Dramatize R effect of pressure alone

[zitaray]

I am a teacher, not an engineer, trying to dramatize for nineth-graders that, even without evaporation, variations of pressure alone can change R effect. We are building a device with H2O in a chamber inside a sealed jacket to be filled with a refrigerant connected to a reservoir under controllable air pressure. We want to effect freezing of the H2O by reducing the pressure on the refrigerant, then remelt the ice by increasing the pressure on the refrigerant. High efficiency would be great, but clarity of principle is our key interest. What amateur-safe refrigerant should we use to keep the pressure within the range of a small air compressor and vacuum pump (preferably between zero psia to 100 psia, but other minimums and maximums are no problem)? We are all excited at making this work. Any other suggestions?

[Gary]

What makes you think you can drop the pressure of the refrigerant and freeze the water without evaporating some of the refrigerant?

[zitaray]

Dear Gary,
Thanks for your query. I don't mean there is no evaporation when I lower the pressure. I just mean low-pressure refrigerant is already absorbing heat even in its liquid state before evaporating. We don't need to have no evaporation in our device (though that would be neat). The main thing is that we can get cooling effect and a warming effect just by changing pressure between low and high pressures. This is just school stuff, but it is extremely enjoyable and I thank you for helping us if you can. Best regards.
zitaray

[Gary]

When you drop the pressure, part of the refrigerant evaporates, absorbing heat in the process. It is the evaporation that causes the refrigeration effect.

Someone please correct me if I am wrong on this.

[zitaray]

I'm sure you're right that evaporation is automatic when we lower pressure. Can you tell me what amateur-safe refrigerant we should use in our device as described in my first message? And at what pressure the refrigerant will effect freezing, and at what pressure it will effect melting? Thanks. zitaray

[Gary]

Probably something like R134a would be best. The freezing/melting point of water being 32F, this would correspond to 27.8 psig using R134a.

I should add that there are no "amateur safe" refrigerants. Like fire or electricity, its as safe or unsafe as the person handling it.

[chemi-cool]

Hi zitaray,

Please be careful with the refrigerants, especialy when little children involved.

Why not ask a pro to help you in class?

Chemi

[RogGoetsch]

I am a teacher, not an engineer, trying to dramatize for nineth-graders that, even without evaporation, variations of pressure alone can change R effect. We are building a device with H2O in a chamber inside a sealed jacket to be filled with a refrigerant connected to a reservoir under controllable air pressure. We want to effect freezing of the H2O by reducing the pressure on the refrigerant, then remelt the ice by increasing the pressure on the refrigerant. High efficiency would be great, but clarity of principle is our key interest. What amateur-safe refrigerant should we use to keep the pressure within the range of a small air compressor and vacuum pump (preferably between zero psia to 100 psia, but other minimums and maximums are no problem)? We are all excited at making this work. Any other suggestions?

I would like to understand the physics of what you propose.

If I remember correctly (and it is frequently obvious that I often do not!) a gas changes temperature slightly when expanded or compressed, due to the net acceleration or deceleration of the gas molecules. I’m not sure such a process occurs within a liquid.

While a refrigerant can be any substance, even water or air, we usually choose substances in order to take advantage of the heat of vaporization accompanying a phase change. One of the key parameters of what we use as refrigerants, therefore, is that the substance evaporate and condense at reasonable pressures.

If you were only interested in heat absorption or rejection from a liquid without phase change, oil should suffice. But it seems to me that you may be misunderstanding the pressure/temperature relationship attributed to refrigerants.

You can change the temperature by changing the pressure because of evaporation and condensation, not because of any special property of refrigerants. Water in a pressure cooker obeys the same laws. We can cook at higher temperature because we trap the gas and allow the pressure to rise.

We could raise the temperature of the cooker by changing only the pressure if we pumped enough water vapor into the vessel to raise the pressure and release enough heat by condensation to heat the vessel to the corresponding temperature/pressure equilibrium point on the H20 chart. The amount of pumping energy needed should equal the amount of heat from the stove which would achieve the same increase.

If you hooked up your vacuum pump to the pressure cooker, you could reduce the pressure, causing the water to boil until its temperature was reduced to the corresponding vapor pressure in the vessel. When vapor pressures of liquid and gas phases are in equilibrium, then and only then do evaporation and condensation not play the predominant role. At this point, the rate of molecules leaving the liquid equals the rate of molecules being reabsorbed by the liquid.

If air were being pumped instead, the mechanism would be different, it seems to me. The air would be heated by compression and then transfer its heat to the water.

Have I misunderstood your experiment?

Rog

[Peter_1]

No Rog, you didn't misunderstood, it's in my opinion indeed impossible.
But your explanation was good lecture for me. I printed it out so that I can read it once.

