proof that compressed air is solar energy

I need to move this blog to a place where formatting tools are provided so I don’t have to send people to download the version that looks right. In the meantime, here is the URL so you can read the correctly formatted version:

http://docs.google.com/fileview?id=0B-DQiVAbbA7WNTA1MWE1YmUtMzQ4Yy00ZjNiLTkwYzktYjY2OWEyN2JkMWVi&hl=en

The associated spreadsheet is at this URL:

http://docs.google.com/fileview?id=0B-DQiVAbbA7WNzU4ZDZkY2ItM2IyYy00OTA3LTkyMzctNWEzNTc0YzA1YjE1&hl=en

Regarding the controversial “lost information” that compressed air can be manipulated to make free energy available. When dreamers and doers come together to shout at each other in the heat of mutual intolerance, they should remember these things so their conversation has some hope of having something to do with compressed air:

The three-part assertion is this:

1. All compression work is lost as heat. (CONTROVERSIAL)
2. After cooling, compressed air still has pressure to expand and do work. (ACCEPTED)
3. This availability of usable energy is accounted for by the fact that the air contained thermal energy before it was compressed. (CONTROVERSIAL)

It is not the textbook writers who find points 1 & 3 above to be doubtful; it is the uneducated including engineers and sciolists who assume they know something about compressed air but are, in reality, effectively uneducated as to compressed air theory in regards to points 1 & 3 above.

1. All compression work is lost as heat.
The wording of the first part of the assertion is critical. It says “compression work”, not “the compressor’s work”. The compressor’s work has three parts: intake, compression, and delivery. It is the compression work—not the intake work or the work of delivering air into the tank—that is, according to the assertion, all lost as heat. “Compression work” (NOT “the compressor’s work”) is the work required to squeeze the air into a smaller volume. During compression work, the compressor cylinder is closed. Nothing is entering or leaving the cylinder. In adiabatic compression, not even heat is entering or leaving the cylinder. The piston is moving; the cylinder is getting smaller. The air is being squeezed. The heat is spontaneous; if the piston were to change direction, allowing the air to expand, the temperature would go down just as spontaneously, just as instantaneously.

So what happens to the work done to draw atmosphere into the compressor’s intake, and what happens to the work done to push the air into the tank at constant pressure?

INTAKE WORK is negative work. It can’t be lost because it is work not done, work canceled. The crankshaft is already moving and atmosphere rushes in behind the moving piston because of its positive pressure, canceling or making up for the work done to keep the crankshaft going around during intake.

DELIVERY WORK is positive work. It is done with the downstream valve open and air flowing out of the cylinder against constant tank pressure. This is the simplified version, assuming that the tank pressure isn’t going up since an equal amount of air is going out of the tank and down the pipe to be used in the air engine. So what is happening to the delivery work? First let’s picture this: imagine there is no air leaving the tank. So the delivery is not at constant pressure, it is at increasing pressure.

Wrong. The compression all takes place inside the cylinder with the valve closed. If tank pressure is ever higher than cylinder pressure, the valve closes. So that scenario is not valid: the delivery part of the power stroke cannot take place against an increasing resistance. Only when cylinder pressure overtakes tank pressure can the air leave the cylinder to enter the tank.

OK, so what happens to the delivery work? The textbook says that it is displacement air, it takes the place of air leaving the tank. At this point I don’t know. But I think it’s important, because depending on the pressure and other factors, delivery work can be more than compression work. The book states firmly that this amount is recovered by a chain of displacement through the system. In reference to air engines.

Just remember that the process of compression is the squeezing of air; the work of an air compressor is more complicated than that, so the two must be dealt with distinctly by theorists.

