Jump to content

Help needed for pv=nrt calculation


Recommended Posts

 

 

 

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.

In general, the piston type compressor is used at the high pressure compression. We can control the compression ratio by reducing the empty space. However, without good sealing, the compressor can not generate high pressure gas output.

Link to post
Share on other sites

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.

Yes. Very good. Scratch my line of thought. I was assuming the expanding gas was doing the pressure-volume work on the piston.

 

I agree, if I understand your interpretation of the system right, Helium would have to be below about 200 K with initial standard pressure to cool, while the nitrogen and oxygen components of air would do so with initial STP.

 

According to this link... for air...

The Joule-Thomson equation for air follows the equation...

 

...that is, for an average pressure of 25 atm, the partial derivative of temperature with respect to pressure equals 0.17 K*atm^-1 at 300 K and 0.29 at 200 K. Thus when air undergoes a Joule-Thomson expansion from 50 atm to some low pressure (~1 atm), we compute change in temperature at approximately -9, if the gas was initially at 300 K, whereas we obtain change in temperature is approximately -14, if the gas was initially at 200 K.

 

JT coefficient of air

So, roughly 1/50th of the pressure amounts to a roughly 9 C drop in temp. That seems quite little. Were you hoping for more of a change?
Link to post
Share on other sites

In general, the piston type compressor is used at the high pressure compression. We can control the compression ratio by reducing the empty space. However, without good sealing, the compressor can not generate high pressure gas output.

 

This would be true. However this so-called "compressor", according to the patent, does not use high pressure compression. Supposedly the air is still at atmospheric pressure AFTER "compression". In other words, it seems, according to the patent that the compressor does not really compressing the air much at all.

 

It mentions that the air, after being delivered to the tank is below freezing. Electric heaters are used to heat the air from the tank for use to run an air motor.

 

The implication, though this is not explicitly stated in the patent, is that the air is being reduced in volume by cooling rather than compression. Water cooling could conceivably bring the temperature down to somewhere above freezing I would suspect, but some additional cooling would have to come from somewhere to get it down below freezing. This "throttling valve" is all I've been able to find that looks like it might have some potential to do that.

 

At any rate, the patent does explicitly state that the pressure does not go above about 15 psi or atmospheric pressure.

Edited by Tom Booth
Link to post
Share on other sites
  • 4 weeks later...

 

 

According to this link... for air...So, roughly 1/50th of the pressure amounts to a roughly 9 C drop in temp. That seems quite little. Were you hoping for more of a change?

 

If this is the case, such a slight drop in temperature seems rather insignificant.

 

There may be another factor involved though: Humidity.

 

I came across this discussion while looking up random references to Joule-Thompson Effect.

 

http://www.clubfrontier.org/forums/f11/throttle-body-icing-74415/

 

It seems that under certain conditions automotive carburetors can ice up at the throttle due to Joule-Thompson Effect resulting in a run away engine - even under the hood where temperatures are rather hot. The result being that some manufacturers have installed heaters to circumvent the problem.

 

One commentator on that thread states: "Carb icing isn't really a hot or cold weather issue, but rather a humidity/dew point issue. so keep in mind that as air flows thru the venturi part of the throttle body, there is a drastic localized drop in temperature that goes along with the drop in pressure."

 

The line along which I'm thinking at this point is that although the temperature drop might be initially rather slight. Just a few degrees, the effect, as the engine continues to operate drawing more and more air through the valve, the effect might be cumulative. That is, a few degrees of localized cooling with the first stroke of the piston, a few more degrees with the second stroke, a few more with the third etc. until the cooling effect becomes significant.

 

In other words, not just the air gets cold but also the valve itself along with all the surrounding metal in the vicinity of the valve, the valve housing, cylinder, top of the piston etc. which might get progressively colder the longer the engine runs. Correspondingly the air entering through the valve would tend to have more and more heat drawn out of it with each cycle.

 

If the temperature drop were only .5 (1/2) degree per cycle, what might it be after 1 minute of operation in an engine running at say 500 RPM ?

