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The compression phase in an air conditioner is approximately adiabatic, so that heat is added to the refrigerant. It then naturally transfers heat to the outside. My question is: since heat was added both during the evaporation phase (from inside the house) and during compression, why should we expect that more heat is lost during condensation than was added by compression?

Secondarily, it seems that in refrigerators, the compression phase turns the gas to a liquid. I assume that air conditioners generally do the same, but then calling the next phase "condensation" doesn't make a lot of sense (since it's already a liquid). What gives?

Edit: http://physics.bu.edu/~duffy/ns549_fall07_notes06/cooling_fridge.html:

  1. The gas is transferred to a compressor, where most of the work is done. The gas is compressed adiabatically, heating it and turning it back to a liquid.
  2. The liquid passes through cooling coils on the outside of the fridge. Because the liquid is now warmer than room temperature, heat is transferred naturally to the room. This is an isobaric compression process.

So it seems that sometimes condensation happens before the "condenser" stage (4).

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    $\begingroup$ Condensation does NOT happen inside the compressor. The compressor is designed to compress a vapor. Since liquids are somewhat incompressible, any liquid inside the compressor would tear it up. Your source is incorrect about this point. $\endgroup$ Sep 19, 2022 at 19:30
  • $\begingroup$ The wording in your link is sloppy thermodynamics. Unfortunately, it is not uncommon. For example, In step 3 when they say “heating it” they actually mean raise its temperature. But the increase in temperature is due to compression work, not heat. $\endgroup$
    – Bob D
    Sep 20, 2022 at 12:06
  • $\begingroup$ The information you got from the link is so awful, I felt compelled to address in an update to my answer. $\endgroup$
    – Bob D
    Sep 20, 2022 at 14:22
  • $\begingroup$ @BobD I really appreciate the help you've provided! I had a major misconception around "heat." I wish I could mark two answers as accepted. My first question may not have been phrased very clearly, but I think the other answer here partially addresses it. I had been assuming that the compression phase was intended to raise the temperature, but now it seems like merely a side effect of increasing the pressure to help it condense. I will do more reading. $\endgroup$
    – A_P
    Sep 20, 2022 at 15:55
  • $\begingroup$ @BobD One more question though: on a standard P-T phase diagram, is "saturated vapor" the line separating liquid/vapor (between the triple and critical points), and "superheated vapor" the gaseous phase? Or is that the wrong way to think about this? $\endgroup$
    – A_P
    Sep 20, 2022 at 15:56

2 Answers 2

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Question 1: Work is done on the refrigerant to increase its pressure enough to condense against ambient air. This work does have to exit the system as heat, but the amount of heat added by compression work is not going to be as much as the amount of heat absorbed by boiling refrigerant inside the evaporator, as this would lead to a VERY low efficiency for the refrigeration system. This can be verified by looking at the coefficient of performance for the particular system that you are dealing with, as shown here.

Question 2: the compressor increases the pressure of the refrigerant enough to allow it to condense at ambient conditions, but compression alone does NOT condense the refrigerant. The high pressure and high temperature refrigerant turns into liquid at the condenser as it releases heat to the environment.

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  • $\begingroup$ Thanks! I've edited my question to show where my confusion in question 2 comes from. $\endgroup$
    – A_P
    Sep 19, 2022 at 17:15
  • $\begingroup$ Hi David. I must confess having just looked at your answer. But do you really want to talk about "the amount of heat added by compression work"? It makes me cringe. The heat exiting the system is the sum of the heat added to the system by the evaporator plus the work done on the system by the compressor. From my experience, that's how it's always framed in a refrigeration cycle. Then you can say the amount of work done on the system by the compressor is not going to be as much as the amount of heat added to the system by the evaporator for a decent Coefficient of Performance (COP). $\endgroup$
    – Bob D
    Sep 20, 2022 at 20:18
  • $\begingroup$ Just a suggestion, no need to reply. $\endgroup$
    – Bob D
    Sep 20, 2022 at 20:18
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It appears you are confusing energy transfer by work with energy transfer by heat, starting with the title:

AC compressor adds less heat than is removed during condensation?

The compressor in a refrigeration cycle does not add heat to the refrigerant. Heat is energy transfer due solely to temperature difference.

The compressor adds energy to the refrigerant by means of work, taking low pressure saturated vapor from the evaporator and delivering high pressure superheated vapor to the condenser.

My question is: since heat was added both during the evaporation phase (from inside the house) and during compression, why should we expect that more heat is lost during condensation than was added by compression?

Again, heat is not added to the refrigerant by the compressor. Energy is added to the refrigerant by the compressor by means of work, not heat. It is important to understand the difference. Heat is energy transfer due solely to temperature difference.

Overall, the stages are

  1. Energy is transferred to the refrigerant from the cool environment in the form of heat by the evaporator.

  2. The compressor transfers energy to the refrigerant by means of adiabatic work converting the refrigerant from saturated low pressure vapor at the evaporator to superheated vapor (high pressure and temperature) at the condenser.

