Learning Refrigeration Cycle: Carnot, Real Life & Reversed

In summary, The regular Carnot cycle does not work in reverse because the gas does not lose heat to the surroundings. The gas is in contact with the hot reservoir and absorbs heat from a cold reservoir.
  • #1
Peter G.
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Hi, :smile:

So I am learning about how refrigerators work. First I have to show the Carnot cycle and explain why it does not work in reverse - from my understanding, it does not.

And then show how it actually works in real life, with the phase transitions, which I am guessing don't exist for ideal gases because the intermolecular forces are ignored - is this correct?

I find understanding the "real life" freezer O.K.. The problem is when it comes to the reverse Carnot cycle...

This is why I believe it does not work, and I would appreciate any help/corrections on what I said above and will write now:

If we look at the regular Carnot Cycle, the first part, that is: Isothermal contraction - the following occurs:

Δ U = 0 Because it is isothermal
Δ W = negative because work is being done on the gas, hence:

ΔQ = 0 - ΔW which means that the ΔQ is negative: Heat is transferred from the gas - heat is lost, and, in order for that to happen, the gas must be contact with a cold source.

Now if we revert the cycle and still keep the cold source where it is, this is what happens:

ΔU = 0
ΔW = positive because the gas is doing work

Therefore ΔQ = positive: The gas absorbs heat from a cold (I am assuming colder) surroundings, which is impossible

Is that it?

Thanks,
Peter G

P.S: I haven't done the Second Law of Thermodynamics yet, so I don't know about entropy. Furthermore, if any of you could explain how there is no temperature change in an isothermal situation even though heat energy is moving in and out? Maybe it is because it loses the heat to the surroundings as shown by:ΔQ = ΔW?
 
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  • #2
Peter G. said:
Hi, :smile:

So I am learning about how refrigerators work. First I have to show the Carnot cycle and explain why it does not work in reverse - from my understanding, it does not.
A Carnot cycle is reversible. This means that the work done in moving heat from the cold reservoir to the hot reservoir in the Carnot refrigerator cycle can be recovered by running a Carnot heat engine between the hot and cold reservoirs.

And then show how it actually works in real life, with the phase transitions, which I am guessing don't exist for ideal gases because the intermolecular forces are ignored - is this correct?
A real life refrigerator uses Joule-Thomson cooling, which involves rapid expansion of an non-ideal gas, which is not reversible.

I find understanding the "real life" freezer O.K.. The problem is when it comes to the reverse Carnot cycle...

This is why I believe it does not work, and I would appreciate any help/corrections on what I said above and will write now:

If we look at the regular Carnot Cycle, the first part, that is: Isothermal contraction - the following occurs:

Δ U = 0 Because it is isothermal
Δ W = negative because work is being done on the gas, hence:

ΔQ = 0 - ΔW which means that the ΔQ is negative: Heat is transferred from the gas - heat is lost, and, in order for that to happen, the gas must be contact with a cold source.
No. The gas is in thermal contact with the hot reservoir which is at an infinitessimally lower temperature than the gas. At this point the gas is undergoing isothermal compression after having been adiabatically compressed. The adiabatic compression is to bring it from the temperature of the cold reservoir up to the temperature of the hot reservoir.

Now if we revert the cycle and still keep the cold source where it is, this is what happens:

ΔU = 0
ΔW = positive because the gas is doing work

Therefore ΔQ = positive: The gas absorbs heat from a cold (I am assuming colder) surroundings, which is impossible

Is that it?
No. In the reversed cycle, the gas is initially in thermal contact with the hot reservoir and has a temperature that is an infinitessimal amount lower than the hot reservoir. It expands as heat flows just enough to keep the temperature constant.
P.S: I haven't done the Second Law of Thermodynamics yet, so I don't know about entropy. Furthermore, if any of you could explain how there is no temperature change in an isothermal situation even though heat energy is moving in and out? Maybe it is because it loses the heat to the surroundings as shown by:ΔQ = ΔW?
That's right.
There is no temperature change because the gas expands and does work PdV = dQ.

AM
 
  • #3
I don't understand for example, how in the regular Carnot Cycle, during Isothermal Compression, how, if heat is lost, is it in contact, as you said, with a hot reservoir? Furthermore, if it is an infinitesimally lower temperature, shouldn't it be a cold reservoir? Or am I confusing the terms
 
  • #4
Peter G. said:
I don't understand for example, how in the regular Carnot Cycle, during Isothermal Compression, how, if heat is lost, is it in contact, as you said, with a hot reservoir? Furthermore, if it is an infinitesimally lower temperature, shouldn't it be a cold reservoir? Or am I confusing the terms
In a refrigerator, heat flows into the system from the cold reservoir and out of the system to the hot reservoir.

In the Carnot refrigerator heat is extracted from the cold reservoir during isothermal expansion. (This reversible isothermal expansion progresses infinitely slowly). Once expansion is complete, the gas undergoes an adiabatic compression until the gas reaches the temperature of the hot reservoir. It is then compressed isothermally and the heat flows to the hot reservoir. The gas is then adiabatically expanded until it is the temperature of the cold reservoir where it is then expanded isothermally and so on.

AM
 
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  • #5
Cool, I think I am almost there (I hope I am not bothering you too much :redface:)

These are my remaining doubts: In the regular Carnot Cycle:

In the "first step", that is, isothermal expansion, what is happening is: As the Gas does work, it's temperature *would* cool down, it therefore creates a temperature gradient and the internal energy is kept constant because it is being supplied with Heat from a Hot reservoir. That is how it gets heat. Correct? But why does the gas need the thermal energy, can't it do work without it? Or is it just part of the purpose of the engine? Transferring heat heat to a cold area?

