Work done by heat engine that uses exhaust from heat pump

• Chemistry
• zenterix
zenterix
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):

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?

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

Replies
4
Views
331
Replies
3
Views
997
Replies
3
Views
230
Replies
1
Views
946
Replies
2
Views
2K
Replies
10
Views
1K
Replies
4
Views
2K
Replies
2
Views
2K
Replies
6
Views
12K
Replies
29
Views
5K