  3. The condenser transfers energy from the refrigerant to the warm environment by means of heat, converting superheated vapor to saturated liquid.

  4. An expansion valve takes saturated liquid from the condenser and delivers a liquid/vapor mixture to the evaporator with no work or heat transfer.

For conservation of energy the energy extracted from the refrigerant by means of heat in the condenser is the sum of the energy absorbed by the refrigerant by means of heat in the evaporator plus the energy absorbed by the refrigerant by means of work by the compressor, or

$Q_{condenser}=Q_{evaporator}+W_{compressor}$

You may find the following article helpful: https://en.wikipedia.org/wiki/Heat_pump_and_refrigeration_cycle

I think I still have some confusion. I understand that heat can be transferred, but can't one also speak of the amount of heat (thermal energy) in a system?

No you can't think about the amount of heat in a system. Heat is strictly energy transfer due to temperature difference. Things don't "contain" heat. The energy contained in the system is properly called the internal energy of the system, which is the sum of the kinetic and potential energies at the atomic and molecular level.

The term "thermal energy" is often used as synonymously with the kinetic energy component of internal energy, but it is often frowned upon because of being confused with temperature and heat.

And if something causes that to grow through work, then although it hasn't "transferred heat," is it not appropriate to say that heat was added in the process?

No, it is not appropriate to say that heat was added to the process. It is appropriate to say that energy was added to the process due to work (energy transfer due to force times displacement) not heat (energy transfer due to temperature difference).

Apart from that, hopefully the rest of my question makes sense.

I will have to reread it since it appears you edited it after I posted my answer.

UPDATE:

This will address your edit and the related link.

  1. The gas is transferred to a compressor, where most of the work is done. The gas is compressed adiabatically, heating it and turning it back to a liquid.

This is at best, sloppy thermodynamics wording and at worst downright wrong. Get yourself a better reference.

First of all, when they say “heating it” they actually mean raise its temperature. But the increase in temperature is due to energy transfer by compression work, not energy transfer by heat. Unfortunately this is not uncommon. Since we are so used to raising the temperature of something by heat transfer, like cooking on a stove, we tend to use the word "heating" as always synonymous with raising temperature. But that is not always the case. Raising the temperature of a gas by compressing it involves no heat transfer.

Secondly, the compressor does not turn the gas "back into a liquid". It takes saturated vapor at the output of the evaporator and raises its temperature and pressure to create superheated vapor for the input to the condenser. See process 1-2 for an ideal (isentropic) compressor in the diagram below taken from the Wikipedia link above. If you Google "Temperature-entropy diagram for refrigeration cycle" to find a multitude of similar diagrams. Nowhere will you see the output of the compressor being a liquid.

  1. The liquid passes through cooling coils on the outside of the fridge. Because the liquid is now warmer than room temperature, heat is transferred naturally to the room. This is an isobaric compression process.

The superheated vapor, not liquid, is passed through the coils first transferring heat to the room to convert the superheated vapor to saturated vapor (lowering temperature and pressure), as shown in process 2-3 in the figure below, then transferring heat to convert the saturated vapor to saturated liquid at constant temperature and pressure (isothermal and isobaric) as shown in process 3-4 in the figure below.

Hope this helps.

enter image description here

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  • $\begingroup$ Thanks. I think I still have some confusion. I understand that heat can be transferred, but can't one also speak of the amount of heat (thermal energy) in a system? And if something causes that to grow through work, then although it hasn't "transferred heat," is it not appropriate to say that heat was added in the process? Apart from that, hopefully the rest of my question makes sense. $\endgroup$
    – A_P
    Sep 19, 2022 at 17:08
  • $\begingroup$ @A_P I've updated my answer to respond to your follow up questions. $\endgroup$
    – Bob D
    Sep 19, 2022 at 17:46
  • $\begingroup$ @A_P, some professions actually do consider the heat content of a system. Chemical engineers assign an enthalpy to a standard state (e.g., ideal gas at 0 K), and measure heat content against that standard state. This means that chemical engineers think of heat and heat transfer a bit differently than physicists do. $\endgroup$ Sep 20, 2022 at 18:35
  • $\begingroup$ @DavidWhite I'm aware of that. Such terms are baked into the vernacular of not only chemists but also some physicists. But from a purely thermodynamicist perspective they are, to put it mildly, inelegant. But here the OP confusion has to do with increasing temperature by compressor work vs heat. There should be no debate about that. And that is my focus here. In no way does the compressor "heat" the gas. $\endgroup$
    – Bob D
    Sep 20, 2022 at 18:54
  • $\begingroup$ @BobD, don't forget about the mechanical equivalent of heat. Also, the OP was asking if it's appropriate to say that heat was added to a process. As "inelegant" as my viewpoint is, the answer is "yes", some professions do think of it in this way. $\endgroup$ Sep 20, 2022 at 20:11

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