# Carnot cycle: Definition, Formula, Pv and Ts Diagram, Derivation [With pdf]

Contents

## What is carnot cycle?

Carnot cycle is ideal cycle consists of four Reversible process, Out of them two are reversible adiabatic & Two are reversible isothermal process.

Carnot cycle has maximum efficiency for engine working between two different temperatures T_{H} (heat source temperature) & T_{L} (Sink temperature).

For Example:- Heat Engine working between source temperature 400 kelvin & Sink temperature 350 Kelvin, For this Carnot efficiency is given by

\eta_{th} =1-\frac{T_{H}}{T_{L}}

\eta_{th} =1-\frac{350}{400}

\eta_{th} = 0.125 Or 125 %

It shows that any engine working between these two temperatures has maximum attainable efficiency of 12.5 %. The efficiency of any heat engine does not exceed Carnot efficiency.

In Carnot cycle, processes are non-flow processes.

## History of carnot cycle:

Carnot cycle is invented by french scientist nicolar leonard sadi carnot in 1824.

## Carnot cycle assumptions:

The assumptions considered in carnot cycle are as follows:-

1. Working fluid is air.
2. Air is considered as ideal gas.
3. No internal heat generation, It means that heat is added & rejected externally.

## Carnot cycle Pv and Ts diagram:

Here,

T_{1} = T_{4} = T_{L}   &

T_{2} = T_{L} = T_{H}

The Processes in carnot cycle are as follows:-

• Reversible Isothermal expansion ( Also known as Reversible  Isothermal heat addition)
• Reversible isothermal compression (Also known as Reversible Isothermal Heat rejection)

Parameters calculation for each process:-

Process (1-2):-

It is reversible adiabatic compression process.

Here temperature of air increases from T_{1} to T_{2} (that is T_{L} to T_{H})

Hence, Internal Energy also increases which is given by,

du_{1-2} = C_{V} ( T_{H}- T_{C} )

Q_{1-2} = 0

Now by first law of thermodynamics,

Q_{1-2} = du_{1-2} + W_{1-2}

0 = C_{V} (T_{H} – T_{L})+ W_{1-2}

W_{1-2} = C_{V}  (T_{H} – T_{L})

Process (2-3):-

It is reversible isothermal expansion process.

For isothermal process the change in a internal energy is zero,

du_{1-2} = 0

And workdone for isothermal process is given by,

W_{2-3} = R × T_{H} ln frac{V_{3}}{V_{2}}

Now, by first law of Thermodynamics,

Q_{2-3} = du_{1-2} + W_{2-3}

Q_{2-3} = R × T_{H} ln frac{V_{3}}{V_{2}}

Process (3-4):-

It is reversible adiabatic expansion process.

Q_{3-4} = 0

In this process temperature of air decreases from T_{3} to T_{4} (that is T_{H} to T_{L})

Hence change in internal energy du_{3-4} is given by,

du_{3-4} = C_{V} ( T_{L} – T_{H} )

Now by first law of thermodynamics

Q_{3-4} = dh_{3-4} + W_{3-4}

0 = C_{V} ( T_{L}-T_{H}) + W_{3-4}

W_{3-4} = C_{V} (T_{H} -T_{L})

Process (4-1):-

It is reversible Isothermal compression or heat rejection process.

In this process temperature is remain constant hence internal energy is not changing.

du_{4-1} = 0

For this process, work done is given by,

W_{4-1} = R × T_{L} × ln frac{V_{1}}{V_{2}}

By first law of thermodynamics

Q_{4-1} = du_{4-1} + W_{4-1}

Q_{4-1} = 0 + R × T_{L} × ln frac{V_{1}}{V_{2}}

Q_{4-1} = R× T_{L} × ln frac{V_{1}}{V_{4}}

All parameters are summarized below:

## Work done in carnot cycle:

The net work done of the carnot cycle is given by,

W_{Net} = Q_{In} – Q_{out}

W_{Net} = Q_{2-3} – Q_{4-1}

W_{Net}=RT_{H}ln(\frac{V_{3}}{V_{2}})-RT_{L}ln(\frac{V_{1}}{V_{4}})

But \frac{V_{1}}{V_{4}}=\frac{V_{3}}{V_{2}}

\therefore W_{Net}=RT_{H}.ln(\frac{V_{3}}{V_{2}})-RT_{L}.ln(\frac{V_{3}}{V_{2}})

W_{Net}=R(T_{H}-T_{L})ln(\frac{V_{3}}{V_{2}})

## Carnot cycle efficiency:

Thermal efficiency of Carnot cycle is highest efficiency of any heat engine working between two temperature T_{L} & T_{H}.

No heat engine will give higher efficiency than a Carnot heat engine.

The efficiency is given by

eta _{th} = frac{W_{Net}}{W_{Supplied}}

Where,  W_{Net} = Q_{In} – Q_{out}

W_{Net} = Q_{2-3} – Q_{4-1}

And the heat supplied is given by,

W_{Supplied} = Q_{2-3}

Put these values in equation of efficiency,

eta _{th} = frac{Q_{2-3} – Q_{4-1}}{Q_{2-3}}

eta _{th} = 1- frac{Q_{4-1}}{Q_{2-3}}

By putting values of Q_{4-1} & Q_{2-3}

eta _{th} = 1- frac{R×T_{L}×frac{V_{1}}{V_{4}}}{R×T_{L}×frac{V_{3}}{V_{2}}}

eta _{th} = 1- frac{T_{L}}{T_{H}} × frac{lnfrac{V_{1}}{V_{4}}}{ lnfrac{V_{3}}{V_{2}}}

Since frac{V_{1}}{V_{4}} = frac{V_{3}}{V_{2}}

Therefore, lnfrac{V_{1}}{V_{4}} = lnfrac{V_{3}}{V_{2}}

Hence, eta _{th} = 1- frac{T_{L}}{T_{H}} × frac{lnfrac{V_{3}}{V_{2}}}{ lnfrac{V_{3}}{V_{2}}}

eta _{th} = 1- frac{T_{H}}{T_{L}}

This is the Carnot cycle efficiency formula.