# How does a refrigerator work?

1. Feb 17, 2014

### rogerk8

Hi!

I wonder how a refridgerator work.

I understand that the Ideal Gas Law

$$pV=N_{mol}RT$$

is used somehow.

This is my blog description of my preliminary understanding:

"Here both N and V is constant. A compressor on the outside of the refridgerator enables high pressure of the gas there. A high pressure valve then lets some of the gas inside the refridgerator. While the pressure inside then is much lower it absorbs heat to get back to its original pressure before it is circulated back into the system(?)"

Thankful for any correction.

Best regards, Roger

2. Feb 17, 2014

### ZapperZ

Staff Emeritus
3. Feb 17, 2014

### Staff: Mentor

Vapor compression refrigeration: http://en.wikipedia.org/wiki/Vapor-compression_refrigeration

This is one of those rare cases where I prefer wikipedia to hyperphysics. Besides being thin on the details for heat pumps/air conditioners, it incorrectly defines the term "heat engine" (saying a heat pump is a heat engine, which is basically a self-contradiction).

4. Feb 18, 2014

### sophiecentaur

The central part of a refrigeration cycle like this is that the expanding gas / vapour has to Do Work, which will lower its internal energy. The nozzle that the gas is allowed to pass through is the thing that does this work in the 'throttling process'. If the gas were allowed to expand without doing any work, there would be no cooling effect.

5. Feb 18, 2014

### Manraj singh

I was wondering, why is it that when we open the refrigirator, we see a 'mist' kind of thing. Is it because of the moisture in the atmosphere condensing when it enters the refrigirator?

6. Feb 18, 2014

### dauto

An ideal gas can be used for refrigeration in a four step process.
1st: The compressor compresses the gas adiabatically. Both pressure and temperature increase.
2nd: Heat is removed from the gas by bringing it to thermal equilibrium with the outside environment (The kitchen)
3rd: The air is allowed to expand adiabatically to a low pressure and low temperature.
4th: Heat is added to the gas by bringing it to thermal equilibrium with the inside environment (heat is removed from the inside of the refrigerator).

Two Important points.

1st: Many gases actually used in refrigeration will actually turn liquid when under high pressure, so using ideal gas equations is completely wrong. Latent heat becomes very important.

2nd: Even for gases, effects such as the Joule–Thomson_effect are very important in most cases. This effect requires treating the gas as a non-ideal gas.

7. Feb 18, 2014

### dauto

That's only true for ideal gases. In the real world the Joule–Thomson effect which requires no external work is the main source of cooling. In the throttling process actually used in most refrigeration systems the amount of work actually performed during the expansion is almost negligible.

8. Feb 18, 2014

### Staff: Mentor

Yes.

9. Feb 18, 2014

### Manraj singh

Thank you.

10. Feb 19, 2014

### jartsa

I still think that in common fridges evaporation is the main source of cooling. I even checked Wikipedia.

High pressure freon flows into low pressure area through a valve which throttles the flow. The freon evaporates, and the pressure of the freon gas decreases.

My intuition says that evaporation is the main cooling effect there.

11. Feb 19, 2014

### Staff: Mentor

It is.

12. Feb 19, 2014

### Andrew Mason

The working fluid does not have to change state, but you have the right idea. Due to strong inter-molecular forces an increase in the separation of molecules will require energy. Since the throttling process is adiabatic and involves no work (free expansion), the only source of this increase in potential energy has to be the internal energy. So molecular kinetic energy has to decrease, hence the temperature decreases.

AM

13. Feb 19, 2014

### Staff: Mentor

The question wasn't whether it has to or can be done a certain way, it was whether it is done that way. In real refrigeration, the phase change absorbs/releases most of the energy.

14. Feb 19, 2014

### Andrew Mason

Not quite. The phase change of the refrigerant does not absorb heat. The throttling process is adiabatic (no transfer of heat). The absorption of heat from the surroundings occurs AFTER the drop in temperature due to the JT effect as the refrigerant gas comes into thermal contact with the space being cooled.

AM

15. Feb 20, 2014

### Staff: Mentor

I'm sorry, but you are really making a mess now and it seems trying to nitpick as well. Most of that is wrong:
1. The phase change in the expansion valve causes latent heat to be converted from sensible heat, dropping the temperature of the fluid as it evaporates:
2. Then in the evaporator, heat is absorbed from the surroundings and the rest of the liquid evaporates, absorbing more latent heat:
And:
For cooling that just uses the Joule-Thompson effect, liquifying gases for cryogenics is one of the few common applications: http://en.wikipedia.org/wiki/Joule–Thomson_effect#Applications
But notice that this is the opposite of what happens in a refrigerator: the gas expands and cools by the Joule-Thompson effect, causing it to liquify. In a refrigerator, the liquid is throttled and allowed to expand, causing it to vaporize.

16. Feb 20, 2014

### sophiecentaur

Also, CO2 from a fire extinguisher causes a change of state as a result of the J-T cooling.

17. Feb 20, 2014

### Andrew Mason

The distinction between evaporative cooling and Joule-Thomson cooling is significant. In evaporative cooling the heat is absorbed from the surroundings during the evaporation process. In JT cooling the energy needed to separate the gas molecules is drawn from the internal kinetic energy of the cooling substance as it is adiabatically throttled. The heat is not absorbed during evaporation. That is the point I was making. I am sorry that you think that is nitpicking or an attempt to mess things up.

