Help needed for pv=nrt calculation

[Tom Booth]

I have been looking over a patent for an air compressor (Bob Neal Compressor).

 

Looking at the illustrations and reading the patent it seems apparent that this "compressor" is not so much compressing air as it is cooling it. Theoretically, by cooling the air significantly it will condense or contract, thereby reducing the volume and in this way effecting "compression". At least that is the theory I'm working with at this point.

 

There is, what seems to me, a rather strange valve arrangement for a compressor. Here is a clip of an image from the relevant illustration from the patent itself:

 

 

The main valve in question (#44 in the illustration at the top of the "compression" cylinder) has no rocker arms or valve lifters to mechanically open the valve. On the contrary it is held closed by, what looks like a rather strong valve spring. This is the air intake. There is also a check valve (#43) for exhaust but this can largely be ignored along with the check valves at the bottom.

 

When the piston (#25) is moving down, it appears that the only way the valve (#44) can open would be by the piston "pulling a vacuum". How much of a vacuum would depend on how strong the spring is (which, unfortunately, is not specified in the patent). The question became, why this apparently strong spring when another simple check valve as on the bottom would due? It appears that the function of this top cylinder chamber is probably cooling by expansion. The piston moves down but the air intake is restricted by the valve spring. The result being, presumably, a certain degree of mechanical expansion of the air that does get through the valve.

 

It can be assumed that the piston is pulling in atmospheric pressure air at ambient temperature.

 

The volume of the chamber into which the air is admitted through the valve is not specified in the patent but we could assign some reasonable arbitrary value.

 

Say, if the valve were missing or "wide open" the piston would draw in 1 pint of air (at one atm ambient temperature) with no resistance. Let's say ambient temperature is 60 degrees Fahrenheit.

 

The air intake is, however, restricted, so I think it can be assumed that something less than 1 pint of air is actually drawn in. The piston however, continues downward, expanding whatever amount of air is admitted to the full 1 pint volume.

 

The actual percentage of the potential 1 pint actually drawn in would depend upon the spring tension. We don't know the spring tension but I think it should be possible to get some rough idea how much COOLING by expansion might be possible with such an arrangement.

 

For example, if just 3/4 pint of 1 atm 60 F air is drawn in and expanded to the full 1pint volume.

 

Or with a stronger spring, if 1/2 pint is drawn in and expanded to the full volume.

 

What would be the resulting temperature if these values are plugged in to pv=nrt ?

 

That is, for the second "strong spring" example, If 1/2 pint of 60 degree ambient atmospheric air is drawn in and expanded to twice its initial volume, what will be the temperature of the gas when the piston reaches the bottom of the cylinder?

 

Do we have enough information? What would be the resulting temperatures for various arbitrary ratios of expansion? (spring strengths or tensions). What would be the temperature reduction ?

 

I've tried to do some calculations using online calculators and am coming up with what seem like ridiculously low temperatures, like -300 degrees F and such, but I find the formula rather confusing and have no confidence that I'm doing it right.

 

Given the above rough figures, starting with ambient air at 1 atm and a potential volume of 1 pint can anyone help me figure out what the resulting temperatures might be after expansion for various arbitrary expansion ratios ?

 

I hope I have made it clear what I'm trying to figure out here. If anyone can help, feel free to use whatever values seem reasonable that will produce some round numbers, I'm just looking to find out what degree of cooling might be possible. Someone is planning to actually reproduce this compressor, but at this point I'm just trying to figure out how much cooling of the air might be possible with such an arrangement.

[Iggy]

Chang in volume, change in pressure, change in temperature, change in amount. If you can say in a single sentence what three of those things are then anyone here could happily solve the fourth. That's pv=nrt.

[Tom Booth]

Chang in volume, change in pressure, change in temperature, change in amount. If you can say in a single sentence what three of those things are then anyone here could happily solve the fourth. That's pv=nrt.

 

Sorry, as I tried to explain, the patent does not provide enough information to give any exact values. Other than; all I know for sure is that there is plain old ambient temperature atmospheric air being drawn into the cylinder, the rest is mostly guesswork or assigning some reasonable arbitrary values.

 

I can say what doesn't change.

 

Assuming 1/2 pint of air enters the chamber (arbitrary), This "amount" (number of air molecules in 1/2 pint of air - moles?) does not change, at least not for the purposes of calculation.