2. After cooling, compressed air still has pressure to expand and do work.

Nobody argues against this. The fact that the sciolist and the engineer who is uneducated on these points will agree readily with point 2 and aggressively deny points 1 & 3 just means that these folks feel that the letters after their names make them omniscient. My viewpoint is that because point 2 is true, points 1 & 3 must be true. But the untapped potential of air cannot speak for itself. Other than weather patterns which are like the forest we can’t see for the trees, it is not obvious to anyone that air carries solar energy; that compressed air in a tank contains none of the compressor’s invested energy. To assume the opposite is natural and therefore commonplace. In our society, the majority rules, and to hell with the math; John Doe, BS, didn’t get his BS for free and he is not going to concede on any point as long as he still has a student loan to pay off. That’s the math he really cares about. It’s personal.

By the time the motivating factors for his early arrogance—the costs of education—become a non-issue, he is a grizzled veteran and nobody questions the craggy mystique of the career’d.

Of course the real winners of that game are the banks, and John Doe, BS and all, is just another puppet, full of himself and thriving in the suburbs, unheralded champion of a fourth mortgage.

3. This availability of usable energy is accounted for by the fact that the air contained thermal energy before it was compressed.

Try getting a skeptic to address, with compressed air math, the fact that the air already contained thermal energy before it was compressed. They won’t do it because they don’t know how. Where is that energy now that the air has been compressed into a tank and cooled? I don’t know what answer they will give because no skeptic has yet faced up to that question or honestly tried to answer it. They will always change the subject, bring up laws and principles and generalizations and accusations that don’t apply, in order to avoid the specific question.

As Simons states in his textbook, the compressor no more put the energy into the air stuffed into a cooled-off air tank than Hoover Dam built itself and put water behind itself. The dam, like the compressor and tank, was built and arranged to be filled by people, and the water, like the air in the tank, was instilled with usable energy by the heat of the sun.

THE MATHEMATICAL PROOF THAT ALL COMPRESSION WORK IS LOST AS HEAT

I can’t stress enough that there are three parts to a compressor’s work, and the part we’re talking about here is the part where the air is trapped in a closed cylinder with a moving piston making its pressure go up while its volume goes down.

That is the work that becomes heat and is all lost by the time the final product is cooled and sitting in a tank.

If we know the initial pressure and volume of air that is going to be compressed, and we know the pressure it will be compressed to, the final volume can be found using the adiabatic equation. Adiabatic compression is when the heat of compression stays in the air, and it is appropriate here because it keeps the system’s energy isolated from outside influences and it produces the maximum possible heat that an air compressor will put into air without having a fire built under it.

Follow along in the Simons textbook and/or the spreadsheet if you want, as the example used is the same. If you know nothing about compressed air math, study Part One of my book Compressed Air Power Secrets, 3rd Edition.

V2/V1 = (P1/P2)1/n

Here the adiabatic equation tells us that as the volume of air is decreased by any method, its pressure increases. The equation also tells us that the process is inverse in general—pressure goes up while volume goes down—but not exactly inversely proportional. If you push air into half its original space, its pressure goes up less than double. The exponent “1/n” defines to what extent the relationship is not exactly inversely proportional.

The index n is the fingerprint of air. Its use in the adiabatic equation is what pinpoints the amount of heat that is generated, and it has been known for maybe 150 years that for air, this index is equal to about 1.406 for adiabatic conditions. Adiabatic condition means no heat exchange with the surroundings.

This is only one way to arrange the adiabatic equation. By simple algebra we can solve it for final volume:

V2 = V1(P1/P2)1/n

We are going to consider the case of compressing one pound of sea level atmosphere from 0 gauge pressure to 75 psig. That’s 14.7 absolute pressure (psia) and 89.7 psia. Ambient conditions will be set at 60° F which is 521° absolute temperature. At this temperature at sea level, one pound of air occupies 13.09 cubic feet. To find the final volume after compressing to 89.7 psia, substitute values into the formula:

V2 = 13.09(14.7/89.7)0.71
V2 = 3.62 cubic feet

Now that we know both initial and final values for both pressure and volume, the adiabatic work equation tells us theoretically how much work is required in the compression of the pound of air. Remember this is not all that a compressor does, but the work of compression during PV change only, while P is getting bigger and V is getting smaller. The adiabatic work equation for the compression phase of a compressor’s work:

W1 = 144(P2V2 - P1V1)/(n -1)

P2V2 is the final energy condition of the air, after compression. P1V1 is the initial energy condition before compression. P2V2 - P1V1 is the difference between the two, which is the basis for the work input by the compressor, the energy added to the intake air by pushing it into a smaller space. But not the whole story. “n - 1” corrects that value to take into account the effect of adiabatic heat buildup. The conversion factor 144 changes the result to ft-lbs since pressure is being stated in psi or pounds per square inch. To get the result to read out correctly in ft-lbs of work, the conversion factor 144 is needed to change the pressure to psf or pounds per square foot.

Substitute values into the formula:

W1 = 144(89.7*3.62 - 14.7*13.09)/0.406
W1 = 46806 ft lbs

Depending on how much rounding off takes place, the result varies by one or two hundred ft-lbs, which is no more than six thousandths of a horsepower if the work is done in one minute. Also the natural constants such as n = 1.406, the absolute zero, and the specific heat capacities of air, are not known exactly. That’s why a slightly different result is to be expected depending on who does the calculations and where they round off their results.

Using another version of the adiabatic equation, we can find out how hot the air got during the compression work:

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

Initial temperature was 521° F abs, so substitute values to find the final temperature:

T2 = 521(89.7/14.7)0.29
T2 = 878.45° F abs

The added temperature due to compression:

T2 - T1 = 878 - 521
T2 - T1 = 357° F.

This adds up to saying that it takes about 47,000 ft-lbs of work to raise the temperature of 1 pound of air 357 degrees by mechanically squeezing the molecules closer together.

But what if we were to heat the same air directly instead of heating it by compression? Use the same temperature differential for 1 pound of air:

BTUs req'd = Cv(T2 - T1)
ft-lbs req'd BTUs * 778

The thermal unit or British Thermal Unit (BTU) is a measurement of thermal energy. One BTU equals 778 ft-lbs of energy or work. Cv is the work required to raise the temperature of one pound of air one degree Fahrenheit. Its value is 0.1689 and it is part of air’s index n:

n = Cp/Cv = 0.2375/0.1689
n = 1.406

Air’s index “n” is thus the ratio of maximum work needed to compress air (adiabatic compression) to the minimum work needed to compress air (isothermal).

Substituting values to find the BTUs and ft-lbs required to do the same heating job with heat instead of compression:

BTUs req'd = 0.1689 * 357 = 60.3 BTUs
ft-lbs req'd = 60.3 * 778 = about 47,000 ft-lbs when rounded off

Without any rounding off and using the most accurate values I have for the natural constants, the work done by compression takes 46,805.84 ft-lbs, and by direct heating the work done on the same pound of air is 46,970.15 ft-lbs. The difference in the results is five thousandths of a horsepower if the work is done in one minute. That is not considered a difference (the difference is negligible) for reasons already mentioned.

So it’s true: the work of compression (PV change) is all turned into heat.

The compressor’s work is NOT stored in the tank. So why is there still usable pressure in the tank when the heat has all dissipated? First of all, how much pressure will there be?

Charles’ Law states that a mass of air at a constant pressure changes volume at a constant increment per change of its increment of temperature. That translates to some very useful generalizations about air, once you transform its mathematical version into relevant forms with simple algebra. One form of Charles’ Law shows that if the volume of the air mass stays constant, pressure and temperature will change proportionately to each other. We’ll use this to see how much pressure is left in the tank after the heat of compression is all gone. Since the tank volume stays constant, V is the same on both sides of the equation and cancels itself out, so is not mentioned in the equation:

P3/P2 = T3/T2

Solving for the unknown new final pressure:

P3 = P2(T3/T2)

T2 is the same as before, the final pressure after compression. T3 is the new final temperature, once again ambient or 521° F abs.

