Maximum work done by a Carnot engine

In summary, the conversation discusses the use of a Carnot engine and its efficiency in a reversible process. The entropy change of the hotter system is equal and opposite to that of the colder system, and the temperature of the reservoirs are changing during the process. Therefore, the efficiency equation cannot be used in this scenario and multiple Carnot cycles must be used to calculate the maximum work done.
  • #1
Pushoam
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Homework Statement


upload_2017-12-31_11-56-44.png


Homework Equations

The Attempt at a Solution

Taking the engine to be a Carnot engine,

## \eta = \frac { T_h – T_c } {T_h} = \frac { W} {Q_h} ## ...(1)

## Q_h = C(T_h – T_f) ...(2)

\\ Q_c = C(T_f – T_c) ...(3)

\\ W = Q_h – Q_c = C(T_h + T_c – 2 T_f) ## ...(4)Solving the above equations,

## T_f = \frac { 2T_c T_h }{T_h +T_c} ## ...(5)This solution does not lead me to any of the given option.

Is this correct?
 

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  • #2
Does ##\eta## remain constant during the process?
 
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  • #3
Algebraically, in terms of ##T_h##, ##T_f##, and the heat capacity C, what is the entropy change of the hot system?
 
  • #4
TSny said:
Does ##\eta## remain constant during the process?
As I understand: ## \eta ## is not defined during a process. ## \eta ## is defined for a process. For a Carnot engine working between temperatures ## T_h ## and ## T_f ##, ## \eta ## is defined as ## \frac W{Q_h} ##. This is what I have used.

I am not getting the intention behind this question. Please give some more hint.
 
  • #5
The entropy change of the hotter system is equal and opposite to that of the colder as it is a reversible process.

## dS = C \frac { dT } T ## ...(1)

## \Delta S_h = - C \ln \frac { T_h}{T_f} = - \Delta S_c = - C \ln \frac { T_f}{T_c} ## ...(2)

## T_f = \sqrt{ T_h T_c} ## ...(3)

## W = Q_h – Q_c = C(T_h + T_c – 2 T_f) ## ...(4)

## W = C(\sqrt{ T_h } - \sqrt{ T_c})^2 ## ...(5}

So, the answer is option (d).Why is using the efficiency equation not a correct step?

The engine is working between the two temperatures through a reversible cycle. So, it is a Carnot engine.

For calculating maximum work done, I have to take maximum efficiency. This is what I did in the OP.
 
  • #6
Pushoam said:
The entropy change of the hotter system is equal and opposite to that of the colder as it is a reversible process.

## dS = C \frac { dT } T ## ...(1)

## \Delta S_h = - C \ln \frac { T_h}{T_f} = - \Delta S_c = - C \ln \frac { T_f}{T_c} ## ...(2)

## T_f = \sqrt{ T_h T_c} ## ...(3)

## W = Q_h – Q_c = C(T_h + T_c – 2 T_f) ## ...(4)

## W = C(\sqrt{ T_h } - \sqrt{ T_c})^2 ## ...(5}

So, the answer is option (d).Why is using the efficiency equation not a correct step?

The engine is working between the two temperatures through a reversible cycle. So, it is a Carnot engine.

For calculating maximum work done, I have to take maximum efficiency. This is what I did in the OP.
The temperatures of the two reservoirs are changing during the process. So you can't use the results for a single Carnot cycle with constant reservoir temperatures. You can't get from the initial state of this system to the final state reversibly with a single Carnot cycle. You need to use multiple Carnot cycles, each with slightly different pair of reservoir temperatures. And, for each of these multiple Carnot cycles, the efficiency is different.
 
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  • #7
Chestermiller said:
The temperatures of the two reservoirs are changing during the process. So you can't use the results for a single Carnot cycle with constant reservoir temperatures. You can't get from the initial state of this system to the final state reversibly with a single Carnot cycle. You need to use multiple Carnot cycles, each with slightly different pair of reservoir temperatures. And, for each of these multiple Carnot cycles, the efficiency is different.
I didn't realize that the temperatures of the reservoirs are changing and the efficiency eqn is valid for constant reservoir temperatures even after solving the question.
Thanks for it.
 

1. What is a Carnot engine?

A Carnot engine is an idealized heat engine that operates on the Carnot cycle, which is a theoretical thermodynamic cycle involving two isothermal processes and two adiabatic processes. It is often used as a benchmark for comparing the efficiency of real-world heat engines.

2. How does a Carnot engine work?

A Carnot engine works by absorbing heat from a high-temperature reservoir, performing work, and then releasing excess heat to a low-temperature reservoir. This process is repeated in a continuous cycle, with the engine's efficiency depending on the temperature difference between the two reservoirs.

3. What is the maximum work done by a Carnot engine?

The maximum work done by a Carnot engine is equal to the difference in temperature between the two reservoirs, divided by the temperature of the high-temperature reservoir. This is known as the Carnot efficiency and is represented by the formula W(max) = (Th - Tl) / Th, where Th is the temperature of the high-temperature reservoir and Tl is the temperature of the low-temperature reservoir.

4. Why is the maximum work done by a Carnot engine important?

The maximum work done by a Carnot engine is important because it represents the theoretical upper limit for the efficiency of any heat engine. This means that no real-world heat engine can ever be more efficient than a Carnot engine operating between the same temperature reservoirs. It also allows for the comparison and evaluation of the efficiency of different heat engines.

5. How can the maximum work done by a Carnot engine be increased?

The maximum work done by a Carnot engine can only be increased by increasing the temperature difference between the two reservoirs. This can be achieved by using a higher temperature heat source or a lower temperature heat sink. However, in reality, it is difficult to achieve large temperature differences and therefore, the Carnot efficiency remains a theoretical limit for the efficiency of heat engines.

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