 

Is it at least possible to calculate, or is it already generally known what "amount" of air (moles?) there would be in 1/2 pint of air at ambient temperature (arbitrarily set at 60 degrees F) at ordinary atmospheric pressure ?

 

Since air is being "throttled" or drawn in and expanded simultaneously it is practically impossible to give an actual starting and finishing value as far as "amount". !/2 pint is not really in the cylinder and then expanded to twice the volume. Initially there is nothing in the cylinder. The air is actually drawn in and expanded simultaneously, but to simplify matters it could be said that X amount of air is drawn in and then expanded to twice or three times or whatever the volume, though in a running "compressor" both would be happening simultaneously.

 

Everything else changes. Volume is increased. Pressure and temperature presumably decrease.

Edited January 26, 2013 by Tom Booth

[Iggy]

Sorry, as I tried to explain, the patent does not provide enough information to give any exact values. Other than; all I know for sure is that there is plain old ambient temperature atmospheric air being drawn into the cylinder, the rest is mostly guesswork or assigning some reasonable arbitrary values.

I did get that. My point is that the inexact values given would only lead to inexact answers. It's hardly worth asking then, eh?

 

What really do you want to know?

[Tom Booth]

I did get that. My point is that the inexact values given would only lead to inexact answers. It's hardly worth asking then, eh?

 

What really do you want to know?

 

The patent for the compressor does not say that this valve or cylinder is for cooling. That's just my theory.

 

The patent does say that air in the tank is below freezing. Heaters are used to bring the temperature back up above freezing to drive a compressed air engine.

 

It seems reasonable to assume then that the compressor is delivering below freezing cold air to the tank. Somehow the air is being cooled. It looks to me like this valve that may be restricting air intake may be the source of the extreme cooling.

 

I'm just trying to figure out if I might be on the right track or not with this theory.

 

Could ordinary ambient air be cooled significantly below freezing by drawing it through a tightly closed valve into a cylinder?

 

That's the basic question. Then, if so, how much cooling might actually be possible ? -10, -100, liquefaction ?

 

It seems some similar process is commonly used for liquefaction of gases, refrigeration, cryogenics etc.

 

In other words, looking at this picture:

 

 

 

 

Could the top cylinder be for cooling by "pulling a vacuum", expanding rather than compressing the air.

 

Does this seem like a reasonable hypothesis given the apparent strong spring on the inlet valve?

 

If so, how much cooling might be possible at various different spring strengths?

 

I thought it might be possible to get some idea using the ideal gas law formula but I don't really know how to use it.

[Iggy]

...Could ordinary ambient air be cooled significantly below freezing by drawing it through a tightly closed valve into a cylinder?

 

That's the basic question...

Yes, if the cylinder expands by some external force.. The temperature by which it drops is proportional to the volume by which it expands. Further detail requires numbers.

 

edit...

 

I think I got what you're thinking. Let me just ask (then I think I can give you a definitive answer) why does the cylinder marked "45, 41, 25" expand? What makes it expand?

Edited January 26, 2013 by Iggy

[Tom Booth]

Yes, if the cylinder expands by some external force.. The temperature by which it drops is proportional to the volume by which it expands. Further detail requires numbers.

 

OK, great!

 

What I imagine happens is the piston is at top. No air is in the cylinder. As the piston moves down it creates a vacuum.

 

When the vacuum is strong enough the valve is forced open slightly by atmospheric pressure outside the cylinder and a little air enters providing some relief, then the valve closes again while the piston continues to the bottom. (or the valve remains slightly open till the piston reaches the bottom then closes).

 

I think, effectively, this would be more or less the same as if there was a little air in the cylinder initially, and this small amount of air was then expanded.

 

So I'm going to say 1/2 pint of air is in the cylinder initially, then the piston moves down expanding this 1/2 pint to twice the volume.

 

The 1/2 pint of air is at 60 degrees Fahrenheit temperature and 1 atmosphere pressure. It is then expanded to twice its initial volume.

[Iggy]

OK, great!

 

What I imagine happens is the piston is at top. No air is in the cylinder. As the piston moves down it creates a vacuum.

 

When the vacuum is strong enough the valve is forced open slightly by atmospheric pressure outside the cylinder and a little air enters providing some relief, then the valve closes again while the piston continues to the bottom. (or the valve remains slightly open till the piston reaches the bottom then closes).

