pressure in a rotating storage tank

A cylindrical tank is filled with air with pressure p at temperature T. It has length l and radius r. There is no air flow into or out of the tank. The cylinder is then rotated in a circle such that one of the faces of the cylinder is at all times pointing directly toward the point of rotation and the other is facing away. The distance from the point of rotation to the near face is R >> r. The tank is rotated with angular velocity ω.

What is the pressure inside of the tank at any distance d perpendicular to the near face?

I know it sounds like a question from my physics textbook that I am trying to trick everyone into giving me an answer to, but I promise it is FRC related. I know that the pressure near the outside face will be greater than the pressure on the inside face, but I have no idea how one would go about calculating this.

EDIT: Assume that the rotation is happening in a horizontal plane so that gravity has no appreciable effect. I am only looking for the pressure as a function of distance from the near face, I don’t care about the miniscule height differences due to gravity. I also don’t believe that the value of r should change anything, but I included it anyway.

Wow, writing completely unambiguous questions like this is much harder than I thought it would be. I probably still missed something.

Appreciate that next time you see something like this on a test…:smiley:

I’m not sure it would be advisable to allow ω to reach a large enough value for this to matter. What are you trying to do? There is almost certainly a simpler way to do it.

I feel like the difference is likely to be negligible (unless omega is very very large), but here it goes anyway:

-Assuming the p is the pressure at the near face, not the far face or the average
-I am going to call d x instead. This reduces confusion when dealing with derivatives (because d could either be a variable or the sign for derivative).

Centripetal acceleration can be found from the rotation rate and distance from the center point:
a = omega^2 * (R + x)

This can give you the force:
F = ma = momega^2 * (R+x)

The mass is the mass of a cylinder of air at distance d, with differential thickness dx.

m = pi * r^2 * dx * rho
F = pi * r^2 * dx * rho * omega^2 * (R+x)

Density is a function of pressure and temperature, which can be determined by the ideal gas law:

P = rho * Rspec * T
rho = P / (RspecT)
F = pi * r^2 * dx * (P(x)/(Rspec
T)) * omega^2 * (R+x)

F is the net force on the differential cylinder, which depends only on the difference in pressure at distance x and distance x+dx. It would also depend on gravity if we accounted for it.

F = (pi * r^2) * (P (x + dx) - P (x)) = (pi * r^2) * ((P(x) + dP) - P(x))
F = (pi * r^2) * dP
Substitute in our expression for F:

F = pi * r^2 * dx * (P(x)/(Rspec*T)) * omega^2 * (R+x) = (pi * r^2) * dP

Divide by pi * r^2:

dx * (P(x)/(Rspec*T)) * omega^2 * (R+x) = dP

Define K = omega^2 / (Rspec*T) to clean up the left hand side:

dx * P(x) * K * (R+x) = dP

This is a linear differential equation for P(x), and can be solved by seperation:

dP/P(x) = dx * K * (R+x)

Integrate both sides:

ln(P(x)) = KRx + K*x^2 / 2 + C (Where C is constant of integration)

Exponential of both sides (define F = e^C):

P(x) = e^C * e^(KRx + Kx^2 /2) = F * e^(KRx + Kx^2 /2)

Use the boundary condition, P(0) = p to solve for f:

p = F * e^(0) * e^(0) = F
F = p

Substitute this back in for F

P(x) = p * e^(KRx + Kx^2 /2) = p * e^(omega^2 / (RspecT) * (R*x+x^2/2))

Using some actual numbers:

p = 120 psi
omega = 10 /s
T = 70 degrees Fahrenheit = 500 degrees Rankine
Rspec (for air) = 53.353 ft * lbf / (lb * degree Rankine)
R = 3 feet
x = 2 feet

P(0) = 120 psi
P(2) = 120.132 psi

So I don’t think you will get any significant pressure rise because of rotation, unless you are rotating extremely fast (10 rad/s is already more than a revolution per second). And rotating a tank very fast would seem to be a safety hazard. I’m curious, what is your application? An arm with the tank on it?

