# Question about centrifugal force and Bernoulli's law.

1. Aug 24, 2006

### RGClark

What I want to do is maintain a gas at very high pressure and temperature for propulsion purposes. The problem is containing a gas at high pressure requires a thick walled tank, and the resulting mass would obviate the advantage of having the gas at high pressure.
So what I'm considering is moving the gas at high speed within the tank. By the Bernoulli principle the tank should see the pressure on its walls as much less.
However, if the gas is moving around a cylindrical tank or a toroidal tank, then by centrifugal force that would seem to indicate it should increase the pressure(!)
Which one wins out?
Here's one possible way to address this problem. You could have the tank consist of long straight portions on either side but be curved at the top and bottom. Then you would only have to keep the portions at the top and bottom to be thick walled to maintain the pressure increase due to centrifugal force. If the straight portions are much longer than the curved parts at top and bottom, you could get a large volume without having to add too much to the tank mass.

Bob Clark

2. Aug 24, 2006

### RGClark

This won't work. The way Bernoulli's law works is that it has to be applied along a streamline. Then an increase in speed along that streamline results in a reduction in the pressure.
What I was envisioning is that if you have a flow at steady speed and it flows horizontally over a surface, while maintaining the same constant speed, then the pressure applied to the surface will be less than the pressure far from that surface. This is not how Bernoulli works. It only applies when the velocity is different between two points in the flow.

Bob Clark

3. Aug 24, 2006

### Clausius2

Don't envisage that, because neither it works as you said. When you are close to the surface, the effect of viscosity is non negligible at all, even if your external flow is extraordinarily fast. Close to the surface there would be the so-called Boundary Layer. Moreover, the pressure gradient ACROSS the Boundary Layer is negligible at first order, so the wall would feel the pressure of the external flow, which would be less than if the fluid is stagnant of course.

4. Aug 24, 2006

### Clausius2

The Bernoulli equation can be still applied to your system. Assume that the flow is steady. There is no radial component of velocity, but there is an azimutal component of value $$u_\theta=\Omega r$$ which increases with r (solid body rotation). The pressure force is counterbalanced with the centrifugal force, such that your new "Bernoulli" equation (which shows up after solving the N-S equations for your flow and it's independent of the Reynolds Number you use), it is:

$$P+\rho\frac{\Omega^2 r^2}{2}=Const$$

if your vessel is closed and the rotation speed is small enough for considering incompressible flow, and you're looking far away from the upper and bottom walls. I'm very afraid the constant should be worked out from the transient regime. I mean, I cannot chose any vessel radial coordinate and set a pressure, because I don't know the pressure inside. If you put gas in the tank at rest at certain pressure, and begin to rotate it with rotation speed $$\Omega$$, the pressure inside will be modified and reordenated spatially until reaching the steady state described by the Bernoulli equation above.

5. Aug 25, 2006

### Andrew Mason

The bottom line is that speeding up the gas flow does not change the energy of the gas molecules. If you confine those molecules withiin a container, they will exert the same average force per unit area on the container whether they are flowing (moving together with the same average energy) or not (moving randomly with the same average energy).

AM

6. Aug 25, 2006

### Clausius2

Wait I am not sure about that. Let's look at it from a reference frame attached to the vessel. The centrifugal force is definitely generating a gradient of pressure. It is a non-equilibrium situation. There is a gradient of a bulk magnitude inside the fluid. I think it definitely increases the pressure towards the vessel walls. Why not?.

7. Aug 25, 2006

### Andrew Mason

You are right. I should have said:the same average pressure throughout the gas.

The centripetal force is provided by collisions with the walls of the container by the individual molecules (ie. pressure). Since the average energy of the molecules is the same, the pressure would be the same unless the number of molecules per unit volume is greater near the container wall, in which case the pressure on the walls would be greater. In any event, it would not be less than the pressure in a static situation, which was really my point. You cannot decrease the pressure by making the gas flow within the container.

AM

Last edited: Aug 25, 2006
8. Aug 25, 2006

### Clausius2

Ok, it seems now I understand you. You mean that despites the pressure is non homogeneous, it's average keeps on being the same initial pressure, as well as the temperature (same average energy). I keep on not being very sure about that. I don't think the average energy remains the same. Where does the power introduced in the vessel because of the rotation go?. There is an external inflow of power into the system, it must be somewhere, and I think it must go to modify the thermodynamic state of the gas, even in the average. I don't think it has sense here to talk about energy in the average for the whole gas. There are small energies in the average, in differentials of r, but not in the whole vessel. With that I mean that we have to employ the Non Equilibrium Kinetic Theory (which by the way gives rise to the N-S equations and to the equation I posted above), because the collision dynamics are not in equilibrium in the whole, and one cannot define a total maxwellian distribution.

What one can do is to assume thin radial slices in which there The Local Equilibrium Hypothesis is almost valid, and do an average inside those thin slices. So I am not sure about your statement about having the same pressure on the average. Is there any way of showing it analitically?. This is an interesting problem...

9. Aug 27, 2006

### Andrew Mason

If you make the gas flow you do not necessarily have to increase its energy. Bernouilli's principle after all is based on conservation of energy: if you speed up the gas flow (without adding energy), its pressure drops. I don't see the point of adding more energy to the gas if what you want to do is lower the pressure.

AM