 

I think, effectively, this would be more or less the same as if there was a little air in the cylinder initially, and this small amount of air was then expanded.

 

So I'm going to say 1/2 pint of air is in the cylinder initially, then the piston moves down expanding this 1/2 pint to twice the volume.

 

The 1/2 pint of air is at 60 degrees Fahrenheit temperature and 1 atmosphere pressure. It is then expanded to twice its initial volume.

 

why does the piston move?

 

Is it because of anything that happens in the cylinder?

 

The reason I ask is because your question can't be answered unless the pressure in the cylinder is known before and after it moves.

[Tom Booth]

why does the piston move?

 

Is it because of anything that happens in the cylinder?

 

The reason I ask is because your question can't be answered unless the pressure in the cylinder is known before and after it moves.

 

The piston is being driven by an air motor (external force). No, the piston does not move because of anything that happens in the cylinder.

 

I'm going to say that the pressure before it moves is 1 atm.

 

This is necessarily fudging things but I don't see any other way to proceed.

 

Reasonably, as the piston moves down the pressure drops along with the temperature. No additional air is admitted.

Edited January 26, 2013 by Tom Booth

[Iggy]

The piston is being driven by an air motor (external force).

 

I'm going to say that the pressure before it moves is 1 atm.

 

This is necessarily fudging things but I don't see any other way to proceed.

 

Reasonably, as the piston moves down the pressure drops along with the temperature. No additional air is admitted.

If that is correct then you are correct. The temp. in the cylinder will be less than ambient. The amount by which it is less depends on the volume and pressure both before and after the cylinder expands. The stronger the spring the lower the temp. Is that what you're after?

[Tom Booth]

If that is correct then you are correct. The temp. in the cylinder will be less than ambient. The amount by which it is less depends on the volume and pressure both before and after the cylinder expands. The stronger the spring the lower the temp. Is that what you're after?

 

Well, that's a good start.

 

It would be nice if it were possible to get some rough idea how much lower the temperature could go.

 

Can we just plug in some arbitrary values?

 

For example, the maximum volume is whatever the cylinder can hold with the piston fully expanded. I'm guessing 1 pint.

 

Start out with 1/2 pint of air in the cylinder (60 F 1 ATM) expand that to double the initial volume.

[Iggy]

Well, that's a good start.

 

I figured as much.

 

It would be nice if it were possible to get some rough idea how much lower the temperature could go.

 

Can we just plug in some arbitrary values?

 

For example, the maximum volume is whatever the cylinder can hold with the piston fully expanded. I'm guessing 1 pint.

 

Start out with 1/2 pint of air in the cylinder (60 F 1 ATM) expand that to double the initial volume.

 

The final temperature will equal the final pressure times the final volume and temperature divided by the initial pressure and volume. If you don't know any one of those things then you can't solve. Like I said before: depends on how strong the spring is.

[Tom Booth]

Well, using various online sources / calculators / charts etc. I've come to the conclusion that for Air, there would be a 10 degree F drop in temperature (approximately) for every loss of 1 psi due to expansion (or pressure reduction).

 

Starting at 14.5 psi at sea level (14.6488 - 14.6950 actually, according to different sources). Would not expanding that to twice the volume (whatever the original volume) result in also cutting the pressure in half ? i.e. to about 7 psi ?

 

So expanding air to twice its volume would result in a drop in temperature of about 7 X 10 or 70 degrees F.

 

So starting at 60 degrees Fahrenheit, in the above example, (60 F 1 atm expanded 2X) the resulting temperature would be about -10 Fahrenheit after expansion to double the volume.

 

I'm not arriving at any of this from pv=nrt directly, though some of the sources used it. Rather this is based on pressure changes due to elevation in the atmosphere.

 

As you climb a mountain the pressure drops. About .5 psi per 1000 ft of elevation. (or 1psi per 2000 ft) and according to the sources, a corresponding temperature drop of about 5 F per 1000 ft.

 

Primary source: http://hwstock.org/adiabat.htm

 

also: http://www.unitarium.com/pressure

 

The thing at this point I'm not really sure about is, am I correct in assuming that doubling the volume of a fixed amount of air would halve the pressure ? Seems to follow from basic logic but...

 

Is it necessary to use "absolute pressure" rather than psi ? Is there a difference ? I'm a little fuzzy there.

