Potential Energy of Compressed Air

[bjones]

Hi,

 

This is my first forum post. Thanks for offering such a nice resource. If my post is in the wrong place, please move it and tell me. Was not sure if here or chemistry would be better. I end up needing science answers from my choices of reading material, but don't have much of a science background.

 

 

Ideally, from your answers, I'm hoping for two things:

 

1. Online source for a graph showing the relationship between pressure (psi, etc.) and potential energy (in some unit of measure) of compressed air. Something along these lines, but with correct axis labeling, etc.: http://www2.chemistr...r_vapor-ans.gif

 

2. I'd like to understand a couple of related concepts. Use as a reference for answers a typical construction site air compressor (30 gallon tank/cylinder; capable of 19 cfm at 125 psi; http://www.constructionequipment.com/jenny-j5a-30p-air-compressor.). I'm not trying to buy one, just using it to understand science concepts. In fact, any compressed air cylinder would be fine for this, but I wanted to provide one to save time.

 

A. Realistic or not, if that system could be taken to 250 psi (double the above-mentioned pressure), would that mean that it would also be capable of double the work (potential energy)?

 

B. Is that a linear relationship in general? If not, what is it? If increased volume of air stored is equivalent to increased potential energy, this website, while not scientific, makes me think the relationship is linear: http://www.spearfish...java/tankv.html

 

C. Does doubling the volume of compressed air stored actually equate to doublling the potential energy?

 

D. Does doubling the pressure in the tank equate to doubling the volume of air stored?

 

E. The specification "30 gallon tank" doesn't seem very meaningful when talking about compressed air. I would expect the capacity to be cubic feet or meters, etc. Please explain.

 

Overall, I'm trying to understand what to expect in terms of increasing potential energy when some compressed air cylinder ("A") is replaced by another cylinder ("B"), which allows an increase in pressure over "A".

 

Thanks in advance for any help.

Edited August 13, 2012 by bjones

[Enthalpy]

1. This is a P vs T graph which doesn't tell how much work you can extract - there are some indirect links.

For water vapour up to a few ten bar, you can take P (atm) = [T (°C) / 100]^4.

 

A & B. 250 psi is realistic. Technology uses commonly hydraulic pressure of 210b, 350b, sometimes 700b, uncommonly 1500b. This pressure is stored in a gas, nitrogen.

Double the pressure less than doubles the maximum available work, unless the pressure increase over 1 atm is very small.

 

B plus. If you pass a liquid through an engine, the maximum available work is volume * pressure (or rather pressure drop, and only as long as this keeps constant). But then the work is obtained from a gas that pushes the liquid: it's not extracted from an energy stored in the liquid itself. In thermodynamics, the liquid keeps its internal energy; its enthalpy is converted to work. Interestingly, enthalpy is defined from the liquid's state, and suffices to compute the available work, but the work comes from an energy stored in part elsewhere, here it was fully in the gas.

 

B more. When expanding a gas, it colds down. The work you can extract depends on if you let it cold down naturally, or inject heat in it so, for instance, its temperature keeps constant - or any other situation. So no direct relation exists between pressure and work - only for some simple cases. If no heat is exchanged, the maximum work is the change of enthalpy; if temperature is constant, it's the change of internal energy.

 

B chatter. Internal energy and enthalpy are strongly linked with the gas' temperature, rather than pressure. In a "perfect gas" (which a real gas may resemble if its density is far less than the liquid's density), internal energy and enthalpy depend only on the temperature.

 

C. Yes.

 

D. Only if the temperature is the same in both cases, and only if density is far less than the liquid's density, the mass stored is propotional to pressure. This mass converts into a volume of air taken at normal atmospheric conditions.

 

E. As a metric integrist I feel imperial units useless in any situation.

 

F. Gas are more complicated because they have a temperature, a pressure and a volume, where one property can be deduced from the two others:

ONLY for perfect gas, it's simply P*V = n*R*T with sensible units, where n is number of moles (29g each for air) and R = 8.3145 J/mol/K.

Since all three can vary when you expand the gas, you must know more about how they vary during the process.

For instance if no heat is exchanged, then P*V^gamma is constant, with gamma~1.4 for air.

 

F more. At the pressure you cite, and reasonable temperatures, air is nearly a perfect gas with gamma~1.4, and then its enthalpy is 3.5 * R * T (K) in Joules per mole of 29g, or nearly 1kJ/kg/K. With no heat exchange, the available work is the change of enthalpy, and the ratio of pressure and the ratio of volume equal the ratio of temperature high 3.5 and 2.5 respectively - put them properly so pressure increases the temperature and decreases the volume. Then the maximum work can be computed with a limited effort. In more complicated cases you need experimental tables.

 

From the power 2.5 and 3.5 you see the relationship with pressure won't be linear. But the tank's volume is, as it defines how many moles you store, since P and T define how much volume each mole takes in the tank.

[bjones]

Thank you very much for your detailed reply. I'll need to chew on that awhile.

 

I knew my graph link was wrong. It was only an example.

 

That said, from what I understand of your reply so far, I can see why I was not able to find any graphs like I desired. That relationship (pressure to potential energy) is much too complicated to depict on one graph as many variables are involved.

 

Any other replies still welcome.

[Enthalpy]

In some special - but realistic - cases graphs exist. If you expand the gas at reasonable speed it won't exchange heat with the outside world, and then the maximum available work (losses exist) is the gas' change of enthalpy. Then, at moderate gas density, the enthalpy (H) depends just on the gas' temperature (T), so diagrams exist, as simple as H versus T.

 

Or rather, H versus T at many different pressures - but this isn't very useful because pressure changes when the gas is exploited.

More commonly, you get diagrams at constant entropy. Or as well, enthalpy versus entropy, very common.

 

Thermodynamics is a bit abstract but interesting. It's one huge basis of physics, but regrettably many physicists are uneasy with it, so learning it would give you an advantage.

I suppose beginning with perfect gas, followed by real gas, by vapour and liquid, and only later generalizing to any system is the best possible approach, so just go on like this. Most books and courses begin with the general case and give gas as a special situation - then it's very abstract and offputting.