Substituting values in the equation:

P3 = 89.7(521/880)
P3 = 53.2 psia
gauge pressure = 53.2 - 14.7 = 38.5 psig

This shows…

• …that it takes 47,000 ft-lbs of compression work to compress 1 pound of air into a tank that has a volume 3.62 cu ft, from atmospheric pressure (sea level at 60° F) to a pressure of 75 psig.
• …that 47,000 ft-lbs of compression work will be lost to heat dissipation when you change the condition of air between these starting and ending conditions.
• …that after the heat has all been lost, the tank will still hold air that is at 38.5 psig.

That is a very typical 38.5 psig, not some trick. It’s business as usual. It can expand down to atmospheric pressure, doing work. To find the new volume it will occupy once it is again part of the atmosphere:

V1 = V2(P3/P1)1/n

V1 is not the same as the V1 we started with, which was the volume occupied by 1 pound of air at sea level and 60° F. This is the final volume after expansion of the same pound of air from the same tank (3.62 cu ft) but starting at ambient temperature. Expanding from 53.2 psia down to 14.7 psia, and with the exponent 1/n taking care of the temperature changes in adiabatic expansion, substitute known values to find how much the same pound of air will occupy when its temperature goes down below ambient during expansion:

V1 = 3.62(53.2/14.7)0.71
V1 = 9.02 cu ft

Now we have all the factors to calculate how much work can theoretically be done by that much air expanding completely under those conditions:

W = 144(P3V2 - P1V1)/(n -1)

Substituting values:

W = 144(53.2*3.62 - 14.7*9.02)/0.406
W = 21,170 ft-lbs

Knowing that the work of compression was 47,000 ft-lbs, we can find out how much work can be done by this air in comparison, once all the work of compression has been lost forever:

W/W1 ratio of expansion work to compression work, adiabatic cond

W/W1 = 21170/47000
W/W1 = 45%

So 45% of the theoretical work of compression—and 100% of that is lost as dissipated heat—is then regained by using the heat of the atmosphere from where it is trapped in an air tank to expand into the atmosphere and do work.

This is not magic, it is not perpetual motion, it is not hard to believe, and it is not taught to engineers anymore. Because it’s changed? Because it’s wrong? Because it’s a conspiracy? None of the above. It’s just not taught. It’s forgotten because it’s not mentioned in textbooks anymore. I don’t believe that there is currently any conspiracy to cover this fact up. Conspiracy isn’t needed when it is assumptions that steer concensus opinion.

It could be a coincidence that the information disappeared from compressed air textbooks at the same time that petroleum fuels were unquestionably on top for the first time in history. I doubt it. But if there was once a conspiracy to change the textbooks, once calculus took over where algebra had always been adequate, textbooks were not used anymore to teach compressed air. Either engineers learn their air routines on the job or they don’t need to, and most engineers don’t make it their hobby to go home after work and study things their professors forgot to tell them about in college. It’s just forgotten. It’s too easy; it’s not cutting edge; it’s low tech.

Compressed air is taught from a practical standpoint in an industry where it is a workhorse taken for granted, and the details of its inner life are not considered worth spending any time on. Too bad.

There’s more:

THE MATHEMATICAL PROOF THAT AIR’S ENERGY IS NOT BASED ON ITS PRESSURE

proof 1:
pressure * area = force
pressure = force/area
force * distance = pressure * volume = energy
pressure = energy/volume
conclusion: pressure ≠ energy

proof 2: Pressure is so often mistaken for energy in the diatribes of those who propound against the fact that energy contains intrinsic energy of its own, that they might as well burn the physics book when it comes time to study the production and use of compressed air.

The internal energy of air is entirely dependent on its thermal energy, the heat it contains. And the internal energy of air is entirely INdependent of its pressure. Pressure is a component of force, and force is a component of work, and work is what energy can do. Pressure is not energy. Here is the proof.

