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# Potential Energy in Compressed Air2

## Potential Energy in Compressed Air

(OP)
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.

### RE: Potential Energy in Compressed Air

ENERGY = PRESSURE x VOLUME

WORK = CHANGE_IN_PRESSURE x VOLUME

POWER = CHANGE_IN_PRESSURE x VOLUME / TIME

Going the Big Inch!
http://virtualpipeline.spaces.msn.com

### RE: Potential Energy in Compressed Air

(OP)
Nice and simple Thanks

What units should I use?

### RE: Potential Energy in Compressed Air

2

If I understood your question, you refer to available energy, A.E.

Assuming it would do reversible work discharging to the surrounding atmosphere (Po =1), neglecting mechanical kinetic and gravitational energies, and air acting as an ideal gas (PV=RT), in kJ/kg:

A.E. = RT[(Po/P1)-1+ln(P1/Po)]

For air R = 0.287 kJ/(kg.K)

P1 = 5, 10, 25 ata;  Po = 1 ata

Examples:

For P1 = 5 ata

A.E. = (0.287)(273+25)[(1/5) -1 + ln(5/1)] = 69.2 kJ/kg

For P1 = 25 ata

A.E. = (0.287)(273+25)[(1/25) -1 + ln(25/1)] = 193.2 kJ/kg

With the densities known one can estimate the A.E. for each given volume.

### RE: Potential Energy in Compressed Air

Unitwise, I would recommend using units of pressure, volume, and energy.  Perhaps Torr, minims, and MeV.

### RE: Potential Energy in Compressed Air

(OP)
Many thanks to member “25362” above. That was very useful and I think I am almost there but I’m getting a funny result.

I’m trying to find out what volume of compressed air I will need to accelerate a human body to around 2m/s (nearly 5mph).  (Assuming body is on a frictionless sled and there are no losses)

Using ½ mV2 and a mass of 80kg I need to store and release about 17 J of energy.

Taking the first example posted by member “25362” above, 5 atmospheres of pressure and a density of air of roughly 1.23 kg/m3 I end up with a volume no bigger than a Rubik’s Cube!

I took Available Energy = 69.2 kJ/kg and I need 17 J so:
mass of air required =  (17/69,200)  =  0.000246 kg
volume of air required  =  (0.000246kg)/(1.25kg/m3)
=  0.000197 m3
which is a cube with sides of 5.8cm

Can is possibly be true that this much compressed air at 5 atmospheres can accelerate a man to nearly 5mph?

### RE: Potential Energy in Compressed Air

I don't think that you evaluated the energy required correctly.  Also, you will need to store energy to overcome any losses as well.

I2I

### RE: Potential Energy in Compressed Air

(OP)
Apologies, correction to the above – but still very strange:

Many thanks to member “25362” above. That was very useful and I think I am almost there but I’m getting a funny result.

I’m trying to find out what volume of compressed air I will need to accelerate a human body to around 2m/s.  (Assuming body is on a frictionless sled and there are no losses)

Using ½ mV2 and a mass of 80kg I need to store and release about 160 J of energy.

Taking the first example posted by user “25362” above, 5 atmospheres of pressure and a density of air of roughly 1.23 kg/m3 I end up with a volume a bit bigger than a Rubik’s Cube!

I took Available Energy = 69.2 kJ/kg and I need 160 J so:
mass of air required =  (160/69,200)  =  0.002312 kg
volume of air required  =  (0.002312kg)/(1.25kg/m3)
=  0.001849 m3
which is a cube with sides of  12.25cm

Can is possibly be true that this much compressed air at 5 atmospheres can accelerate a man to nearly 5mph?

### RE: Potential Energy in Compressed Air

That's assuming that the man is the piston.  You have to account for the accelerating the mass of the piston, the friction of the seals, etc.

