# Gas flow rate, Bernoulli

1. Jun 18, 2004

### Locrian

My statistical dynamics and other useful areas of physics was sorely lacking in my undergrad education. Take this for instance, at work I need to calculate the gas flow rate from a tube into a very large chamber. I know the diameter of the tube and pressures of the gas.

Bernoulli's equation seems suited for this

(p + (rV^2)/2) = constant

and so should be equal to each other on either side of the tube connection. However for some reason that didn't get me a useful equation, mostly I think because I don't know what basic assumptions to apply. Note that I only need the rate the air is coming through, which is a large effect, and can ignore details that would actually make this problem hard.

I know the density isn't the same on either side, so can I assume the velocity is the same?

If someone could get me started I would greatly appreciated, I get the feeling htis is going to be awful simple when it is laid out correctly

2. Jun 18, 2004

### enigma

Staff Emeritus
Hi Locrian,

Welcome to PF!

Can you list exactly what you do know?

If the densities are different, then Bernoulli doesn't apply. It makes the assumption of incompressible fluid.

3. Jun 18, 2004

### Gokul43201

Staff Emeritus
If your chamber is so large that adding gas to it doesn't change its pressure, then the flow rate is dominated by the pressure at some point inside the tube, and the impedance of the length of tube from that point out.

At what point is the pressure of the gas specified ? Do you have geometric info on the tube ? Is the chamber really "large" ? If the last condition is not true, you will have a decaying flow rate, that goes to zero as the pressure difference narrows.

4. Jun 18, 2004

### Locrian

Thank you very much enigma and gokul for your replies!

I think you both are asking me for more detailed information. The problem at hand is very simple (enough so I'm a bit embarassed).

The air coming through the tube will be at 80psi approximately. The tube is 1.5" diameter. I call it a chamber (it is a microwave cavity, we are doing diamond growth) but this part of it is open at the top, so the pressure inside the microwave cavity will always be 14psi, it should never go higher. If it does, it means the chamber below it exploded catastrophically and the gas flow rate in the room will vary greatly from calculations

All i could do is hook a baloon onto it, see how mcuh air enters the baloon over a short time and measure the flow rate by that... but I'd really rather have an equation I can employ to determine changes that need to be made. I've actually found some places ont he web that calculate this, but they don't offer me enough control over the problem.

Kyle

Edit: My real goal is less to solve this particular problem, and more figure out why I can't solve the problem. Would this be covered in a statistical dynamics book?

Last edited: Jun 18, 2004
5. Jun 19, 2004

### arildno

Bernoulli's equation (or rather, the "energy integral" over the streamline) is readily generalized to stationary, inviscid flow of an isentropic, compressible gas.
Is this what you're seeking?

6. Jun 19, 2004

### Clausius2

Well, it seems to me I have not understood your problem at all. Please check my assumptions:

There is a gas flow through a pipe, which discharges into a chamber with a static discharge pressure of 14 Psi. You want to know what is the mass flow, isn't it?. And you said total pressure of the inlet flow is 80 Psi.

If this is true, the static pressure of the gas inside the pipe is 14 psi along it. As enigma said, the "usual" Bernoulli equation is only for incompressible flow. But he knows too that this equation is valid for gas flow at low Mach numbers. I have not made the calculus, but if you suppose enough supersonic flow (Ma<0.6) then:

Pt=Pc+rho*v^2/2.

where Pt=80 psi^
Pc=14 psi
rho= average density
v=flow velocity.

If you seek for the velocity, then you will have the mass flow.

PD: It would be a good time for learning Latex. I'm too lazy :zzz: .

7. Jun 19, 2004

### arildno

Let us consider an inviscid fluid with a density profile $$\rho$$ given by:
$$\rho=\rho(p)$$
where p is the pressure.
Then, the equations of motion in the stationary case with only gravity as the external force reads:
$$\vec{v}\cdot\nabla\vec{v}=-\frac{1}{\rho}\nabla{p}-\nabla(gz)$$
Multiplying with the differential streamline curve element $$d\vec{s}$$ and integrate from $$\vec{x}_{0}$$ to $$\vec{x}_{1}$$ then you gain:
$$\frac{1}{2}\vec{v}_{1}^{2}+gz_{1}+\int_{p_{0}}^{p_{1}}\frac{dp}{\rho(p)}=\frac{1}{2}\vec{v}_{0}^{2}+gz_{0}$$
This is the proper generalization of Bernoulli's equation for fluids with variable density.

8. Jun 20, 2004

### Gokul43201

Staff Emeritus
I believe $$\rho(p)=const*p$$ will give you a good approximation

9. Jun 21, 2004

### Locrian

Okay, I'm seeing where to go from here.

Thank you all very much for your help, I greatly appreciate it!