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Centrifugal pressure

by dushak
Tags: centrifugal, pressure
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dushak
#1
Jan10-06, 06:53 PM
P: 3
(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 + (k-1)*Mv^2/(2*R*T0) } ^ [k/(k-1)]
,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 ready-for-use formula?
Attached Files
File Type: doc There is an ideal gas in the centrifuge.doc (27.0 KB, 10 views)
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psi
#2
Dec21-06, 07:59 AM
P: 2
Quote Quote by dushak View Post
(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 + (k-1)*Mv^2/(2*R*T0) } ^ [k/(k-1)]
,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 ready-for-use formula?
i am interested in the solution, did you find one?

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.
AlephZero
#3
Dec21-06, 04:04 PM
Engineering
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HW Helper
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P: 6,932
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.

psi
#4
Dec21-06, 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?
AlephZero
#5
Dec22-06, 05:03 AM
Engineering
Sci Advisor
HW Helper
Thanks
P: 6,932
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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