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Centrifugal pressure 
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#1
Jan1006, 06:53 PM

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(see also attached .doc)
There is an ideal gas in the centrifuge; no flows, everything is steady (not very steady in fact  process is adiabatic). If gas is incompressible, it is very easy: P=0.5 rho (w*r)^2 But my gas is compressible, and it's a trouble. I started from distribution law of Bolzman (with C = Bolzman constant): n=n0 exp(mgx/CT) ,hence d(ln n)=(M * w^2 * r) dr / RT ,where M is molar mass of gas after integrating this equation (also used PV/T=const and PV^k=const) I got: P/P0 = { 1 + (k1)*Mv^2/(2*R*T0) } ^ [k/(k1)] ,where k=Cp/Cv v=w*r And now goes the trouble. At low velocities (and low compression ratios) the last equation must transform into the first: deltaP=0.5 rho v^2 / 2 But it doesn't! Using the rule (1+x)^a = 1+a*x ,when x<<1 we have: P/P0 = 1 + k*Mv^2 / 2RT0 deltaP/P0 = k*Mv^2 / 2RT0 deltaP = k*rho*v^2 / 2 ,which is k time larger... So what is wrong? Or anybody have a readyforuse formula? 


#2
Dec2106, 07:59 AM

P: 2

FWIW i don't see what Boltzmann's dist has to do with it, can't you just integrate the ideal gas law with an acceleration correction? i think it makes a lot of difference if the centrifuge is open or closed at the top, having the constraint of fixed total mass will probably change the form of the solution. 


#3
Dec2106, 04:04 PM

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If this is a steady state situation, isn't the temperature constant? Sure there is a transient adiabatic temp change when the centrifuge starts up, but conduction in the gas will get rid of that in the steady state.
So I think you have an ideal gas, with pressure and density varying with radius, and a force dm.r.omega^2 = rho.A.dr.r.omega^2 on the gas between radius r and r+dr. 


#4
Dec2106, 06:21 PM

P: 2

Centrifugal pressure
i think you must have meant to substitute the ideal gas law where one on the centrifugal forces is, as you have it the r.omega^2's just cancel out.
another way to say it, the difference in pressure difference across the layer must be caused by the force on the layer (area cancels, i think) so: dP=w.w.r.dm=w.w.r.dr.P/RT integrate over range , P;Po to P and r;0 to r, so P/Po=K.Exp(w.w.r.r/2RT) and with w=0, P=Po so K=1 so P/Po = Exp(w.w.r.r/2RT) does this seem OK? 


#5
Dec2206, 05:03 AM

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I was just writng the mass in 2 different ways  sorry if that was confusing.
To be pedantic A.dP = w.w.r.dm (what you wrote says "pressure = force" which isn't right) and using dm = rho.dV = rho.A.dr the area cancels out and dP = w.w.r.dr.P/RT 


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