TTFN

### RE: Potential Energy in Compressed Air

Swetenham,

the principle behind your calculation is OK. The work to be done on the body without friction to accelerate it horizontally from rest (vinit=0) to 2 m/s is indeed 1/2 mv2, e.g., the change in its kinetic energy.

However, as IRstuff pointed out, friction, μ, would have to be considered, and the force to overcome this friction would be

μ(80 kg)(9.8 m/s2)= 784μ N

Assume the kinetic friction coefficient is 0.3, the energy needed to overcome this friction over 10 m length would be 0.3×784×10 = 2,352 J, about 15 times the previously estimated energy !

Once a teacher of physics said to us: "remember, the brakes stop only the wheels; it's road friction that stops the car."

### RE: Potential Energy in Compressed Air

To get an idea why hydraulic pressure tests are preferred to air pressure tests look at the energies involved.

100 kJ/kg in a container with 10 kg compressed air, when it explodes the release of energy is equivalent to

100 kJ/kg×10 kg÷4270 kJ/kg TNT = 0.234 kg TNT

### RE: Potential Energy in Compressed Air

Swetenham,

Do you really want to consider the person to be the piston, or should you consider the person to be an object in the path of the expanding air and use a drag coefficient approach?

The probability that a person is laying up against a container of compressed air *may* be pretty small, or it may not.  If this is for a fabrication shop and they have to mark any leaks, it would not make sense.  If it does make sense to your application, then a rule about making walkways and normal traffic access X feet from a pressure containing component may benefit your approach.  In the plants I work in, it would not make sense just from the volume of pressure containing components there.

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)?

Good luck,
Latexman

### RE: Potential Energy in Compressed Air

The original formula for work on expansion = A.E. = PoΔV -ToΔS

Since ΔV = RTo (1/P1 - 1/Po)
and ΔS = R ln (P1/Po)

A.E. = RTo[(Po/P1)-1] +RTo ln (P1/Po) = RTo(ln P1/Po -1 + Po/P1)

### RE: Potential Energy in Compressed Air

If the person isn't the piston, then you need to put more mass in the equation, since the air must accelerate the piston as well.  Of course, trying to squish a person to squish air to 70 psi might also be a challenge.

TTFN

### RE: Potential Energy in Compressed Air

25362,

They may be equivalent, but I approach it differently.

The work the expansion can do = ∫PsystemdVsystem.  Here you have to model the expansion as isentropic, adiabatic, or isothermal, if an analytical solution is wanted because P and V vary during the expansion.  A "real" solution requires a numerical or graphical solution.  Since the fluid is air at low reduced pressures and high reduced temperatures the analytical solution is likely close enough.  Isothermal is the easiest, but adiabatic is probably closer to reality.  Please pardon me from supplying an analytical equation now as my references are at the office.

The work done on the surroundings = ∫PsurroundingsdVsurroundings.  This one is very straight forward.  The work done on the surroundings = Patmosphere ΔV.

The work to accelerate the piston (person) = M/gc∫u du.  This one is very straight forward too.  The work done to accelerate the piston (person) from rest = Mv2/2gc.

This all rolls up to ∫PsystemdVsystem - Patmosphere ΔV = Mv2/2gc.

I'm still not sure you've accounted for the expansion having to push the surrounding atmosphere bach.

Good luck,
Latexman

### RE: Potential Energy in Compressed Air

IRstuff,

If the person isn't the piston, then you would have to use the entire mass of the failed containment vessel for a total loss of containment (that doesn't seem credible) or the fraction that is involved in the hypothetical failed area of a credible scenario.  Right?

Good luck,
Latexman

### RE: Potential Energy in Compressed Air

No, the person would simply be sitting on the atmosphere side of the piston, but both the piston and the person would need to be accelerated, thus incurring a larger energy consumption.

There may be some confusion about the OP's actual desires.  His most recent postings involve propelling a person on purpose and not about a failed containment vessel.

TTFN

### RE: Potential Energy in Compressed Air

The work in the expression I brought is indeed Workmax- Worksurr.

