Carnot cycle: Definition, Efficiency, PV TS diagram, Formula

What is the Carnot cycle?

The Carnot cycle is an ideal cycle that consists of four reversible processes, Out of them two are reversible adiabatic processes & remaining two are reversible isothermal processes.

This cycle was invented by french scientist Nicolas leonard sadi carnot in 1824.

Carnot cycle has maximum efficiency for engine working between two different temperatures `T_{H}` (heat source temperature) & `T_{L}` (Sink temperature). The efficiency of any heat engine does not exceed Carnot’s efficiency.

In the Carnot cycle, processes are non-flow processes, which means that this cycle never exchanges mass with the surrounding.

Carnot cycle assumptions:

The assumptions considered in Carnot cycle are as follows:-

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

Carnot cycle schematic diagram:

Carnot cycle schematic diagram

Carnot cycle PV and TS diagram:

Carnot cycle Pv and Ts diagram

Here,
`T_{1}=T_{4}=T_{L}`   &
`T_{2}=T_{3}=T_{H}`

The Processes in the Carnot cycle are as follows:-

  • Process 1-2:- Reversible adiabatic compression
  • Process 2-3:- Reversible Isothermal expansion (Also known as Reversible Isothermal heat addition)
  • Process 3-4:- Reversible adiabatic expansion
  • Process 4-1:- Reversible isothermal compression (Also known as Reversible Isothermal heat rejection)

Parameters calculation for each process:

Process (1-2):- 

It is a 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_{2}- T_{1})`

As `T_{2}=T_{H}` and `T_{1}=T_{L}`

`\therefore \mathbf{du_{1-2}=C_{V}(T_{H}- T_{L})}`

For adiabatic process, 

`\mathbf{Q_{1-2}= 0}`

Now by the first law of thermodynamics, 

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

`0=C_{V}(T_{H}- T_{L})+W_{1-2}`

`\mathbf{W_{1-2}=C_{V}(T_{L}- T_{H})}`

Process (2-3):-

It is a reversible isothermal expansion process.

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

`\therefore \mathbf{du_{2-3}}` = 0

And work done for an isothermal process 2-3 is given by, 

`W_{2-3}= RT_{2} ln frac{V_{3}}{V_{2}}`

But as `T_{2} = T_{H}`,

∴`\mathbf{W_{2-3}= RT_{H} ln frac{V_{3}}{V_{2}}}`

Now, by the first law of thermodynamics, 

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

`Q_{2-3}=0+RT_{H} ln frac{V_{3}}{V_{2}}`

`\mathbf{Q_{2-3}=RT_{H} ln frac{V_{3}}{V_{2}}}`

Process (3-4):-

It is a reversible adiabatic expansion process.

For adiabatic process, 

`\mathbf{Q_{3-4}=0}`

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

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

`du_{3-4}=C_{V}( T_{4}-T_{3})`

As `T_{3}=T_{H}` and `T_{4}=T_{L}`

`\therefore \mathbf{du_{3-4}=C_{V}( T_{L}-T_{H})}`

Now by the first law of thermodynamics,

`Q_{3-4}=du_{3-4}+W_{3-4}`

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

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

Process (4-1):-

It is reversible Isothermal compression or heat rejection process.

In this process, as the temperature remains constant, thus the internal energy is not changing.

∴`\mathbf{du_{4-1}=0}`

For this isothermal process, work done is given by, 

`W_{4-1}=RT_{4}ln frac{V_{1}}{V_{4}}`

But as `T_{4} = T_{L}`,

∴`\mathbf{W_{4-1}=RT_{L}ln frac{V_{1}}{V_{4}}}`

By the first law of thermodynamics,

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

`Q_{4-1}=0+RT_{L}ln frac{V_{1}}{V_{4}}`

`\mathbf{Q_{4-1}=RT_{L}ln frac{V_{1}}{V_{4}}}`


All parameters are summarized in below table:

carnot cycle formulas

Work done in Carnot cycle:

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

`W_{\text{Net}}=Q_{\text{Net}}`

`W_{\text{Net}}=Q_{\text{In}}+Q_{\text{out}}`

`W_{\text{Net}}=Q_{2-3}+ Q_{4-1}`

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

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

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

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

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

Carnot cycle efficiency derivation:

The thermal efficiency of the Carnot cycle is the highest efficiency of any heat engine working between two temperatures `T_{L}` & `T_{H}` (`T_{H}>T_{L}`).