 

I'm also assuming that pressure and temperature drop due to elevation would correspond to temperature and pressure drop due to mechanical expansion, which may be wrong but I have little else to go by at this point.

 

Anyway, if anyone can come up with a refutation, correction or improvement in regard to these basic conclusions of rough estimates I would certainly appreciate any additional help or insight.

 

Thanks!

[Iggy]

Well, using various online sources / calculators / charts etc. I've come to the conclusion that for Air, there would be a 10 degree F drop in temperature (approximately) for every loss of 1 psi due to expansion (or pressure reduction).

 

Starting at 14.5 psi at sea level (14.6488 - 14.6950 actually, according to different sources). Would not expanding that to twice the volume (whatever the original volume) result in also cutting the pressure in half ? i.e. to about 7 psi ?

 

So expanding air to twice its volume would result in a drop in temperature of about 7 X 10 or 70 degrees F.

 

So starting at 60 degrees Fahrenheit, in the above example, (60 F 1 atm expanded 2X) the resulting temperature would be about -10 Fahrenheit after expansion to double the volume.

 

I'm not arriving at any of this from pv=nrt directly, though some of the sources used it. Rather this is based on pressure changes due to elevation in the atmosphere.

 

As you climb a mountain the pressure drops. About .5 psi per 1000 ft of elevation. (or 1psi per 2000 ft) and according to the sources, a corresponding temperature drop of about 5 F per 1000 ft.

 

Primary source: http://hwstock.org/adiabat.htm

 

also: http://www.unitarium.com/pressure

 

The thing at this point I'm not really sure about is, am I correct in assuming that doubling the volume of a fixed amount of air would halve the pressure ? Seems to follow from basic logic but...

 

Is it necessary to use "absolute pressure" rather than psi ? Is there a difference ? I'm a little fuzzy there.

 

I'm also assuming that pressure and temperature drop due to elevation would correspond to temperature and pressure drop due to mechanical expansion, which may be wrong but I have little else to go by at this point.

 

Anyway, if anyone can come up with a refutation, correction or improvement in regard to these basic conclusions of rough estimates I would certainly appreciate any additional help or insight.

 

Thanks!

No, you can't assume that doubling the volume cuts the pressure in half. It depends on the strength of the spring. Imagine a strong spring that admits no air at all. The temperature would drop drastically. Imagine then a soft spring that lets in any air. The temperature wouldn't change. It depends entirely on the strength of the spring. The change in pressure can't be avoided. Without it you can't solve.

[John Cuthber]

"what looks like a rather strong valve spring."

Why do you think that?

[Tom Booth]

No, you can't assume that doubling the volume cuts the pressure in half. It depends on the strength of the spring. Imagine a strong spring that admits no air at all. The temperature would drop drastically. Imagine then a soft spring that lets in any air. The temperature wouldn't change. It depends entirely on the strength of the spring. The change in pressure can't be avoided. Without it you can't solve.

 

By "doubling the volume" I'm just adapting an arbitrary scenario.

 

The cylinder holds X volume of air. It admits only 1/2 that potential due to the spring being "medium" strength.

 

Yes that depends on the strength of the spring for sure. However, this is purely hypothetical so I could make the spring tension adjustable and turn the tension wing nut or whatever until that's the way it is. I could increase the tension and then say tripling the volume and work it out from there. In that case the even stronger or tighter valve would only allow 1/3 X into the cylinder.

 

I'm treating this AS - IF the amount of air entering the chamber were controllable and the action of the valves opening and closing were controllable. In that case, you open the valve fully. Draw the piston down 1/2 way then close the valve completely and draw the piston down the rest of the way. Now you have expanded the air to twice the volume.

 

I think it can be assumed that under such hypothetical controlled circumstances, doubling the volume of a gas would halve the pressure.

 

"what looks like a rather strong valve spring."

Why do you think that?

 

By looking at it. (in the patent drawings above)

 

By comparison, the spring on the inlet valve (#44) is at least 4 times bigger than the springs on the other valves (#43).

 

Bigger does not necessarily mean stronger but I think it's a fairly safe assumption.

 

This is a double acting compressor. The other three valves are all regular "check valves". Check valves are generally designed to offer as little resistance as possible so have very weak springs, just enough to hold the valve closed. In a normal compressor, there would be no need for anything else.