Your point that in a refrigerator there is some evaporative cooling is well taken (the liquid portion of the refrigerant that remains after throttling may evaporate during the heat exchange process).

AM

Last edited: Feb 20, 2014
18. Feb 26, 2014

### rogerk8

Hi!

I have thought some more regarding how a refrigerator might work.

I have read all your nice explanations but the most important part of understanding came from my collegue.

Now I only need to know how a compressor actually works. At least I think so.

Anyway, I am attaching a drawing of my current understanding.

Best regards, Roger

#### Attached Files:

• ###### cooler.PNG
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4.5 KB
Views:
246
19. Feb 26, 2014

### Staff: Mentor

Still a lot you aren't getting, roger:
1. A compressor increases pressure to move/circulate the fluid.
2. The valve creates a pressure drop.
3. The fluid isn't always a gas, it switches back and forth between gas and liquid.

Have you not looked at any of the links/information provided?

20. Feb 26, 2014

### Andrew Mason

You have it backwards. The compressor heats up the fluid. This is simply the first law: Q = ΔU + W where W = work done BY the gas. If work is done on the gas (W<0) and no heat is allowed in or out, then ΔU must be positive ie. T increases. The key to cooling is the expansion. PV=nRT just tells you that T is proportional to V if P is constant. When you compress the gas, the work done in compressing increases the internal energy so T increases unless heat flows out (Q<0).

If you run any thermodynamic engine cycle backward you can move heat the other way i.e from a cooler reservoir to a warmer reservoir by doing work. Work is done on the gas by compressing it, letting heat flow out of it and then letting it expand against an external pressure until it reaches a temperature just below the temperature of the cold reservoir (W > 0 means ΔU < 0 so T decreases).

But the reverse engine cycle can be slow in creating a temperature difference. Modern refrigeration requires creating that cold temperature quickly. That is where the Joule-Thomson effect comes in. A gas with a high Joule-Thomson coefficient is compressed and then allowed to expand freely and adiabatically through a throttle (narrow aperture) valve. This results a significant temperature drop but no (very little) change in internal energy. That cold refrigerant is then placed in a heat exchanger to cool the space to be refrigerated. The cold fluid absorbs heat rapidly due to the significant temperature difference. The Joule-Thomson effect is at the heart of modern refrigerators and cooling systems.

AM

21. Feb 27, 2014

### Staff: Mentor

You need to stop saying this because it just isn't true. The refrigerant is not a gas before going through the expansion valve, it is a liquid. The fact that it is a liquid at that point is the critical issue. And the temperature of the liquid doesn't drop due to the Joule-Thompson effect, it drops because its boiling point drops and heat is carried away by its boiling. The new temperature after the expansion valve is determined by the pressure and boiling point at that pressure, not Joule-Thompson cooling.

Think of it this way:
If you have a volume of liquid, there is no gas in it to undergo Joule-Thompson cooling. If you suddenly expand the container, you would end up with a vacuum above the liquid, but to avoid that, a certain amount of the liquid flashes to vapor, which converts sensible heat to latent heat and lowers the temperature of the entire mixture. Any joule-thompson cooling on the liquid is negligible because liquids are not very compressible.

Similarly, a can of "compressed air" is actually a liquid that is a lot like a refrigerant. There is some gas above it that blows-out when you use it. This gas cools a little, but the liquid left inside cools a lot due to the fact that it starts to boil. The liquid inside the compressed air can can't be getting colder due to the Joule-thompson effect because it starts and ends at the same pressure and never goes through the expansion valve.

If you do a sample problem, you'll see that you don't use the Joule-Thompson effect in the calculation of the new temperature after the expansion valve.

The joule-Thompson effect is relevant only at the compressor, where the refrigerant is a gas on both sides of the process.

22. Feb 27, 2014

### sophiecentaur

Air cannot be compressed to a liquid at room temperature, afaik. Even cans of Compressed CO2 are full of gas - and the triple point of CO2 is only about -60C. Are you sure you meant that?

23. Feb 27, 2014

### sophiecentaur

I think we have a dialogue of the deaf going on here. There are two distinct forms of refrigeration, it seems and the proponents of each of the two on this thread are doomed not to agree. An ordinary 'refrigerator' uses change of state in the 'evaporator' to take heat out of the fridge (or hot room). Cryogenic refrigeration doesn't start with a liquid; it can't. The liquid is only there when the JT cooling has reduced the temperature far enough to produce a liquid. The Wiki article seems to be over-emphasising JT.
The secret behind domestic refrigerators was to find a suitable fluid with an appropriate boiling point. CFCs seem to fill the bill. Unfortunately they are a bit nasty for the environment. The original Ammonia based fluid now seems to be restricted to the 'Absorption' type cycle, which is very inefficient (loose term).

24. Feb 27, 2014

### Staff: Mentor

"A can of 'compressed air'" is a colloquialism people use to describe dusting products. I have a can of Endust Multi-Purpose Duster on my desk, which is compressed liquid tetrafluoroethane:

http://www.gjfood.com/pdf/msds/116_821480.pdf

There was one small error in my previous post: after you use some and it cools, it is at a lower pressure.

Last edited by a moderator: Feb 27, 2014
25. Feb 27, 2014

### Staff: Mentor

That's all fine except that the title and OP say that the thread is about ordinary refrigerators. So it is my perception that Andrew mistakenly believes what he is saying applies here. I agree that the wiki overemphasizes it by doing the same thing. They say joule-thompson is at the heart of modern refrigeration, then go on to describe only vapor-vapor processes.