Wow, I didn’t think that someone was going to work through the entire thing themselves. Thank you John. I was hoping someone might give me a helpful formula to start out with, but you just went and did the entire thing using Newton’s 2nd. Kudos to you. I tried a similar approach but thought that there must be a simpler way to do it so I quit. I’m going to retry that approach now. I will post again with context and more thoughts after.

By the way, I had never even heard of the Rankine temperature scale until right now.

Good theoretical exercise, but eminently impractical. But you knew that already.

The conversion between Fahrenheit and Rankine is approximately R = F + 460 (I think that’s easier to remember than K = C + 273), so the correct temp to use is 530 degrees R – not that it makes much difference to the problem.

Yeah, you’re right. I think i made another small error: it should be 120 psi gauge, so 135 or so absolute. Doesn’t affect the final pressure gradient much though.

John, I worked through the math myself and came to the exact same result as you did:
P(x) = P(0)e^(ω^2(Rx+(x^2)/2)/(TRspec))
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) = 2lp*sqrt(k)/sqrt(pi)erfi(sqrt(k)l)
l is the length of the cylinder
p is the initial pressure
k is shorthand for ω^2/(2
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.

finding P(0).docx (14 KB)

finding P(0).docx (14 KB)

Well, you’ve certainly piqued our curiosity about what you’re trying to accomplish.

Can we assume that rotational speed is low enough, that we don’t need to account for the different gases in air?

(while not the normal implementation, doesn’t this start looking like a gas centrifuge (above some rotational speed)?)


This certainly has my interest… I might have to cross the street at the end of March to see what you put together! If nothing else, it sounds like you’re thinking of giving MN’s newest LRI an interesting robot to look at :slight_smile:

Now that I have everyone’s attention ;), I’ll talk about the potential application. I was curious to see if some kind of “super goalie” would be plausible with this setup. Let me explain:
According to the rules, a robot is only allowed to have one 6" diameter vertical extension. However, there is nothing to say that there can’t be more than one possible extension (this is even shown in the description of the rule in one of the pictures). It would be interesting to see a robot that plays the goalie position with 2 pistons (or equivalent devices) that rapidly extend and retract out of phase with each other such that only 1 is above 60" at any given time. If they alternate quickly enough, this goalie has the potential to block an effective area with a width much larger than 6". I am still hoping that someone will do this.

Another similar idea would be for the extension to be moving relative to the robot. A non-articulating 2’ vertical metal bar that spins quickly in a 16" radius circle about the center of the robot would be effectively blocking a 32" wide area of the HIGH GOAL. I would also love to see a team do that for this year. The trade-off with this design though is that it confines said robot to their own GOALIE ZONE for the entire match. A piston instead of a metal rod could potentially solve this problem. However, running wires and/or pneumatic tubing to something spinning at this speed is nontrivial.

Hence the above question. I was wondering if a piston attached like this could be triggered just due to the difference in air pressure that results from the spinning. The bottom of the piston would be connected to storage tanks that are spinning near the center of rotation. The top of the piston would be open to the surrounding air. In this way, the spinning motion would automatically trigger the extension of the piston. When the motor slows down/stops, the piston would retract.

The cylinder in the above problem was just there to simplify it. At any point of distance d from the center of rotation, the pressure should be the same as that of any other point of distance d from the center of rotation. Thus, this scenario can be modeled by using an R of the closest point in the pneumatic system to the point of rotation, and an x of the distance from R to the furthest point in the pneumatic system.

I was fairly confident that this would not work without a massive volume of stored air, but I wanted to quantify just how much. There are probably other ways to accomplish this same objective, but this one has the potential be more elegant than the others since only a single controlled motor is required after the rest of the system is set up. The rule that will probably get in the way regardless is R78. If this system is considered a pneumatic system, then this whole setup would be invalid.

My team will not be prototyping this anytime soon, so if anyone out there is interested in prototyping this, please post video here. I would love to see this kind of system see at least a bit of action.