 

The one valve on top however is obviously different. Much bigger with a much larger spring. I think it can be assumed that this one valve serves some special purpose.

 

Of course I can't take it out of the patent drawing and test it to see if it is stronger than the other springs but it sure looks stronger to me judging by what I can see in the drawings. Don't you agree ?

[Iggy]

By "doubling the volume" I'm just adapting an arbitrary scenario.

 

The cylinder holds X volume of air. It admits only 1/2 that potential due to the spring being "medium" strength.

 

Yes that depends on the strength of the spring for sure. However, this is purely hypothetical so I could make the spring tension adjustable and turn the tension wing nut or whatever until that's the way it is. I could increase the tension and then say tripling the volume and work it out from there. In that case the even stronger or tighter valve would only allow 1/3 X into the cylinder.

 

I'm treating this AS - IF the amount of air entering the chamber were controllable and the action of the valves opening and closing were controllable. In that case, you open the valve fully. Draw the piston down 1/2 way then close the valve completely and draw the piston down the rest of the way. Now you have expanded the air to twice the volume.

 

I think it can be assumed that under such hypothetical controlled circumstances, doubling the volume of a gas would halve the pressure.

 

If you double the volume and half the pressure then there would be no change in temp. The equation is p*v/t = P*V/T where lowercase are the initial values and uppercase are final values.

 

The final temperature equals the initial temperature times the final pressure times the final volume divided by the initial pressure divided by the initial volume. T=t*P*V/(p*v)

[John Cuthber]

It's not really possible to tell from the diagram but, for a given wire size, a longer spring is weaker (i.e. deflects more under a given load) and I think a spring with a larger diameter is also weaker.

 

More importantly, I don't think the drawings are "to scale" in that way.

 

The inlet valve is bigger because it's handling a bigger volume of gas (since it hasn't been compressed yet).

[Tom Booth]

If you double the volume and half the pressure then there would be no change in temp. The equation is p*v/t = P*V/T where lowercase are the initial values and uppercase are final values.

 

The final temperature equals the initial temperature times the final pressure times the final volume divided by the initial pressure divided by the initial volume. T=t*P*V/(p*v)

 

Really ? I was actually getting that result at one point using an online calculator, but I thought, that couldn't possibly be right.

 

Increasing pressure raises temperature right ? So doing the opposite should result in cooling.

 

Then I was reading how if you have a balloon with air in it and you carry it up a mountain, the balloon will expand and the air inside will cool. Bring it down and the opposite happens.

 

Somehow this just doesn't seem like it could be right. Like a violation of the first law of thermodynamics not right.

 

If I expand a gas to twice the original volume and its stays the same temperature, then what happens when I compress it back down again ? It would get hot right. If I just kept expanding and compressing the same air I could make it as hot as I wanted.

 

BTW, I'm not doubling the volume by adding more air.

 

Just to be absolutely clear on this and make sure we are both thinking in the same terms.

 

What I'm talking about is this.

 

 

Words can be tricky sometimes. Like by "pressure" I could mean inside pressure or outside. "Volume" could mean amount or size etc. so just to be clear. these are the conditions:

 

The cylinder contains a piston with an air tight seal. No air can get in or out.

 

There is a fixed amount of air in the cylinder.

 

Press down on the handle to raise the piston.

 

If the starting conditions are known, is it impossible to determine the result ?

 

The gas inside the cylinder is 1 atmosphere, (about 14.5 Psig) and 60 F. call it one pint.

 

That is before pushing down on the handle to expand the air to twice the volume.

 

There is no air being added. Just mechanically expanded to increase the volume.

 

Does it really stay the same temperature ? Just double checking.

 

It's not really possible to tell from the diagram but, for a given wire size, a longer spring is weaker (i.e. deflects more under a given load) and I think a spring with a larger diameter is also weaker.

 

More importantly, I don't think the drawings are "to scale" in that way.

 

The inlet valve is bigger because it's handling a bigger volume of gas (since it hasn't been compressed yet).

 

Well, none has been compressed yet in the other half of the cylinder either. There, just check valves are used.

 

The compressor is double acting NOT two stage. That is it compresses air on both the up and down stroke. but always new air.

 

Anyway, this one valve is accessible, The check valves are not. So, it could be made adjustable, or could be replaced. Perhaps the inventor built it that way so he could experiment with different spring strengths or make adjustments while the engine was running.