Zitaray, suppose water in an R134a cylinder at room temperature.
The water will be in liquid state, not frozen.
If you release some of the the gas, then heat will be absorbed and pressure (=temperature) will come back to it's equilibrium pressure.
The only way you can freeze the water is release so fast the gas so that the heat can not be transfered fast enough in pressure again.
And how will you visualize this? A see-through cylinder?

A friend of me is a refrigeration teacher and he got from Daikin Ostend a didactic stand: a Plexiglas evaporator and Plexiglas condenser. So you could (!) see the thermodynamic process with you own eyes.
But on a day, the condenser exploded in the face of a student. He never used it anymore and he even had to go to the court for this.
What he still uses is a small tube of Plexiglas tube and fill it with water. He hen attach a vacuum pump on it and you can see that the water boils and evaporates and 'disappears' very fast.

[zitaray]

Dear Rog,
Thanks for a great analysis! You are very clear in your presentation of principles. And your examples are beautiful. Also, one thing that really caught my imagination was your point: "If you were only interested in heat absorption or rejection from a liquid without phase change, oil should suffice." That actually is the main thing I am hoping to find out, and you put it better than I did. Would it really be possible for some kind of oil at say room temperature, under enough vacuum, to absorb enough heat from the H2O in our chamber to effect freezing? Conversely, would it be possible for that same room-temperature oil, under enough pressure, to reject enough heat back into the ice to melt it? All this without state change in the oil, of course. I think we need to insulate our jacket and also somehow stop our H2O chamber so that heat doesn't enter or exit from the top. If all this is possible, it might lend my claim some validity, i.e. that there is some heat exchange--maybe even as much as we are looking for in this device--effected by pressure change alone. I have always understood there is a period of time in any refrigeration system just before the refrigerant is still liquid under high pressure and not yet changed to vapor when heat flows out the walls of the tubing because of the pressure. In other words, I understand pressure causes the molecules to accelerate, even on a liquid. But maybe I'm wrong on this point, and I have been wrong about fifty million times before, even on basic things like this.
If this oil suggestion could work, it might also solve another potentially serious problem. Several respondents advised me that there are no amateur-safe refrigerants, and that I should be wary about letting kids (or their teacher) near any of them without a pro overseeing the whole thing. I'm game to bring a pro in, but I'd like to look into this oil possibility first. I have done enough with pressure generation and control in my life, but I have no hands-on experience with refrigerants. Also, as you obviously sensed in reading my initial question, even my theoretical experience is far from perfect.
So, Rog, thanks again. I hope you can take the time to answer. All the best.
zitaray

[Peter_1]

Zitaray, sorry I intrude your chat with Rog but the answer to your question seemed on the first sight so simple but it isn't simple to explain.
What is in fact the meaning of your experiment if I may ask you?

[zitaray]

Dear Peter_1,
Thanks for your interest. I saw your first message about your friend the teacher and I wanted to respond but I couldn't find how to direct my response to you. I am a retired teacher trying to guide some 14-year-olds into an understanding of how pressure and temperature work together in causing an R-effect. Specifically, we are trying to build a device for visualizing only the pressure factor in the thermodynamic process, to paraphrase your words. I answered Rog's analysis and suggestion that maybe oil would suffice in place of a usual refrigerant. I have copied here my response to him:

"Dear Rog,
Thanks for a great analysis! You are very clear in your presentation of principles. And your examples are beautiful. Also, one thing that really caught my imagination was your point: "If you were only interested in heat absorption or rejection from a liquid without phase change, oil should suffice." That actually is the main thing I am hoping to find out, and you put it better than I did. Would it really be possible for some kind of oil at say room temperature, under enough vacuum, to absorb enough heat from the H2O in our chamber to effect freezing? Conversely, would it be possible for that same room-temperature oil, under enough pressure, to reject enough heat back into the ice to melt it? All this without state change in the oil, of course. I think we need to insulate our jacket and also somehow stop our H2O chamber so that heat doesn't enter or exit from the top. If all this is possible, it might lend my claim some validity, i.e. that there is some heat exchange--maybe even as much as we are looking for in this device--effected by pressure change alone. I have always understood there is a period of time in any refrigeration system just before the refrigerant is still liquid under high pressure and not yet changed to vapor when heat flows out the walls of the tubing because of the pressure. In other words, I understand pressure causes the molecules to accelerate, even on a liquid. But maybe I'm wrong on this point, and I have been wrong about fifty million times before, even on basic things like this.
If this oil suggestion could work, it might also solve another potentially serious problem. Several respondents advised me that there are no amateur-safe refrigerants, and that I should be wary about letting kids (or their teacher) near any of them without a pro overseeing the whole thing. I'm game to bring a pro in, but I'd like to look into this oil possibility first. I have done enough with pressure generation and control in my life, but I have no hands-on experience with refrigerants. Also, as you obviously sensed in reading my initial question, even my theoretical experience is far from perfect.
So, Rog, thanks again. I hope you can take the time to answer. All the best.
zitaray"
I think this explains why I asked for help from professionals like yourselves. The response has been better than I expected. Some warned me there are no "amateur-safe refreigerants" and I take such advice humbly. I will bring in a pro at a certain point, but, as I wrote to Rog, I want to experiment with oil as he suggested, if I can obtain sufficient absorbtion and rejection of heat through pressure alone.
You guys are great! Thanks for everything so far and for any other suggestions you have! All the best, Peter_1.
zitaray