 

This is, of course, mere speculation, but if a carburetor in a car engine, under the hood, in a relatively hot environment with what could conceivably only be a very slight pressure drop at the throttle could freeze up the entire throttle body I would assume this would have to be due to a cumulative effect in a running engine. The temperature drop with each cycle might be nearly imperceptible but the cumulative effect over time could be significant.

 

So, was I hoping for more of a change?

 

I suppose. I'm mostly just trying to find some logical explanation for how this so-called "compressor" could operate as described in the patent. I would tend to simply dismiss the whole thing as manifestly impossible, if not for the fact that in an interview with the inventors son, he stated that his father did indeed have a working prototype as well as a small working model which he saw operating when he accompanied his father to the patent office.

 

This Wiki reference was also cited in that thread:

 

http://en.wikipedia.org/wiki/Carburetor_icing

 

What I'm getting at is that in most scientific discussions about Joule-Thompson, such as you have cited above from "Chemical Thermodynamics" the discussion is limited to a strictly one time event. That is, air being passed through some throttling device from one chamber to another where there is no possibility for any cumulative effect as might be seen in a running machine where the air flows through the valve continuously or repeatedly in quick succession.

Link to post
Share on other sites
  • 3 years later...

On second thought, a 9 degree change on the C scale is more like 15 on the F. That seems a bit more significant somehow.

 

Especially when taking into account the humidity factor along with the possible cumulative effect of repeated iterations in a running engine addressed above, it seems to me that the cooling effect could be quite pronounced.


This video helps to explain the process of "Throttling" and why it results in cooling. Essentially kinetic energy is reduced and potential energy is increased as the air molecules are drawn further and further apart.

 

Edited by Tom Booth
Link to post
Share on other sites

Your problem here is that you are faced with a process where four variables (P, V, T, and to a certain extent n) are changing simultaneously and we only have the tools to cope with two variables at a time.

 

The general approach to these problems is to consider changes between two different states not by the direct route, but by a combination of simpler steps where only two variables are in play at the same time. So long as the initial and final states are the same, it doesn't matter which path we choose - the overall thermodynamic behaviour remains the same (google 'state variables' for background).

 

When your piston is top dead centre, some air remains at the top of the cylinder. It will be around the piston discharge pressure (lets call it 5 bar absolute) and its elevated compression temperature (let's guess 400 K). Lets also assume the volume is 5% of the piston swept volume.

 

We will also be assuming that the overall process is adiabatic (although you will see some interesting wrinkles).

 

The poppet will not open until the cylinder pressure has fallen a little below atmospheric - lets say 0.95 bar absolute.

 

Now the first wrinkle, We will invent an imaginary constant volume cooling process that will reduce the air pressure to 0.95 bar. Only pressure and temperature are in play so the Pressure Law applies - pressure is directly proportional to temperature.

 

So temperature reduces to 400 *0.95/5 = 76 K (not in reality!!) involving a heat removal of 5 somethings times the heat capacity at constant volume (Cv) * (400-76) Joules.

 

As the overall process is assumed adiabatic, we must return this heat energy to the air (second wrinkle), but this time at constant pressure causing the air to expand in accordance with Charles's Law - volume is directly proportional to absolute temperature. This time we use the heat capacity at constant pressure (Cp) and solve for

 

5*Cv*(400-76) =5*Cp*(T-76), The 5s drop out and Cp/Cv is the ratio of specific heats (1.4 or thereabouts) which gives a temperature of 307 K (which for me in Lagos is ambient).

 

So by Charles's Law our 5% of stroke has become 5*307/76 = 20% (-ish) of stroke by the time our poppet valve opens so 15% of piston travel is doing nothing.

 

But at least the poppet valve is open, and air will enter the cylinder as the piston drops to BDC at essentially constant pressure and temperature. There is no reason to treat the air flow through the poppet valve as anything but an isothermal process. There are no extreme conditions here.

 

Hope this helps.

Edited by sethoflagos
Link to post
Share on other sites

Btw significant J-T cooling of a gas requires the high pressure state to be getting somewhere towards the vicinity of its critical point. 50 bar air at 200 K will give significant J-T cooling on depressurisation. Warm high pressure air, if anything tends to go a little the other way (Z>1).

 

In the system under consideration, J-T is a herring of a deep red hue.

Edited by sethoflagos
Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.