# Heat Engine problem in Thermophysics

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1. Oct 16, 2016

### reh

1. The problem statement, all variables and given/known data

Im working on the following problem and could need some help in answering them:

Work is being produced from a cycle. In order to produce this work, energy is being taking from a high temperature sources at a ratio of 1000 kJ/kg and the extra energy is being deliver to a cold source in a ratio of 500 kJ/kg, the low temperature reservoir is 250 K. a) What is the maximum temperature that the high temperature source could be? b) What is the change in entropy of the system?

2. Relevant equations

3. The attempt at a solution
a) I assume that the system consist of the both heat sources and the device/engine. The total entropy in any process is always:

ΔStotal ≥ 0 where equality holds only for a reversible process.

The entropy change for the sorroundings is 0, because only work is done on the sourrundings, and the entropy change for the device is 0 because the process in the device is cyclical. The entropy change is then:

ΔStotal = -qH /TH + qL /TL ≥ 0

Rearranging to get the Temperature at the highest source:

TH ≥ TL * qH / qL

where TH = TL * qH / qL if the cycle is reversible and
TH > TL * qH / qL if it is irreversible. but then it would have to maximum value?

b)
For Reversible process the entropy change for the system would be 0, and if TH →∞ then the maximum entropy that can be made in the system is ΔS = qL /TL?

2. Oct 16, 2016

### Staff: Mentor

If the system is working in a cycle, $\Delta S=0$ for the system (i.e., the working fluid).

3. Oct 16, 2016

### reh

ok, thank you for answering. They ask for the maximum temperature in a). As i have understood it, there wont be any upper limit to the temperature for the high temperature source, and the lowest temperatur is 500 K where the cycle is reversible. That means that temperatures above 500 K gives a irreversible cycle and the total entropy change approaches qL /TL > 0 and temperatures below 500 K would give a negative change for the total entropy which would violate the second law, so there is no maximum temperature. Does my answer seem reasonable?

I have also been thinking that if the cycle is reversible, the maximum efficiency of the engine would be 0.5. From the given data, the efficiency is equal to 0.5, and that would imply that the engine runs as a reversible heat engine, so the temperature of the high temperature source should be 500 K and the total change in entropy is 0.

4. Oct 16, 2016

### Staff: Mentor

Well, if we do an entropy balance on the working fluid, we have $$S_{in}-S_{out}+\sigma=\Delta S =0$$where $\sigma \geq 0$ is the irreversible entropy generated, $S_{in}=Q_{H}/T_{H}$, and $S_{out}=Q_C/T_C$. So,$$\frac{Q_C}{T_C}-\frac{Q_H}{T_H}=\sigma \geq 0$$So, $$\frac{Q_C}{T_C}\geq \frac{Q_H}{T_H}$$So,$$T_H \geq \frac{Q_H}{Q_C}T_C$$

I agree with your assessment. But, a temperature less than 500 K means negative entropy generation in the system, which is impossible.

5. Oct 18, 2016

### rude man

No, not really. It's the universe the entropy of which can never go negative.

6. Oct 18, 2016

### Staff: Mentor

While it is true that the entropy change of the universe must never be less than zero, it is also true that the entropy generated within the system must likewise never be less than zero. For a closed system experiencing an arbitrary process, the entropy change of the system is comprised of the sum of two distinct parts:

1. The net entropy entering through the boundaries of the system as a result of heat flow into and out of the system.
2. The entropy generated within the system as a result of irreversibilities such as viscous dissipation and finite heat conduction

The sum of these two contributions add up to the total entropy change of the system.

In the present problem, the total entropy change of the system in passing through the cycle is zero, so the net entropy entering through the boundaries of the system as a result of irreversibilities must be negative. Since the hot and cold reservoirs are assumed to be ideal, their combined entropy change is given by:
$$\Delta S_{surroundings}=\frac{Q_C}{T_C}-\frac{Q_H}{T_H}$$But, from the equations I developed in post #4, this is just equal to $\sigma$, the heat generated within the system. So the entropy change of the surroundings is positive, while the entropy change of the system is zero. And we have:$$\Delta S_{universe}=\Delta S_{system}+\Delta S_{surroundings}=0+\sigma$$ So in this cyclic process, any entropy generated within the system is transferred to the surroundings. And the entropy change of the universe is just equal to the entropy generated within the system.
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