# Air Pressure in PVC pipe

1. Aug 14, 2014

### skeeve

Consider the following:

I have 2 PVC pipes, one is small enough to fit into the other with a little extra space. A rubber ring is placed around the smaller one to fill up the empty space, making a good seal to push air in and out of the larger one. Ignore any friction caused by the ring on the inside of the large pipe. The small pipe also has a strong stopper on the side that is plugged into the large pipe.

The small pipe is pushed slightly up into the big one, and on the other end of the big one there is a stopper, preventing air from escaping. Imagine that it's perfectly attached and won't blow off with any amount of force.

The small pipe is standing on the ground, slightly inside the large pipe, and some amount of weight is placed on top of the large pipe, say, 300 lbs. (I would use metrics, but there would be too much conversion for my situation)

Question:
Assuming I'm at sea level, and using your every day air, how is the pressure (psi) on the inside of the large pipe determined when you know how much weight is pushing it down?

Thanks.

2. Aug 15, 2014

### 256bits

Try
PV = constant, assuming the temperature before and after compression is the same

3. Aug 15, 2014

### gmax137

Once the assembly comes to rest, the applied weight is balanced by the pressure inside acting on the cross-sectional area at the top of the large pipe. So, you need to know the weight and the inside area.

Using 256bits hint, you can predict how far down the large pipe will slide for your given weight and pipe size.

4. Aug 15, 2014

### skeeve

All right, let's try adding some numbers here.

Large pipe length minus inside stopper length: 5 ft
Small pipe: 5 ft
(pretend stopper doesn't expand beyond the top of the pipe)
Weight on top: 300 lbs
Average external air pressure: 14.7 psi

Please include the basic equation(s) that you would use to solve this, as it's been a while since I've gone over this kind of thing, and I don't have the textbook that I once had.

Thanks!

The

5. Aug 15, 2014

### .Scott

To be clear, only the outside diameter of the inside pipe is important.
Since you are only interested in the final pressure, temperature changes are unimportant.

6. Aug 15, 2014

### .Scott

I'll estimate the outside diameter of the inside pipe to be 2". So its area will be 3.14 square inches.
Your resulting gauge pressure will be 300 pounds per 3.14 square inches = 95.5psi.

Since you are mentioning the external air pressure, we can compute the absolute pressure in the pipe as:
95.5 + 14.7 psi = 110.2psi (absolute).

7. Aug 15, 2014

### skeeve

Oops. I forgot to put in the diameters. id=inside diameter, od=outside diameter

Big Pipe od: 2.375 in
Big Pipe id: 2.067 in

Small od : 1.900 in
Small id : 1.593 in

8. Aug 15, 2014

### skeeve

I've probably put more information then necessary, but it's been a while since I've taken a Physics class (which I love) and can't find my old textbook, so I didn't want to risk anything. I'd appreciate any recommendations for a good website that talks about this concept and includes some equations that would help me understand and calculate this question.

And if there does happen to be something that I should have added to my information, let me know. And please include the basic equation(s) that you used to solve this.

Thanks to everyone helping with this.

9. Aug 15, 2014

### .Scott

My guess for the inside pipe OD was only off by 0.1"! So:
Gauge pressure: $P_g = 300lb/(\pi (1.9/2)^2)psi = 105.8psi$
Absolute pressure: $P_a = P_g + 14.7psi = 120.5psi$

There's actually a bit of ambiguity here. The exact amount depends on how that gasket (the rubber ring) is supported. It's against stretched against the inner pipe, but is also pressing against the outer pipe. So which pipe is keeping it from blowing free?

If it was firmly attached to the inside pipe, then we would use the inside diameter of the outside pipe in our calculation (2.067"). If it was firmly attached to the outside pipe, then we would use the outside diameter of the inside pipe (1.900"), as we did in the calculations above.

Let's assume that it is a shared task with the inside pipe providing support for 2/3 of the rings width and the outside ring holding 1/3 of that ring.

Then the diameter we will use is:
Effective piston diameter: $D_p = (2/3)1.900 + (1/3)2.067 = 1.956 inches$

The area we are applying the force against is:
Effective piston cross section: $A_p = \pi (D_p/2)^2 sq. in. = 3.004 sq. in.$

The force we are applying:
Force against this cross section: $F_p = 300 lb.$

The resulting pressure:
Gauge pressure: $P_g = F_p/A_p = 300lb/(3.004 sq. in.) = 99.9psi$
Absolute pressure: $P_a = P_g + 14.7psi = 114.6psi$

Last edited: Aug 15, 2014
10. Aug 15, 2014

### skeeve

.Scott,

I'm not sure how exactly this will affect the situation, but the smaller tube will have a small groove cut into it to hold the rubber ring, probably a bit of glue to try and hold it. To make it simple, there will be some kind of oil on the inside of the large tube to take away as much friction as possible.

11. Aug 16, 2014

### .Scott

In that case, the ring is fully supported by the inside tube, so the diameter that should have been used is the inside diameter of the outside pipe (2.067"). I'll let you post the arithmetic this time.

12. Aug 24, 2014

### skeeve

Dpiston=inside big pipe=2.067in
Ap=∏(Dp/2)2=∏(2.067/2)2=3.356in2
Fp=300lb

Pg=Fp/Ap
Pg=300lb/3.356in2=89.39psi

13. Aug 25, 2014

### skeeve

With all of these previous pieces of information, I have one more question. How is a new weight calculated of an object that is has been falling?

I'm going to add a couple things here. Let's assume the small cylinder weighs 5 lbs, and the large cylinder weighs 10 lbs.

The smaller cylinder in the piston is on the ground and kept up by some metal feet, the large cylinder is being pulled down by gravity (9.8 m/s2). How far would the large cylinder go until the air pressure inside the piston is equally opposite to the gravitational force?

Next, a heavy object falls down on top of the large cylinder. Let's say it's the 300 pound weight again. If the large cylinder is weaker in pressure capability, and it's max is 350 PSI, how high can the 300 lbs weight fall from to match the 350 PSI?

Since we already know that P=F/A, we can decide that F=P*A.
So the max force, F = 350 PSI * 3.356 in2 = 1174.6 lbs

How high can we drop a 300 lbs object from to get a max pressure of 1174.6 lbs?