# Carnot cycle - Zero Power Extremes

• Joshua Pham
In summary: Starting to become clearer. How did you obtain ##T_{LC} = \frac{(T_H + T_C)}{2}##...By solving for ##T_H## and ##T_C## using the equation for heat transfer between two bodies in thermal equilibrium: By solving for ##T_H## and ##T_C## using the equation for heat transfer between two bodies in thermal equilibrium: T_H = Q_H - Q_CT_C = W - Q_H
Joshua Pham
Hey guys,
I ran into this paper talking about the Maximum power you can obtain from a Carnot cycle: http://aapt.scitation.org/doi/abs/10.1119/1.10023 From what I understood, there are two extremes. To achieve maximum efficiency you have to make sure that the temperature of the system is never more than infinitesmally out of thermal equilibrium with its thermal reservoirs. So during the isothermal expansion, you have heat transfer coming in albeit at a very slow rate because of the negligible temperature difference between the hot reservoir and the system. It then isentropically expands to its low temperature which is infinitesmally higher than the cold reservoir and you get heat transfer out.

The other extreme is when you set the system temperature so that it does not change during the isothermal processes. From the paper this leads to all of the heat going straight from the hot sink to the cold sink and 0 power output. I don't quite understand this bit. Why does all of the heat go straight from the hot to cold sink?

Joshua Pham said:
Hey guys,
I ran into this paper talking about the Maximum power you can obtain from a Carnot cycle: http://aapt.scitation.org/doi/abs/10.1119/1.10023 From what I understood, there are two extremes. To achieve maximum efficiency you have to make sure that the temperature of the system is never more than infinitesmally out of thermal equilibrium with its thermal reservoirs. So during the isothermal expansion, you have heat transfer coming in albeit at a very slow rate because of the negligible temperature difference between the hot reservoir and the system. It then isentropically expands to its low temperature which is infinitesmally higher than the cold reservoir and you get heat transfer out.

The other extreme is when you set the system temperature so that it does not change during the isothermal processes. From the paper this leads to all of the heat going straight from the hot sink to the cold sink and 0 power output. I don't quite understand this bit. Why does all of the heat go straight from the hot to cold sink?
Did you mean "during the isentropic process" rather than "during the isothermal process" in the last paragraph?

Chestermiller said:
Did you mean "during the isentropic process" rather than "during the isothermal process" in the last paragraph?

No the paper says isothermal processes. I'm still trying to decipher what this means.

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The author is talking about carrying out the cycle irreversibly, so that there is a finite temperature difference between the working fluid and the hot reservoir, and an equal finite temperature difference between the working fluid and the cold reservoir (so that, in both isothermal steps of the cycle, the working fluid is half way between the hot reservoir and the cold reservoir temperature). Under these circumstances, the heat transferred from the hot reservoir to the working fluid is equal to the heat transferred from the working fluid to the cold reservoir. So, no net work is done, and the net effect is a transfer of heat from the hot reservoir to the cold reservoir.

Chestermiller said:
The author is talking about carrying out the cycle irreversibly, so that there is a finite temperature difference between the working fluid and the hot reservoir, and an equal finite temperature difference between the working fluid and the cold reservoir (so that, in both isothermal steps of the cycle, the working fluid is half way between the hot reservoir and the cold reservoir temperature). Under these circumstances, the heat transferred from the hot reservoir to the working fluid is equal to the heat transferred from the working fluid to the cold reservoir. So, no net work is done, and the net effect is a transfer of heat from the hot reservoir to the cold reservoir.

Thanks Chestermiller! My lecturer actually drew a picture like this when he attempted to explain it. Where THC and TLC are the high and low temperatures of the system respectively. "To maximise the heat transfer rate in, let THC = TLC. But then Qin = Qout." I don't really understand this statement and I'm still struggling to understand what the paper is saying

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The heat transfer rate into the working fluid during the expansion is ##k(T_H-T_{HC})## where k is a constant of proportionality; the heat transfer rate out of the working fluid during the compression is ##k(T_{LC}-T_C)##. If ##T_{HC}=T_{LC}=\frac{(T_H+T_C)}{2}##, then ##Q_H=Q_C## and ##W=Q_H-Q_C=0##

Joshua Pham
Chestermiller said:
The heat transfer rate into the working fluid during the expansion is ##k(T_H-T_{HC})## where k is a constant of proportionality; the heat transfer rate out of the working fluid during the compression is ##k(T_{LC}-T_C)##. If ##T_{HC}=T_{LC}=\frac{(T_H+T_C)}{2}##, then ##Q_H=Q_C## and ##W=Q_H-Q_C=0##

Starting to become clearer. How did you obtain ##T_{LC} = \frac{(T_H + T_C)}{2}## though?

Joshua Pham said:
Starting to become clearer. How did you obtain ##T_{LC} = \frac{(T_H + T_C)}{2}## though?
I just assumed the expansion and compression temperatures are the same, and half way between the hot and cold reservoir temperatures, for illustrative purposes.

Joshua Pham

## What is a Carnot cycle?

A Carnot cycle is a theoretical thermodynamic cycle that describes the most efficient way to convert heat into work. It consists of two isothermal (constant temperature) processes and two adiabatic (no heat exchange) processes, and operates between two heat reservoirs at different temperatures.

## What are the zero power extremes of a Carnot cycle?

The zero power extremes of a Carnot cycle refer to the points at which the cycle operates with zero power output. At the upper extreme, the cycle operates with zero power output when the heat source is at its maximum temperature and the heat sink is at its minimum temperature. At the lower extreme, the cycle operates with zero power output when the heat source is at its minimum temperature and the heat sink is at its maximum temperature.

## Why is the Carnot cycle considered the most efficient?

The Carnot cycle is considered the most efficient because it operates at the maximum possible efficiency for a heat engine, known as the Carnot efficiency. This efficiency is based on the temperature difference between the heat source and heat sink, and is unaffected by the specific properties of the working fluid used in the cycle.

## Can the Carnot cycle be achieved in practice?

No, the Carnot cycle is a theoretical concept and cannot be achieved in practice. This is because it requires reversible processes, which are not possible in real-world systems. However, the Carnot cycle serves as a benchmark for the maximum possible efficiency of actual heat engines.

## What are some real-world applications of the Carnot cycle?

The Carnot cycle has been used to model the operation of heat engines such as steam turbines, gas turbines, and refrigeration systems. It is also used in the study of thermodynamics and serves as a theoretical basis for understanding the limitations of actual heat engines.

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