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Unread 13-01-2014, 01:53
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Re: pressure in a rotating storage tank

John, I worked through the math myself and came to the exact same result as you did:
P(x) = P(0)*e^(ω^2*(R*x+(x^2)/2)/(T*Rspec))
where:
P(0) is the pressure at the near face
ω is the angular velocity of the system
R is the distance from the center of rotation of the system to the near face
x is the distance from the near face
T is the temperature
Rspec is the specific gas constant of the gas

I hadn't expected the result to be nonlinearly dependent on temperature since temperature and pressure are directly proportional in the ideal gas law. Interesting…

Since I am also interested in finding the pressures when I only know the starting requirements, I solved for P(0) in the case where R=0. The integral otherwise is not nice at all and is probably unsolvable. When R=0, P(0) = 2*l*p*sqrt(k)/sqrt(pi)*erfi(sqrt(k)*l)
where:
l is the length of the cylinder
p is the initial pressure
k is shorthand for ω^2/(2*T*Rspec)
erfi is the imaginary error function (Didn't even know that this existed until right now)

This was assuming that Boyle's law is true for a non-constant pressure environment. That is, that the sum of the pressures at every miniscule volume element is constant in a sealed system even if these elements do not all have the same pressure. I think that this is true? Someone should confirm that though. If anyone is interested, my work is in the attached word doc since integrals aren't nice in plain text. I'll post some context tomorrow, but I am tired now, so good night.

EDIT: The first step in the attached document should have been cancelling the pi*r^2 since it appears on both sides of the equation.
Attached Files
File Type: docx finding P(0).docx (14.0 KB, 7 views)

Last edited by Caleb Sykes : 13-01-2014 at 14:52.