Chemistry Work done by heat engine that uses exhaust from heat pump

AI Thread Summary
The discussion centers on the efficiency calculations of a heat pump and a heat engine, with specific temperatures for the hot and cold reservoirs. The efficiency of the heat pump is calculated to be 9.48, while the heat engine's efficiency is 0.753, leading to a calculated work output of 12.39 kJ, which differs from the expected 11.2 kJ according to EdX. Participants debate the feasibility of using a heat pump to power a heat engine and the implications of the second law of thermodynamics, emphasizing that energy conversion requires a temperature difference. The conversation also touches on the fundamental principles of thermodynamics and the impossibility of perpetual motion machines.
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Homework Statement
Suppose an inventor claims that he/she can do the following.

Using only the heat exhaust from a refrigerator and an unlimited supply of liquid nitrogen, he/she can generate more energy than his refrigerator uses to operate.

The refrigerator is an ideal heat pump that keeps its contents at 275K by pumping out exhaust at 304K.

The inventor uses an ideal heat engine to generate work from the refrigerator's exhaust.

Liquid nitrogen at 75K is used to cool the heat engine.
Relevant Equations
Calculate

a) refrigerator efficiency

b) heat engine efficiency and ##|W_{eng}|## output by the engine if the refrigerator uses 1570J to run

Note that there are other further items in the original problem but my question is about the calculation of engine work in part b).
The efficiency of a heat pump is

$$\mathcal{\epsilon}_{ref}=\frac{T_C}{T_H-T_C}=\frac{275}{29}=9.48$$

where ##T_H=304K## is the hot reservoir and ##T_C=275K## is the cold reservoir.

The efficiency of the heat engine is

$$\mathcal{\epsilon}_{eng}=1-\frac{T_C}{T_H}=1-\frac{75}{304}=0.753$$

where now we have a hot reservoir at 304K (the exhaust from the refrigerator) and a cold reservoir at 75K (the liquid nitrogen).

Schematically, here is the reverse Carnot cycle that represents the heat pump (ie, the refrigerator):

1719978892673.png


Note that I am naming the individual processes as -A, -B, -C, and -D because the corresponding heat engine has processes A, B, C, and D.

The refrigerator uses 1570J to run. This is the work required in order to obtain a certain amount of heat from the refrigerator's cold reservoir.

Let's denote this certain amount of heat as ##|Q_{-C}|##. This is the absolute value of the heat flow in process -C in the picture above.

Then

$$\mathcal{\epsilon}_{ref}=9.48=\frac{|Q_{-C}|}{1570\text{J}}$$

$$\implies |Q_{-C}|=9.48\cdot 1570=14.8\text{kJ}$$

As far as I understand, this is not the heat exhausted by the refrigerator. This is the heat removed from the cold reservoir.

The heat exhausted is the heat in process -A, namely, the heat that is transferred to the hot reservoir.

1570J represents the total work in the cyclic process.

$$|W_{pump}|=1570\text{J}=|W_{-C}+W_{-A}|=|-(Q_{-C}+Q_{-A})|$$

We can calculate ##W_{-C}## and ##W_{-A}## and show that ##W_{-C}+W_{-A}=-(Q_{-C}+Q_{-A})>0##. Thus, we can remove the absolute values and so

$$Q_{-A}=-Q_{-C}-W_{pump}$$

$$=-9.48\cdot 1570 - 1570=-10.48\cdot 1570$$

$$=16.45\text{kJ}$$

This is the heat exhausted by the refrigerator.

The work done by the engine is then

$$|W_{eng}|=0.753\cdot 16.45\text{kJ}=12.39 kJ$$

This problem is from EdX and has automated grading. The answer for this last calculation is, according to EdX, 11.2 kJ.

I am wondering where the discrepancy comes from.

More importantly, I am wondering if the reasoning I used above is correct?
 
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If you had an unlimited supply of liquid ##N_2## and a Carnot engine you could dispense with the refrigerator and just run the Carnot engine between the hot reservoir (ambient temperature 304K) and the liquid ##N_2##. So long as the Sun maintained the temperature of the hot reservoir it would keep going as long as you maintained your supply of liquid ##N_2##.