The maximum specific heat capacity of air is not 0.1689 BTU per lb per °F as pertains to heating air at constant volume. The higher quantity 0.2375 BTU refers to heating air at constant pressure, which is a condition in which the heated air is allowed to expand instead of being confined in a constant volume. The work available from air in expansion and the work needed to compress air are therefore going to be related to the proportions between the two specific heat capacities of air, as well as the arithmetical difference between them. I have gone into this in detail in my book Compressed Air Power Secrets.

One pound of air at atmospheric pressure and 60° F or 521° F abs contains an amount of energy proportional to the specific heat capacity of air, or Cp:

Internal Energy = 521 * 0.2375
Internal Energy = 123.7375 BTUs
123.7375 BTUs * 778 = 96,268 ft-lbs of work

Work is what energy can do. Work and energy are measured in the same units, in my case foot-pounds (or ft-lbs) since I’m too old to switch over to SI units and the older textbooks I read are not going to changeover with me if I do.

A pound of atmosphere at the stated conditions contains energy that could do 96,268 ft-lbs of work if it could expand down to a vacuum. But atmospheric pressure is generally unavailable for doing work because the back pressure of the atmosphere resists it and it can do no work without added energy.

If the circumstances were right, then some of that energy could be used. The same air, containing the same energy, could do some work with part of that energy if the air were at a higher pressure than ambient, so it could expand and give up some of its heat or internal energy.

Int. energy = Cp * T1 in BTUs
internal energy of 1 pound of air in ft-lbs

Substituting values for air at sea level and 60° F:

internal energy of 1 pound of air = 0.2375 * 521
internal energy of 1 pound of air = 123.7 BTUs
intern al energy of 1 lb air * 778 = 96,268 ft-lbs, but none is available

If the same pound of air is in a tank compressed to 100 psig, the adiabatic equation can be used to find the air’s final temperature T1 after expanding to 0 psig.

T1 = T(P1/P)n-1/n

Substituting values:

T1 = 521(14.7/287.18)0.29
T1 = 287.8 °F abs
287.8 - 461 = -173.18°

The difference in temperature T - T1 = 521 - 287.8 = 233.18° F

The difference in temp 233.18 * 0.2375, the specific heat capacity of air at constant pressure, gives the work of expansion:

233.18 * 0.2375 = 55.38 BTUs

Converting to ft-lbs by the multiplier 778:

55.38 * 778 = 43,086 ft-lbs.

So the theoretical usable work available from a pound of 100 psi air is 43,086 ft-lbs.

Now to find out the total energy of a pound of air, usable or not. The absolute temperature remaining in the air * Cp:

287.8 * 0.2375 = 68.36 BTU

Convert to ft-lbs:

68.36 * 778 = 53,182 ft-lbs still in the air but unavailable since ambient pressure has already been reached.

Unavailable energy still in air after expansion to ambient pressure: 53,182 ft-lbs
Available energy already used expanding to ambient pressure: 43,086 ft-lbs
Total energy of 1 pound of air starting at 100 psig: 96,268 ft-lbs

From above: total energy of 1 pound of atmosphere: 96,268 ft-lbs

Let the doubters doubt, but better yet, let them break their pattern and present some evidence:

COMPRESSED AIR IS SOLAR ENERGY!

We are living in a gigantic tank of compressed air heated by the sun…now what are we going to do about it?

THE LAST WORD FOR TODAY: IS THERE A GLITCH IN MY THEORY?

Because of this chapter from Simons and the less detailed sections from similar textbooks, I believe that some of the people who claim they can make an air car operate its own compressors are telling the truth. But I have to be skeptical and question everything.

Simons seems to contradict himself on one point. In two different parts of his chapter he refers to the energy stored in atmosphere as it relates to compressed air made by compressors and used by air engines.