[RogGoetsch]

Zitaray,

Your question reminds me of the early work, by Newton, I think, where energy and temperature change relationships were established by using an insulated container of water with a paddle connected to a weight suspended outside the box. The known weight was allowed to descend a measured distance and the temperature rise in the known volume of water due to the agitation by the paddle gave the first value for the thermal coefficient of water (1 BTU/Lb-°F).

You might test your theory by compressing a mass of water, in a hydraulic cylinder perhaps, with a known weight and measuring the temperature change, if any.

I have always understood there is a period of time in any refrigeration system just before the refrigerant is still liquid under high pressure and not yet changed to vapor when heat flows out the walls of the tubing because of the pressure. In other words, I understand pressure causes the molecules to accelerate, even on a liquid.

The pressure and temperature are highest at the discharge of the compressor. The high-pressure gas is cooled in the condenser and warm high-pressure liquid exits the condenser. Any measurement downstream of the condenser will show lower pressure and temperature the farther you go.

If the temperature stays high and the pressure falls below the saturation point, as it might if the line rises, some liquid will evaporate, creating bubbles in the line, but cooling the remaining mass to the saturation point. If the line subsequently cools, these bubbles will recondense. The fluid always obeys the vapor pressure law.

Regarding heat flow, energy in equals energy out, so if you draw a box around the part of your process in question, any energy you measure going in must be balanced by a temperature or phase change within the box or by energy leaving the box another way.

Hope this helps.

Rog

[Dan]

I never felt this demonstration dangerous. I used a half-pint mayonaise jar with a schrader valve soldered into the lid. I caulked the lid with permagum - nothing fancy. I used a refrigeration grade vacuum pump. I could easily boil water in the jar at room temperature, letting the students hold "boiling water" in their hands.

If you attach a vacuum gauge, you can show and predict at what pressures a given temperature of water will begin to boil.

[zitaray]

Dan, Rog, Marc, Peter and others kind enough to help me:
The water boiling in the mayonaise jar is beautiful. I will try it. Thanks much. But isn't that a demo only of how pressure changes the point of fusion and vaporization, not a demo of heat absorbtion and rejection under pressure alone? Am I wrong in believing a liquid, keeping the temperature constant, absorbs heat as the pressure lowers and rejects heat as the pressure increases? I am going to try Rog's suggestion of placing H20 in a cylinder under pressure and measure the H20 temperature.
I'll be back when I have some results. Again thanks to one and all. zitaray

[Carlo Hansen]

Hi to all and a happy new year.

It is difficult to explaine in text what goes on in a cooling/heating
process. But here is a link, witch i think Zitaray can use.

http://www.sfsb.hr/test/testhome/vtA...on/index1.html


Best regards
Carlo Hansen

[RogGoetsch]

The constant temperature lines of a liquid are almost at 90 degrees to the lines of pressure on a PH chart. The higher the pressure on a liquid, without enthalpy change, the higher the temperature. However, without having a PH chart in front of me right now, this temperature change might be just 2 to 5 deg celcius over a pressure range of say 30 Bar (approx 30 atmospheres).

Good one, Marc. I hadn't thought of that. I show a very slight change in temperature for ammonia, about 1°F per 300psi increase. For R-11, R-12 & R-22, the temperature lines for the liquid phase are completely vertical, so no measurable temperature change with increasing pressure.

Zitaray, these properties are shown on what are called Mollier Charts or Pressure-Enthalpy Diagrams.

Rog

[Peter_1]

Zitaray, have a look here http://www.hampden.com/pdf/rst-2.pdf

[zitaray]

Peter 1
Thanks for that Hampden Engineering link. It looks super. Plus it is only about 90 minutes from where I am. I'll be calling them Monday and see where it goes. Thanks again.
zitaray

[zitaray]

Rog,
Thanks for steering me to enhalpy and for your readings. I'l be back when I get on top of these things. Thanks again.
zitaray

[zitaray]

Carlo,
Thanks for the super VT link. It is brilliant, and useful beyond words. It's amazing how helpful so many of you guys have been to a mere amateur. I can't tell how great I feel. Thanks again. zitaray

[zitaray]

Marc,
Thanks for really solid info. I stand corrected and instructed. Plus I have a lot of stuff to work on. Best regards. zitaray

[Peter_1]

Just 28 minutes in the Newyear, so happy Newyear Zitaray.
Yeah.... we all know, we're all great guys
LOL.
Something complete different than alt.hvac, isn't it?