# Heat Engine problem in Thermophysics

• reh
In summary, the maximum temperature that the high temperature source could be is 500 K, and the entropy change for the system is negative.
reh

## Homework Statement

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?

## 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?

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

ok, thank you for answering. They ask for the maximum temperature in a). As i have understood it, there won't 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.

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.

Chestermiller said:
But, a temperature less than 500 K means negative entropy generation in the system, which is impossible.
No, not really. It's the universe the entropy of which can never go negative.

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

## 1. What is a heat engine?

A heat engine is a device that converts thermal energy into mechanical work. It operates by harnessing the difference in temperature between a high-temperature reservoir and a low-temperature reservoir.

## 2. What is the Carnot cycle?

The Carnot cycle is a theoretical thermodynamic cycle that describes the most efficient way to convert heat into work. It involves four steps: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.

## 3. How does a heat engine work?

A heat engine works by using a working substance, such as a gas or liquid, to absorb heat from a high-temperature source and then release some of that heat to a lower-temperature sink. This difference in temperature creates a pressure difference, which can be used to perform mechanical work.

## 4. What is the efficiency of a heat engine?

The efficiency of a heat engine is the ratio of the work output to the heat input. In other words, it is the percentage of the heat energy that is converted into useful work. The maximum possible efficiency for a heat engine is given by the Carnot efficiency.

## 5. What factors affect the efficiency of a heat engine?

The efficiency of a heat engine is affected by several factors, including the temperature difference between the hot and cold reservoirs, the type of working substance used, and the design and operation of the engine. Generally, higher temperature differences and more efficient working substances lead to higher efficiencies.

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