A Bernoulli Bet (experts please)

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I have a bet with one of my friends,

it has to do with the bernoulli equation that says the higher the velocity of a gas, the lower the pressure.

I came up with an invention idea that involves a gas cylinder, the gas cylinder has a maximum pressure of 4000 PSI and the gas inside exerts that maximum pressure.

in order to increase the volume, a spinning rod is placed in the center of the canister, and spins the gas so that the gas at the edge of the canister moves at a high velocity relative to the wall of the canister,

My hypothesis is that this velocity forces the pressure differential of the canister walls to decrease thus allowing for a greater volume of air inside the canister.

My friend said that it won't work, because he said that the hydrogen gas at that pressure is incompressible, and buy putting more air into the can you'd have to increase the density,

he said to imagine water spinning around in a can there are no air bubble inside the can, it is filled to the brim, the faster you spin it, (neglecting the centrifugal force) the pressure decreases on the sidewall of the can, but if you were to put more air inside, it would be as if the liquid were solid, it might not shoot out of the can, it would just stay there, and if you were to push more water into you couldn't because you'd have to increase the density, and that requires a huge amount of energy.

So I said, no that's wrong

and now we are in a bet for twenty five cents
 

Answers and Replies

  • #2
rcgldr
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When a fluid is sped up, the pressure is only reduced if the fluid is sped up with little or no work done. Using a rod or impeller to speed up the fluid is going to peform a lot of work on the fluid, making a significant change in the total enery and the total pressure energy (static + dynamic). Then there is viscosity between fluid and walls of container, but I guess you could assume a friction less container.

In the case of a gas, you could have the vortex effect, higher pressure hotter gas on the outside, lower pressure colder gas on the inside. I'm not sure about the pressure, but the gas is colder on the inside and hotter on the outide, and a vortex tube can be used to generate a stream of cold and hot air:

http://en.wikipedia.org/wiki/Vortex_tube
 
  • #3
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"My hypothesis is that this velocity forces the pressure differential of the canister walls to decrease thus allowing for a greater volume of air inside the canister."

What does 'pressure differential of the canister walls' mean?
 
  • #4
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When a fluid is sped up, the pressure is only reduced if the fluid is sped up with little or no work done. Using a rod or impeller to speed up the fluid is going to peform a lot of work on the fluid, making a significant change in the total enery and the total pressure energy (static + dynamic). Then there is viscosity between fluid and walls of container, but I guess you could assume a friction less container.

In the case of a gas, you could have the vortex effect, higher pressure hotter gas on the outside, lower pressure colder gas on the inside. I'm not sure about the pressure, but the gas is colder on the inside and hotter on the outide, and a vortex tube can be used to generate a stream of cold and hot air:

http://en.wikipedia.org/wiki/Vortex_tube
I'm afraid that doesn't settle the bet.

I am not too concerned with the differential pressure inside the can, what I want to know is whether the fast moving pressurized air in the outer interior of the canister can create a bernoulli effect with the outside air greater than centrifugal force of the rotating air mass, decreasing the overall pressure of the air inside the canister and therefore whether as a result it would be possible to increase the maximum molar volume.
 
  • #5
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"My hypothesis is that this velocity forces the pressure differential of the canister walls to decrease thus allowing for a greater volume of air inside the canister."

What does 'pressure differential of the canister walls' mean?
What I meant by the pressure differential of the canister walls is the pressure difference between the outer interior airmass and the exterior.
 
  • #6
russ_watters
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You've just created a centrifugal pump/fan: you're increasing the static pressure on the cylinder wall, not decreasing it.
 
  • #7
vanesch
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I would place the canister on a fast-moving train, then the speed of the gas is also higher, and hence the pressure lower :smile:

No, seriously, Bernouilli is often mis-understood, as is the case here.

Bernouilli's law tells you the following:
- in a lossless flow (no viscosity), there is a conservation law *along a fluid line* in a given stationary flow. That means that, along the line, you add 3 quantities, which correspond to velocity (kinetic energy), pressure, and eventually height (potential gravitational energy), and along the streaming, one can convert in another, but the balance is conserved.

