Potential Energy in Compressed Air 2

[Swetenham]

How would one calculate the potential energy content of compressed air at 25 degrees C (~room temperature) for a range of volumes and pressures? For example:

0.25, 0.5 and 1 cubic meter at 5, 10 and 25 Bar?

Please ignore all losses e.g. heat loss when pressurising.

 

[Latexman]

25362,

Where was I?

[∫]PdV = (P[sub]2[/sub]V[sub]2[/sub] - P[sub]1[/sub]V[sub]1[/sub])/(1 - k) Balzhiser, Samuels, and Eliassen agree with you.

P[sub]atmosphere[/sub] = P[sub]2[/sub]

[Δ]V = V[sub]2[/sub] - V[sub]1[/sub]

(P[sub]2[/sub]V[sub]2[/sub] - P[sub]1[/sub]V[sub]1[/sub])/(1 - k) - P[sub]2[/sub](V[sub]2[/sub] - V[sub]1[/sub]) = Mv[sup]2[/sup]/2g[sub]c[/sub]

V[sub]1[/sub] = 250 liters
P[sub]1[/sub] = 4.935 atm (5 bar abs)
P[sub]2[/sub] = 1 atm
k = 1.4
V[sub]2[/sub] = V[sub]1[/sub](P[sub]1[/sub]/P[sub]2[/sub])[sup]1/k[/sup] = 782 liters

Mv[sup]2[/sup]/2g[sub]c[/sub] = ((1)(782) - (4.935)(250))/(1 - 1.4) - (1)(782 - 250)

Mv[sup]2[/sup]/2g[sub]c[/sub] = 597 liter.atm

liter.atm x 101 = N.m

Mv[sup]2[/sup]/2g[sub]c[/sub] = 60335 N.m

M = 80 kg

g[sub]c[/sub] = 1 kg.m/N/sec[sup]2[/sup]

v = ((60335 N.m)(2)(1 kg.m/N/sec[sup]2[/sup])/(80 kg))[sup]1/2[/sup]

v = 39 m/sec = 87 mph

Wow, that does seem unrealistic!

Using your equation:

Mv[sup]2[/sup]/2g[sub]c[/sub] = ((4.935)(250)/0.4)(1-(1/4.935)[sup]0.4/1.4[/sup]

Mv[sup]2[/sup]/2g[sub]c[/sub] = 1129 liter.atm Hmmmmm, yours is different from mine by P[sub]2[/sub](V[sub]2[/sub] - V[sub]1[/sub]) or (1)(782 - 250) or 532 liter.atm

1129 - 532 = 597

I think you are wrong. Your equation does not account for the work to push back the surroundings.


Good luck,
Latexman

 

[25362]

Latexman, please consider, for a moment, as the control volume only the compressed air in its well-insulated cylindrical container, and assume it pushes a piston until it reaches atmospheric pressure. Now consider the final and initial conditions of the air all the time inside the container.
We both agree on that the work the air has done against the piston on its adiabatic expansion = (P1V1-P2V2)/(k-1)

Where, using your figures:

P are pressures: P1=4.935 ata, and P2=1 ata,
V are the volumes: V1=250 L, and V2=782 L. Please note the gas is cooler.
K=1.4

W = [(4.935)(250)-(1)(782)]/0.4 = 1129 L-atm​

Am I missing something ?

 

[Latexman]

25362,

Maybe.

The original post asked only about the potential energy content of compressed air, which your available energy equation answered very nicely.

Later, the original poster (OP) further explained that he was calculating how much compressed air was needed to accelerate a human body to a certain velocity. With this extra information on what the OP was doing, it became possible to progress and refine an answer, which I was working on. In my weekend post I asked, “Does the -1 in the A.E. equation account for the work done by the expanding gas pushing back the surrounding air (not the person)?”

You said, “The work in the expression I brought is indeed Work[sub]max[/sub]- Work[sub]surr[/sub].”.

After I referred to my copy of Balzhiser, Samuels, and Eliassen at the office, I finished looking at the OP’s smallest volume of air at the lowest pressure in the original post to accelerate the human body. While doing this, it became obvious that the A. E. equation “does not account for the work to push back the surroundings”.

Your equation is correct for the potential energy content. To calculate the final velocity of the body (piston), the work for the piston to push back the surrounding atmosphere must be subtracted first.

The adding of additional information may have put us on different pages.

Good luck,
Latexman

 

[25362]

Latexman, you are right. In my first answer, with examples, I used the Gibbs free energy function: G = H-TS = U+PV-TS.

Although I neglected potential and kinetic energies, I took
W[sub]max[/sub]=U[sub]1[/sub]+P[sub]o[/sub]V[sub]1[/sub]-T[sub]o[/sub]S[sub]1[/sub] and substracted W[sub]surr[/sub]=U[sub]o[/sub]+P[sub]o[/sub]V[sub]o[/sub]-T[sub]o[/sub]S[sub]o[/sub], considering the compressed air should overcome the resistance presented by the surrounding air's pressure P[sub]o[/sub].

The subindex o means the surrounding air, the subindex 1 refers to the compressed air. Since both U's are equal they cancel out, I finally got the expression:

A.E.=RT[sub]o[/sub] [ln(P[sub]1[/sub]/P[sub]o[/sub])-1+(P[sub]o[/sub]/P[sub]1[/sub])]​

which I thought (probably wrongly) it would include the energy needed to compress the surrounding air.

 

[magicme]

i hope this response does not violate any forum rules.... if it does, my apoplogies and please trash it.

this question is not very clear to me as the numbers are flying about fast and furious and i'm not sure i got the knowns and unknowns straight in my mind.

but i recently posted a javascript calculator on my thermo course web page for polytropic expansion / compression and it calculates the work involved in the process. which of course depends on how you do the expansion / compression.

the calculator is for students to compare to their homework answers.

possibly it is useful here (only in terms of work / energy in a mass of compressed gas).... possibly it is not.

the link to the page is here:

http://myweb.wit.edu/leod1/MECH240/polytropic.html

i hope this can be helpful, but again, the known and unknown variables in this problem are not clear to me.

daveleo