Also, I would question the correctness of the assumption that the refrigerator can keep pumping 1570 J per cycle indefinitely without lowering the temperature of the cold reservoir and, therefore, the COP of the refrigerator.

AM
 
1743260666079.png

in this formula if Tc is very close by Th, then Th-Tc->0
and the result goes to infinity. isn't that weird?
 
Pisica said:
View attachment 359175
in this formula if Tc is very close by Th, then Th-Tc->0
and the result goes to infinity. isn't that weird?
No, it's not. Why would you think it is? It's saying it gets easier to move heat if you have a smaller temperature difference. (Heat pump efficiency).

If you're looking for perpetual motion in such equations you won't find it. A lot of people have looked.
 
Pisica said:
View attachment 359175
in this formula if Tc is very close by Th, then Th-Tc->0
and the result goes to infinity. isn't that weird?
I Carnot agree with that.
 
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russ_watters said:
If you're looking for perpetual motion in such equations you won't find it. A lot of people have looked.
Please give me a single example of a single man who is looking for a perpetuum mobile or free energy in this equation!
 
Pisica said:
Please give me a single example of a single man who is looking for a perpetuum mobile or free energy in this equation!
Weird request, but the idea has come up a bunch of times in this forum. Since they always fail there are of course no famous examples, but if you Google "heat pump to power heat engine" you'll find a bunch of examples, including from PF.

[Edit] Lol, such as this thread, which you piggybacked on.
 
Last edited:
russ_watters said:
but if you Google "heat pump to power heat engine" you'll find a bunch of examples, including from PF.
Yes, that is true.
Well, let's:
1. I'm waiting for arguments why something like this wouldn't work? "heat pump to power heat engine"

In other words, why can we not transform heat from the external environment into mechanical work without a temp. difference?

This is a postulate, that is, an unproven physical law
A monothermal transformation is impossible because it would generate perpetuum mobile.

2. Why do we study thermodynamics?
To improve thermal machines.
Ideas?
Is there another reason why we study thermodynamics?
 
Pisica said:
1. I'm waiting for arguments why something like this wouldn't work? "heat pump to power heat engine"
You have received them. If the fact that the efficiency equations are inverses of each other* doesn't satisfy you, what sort of answer are you looking for?
Pisica said:
In other words, why can we not transform heat from the external environment into mechanical work without a temp. difference?

This is a postulate, that is, an unproven physical law
This question is almost too basic to have a detailed answer. Why can't you run a hydroelectric generator in a swimming pool? There is no energy to harness in an isolated reservoir. The energy you are trying to harness comes from the flow between reservoirs. This is a critical and fundamental concept in thermodynamics. It is, in essence, what thermodynamics is.
Pisica said:
2. Why do we study thermodynamics?
To improve thermal machines.
Ideas?
Is there another reason why we study thermodynamics?
Well, the first reason anyone should study anything is to learn it. You cannot skip that step and expect to succeed in improving it.

*Caveats/details discussed in a separate thread.
 
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Pisica said:
1. I'm waiting for arguments why something like this wouldn't work? "heat pump to power heat engine"
This sounds a little backwards as heat engines produce work and you can use that work to power the heat pump. When you look into the details, though, you find that the work output from a real heat engine isn't enough to run a real heat pump to completely undo the effect of the heat engine.

In principle, if you had a Carnot engine and a Carnot refrigerator working between the same two temperature reservoirs, the output from the engine would be exactly the amount needed to run the refrigerator. There'd be no extra work left over to do anything with.

Pisica said:
In other words, why can we not transform heat from the external environment into mechanical work without a temp. difference?

This is a postulate, that is, an unproven physical law.
To be more specific, it's the second law of thermodynamics. Note that in science, one generally does not prove things, so when you say the second law is unproven, my response would be "well, duh!"

The second law is, however, a statement describing what's we observe. In one formulation, it says that if heat is allowed to flow spontaneously, it always flows from hotter objects to cooler objects. If you're positing that the second law is wrong, you're essentially saying you can devise situations where heat spontaneously moves from a cold object to a hot object.
 
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