No heat engine gives higher efficiency than a Carnot heat engine.

The heat supplied `Q_{\text{In}}` during the process 2-3 is given by,

`Q_{\text{In}}= Q_{2-3}=RT_{H}ln(\frac{V_{3}}{V_{2}})`

The heat rejected during the process 4-1 is given by,

`Q_{\text{Out}}= Q_{4-1}=RT_{L}ln(\frac{V_{1}}{V_{4}})`

The net heat transfer is given by,

`Q_{\text{Net}}=Q_{\text{In}}+Q_{\text{Out}}`

`Q_{\text{Net}}=RT_{H}ln(\frac{V_{3}}{V_{2}})+RT_{L}ln(\frac{V_{1}}{V_{4}})`

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

Now the efficiency is given by,

`\eta _{th}= frac{W_{\text{Net}}}{Q_{\text{In}}}`

For the cycle `W_{Net}=Q_{\text{Net}}`

`\therefore \eta _{th}= frac{Q_{\text{Net}}}{Q_{\text{In}}}`

`\eta _{th}= frac{RT_{H}ln(\frac{V_{3}}{V_{2}})-RT_{L}ln(\frac{V_{4}}{V_{1}})}{RT_{H}ln(\frac{V_{3}}{V_{2}})}`

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

`\therefore \eta_{th}= frac{RT_{H}ln(\frac{V_{3}}{V_{2}})-RT_{L}ln(\frac{V_{3}}{V_{2}})}{RT_{H}ln(\frac{V_{3}}{V_{2}})}`

`\therefore \eta_{th}=\frac{T_{H}-T_{L}}{T_{H}}`

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

This is the equation to find the efficiency of the Carnot cycle working between the temperature `T_{H}` and `T_{L}`.

Carnot cycle examples:

1] Find the maximum efficiency of the heat engine working between source temperature 400 kelvin & Sink temperature 350 Kelvin.

Given:
`T_{H}=400\ K`
`T_{L}=350\ K`

Solution:

The maximum efficiency attained by any heat engine is equal to the efficiency of a Carnot cycle which is given by,

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

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

`\eta_{th}` = 0.125 OR 12.5 %

It shows that any engine working between these two temperatures has maximum attainable efficiency of 12.5%.


2] A Carnot cycle engine working between temperatures 300K and 400K gives a net work output of 70 Watt. Find the amount of heat supplied to the Carnot engine.

Given:
`T_{H}=400\ K`
`T_{L}=300\ K`
`W_{\text{Net}}= 70\ \text{watt}`

Solution:

The Carnot engine’s efficiency is given by,

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

`\eta_{th} =1-\frac{300}{400}`

`\eta_{th}=0.25`

The heat engine’s thermal efficiency is determined by,

`\eta _{th}= frac{W_{\text{Net}}}{Q_{\text{In}}}`

`0.25= frac{70}{Q_{\text{In}}}`

`\mathbf{Q_{\text{In}}=280\ \text{Watt}}`

FAQs:

  1. What four processes make up a Carnot cycle?

    The Carnot cycle is made up of four reversible processes.
    Process 1-2: Reversible adiabatic compession
    Process 2-3: Reversible Isothermal expansion
    Process 3-4: Reversible adiabatic expansion
    Process 4-1: Reversible isothermal compression.

  2. What is the efficiency of a Carnot engine?

    It is given by,
    ηₜₕ = 1 – (T🇱​​​​​/T🇭​​​​​).

  3. What factors influence the efficiency of a Carnot cycle?

    The efficiency of the Carnot cycle is influenced by the absolute temperatures of the source and sink (T🇭, T🇱​​​​​).

  4. Which thermodynamic cycle is the most efficient?

    The Carnot cycle provides maximum efficiency among all cycles working between the same source and sink temperatures.

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