See the wiki on the subject for some more information:
http://en.wikipedia.org/wiki/Bernoulli's_principle

So your reasoning doesn't hold for different reasons. The first one is that Bernouilli's law is not valid when comparing two different flows. It is within the same flow that it works out, but you change from one flow to another. The second reason is that the boundary conditions are not stationary (the "walls" are moving in a pump). And the third reason is that along a single (circular) flow line, the velocity doesn't change, so nor does the pressure.
 
  • #8
Lok
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The centrifugal force of the spinning gas will be greater than any Bernoulli effect, so the vessel will experience more pressure. And hydrogen can be compressed further than that.
 
  • #9
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So your reasoning doesn't hold for different reasons. The first one is that Bernouilli's law is not valid when comparing two different flows. It is within the same flow that it works out, but you change from one flow to another. The second reason is that the boundary conditions are not stationary (the "walls" are moving in a pump). And the third reason is that along a single (circular) flow line, the velocity doesn't change, so nor does the pressure.
I had read that wikipedia article and I was expecting an answer like this. Perhaps it doesn't follow Bernoulli's Principle but I do expect there will be an effect and here is my theoretical experiment that makes me think so.

say you have a tube of an incompressible fluid, each one with a separate diameter, like in bernoulli's experiment, a barometer is placed with one end in the place with the larger diameter and the other end in the place of the smaller diameter. then you cut the flexible barometer in two and compare the pressures with the outside atmosphere.

my belief is it would be impossible for the two pressures to be the same by virtue of not being in reference to one another (there will be a difference in pressure and that difference will be exactly as measured by the difference in pressures between the two areas of tube.
 
  • #10
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Oh I understand it now, in my little theoretical experiment I would get the result, but it's not because there exists some inverse relationship between velocity and pressure, but because the pressure inside the tube (which is created by the different flow rates, the pressure in the area of the larger tube builds because the flow rate in the larger diameter tube is greater than the smaller one. Then once the pressure becomes high enough, the energy of the pressure is converted to kinetic energy. The system reaches equilibrium probably precisely at Bernoulli's equation.

My spinning air canister therefore has no effect at all and I am down 25 cents

A bad day indeed.
 
  • #11
russ_watters
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So your reasoning doesn't hold for different reasons. The first one is that Bernouilli's law is not valid when comparing two different flows. It is within the same flow that it works out, but you change from one flow to another. The second reason is that the boundary conditions are not stationary (the "walls" are moving in a pump). And the third reason is that along a single (circular) flow line, the velocity doesn't change, so nor does the pressure.
I don't think I'd go that far - the misunderstanding isn't quite so fundamental.

Bernoulli's equation can't be applied to separate scenarios in a single pass of course, but if there is enough common ground, you can construct two separate Bernoulli relations and set them equal to each other.

A typical problem presented in Fluids 101 uses a large pressurized vessel with a fluid in it (doesn't matter if it is liquid or gas) and a short length of pipe with a closed valve at the end on the bottom. The valve is opened, what is the velocity and static pressure in the pipe? The usual Fluids 101 simplifying assumptions apply and the key is that the vessel is big enough that when the valve is opened, the pressure doesn't change quickly. So you have two separate cases:

SP+VP=TP
Case 1: SP1+0=TP1
Case 2: 0+VP2=TP2

Total pressure is the same in both cases, so you have:
SP1=VP2
Then you solve for the velocity. And the static pressure inside the pipe is in this theoretical case 0 - in reality it isn't due to losses, but the sides of the wall do "feel" a significantly lower pressure with the valve open than with it closed.

The logic of the OP is based on that concept and it is a good thought, but it just happens to not work in this case. The reason it doesn't work is that you can't equate the two total pressures, but the "why" isn't that you can't do that in general, it is the specifics of this problem that interfere. And you can construct separate Bernoulli relations and just compare them. So I don't think your first objection works as a blanket statement. For the second, my read of the OP says that the walls are stationary, ie the flow is moving with respect to the walls - this isn't like putting the tank on a train at all. The third objection I don't quite understand, but it seems about the same as the first.

The OP did note in the OP what the flaw might be and frankly, though I think the pressure increase due to the centrifugal force is greater than the static pressure drop along the wall due to the motion, I don't know it for sure and it is something I don't care to integrate! Perhaps a fan curve at stagnation would give that answer....hmm (I'll see if I can find one that applies here).
 
