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.

 

[BigInch]

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

 

[Swetenham]

Nice and simple ) Thanks

What units should I use?

 

[25362]

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

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

A.E. = RT[(P_o/P₁)-1+ln(P₁/P_o)]​

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

P₁ = 5, 10, 25 ata; P_o = 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.

 

[ivymike]

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

 

[Swetenham]

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?

 

[insult2injury]

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

 

[Swetenham]

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?

 

[IRstuff]

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

 

[25362]

Swetenham,

the principle behind your calculation is OK. The work to be done on the body without friction to accelerate it horizontally from rest (v_init=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)= 784[μ] 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."

 

[25362]

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​

 

[Latexman]

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

 

[25362]

The original formula for work on expansion = A.E. = P_o[Δ]V -T_o[Δ]S

Since [Δ]V = RT_o (1/P₁ - 1/P_o)
and [Δ]S = R ln (P₁/P_o)

A.E. = RT_o[(P_o/P₁)-1] +RT_o ln (P₁/P_o) = RT_o(ln P₁/P_o -1 + P_o/P₁)

 

[IRstuff]

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

 

[Latexman]

25362,

They may be equivalent, but I approach it differently.

The work the expansion can do = [∫]P_systemdV_system. 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 = [∫]P_surroundingsdV_surroundings. This one is very straight forward. The work done on the surroundings = P_atmosphere [Δ]V.

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

This all rolls up to [∫]P_systemdV_system - P_atmosphere [Δ]V = Mv²/2g_c.

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

Good luck,
Latexman

 

[Latexman]

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

 

[IRstuff]

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

 

[25362]

The work in the expression I brought is indeed Work_max- Work_surr.

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 g_c 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​

 

[israelkk]

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.

 

[moltenmetal]

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.