# Homework Help: Centrifugal pressure

1. Jan 10, 2006

### dushak

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... :surprised
So what is wrong? Or anybody have a ready-for-use formula?

#### Attached Files:

• ###### There is an ideal gas in the centrifuge.doc
File size:
27 KB
Views:
99
2. Dec 21, 2006

### psi

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.

3. Dec 21, 2006

### AlephZero

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. Dec 21, 2006

### psi

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. Dec 22, 2006

### AlephZero

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