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

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 schematic diagram:

Carnot cycle diagram

Carnot cycle Pv and Ts diagram:

Carnot cycle Pv and Ts diagram


`T_{1}` = `T_{4}` = `T_{L}`   &

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

The Processes in carnot cycle are as follows:-

  • Reversible adiabatic compression
  • Reversible Isothermal expansion ( Also known as Reversible  Isothermal heat addition)
  • Reversible adiabatic expansion
  • 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}` )

For adiabatic process, 

`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.

For adiabatic 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:

carnot cycle formulas

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}`


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}})`


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, ln`frac{V_{1}}{V_{4}}` = ln`frac{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.

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