# Potential Energy of a compressed gas cylinder

1. Oct 18, 2007

### Sean Powell

Hello,

This is not a homework problem, but an engineering office problem.

I’m looking to replace a steel compression spring with a gas cylinder. The spring is presently designed to absorb a known amount of kinetic energy which it ideally converts to potential energy (neglecting friction, yadda yadda) and compression is solvable by a simple formula. PE = ½*K*dL assuming the spring was fully relaxed to start.

When I impact a closed gas cylinder the force response is governed by PV=nRT and we can generally assume no major temperature changes during impact. That means force doubles at half stroke, quadruples at ¾ stroke, is 10 times at 90% stroke etc. I know enough to subtract ambient air pressure to get a correct baseline. My problem is the force is so low for the first half of the stroke and climbs so rapidly for the last 20% or so that I’m having a tough time predicting the response of stopping a rolling vehicle impact.

Ideally I need a formula for potential energy in a compressed gas cylinder if I know: Diameter, L1 (usually free length), P1 (usually atmospheric), L2 (new compressed length) and some basic assumptions like human breathable atmosphere not too far from sea level.

From here I think I can calculate peak force so I know how to design the ratchet mechanism so the car isn’t rebounded. Yes, I know I could probably do this better with a dampened spring but there are other design issues (and patents to get around).

Sean

2. Oct 18, 2007

### rbj

whether it's a spring or an isothermal (or adiabatic) gas cylinder being compressed, the general formula for work done in compressing is:

$$W = \int_{x_1}^{x_2} F(x) dx$$

which for a relaxed spring is

$$W = \int_{x_1}^{x_2} K \cdot (x-x_1) dx = \frac{1}{2} K (x_2-x_1)^2$$.

for the gas cylinder, you have to define what the dimensions of it are (the radius of the cross section, call it $r_0$, and its length $L$. i'm setting $x_1$ to zero, with no loss of generality. the ambient pressure is $P_0 = nRT/(\pi r_0^2 L)$

$$W = \int_{0}^{X} F(x) dx = \int_{0}^{X} \frac{P(x)-P_0}{\pi r_0^2} dx = \int_{0}^{X} \frac{nRT/\left(V(x)\right) - P_0}{\pi r_0^2} dx$$

$$= \int_{0}^{X} \frac{nRT/\left(\pi r_0^2 \cdot(L-x)\right)-P_0}{\pi r_0^2} dx$$

someone's gotta blast out that integral for me. it has a log() in it, that's all that i know. everything other than x is constant.

Last edited: Oct 18, 2007
3. Oct 18, 2007

### Shooting Star

When a volume of gas is compressed suddenly, it is generally an adiabatic process, and no heat is exchanged with the environment. But the temperature will change, and that has to be taken into account.

I have given a simple and rough treatment for another situation much like this in another thread. It's nothing but work done during adiabatic expansion. Perhaps you can have a look at it.

Last edited: Oct 18, 2007
4. Oct 22, 2007

### Sean Powell

Hello Shooting Star.
Thanks for the input. You are correct, while system transfers no heat to the surroundings there probably is a sudden increase in temperature. That was something I did not consider.

****
Since the process is quick, we’ll consider it to be adiabatic. Then, PV^g=K, where I am writing ‘g’ for gamma. For air, g=7/5. The value of K can be obtained by putting in the initial values of P and V, which you had given.

Work done= integral (P-Pa)dV between V1 and V2, where V1 is the volume of the compression chamber and V2 is (V1+volume of barrel), and Pa is the outside pressure.

After integrating, W=K[V2^(-g+1)-V1^(-g+1)]/(-g+1) – Pa(V2-V1).
****
Which I think I follow fairly well and have all of the relavent information to fill out and calculate... but the units don't seem to be working out for me. Assuming gamma is unitless, P is N/m^2 & V is m^3 then K has the units of N*m^21/10??? The (-g+1) term becomes -2/5 and after a bunch of crunching I see N*m^1.5 for the first chunk of the equation while the right hand portion is N/m^2*m^3 => N*m (please excuse my sloppy shorthand as I convert my hand scribbles into a post) Was the first part of the equation supposed to condense to N*m as well?

Am I making a math error or am I missing something more important here?

Sean

5. Oct 22, 2007

### Shooting Star

If you follow a consistent sytem of units, then there is no need to worry about the unit of K -- both sides will ultimately give you the dimension of energy. Just find the values, after converting everything to same units like kg, m ,s.

You have done some math error. I am getting that K has unit of Nm^(11/5). On the RHS inside the bracket it is v2^(-2/5). After multiplying by K, the unit is simply Nm. Both sides should have units of N*m, since that is the unit of energy.