# Carnot Cycle

The ideal thermodynamic cycle

### Key Facts

**Gyroscopic Couple**: The rate of change of angular momentum () = (In the limit).

- = Moment of Inertia.
- = Angular velocity
- = Angular velocity of precession.

## Overview

The Carnot Cycle is an entirely theoretical thermodynamic cycle utilising reversible processes. The thermal efficiency of the cycle (and in general of any reversible cycle) represents the highest possible thermal efficiency (this statement is also known as Carnot's theorem - for a more detailed discussion see also Second Law of Thermodynamics ). This ultimate thermal efficiency can then be used to compare the efficiencies of other cycles operating between the same two temperatures. The thermal efficiency of any engine working between the temperatures of*T*

_{1}and

*T*

_{2}is: From equation (2) it can be seen that in order to improve the thermal efficiency of an engine, we should basically increase the value of (

*T*), i.e. increase the temperature difference under which the engine works.

_{2}- T_{1}## Thermal Efficiency

The thermal efficiency of a cycle, also denoted by , is a measure of the ability to convert heat energy into work. Therefore, the thermal efficiency can be defined as: where*W*is the work output, and

*Q*

_{S}the heat energy supplied. Replacing

*W*with the heat supplied minus the head rejected, then equation (3) becomes: from which: The cycle with the highest possible thermal efficiency is the Carnot cycle (diagramed on a plot in Figure 8).

##### MISSING IMAGE!

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- a reversible adiabatic (i.e. isentropic) compression of the gas from temperature
*T*_{1}to*T*_{2}(step 1-2), - followed by an isothermal heating with expansion (step 2-3),
- then a reversible adiabatic (isentropic) expansion of the gas from
*T*_{2}to*T*_{1}(step 3-4), - and ending with an isothermal cooling with compression which reverts the system back to its initial state (step 4-1).

**TS**diagram (see Figure 9), which is useful for calculating the Carnot Cycle efficiency.

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*Q*

_{S}and

*Q*

_{R}terms (see Eq. 5). The heat supplied

*Q*

_{S}during step 2-3 can be calculated on a

**TS**diagram as the area under the cycle beneath the

*T*

_{2}line (the blue shaded area in Figure 9). The area of this rectangle can also be calculated as: On the other hand, the heat rejected

*Q*

_{R}during step 4-1 of the Carnot cycle can be calculated on a

**TS**diagram as the area under the cycle beneath the

*T*

_{1}line (the blue shaded area in Figure 9C).

##### MISSING IMAGE!

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*T*, the temperature of the cold reservoir, and

_{1=T}_{C}*T*, the temperature of the hot reservoir) are expressed on an absolute scale, such as the Kelvin scale. On the right side we provide calculators for the Carnot efficiency where you can input the temperatures in degrees Fahrenheit or degrees Celsius as well (the conversions are computed automatically).

_{2=T}_{H}Example:

[imperial]

##### Example - Heat supplied to a steam engine

Problem

Consider a steam engine for which the steam is supplied at and condensed at . If the thermal efficiency of the steam engine is of the Carnot efficiency, find the heat required (expressed in ) to produce a work output of 1 horsepower () for 1 minute.

Workings

We know that the Carnot efficiency of an engine working between temperatures and is given by:
where and are absolute temperatures expressed in kelvins ().
Therefore, in order to calculate the Carnot efficiency of the steam engine from the hypothesis, we first have to convert the temperatures into absolute temperatures. As we have that:
and
we get the Carnot efficiency of the steam engine:
from which we obtain:
or, expressed as percentage:
As the actual efficiency of the steam engine is of the Carnot efficiency, we can calculate the actual efficiency as:
from which we obtain:
or, expressed as percentage:
We know that the thermal efficiency of an engine is given by:
where is the work output, and the heat supplied. Therefore, the heat supplied in order to produce a work output of , while working at an efficiency of can be written as:
From the hypothesis we have that the required work output is for . Taking into account that , the work can also be expressed in as:
from which we obtain:
By using (13) and (8) in equation (11), and also considering that , we get the heat required (expressed in ) to produce for as:
from which we obtain:

Solution