In the first quote below he seems to be saying that the contribution of energy by the atmosphere to the operation of the compressor is canceled out by back pressure of the atmosphere against the exhaust of an air engine later. (?I thought that the “contributed” or “negative” intake work in the compressor was already canceled out by the fact that the compressor had already used an equivalent amount of energy to resist atmospheric pressure which would otherwise pull back on the progress of the crankshaft in its rotation?)

In the second quote he seems to contradict the earlier statement by saying that the air in the tank is going to be used, it can do work, and it wouldn’t be there if not for the energy of the atmosphere.

Here is the background I see for this apparent contradiction. A closer look is all that’s needed, but so far I don’t understand how both his statements can be true.

In the first quote (see below) he is referring to a specific amount of energy at a specific time: atmospheric pressure entering and leaving a compressor. He states that the atmospheric pressure, or the energy it represents, is stored up in the compressor’s discharge air. Then, we can infer, the air is pushed into the tank through the delivery part of the compressor’s power stroke. According to the later quote from the same chapter (see below), the air cools and when the air is to be used, it contains useful energy that was in it before it was compressed. But according to the first quote, even later than that the air will be used in an engine, and when the engine’s exhaust valve opens, the same atmospheric pressure is now OUTSIDE THE SYSTEM resisting its exhaust. So the atmospheric pressure in the engine that is part of its exhaust must now cancel out or resist the outside air in order for the air to move. All air engines that have an exhaust must operate against a back pressure of outside atmospheric air, minimum.

It needs to be figured out, because both his statements are true but he was just unclear enough about the context of each that we have to figure out how and when the two seemingly contradictory statements could both be true. I will save that for next time: here are the quotes:

First quote:

It may be asked: What becomes of the energy contributed by the atmospheric air toward compression and delivery…? This energy is actually stored up in the compressed air when the latter leaves the compressor. It could do useful work if it were practicable to exhaust the air from the engine into a vacuum. But since we must exhaust against atmospheric pressure, the energy is consumed in the process of exhaustion and is therefore not available for useful work. It is not included in the formulas expressing power to be furnished by the compressor because it is furnished gratis by the atmosphere; and it is not included in the formulas expressing the useful work which a volume of compressed air can perform, because it is not available for such work.

Second quote:

The answer to the question, why energy still remains in the compressed air after all the heat of compression has been dissipated, is that a certain capacity for work resides in the air which is due to the latter’s ability to expand when the proper conditions prevail.

Such conditions could be brought about by confining a volume of atmospheric air in a cylinder under a piston and then create a partial vacuum on the other side of the piston; the atmospheric air in the cylinder would expand and push out the piston, that is, perform work. But creating a vacuum requires extra work, and is therefore not of practical application in air engines.

As a matter of fact, after all the heat generated during compression of a volume of air has been dissipated, the compressed air possesses no more energy than it did before compression, but part of the energy which it did possess has, by mechanical compression, been made available for doing useful work.

To do work, however, the air requires energy in the form of heat and while expanding, it consumes heat that was contained in its mass before compression. As a consequence the temperature of the expanded air falls below that of the surrounding atmosphere. The amount of heat consumed is equivalent to the amount of work performed and equal to the amount of heat that would be generated in compressing this air from the pressure at which it exhausts from the air engine to the pressure at which it enters the same.

The consumption of heat from the mass of the expanding air is manifested by the cold created in and around the cylinders of an engine using air expansively. Theoretically this is exactly the reverse of the generation of heat in the air cylinders of a compressor.

BIBLIOGRAPHICAL INFO ON SOURCE “SIMONS” AND WHERE IT CAN BE DOWNLOADED

This math is not mine, it is from a compressed air textbook:
Compressed Air by Theodore Simons, 1921, p. 113-123, articles 117-120
"Effect of Loss of Heat, Generated During Compression, on the Ultimate Useful Energy Residing in a Given Quantity of Compressed Air"
The book can be downloaded from my site:
http://www.aircaraccess.com

and is also reproduced in my book Compressed Air Power Secrets, 3rd Edition.

Luther
April 18, 2010