[John Cuthber]

Or perhaps the artist who drew it didn't understand that the thickness of the wire was important.

 

Nobody said it was a two stage pump.

 

The air that enters the cylinder from the atmosphere hasn't been compressed, So it still has quite a large volume. so it needs a big hole to go through quickly so it needs a big valve.

 

The essence of this thread seems to be

"The main valve in question (#44 in the illustration at the top of the "compression" cylinder) has no rocker arms or valve lifters to mechanically open the valve. On the contrary it is held closed by, what looks like a rather strong valve spring. This is the air intake.".

 

It may look that way to you, but it doesn't to me.

Edited January 29, 2013 by John Cuthber

[Iggy]

Really ? I was actually getting that result at one point using an online calculator, but I thought, that couldn't possibly be right.

 

It was right.

 

Increasing pressure raises temperature right ? So doing the opposite should result in cooling.

 

Depends on volume. Let me get clearly across: temp. pressure, and volume are a menage a trois for which one cannot hope to solve one without dealing with the other two. If the amount stays the same then you have to know the other two to guess the third.

 

 

Then I was reading how if you have a balloon with air in it and you carry it up a mountain, the balloon will expand and the air inside will cool. Bring it down and the opposite happens.

 

Somehow this just doesn't seem like it could be right. Like a violation of the first law of thermodynamics not right.

 

If I expand a gas to twice the original volume and its stays the same temperature, then what happens when I compress it back down again ? It would get hot right. If I just kept expanding and compressing the same air I could make it as hot as I wanted.

 

BTW, I'm not doubling the volume by adding more air.

 

Just to be absolutely clear on this and make sure we are both thinking in the same terms.

 

What I'm talking about is this.

 

 

Words can be tricky sometimes. Like by "pressure" I could mean inside pressure or outside. "Volume" could mean amount or size etc. so just to be clear. these are the conditions:

 

The cylinder contains a piston with an air tight seal. No air can get in or out.

 

There is a fixed amount of air in the cylinder.

 

Press down on the handle to raise the piston.

 

If the starting conditions are known, is it impossible to determine the result ?

 

The gas inside the cylinder is 1 atmosphere, (about 14.5 Psig) and 60 F. call it one pint.

 

That is before pushing down on the handle to expand the air to twice the volume.

 

There is no air being added. Just mechanically expanded to increase the volume.

 

Does it really stay the same temperature ? Just double checking.

 

Yes. To be a violation of that case would be, as you suggest, a violation of the laws of thermodynamics. You don't get perpetual heat from the temperature of a constantly expanding and contracting gas. The heat you get is from friction. An adiabatic process doesn't change temp, as the learned would say.

 

 

EDIT:::

 

By the way, John is making an extraordinary amount of sense.

Edited January 29, 2013 by Iggy

[Tom Booth]

Or perhaps the artist who drew it didn't understand that the thickness of the wire was important.

 

Nobody said it was a two stage pump.

 

The air that enters the cylinder from the atmosphere hasn't been compressed, So it still has quite a large volume. so it needs a big hole to go through quickly so it needs a big valve.

 

The essence of this thread seems to be

"The main valve in question (#44 in the illustration at the top of the "compression" cylinder) has no rocker arms or valve lifters to mechanically open the valve. On the contrary it is held closed by, what looks like a rather strong valve spring. This is the air intake.".

 

It may look that way to you, but it doesn't to me.

 

"Nobody said it was a two stage pump."

 

OK, my mistake. I thought that was what you were implying. Because the valve on the lower left of the bottom part of the cylinder is also a (fresh air) inlet valve.

 

"It may look that way to you, but it doesn't to me."

 

No problem. you may very well be right. I appreciate your input. Thanks!

 

It was right.

 

 

Depends on volume. Let me get clearly across: temp. pressure, and volume are a menage a trois for which one cannot hope to solve one without dealing with the other two. If the amount stays the same then you have to know the other two to guess the third.

 

 

 

Yes. To be a violation of that case would be, as you suggest, a violation of the laws of thermodynamics. You don't get perpetual heat from the temperature of a constantly expanding and contracting gas. The heat you get is from friction. An adiabatic process doesn't change temp, as the learned would say.

 

 

EDIT:::

 

By the way, John is making an extraordinary amount of sense.

 

OK, thank you all very much.

