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July 6, 2007, 09:28 |
gauge pressure
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#1 |
Guest
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Hi.! I noticed that there are two kinds of pressure in Fluent, gauge and operational. I've read the user guide, but still I dont really understand about the difference.
I conduct simulation on M=0.13 and M=0.044. I want to compare with an experiment. The problem is that the description of this experiment is only that the q (dynamic pressure) is =1192 Pa which corresponded to M=0.13. the experiment conducted in an area with atmospheric pressure of 101325 Pa (only 3.4m above sea level). So, is this 1192 is the gauge or operational pressure? Thanks a lot |
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July 6, 2007, 10:58 |
Re: gauge pressure
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#2 |
Guest
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sheila,
Pgauge=Pabs-Pop where: Pop=Patm Pabs=absolute pressure and for compressible fluid: Pt=Ps(1+(gamma-1/2)*M^2)^(gamma/gamma-1) where: Ps=static pressure Pt=total pressure M=Mach # gamma=cp/cv |
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July 6, 2007, 12:06 |
Re: gauge pressure
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#3 |
Guest
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sheila remember that fluent wo
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July 6, 2007, 12:13 |
Re: gauge pressure
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#4 |
Guest
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sheila remember that fluent woring with the guage pr. and as you know Pr.abs=Pr.g+Pr.oper
and the operation pr. mean the atmospher pr. at the certain location so if you have guage pr you can compare it directly with the results of fluent and if you have absolute pr. you must convert it to guage to be abale to compare with the results of fluent I hope this helpful for you |
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July 6, 2007, 12:34 |
Re: gauge pressure
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#5 |
Guest
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hi nick and rosol. thank you for your immediate response.
for nick: I want to ask other question now (sorry if this sound dummy...) since u gave me the total pressure equation. is the total pressure the same with absolute pressure? is gauge pressure similar to dynamic pressure? data that I have are dynamic and operational pressure. for rosol: so, If I use the gauge pressure in the initialisation, I can compare the result directly? if not, then the pressure -> force -> coefficient in Fluent is not directly comparable to experiment? Thanks a lot |
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July 6, 2007, 13:28 |
Re: gauge pressure
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#6 |
Guest
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sheila, absolute pressure it's (gauge) local pressure (static or total) + Patm (or Pop). Gauge pressure it's local pressure without Patm contribution. Dynamic pressure is due to the motion of the fluid. For incompressible flow is Pdyn=.5*density*v^2 and Ptot=Pst+Pdyn. For compressible flow use the formula in my previous message.
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July 8, 2007, 23:15 |
Re: gauge pressure
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#7 |
Guest
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Thanks nick.
Now I understand the difference between gauge, operational and absolute pressure. The problem is that in the paper that I refer to, there are only dynamic pressure and mach number as the experiment description. How can I find out the gauge, operational and absolute pressure. Thanks again |
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July 9, 2007, 02:35 |
Re: gauge pressure
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#8 |
Guest
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If u r getting too confused u can set the operating pr.=0 in operating conditions panel and use a 3ddp solver... since ur mach number is >0.01 this approch is going to be accurate and not result in any error.
then all the pressure r absolute pr. |
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July 9, 2007, 03:24 |
Re: gauge pressure
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#9 |
Guest
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hi Mayur,
thanks for ur reply. I tried this before, but the result could not converge. rgrds, |
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July 9, 2007, 05:13 |
Re: gauge pressure
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#10 |
Guest
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u can get the solutioin converged by changing the URf's
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July 9, 2007, 05:16 |
Re: gauge pressure
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#11 |
Guest
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if u r talking of dynamic pr. then it does not depend on operating pr. Pdyn=Ptot-Pstat ... it only depends on the maich number and the velocity I think.
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July 9, 2007, 12:52 |
Re: gauge pressure
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#12 |
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the dynamic pressure is Pdyn=0.5*density*v^2=0.5*gamma*Pst*M^2, then you can get Pst(gauge). I assume you have Pop=Patm which you can add to Pst to get Pabs.
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July 9, 2007, 13:05 |
Re: gauge pressure
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#13 |
Guest
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hi nick!
thanks for your answer. the paper that I refer to, stated the Pop, q and M, so I know that the rho is 1.225 kg/m3. i have tried to use the p gauge (from your formula) and p operation (known). the display result from fluent showed that the rho in pressure far field (around the object) is 2.3 kg/m3. If i enter the right Pop, P gauge and V; can u help explain to me about this result. thanks |
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July 9, 2007, 13:53 |
Re: gauge pressure
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#14 |
Guest
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1.225 kg/m3 is the value of air density at Patm and Tamb. the formula for density is: ro=(Pgauge+Pop)/(R*T/MW) where Pgauge is your input and Pop=101325Pa is the default value in the operating condition panel.For M=0.13,T=300K,q=1192Pa,gamma=1.4, you get Pst=100760Pa. Thus Pgauge+Pop=202000Pa, and ro=2.3kg/m3.
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July 9, 2007, 23:03 |
Re: gauge pressure
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#15 |
Guest
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thanks nick...
now, it's crystal clear. thanks again |
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July 9, 2007, 23:30 |
Re: gauge pressure
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#16 |
Guest
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nick, if you could tell me...
from where are these equations? i have read some fundamental aero books, but never found it. or maybe it was just me who skipped it. thanks |
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