Studies have confirmed that with exploding vessels the work done by the expanding gas is associated with adiabatic expansions. The general expression being:

W = ∫ PdV = (P2V2-P1V1)] / (1-n)

where n = k = Cp/Cv assumed constant, for adiabatic expansions, the subscripts 1 and 2 represent initial and final states. Rewriting the expression and replacing V2, the result is:

W = [P1V1/(k-1)] [1-(P2/P1)(k-1)/k]

n, a bit smaller than k, for a polytropic expansion; for an isothermic expansion, W = P1V1 ln(P1/P2)

As for KE one doesn't need the conversion factor gc when using mass in kg, velocity in m/s, since the resulting kg.m2/s2 = J.

BTW, if the depressurizing is done adiabatically (thermally insulated containers) the remaining air in the containers would cool down by

T2/T1 = (P2/P1)(k-1)/k

### RE: Potential Energy in Compressed Air

I is not just the volume of air. The time to transfer it is also an issue. If you want to use a piston to create the force to push/eject an object then you have to take into account the flow rate into the piston. From my experience this may be the main concern. I have designed an ejecting system using a fabric/rubber sack as a mass less piston. The gas to fill the sack came from a pressure vessel. It appears that the object speed may be sometimes faster then the filling process of the sack so the pressure in the sack is variable. To solve such a system you have to write a set of differential equations that accounts for the gas dynamics and thermodynamics coupled with the dynamic equations of the moving objects and solve them numerically.

### RE: Potential Energy in Compressed Air

Note the significant difference between the quantity of work you can get out of a given volume of stored compressed air depending on whether the expansion is adiabatic or isothermal (i.e. absorbing heat from the surroundings).  If you can do the ideal, reversible expansion under isothermal conditions, you can get a great deal more work out of a given stored volume of gas than if you do it adiabatically.  Unfortunately, you need the time and area (and potential frictional loss) associated with the required heat transfer area to get the benefit of that additional work, which in most devices propelled by compressed air is just simply not used.  This reality, combined with the realities of the compression side of the equation using real compression equipment, limits the energy efficiency of using compressed air as an energy storage medium.

### RE: Potential Energy in Compressed Air

25362,

Where was I?

∫PdV = (P2V2 - P1V1)/(1 - k)  Balzhiser, Samuels, and Eliassen agree with you.

Patmosphere = P2

ΔV = V2 - V1

(P2V2 - P1V1)/(1 - k) - P2(V2 - V1) = Mv2/2gc

V1 = 250 liters
P1 = 4.935 atm (5 bar abs)
P2 = 1 atm
k = 1.4
V2 = V1(P1/P2)1/k = 782 liters

Mv2/2gc = ((1)(782) - (4.935)(250))/(1 - 1.4) - (1)(782 - 250)

Mv2/2gc = 597 liter.atm

liter.atm x 101 = N.m

Mv2/2gc = 60335 N.m

M = 80 kg

gc = 1 kg.m/N/sec2

v = ((60335 N.m)(2)(1 kg.m/N/sec2)/(80 kg))1/2

v = 39 m/sec = 87 mph

Wow, that does seem unrealistic!

Mv2/2gc = ((4.935)(250)/0.4)(1-(1/4.935)0.4/1.4

Mv2/2gc = 1129 liter.atm  Hmmmmm, yours is different from mine by P2(V2 - V1)  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

### RE: Potential Energy in Compressed Air

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)

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 ?

### RE: Potential Energy in Compressed Air

25362,

Maybe.

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 Workmax- Worksurr.”.

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

### RE: Potential Energy in Compressed Air

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
Wmax=U1+PoV1-ToS1 and substracted Wsurr=Uo+PoVo-ToSo, considering the compressed air should overcome the resistance presented by the surrounding air's pressure Po.

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.=RTo [ln(P1/Po)-1+(Po/P1)]

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

### RE: Potential Energy in Compressed Air

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

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