011, 08:58 |
static pressure negative
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#3 |
New Member
Juan
Join Date: Dec 2010
Posts: 18
Rep Power: 15 |
Thank you Amir
According to your reply I am not sure if i should change my operating pressure because the head losses i have got are very similar to experimental values i have measured. Should I change my operating pressure or maybe it is not important? I have used velocity inlet and outflow as boundary conditions Thank you |
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August 5, 2011, 09:13 |
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#4 | |
Senior Member
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Quote:
you are talking about head loss which means difference between heads, so if you have set any value for your operating pressure, you'd obtain same result; because: pressure_head_difference=(p1-p2)/(rho*g)=((p1+C)-(p2+C))/(rho*g) C=any arbitrary value this discussion is valid only for incompressible flow. Amir |
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August 5, 2011, 09:51 |
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#5 |
New Member
Juan
Join Date: Dec 2010
Posts: 18
Rep Power: 15 |
Thank you
So if I would want to know the real static pressure I should define other boundary conditions such as pressure inlet or outlet because if I introduce velocity inlet and outflow fluent does not compute the real satic pressure. Is it right?. How can I compute the real static pressure in to the pipe? Thank you again |
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August 5, 2011, 10:04 |
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#6 |
Senior Member
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No, it's not right!
Even when you are using pressure conditions, you have to set gauge pressure not absolute one. In incompressible flow, if you want to find absolute pressure, you would need pressure of one reference point, othrewise, it's not a big problem; after simulation, you can shift your results in order to coincide reference pressure. Bests, Amir |
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