  • #12
russ_watters
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say you have a tube of an incompressible fluid, each one with a separate diameter, like in bernoulli's experiment, a barometer is placed with one end in the place with the larger diameter and the other end in the place of the smaller diameter. then you cut the flexible barometer in two and compare the pressures with the outside atmosphere.

my belief is it would be impossible for the two pressures to be the same by virtue of not being in reference to one another (there will be a difference in pressure and that difference will be exactly as measured by the difference in pressures between the two areas of tube.
I'm not clear on what you are getting at with this. Are the fluds moving in these tubes? Is it one tube with just different diameters (ie, a venturi tube?)? And are you talking about a U-tube manometer as opposed to a barometer? Two different tubes may or may not have the same pressures in them...
Oh I understand it now, in my little theoretical experiment I would get the result, but it's not because there exists some inverse relationship between velocity and pressure, but because the pressure inside the tube (which is created by the different flow rates, the pressure in the area of the larger tube builds because the flow rate in the larger diameter tube is greater than the smaller one. Then once the pressure becomes high enough, the energy of the pressure is converted to kinetic energy. The system reaches equilibrium probably precisely at Bernoulli's equation.
Now you're really barking up the wrong tree for sure. Bernoulli's equation says nothing whatsoever about flow rate. It is about velocity. Remember, along a streamline, the mass and volumetric flow rate is constant in a standard (incompressible) Bernoulli situation. Ie, in a venturi tube, the area gets smaller, the velocity goes up, and the volumetric flow rate stays the same.
My spinning air canister therefore has no effect at all and I am down 25 cents
Well, you are almost certainly wrong, but you might try weaseling your way out of it by saying he's wrong too. His objection, as you worded it, is irrelevant. If what you speculated had been correct, adding more gas wouldn't necessarily result in you ending up being wrong - as long as you didn't add too much!
 
  • #13
vanesch
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Then you solve for the velocity. And the static pressure inside the pipe is in this theoretical case 0 - in reality it isn't due to losses, but the sides of the wall do "feel" a significantly lower pressure with the valve open than with it closed.
Yes, but you don't use any pressures from the "closed valve" problem to solve the "open valve" problem, except for the *given* pressure in the vessel which is equal in both cases here. In other words, the solution of the "open valve" problem doesn't need the "closed valve" situation.

The logic of the OP is based on that concept and it is a good thought, but it just happens to not work in this case. The reason it doesn't work is that you can't equate the two total pressures, but the "why" isn't that you can't do that in general, it is the specifics of this problem that interfere. And you can construct separate Bernoulli relations and just compare them.
What I meant was: the situation of a static gas in a bottle is not the starting point of the calculation of the pressure in the rotating gas in the bottle. The gas doesn't go, along a single fluid line, from "being static" to "rotating". This is different in your example, where in BOTH cases, the starting point was the same (namely an essentially static gas in a vessel).

For the second, my read of the OP says that the walls are stationary, ie the flow is moving with respect to the walls - this isn't like putting the tank on a train at all.
I meant the moving parts of the pump or the rotor that is going to impart the rotation movement to the fluid. Bernouilli doesn't work (unless you add an extra term) when work is done on a fluid by a pump or any other moving thing.

The third objection I don't quite understand, but it seems about the same as the first.
In a way, yes, it is the same as the first: in the rotational movement of the fluid, there is no point where you can say "velocity = zero and pressure is the previous, static, pressure" and from that point onward, follow the line and say "now the fluid has sped up, so the pressure must be lower now".

In other words, what I considered as the fundamental flaw in the reasoning was a misuse of Bernouilli, thinking that "a gas that has a velocity has a lower pressure than a static gas" in all generality. This is only true in a single "problem" (not between two different situations), along a single fluid line, and if on top of that, the gas didn't receive any "help from the outside" (pump, moving wall,...). It is only in that case that the gas "pays pressure to build up speed".

This is almost like saying: when a ball is higher, it moves slower. So can we decrease the velocity of a car by putting the highway on poles ? This is only true when the car came from a lower altitude with a certain velocity, and then moved up the hill without using its engine. Not between a highway on the floor, and another highway on poles.
 
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