# The Big Air Piston

1. Aug 6, 2011

### FreakMath

Hi all I have a question which came up in one of our physics class and thought you can help out :)

It is not homework!

The objective is to calculate the volume and pressure of the air coming out of the following experiment:

Imagine a huge piston and cylinder. The surface of the piston is 10000sq ft and it is moving up in the cylinder at a rate of 1 ft per hour. As a result, air is being pressured to exit through a 1sq ft opening at the top.

1- what will be the pressure and airflow volume exiting the opening.

If any one has a basic equation for this it will be wonderful.

Happy day
Ariel

2. Aug 8, 2011

### gsal

it all depends on the opening...if it is a nice circular opening and it is just a thin wall between the inside and the outside, then, it can be considered as an orifice. There should be some webpage outthere that lists loss coefficients in velocity heads...then, I think, the pressure inside should simply be the pressure loss through the orifice, given the speed of the air through it.

3. Aug 8, 2011

### xts

The pressure will be innoticeably higher than ambient one, and air flow will be 10000/3600 (~3) cubic feet per second.

4. Aug 8, 2011

### Bill_K

FreakMath, There's a standard approach to these "hole in the wall" problems. An accurate solution of course is quite complicated, since you have to solve a fluid flow which depends on the viscosity of the fluid, details of the geometry, and so on. But to get an approximate answer, here's what to do, based on the particle current from kinetic theory.

Imagine that the fluid flows straight out from the hole. After one second the outflow will occupy a cylindrical shape with a volume A times vav, where vav is the average velocity of the gas particles. (Actually if you do it more carefully, it comes out A vav/4.) The average particle velocity can be calculated from the Maxwell-Boltzmann distribution, and is like √(kT/m) where m is the mass per particle. (Again, there's a small numerical factor which I'm ignoring.) So the outflow is approximately A √(kT/m) and from that you can easily figure out how the other variables are changing.

5. Aug 8, 2011

### xts

Bill_K, common sense approximate solution is that if air flows at the speed of 3 ft/s (1 m/s) through a hole as large as 1 sq.ft (e.g. round hole 40 cm in diameter) there is absolutely no noticable difference in pressure.
This is a case of quite mild wind through an open window.

6. Aug 8, 2011

### Bill_K

Ok, no pressure difference. Now calculate the pressure required to maintain that rate of flow.

7. Aug 8, 2011

### xts

In this case:
$$P=\frac{v^2\rho}{2}\approx0.5\,{\rm Pa} = 5\cdot 10^{-6}\,{\rm atm} = 50 \mu {\rm m \hbox{ of water}} = 4 \mu {\rm m \hbox{ of mercury}}$$
Or if you prefer - that is a change of barometric pressure as you climb two inches.

Last edited: Aug 8, 2011
8. Aug 8, 2011

### Bill_K

Very good! See, it's not so small after all.

9. Nov 5, 2011

### FreakMath

Thank you guys all for your help and sorry I could not get a reply to you sooner :)

back to our subject:

Lets turn this into a more predictable scenario and say that this piston has a long shaft attached to it measuring 1sqf and it is pushing upwards.

The push force is 10000lb.

based on the size of the piston will it be safe to say that the force that the shaft is pushing at is 10000 lb per sq f?

or I am getting this wrong?