 

An adiabatic process doesn't change temp, as the learned would say.

 

I'll accept your judgements. What I don't get though is then why this elaborate description of adiabatic cooling on expansion from this website:

 

http://hwstock.org/adiabat.htm

 

And I think I've read other references similar to this over the years.

 

Could it be because Air is not an "Ideal Gas" ?

Edited January 30, 2013 by Tom Booth

[alpha2cen]

"Nobody said it was a two stage pump."

 

OK, my mistake. I thought that was what you were implying. Because the valve on the lower left of the bottom part of the cylinder is also a (fresh air) inlet valve.

 

"It may look that way to you, but it doesn't to me."

 

The main compression is occuring at the upper part of the machine. Botton part sealing state is not so good.

To maintain a good sealing state , suitable spring pressure is required.

Sealing is one of the important factor in the compressor design.

Edited January 30, 2013 by alpha2cen

[Iggy]

I'll accept your judgements. What I don't get though is then why this elaborate description of adiabatic cooling on expansion from this website:

 

http://hwstock.org/adiabat.htm

 

And I think I've read other references similar to this over the years.

 

Could it be because Air is not an "Ideal Gas" ?

 

Uh... yes. You are right. Halving volume and doubling pressure isn't adiabatic, and adiabatic expansion does lower temp. I think.... Yes. Because the big bang is adiabatic and obviously the temp of the universe is getting lower. I misspoke / sleepwalked through that post.

 

Nevertheless, the equation I gave works and the conclusions are the same. If amount stays the same you need to know change in two of three (pressure, volume, temp) to solve the other one. It can't be sidestepped.

[Tom Booth]

The main compression is occuring at the upper part of the machine. Botton part sealing state is not so good.

To maintain a good sealing state , suitable spring pressure is required.

Sealing is one of the important factor in the compressor design.

 

Are you suggesting then that the bottom intake valve is stronger to compensate for the poor seal around the connecting rod, and that the top has no connecting rod so can make use of a weaker spring.

 

If so, I would say that makes no sense. A stronger spring on the bottom chambers intake would only create more vacuum aggravating the problem of air leakage through the connecting rod packing.

 

I've work as an engine mechanic most of my life. In my experience, check valves using a little ball have very weak springs that can be easily compressed between the fingers. The valve at the top looks exactly like a typical automotive type or internal combustion type valve. Typically these valve springs are so strong a special valve spring compressor tool is required and they can be rather dangerous if they slip out of the valve spring compressor. One could put a hole in the ceiling or put out an eye.

 

If you are saying that a stronger spring is needed at top because this is the main compressor and a better seal is needed to prevent leakage back through the valve. That would make more sense.

 

What I'm generally looking at is references such as:

 

"The throttling of the gas at this stage produces a forced expansion of the gases and condensable vapors, and simultaneously lowers the temperature in proportion thereof. The fall in temperature and pressure causes the entrained vapors to condense to liquid, and collect this condensable matter, the gases are caused to discharge into the separator b through nozzle c."

 

From "Industrial fuels" by Joseph Stephenson pg 33.

 

Or this extract from Wiki:

 

"

Helium and hydrogen are two gases whose Joule–Thomson inversion temperatures at a pressure of one atmosphere

are very low (e.g., about 51 K (−222 °C) for helium). Thus, helium and

hydrogen warm up when expanded at constant enthalpy at typical room

temperatures. On the other hand nitrogen and oxygen,

the two most abundant gases in air, have inversion temperatures of 621 K

(348 °C) and 764 K (491 °C) respectively: these gases can be cooled

from room temperature by the Joule–Thomson effect.^[1]

For an ideal gas, is always equal to zero: ideal gases neither warm nor cool upon being expanded at constant enthalpy."

 

In other words, for an "ideal gas" which is what pv=nrt applies to, there would be no change in temperature from such "throttling" or forced expansion through a valve. Some gases, like helium would actually heat up but regular air being mostly nitrogen and oxygen would be cooled down.

 

I was hoping there might be some way to get an idea how much cooling would be possible, or how much cooling could reasonably be expected, but it seems pv=nrt isn't going to be of any help as air is not an ideal gas.

 

Perhaps there is some other formula or equation that would be applicable from some HVAC manual but I'm guessing that any equation relating strictly to an "Ideal Gas" is going to show no change in temperature.