Secondly, I don't understand the purpose of the adiabatic parts. Furthermore, when work is being done on the object all the work is used to raise the temperature right?

And last but not least (and I really hope this is not confusing - if it is, ignore it). I was watching this video and the guy explained the Carnot Cycle using as an example a gas container, containing an ideal gas and with rocks on top of the the container's piston. (Using the regular Carnot Cycle) During the process of expansion of the gas, being it adiabatic or isothermal, rocks were removed, allowing for an increase in volume and rocks were added during the compressions. I was trying and think in terms of this example, how work is done. So, this is my thought: Let's say the rocks start at a height A. During the period when the Gas is doing work, the rocks are raised to height B, which is well above A. Work however, is then done by us, compressing the gas, lowering the rocks to a height C. Height C is lower than B but higher than A, hence work is done.

Thanks,
Peter G.
 
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  • #6
Peter G. said:
Cool, I think I am almost there (I hope I am not bothering you too much :redface:)

These are my remaining doubts: In the regular Carnot Cycle:

In the "first step", that is, isothermal expansion, what is happening is: As the Gas does work, it's temperature *would* cool down, it therefore creates a temperature gradient and the internal energy is kept constant because it is being supplied with Heat from a Hot reservoir. That is how it gets heat. Correct? But why does the gas need the thermal energy, can't it do work without it? Or is it just part of the purpose of the engine? Transferring heat from a hot to cold area?
The purpose of the Carnot cycle is to create the most efficient engine, not the most practical. The system and surroundings are at equilibrium (ie an infinitessimal amount out of equilibrium) at all times. All heat flow occurs with the system and the reservoir it is in contact with are at the same temperature.

Secondly, I don't understand the purpose of the adiabatic parts. Furthermore, when work is being done on the object all the work is used to raise the temperature right?
In order to have heat flow isothermally to and from the system but also operate between two different temperatures (which is essential if you are to get work out of heat flow) you need to cease heat flow when the system changes from being in thermal contact with one reservoir to being in thermal contact with the other (different temperature) reservoir. This is done through adiabatic compression or expansion.
And last but not least (and I really hope this is not confusing - if it is, ignore it). I was watching this video and the guy explained the Carnot Cycle using as an example a gas container, containing an ideal gas and with rocks on top of the the container's piston. (Using the regular Carnot Cycle) During the process of expansion of the gas, being it adiabatic or isothermal, rocks were removed, allowing for an increase in volume and rocks were added during the compressions. I was trying and think in terms of this example, how work is done. So, this is my thought: Let's say the rocks start at a height A. During the period when the Gas is doing work, the rocks are raised to height B, which is well above A. Work however, is then done by us, compressing the gas, lowering the rocks to a height C. Height C is lower than B but higher than A, hence work is done.
This is how you would reverse the Carnot cycle. Store the work and then reverse the cycle using the stored work to drive the heat flow back to reach the initial starting condition.

AM
 
  • #7
So the adiabatic process is there because when changing from one reservoir to the other we need to cease heat flow, as simple as that?

And I still don't quite get why thermal energy moves in during the beginning of the cycle (Isothermal Expansion). Can't the gas do work without it? I think I have this doubt because I don't understand the function of the heat engine...
 
  • #8
Peter G. said:
So the adiabatic process is there because when changing from one reservoir to the other we need to cease heat flow, as simple as that?

And I still don't quite get why thermal energy moves in during the beginning of the cycle (Isothermal Expansion). Can't the gas do work without it? I think I have this doubt because I don't understand the function of the heat engine...
You have to get heat to flow into the engine. The heat engine uses natural heat flow to generate work. Heat flow naturally occurs from hotter to colder. So you need a slight temperature difference between the hot reservoir and system so that the heat flows into the system from the hot register, isothermally.

This heat flow causes the engine to do work: dW = dQ (dU = 0). The system gas then expands adiabatically doing further work. At the point that system temperature reaches the cold reservoir temperature, thermal contact is made with the cold reservoir and heat flows out of the gas. Work is done on the gas compressing it letting the heat flow out of the gas isothermally to the cold reservoir. It is then compressed adiabatically to get back to the starting position. You can show on a PV diagram that less work is required to do the compression phase than was done by the gas in the expansion phase. The difference is the work produced by the engine.

AM
 
  • #9
Thank you A LOT for your help and patience. I got it now
 

1. What is the Carnot refrigeration cycle?

The Carnot refrigeration cycle is a theoretical thermodynamic cycle that describes the most efficient way to transfer heat from a cold reservoir to a hot reservoir. It consists of four processes: isentropic compression, constant temperature heat rejection, isentropic expansion, and constant temperature heat absorption.

2. How does the Carnot refrigeration cycle differ from a real-life refrigeration cycle?

In a real-life refrigeration cycle, the processes are not perfectly reversible or isentropic, and there are energy losses due to friction and other inefficiencies. This results in a lower efficiency compared to the Carnot cycle.

3. What is the purpose of reversing the refrigeration cycle?

Reversing the refrigeration cycle allows for the extraction of heat from a cold source and transferring it to a hotter source, which is the opposite of the natural flow of heat. This is useful for air conditioning and refrigeration applications.

4. How does the coefficient of performance (COP) relate to the efficiency of a refrigeration cycle?

The coefficient of performance (COP) is a measure of the efficiency of a refrigeration cycle. It is defined as the ratio of the desired output (cooling) to the required input (work). The higher the COP, the more efficient the refrigeration cycle is.

5. What are some real-life applications of the refrigeration cycle?

The refrigeration cycle is used in a wide range of applications, including air conditioning systems, refrigerators, freezers, and heat pumps. It is also used in industrial processes such as food and beverage production, chemical and pharmaceutical manufacturing, and cold storage.

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