Question

Derive the expression for Carnot engine efficiency.

Answer 1

Efficiency of a Carnot engine: Efficiency is defined as the ratio of work done by the working substance in one cycle to the amount of heat extracted from the source.

η = `"work done"/"Heat extracted" = "W"/"Q"_"H"` .............(1)

From the first law of thermodynamics, W = Q_H − Q_L

η = `("Q"_"H" - "Q"_"L")/"Q"_"H" = 1 - "Q"_"L"/"Q"_"H"` ............(2)

Applying isothermal conditions, we get,

`"Q"_"H" = µ"RT"_"H" ln ("V"_2/"V"_1)`

`"Q"_"L" = µ"RT"_"L" ln ("V"_3/"V"_4)` ........(3)

Here we omit the negative sign. Since we are interested in only the amount of heat (Q_L) ejected into the sink, we have

∴ `"Q"_"L"/"Q"_"H" = ("T"_"L" ln ("V"_3/"V"_4))/("T"_"H" ln ("V"_2/"V"_1))` ........(4)

By applying adiabatic conditions, we get,

`"T"_"H""V"_2^(γ - 1) = "T"_"L""V"_3^(γ - )`

`"T"_"H""V"_1^(γ - 1) = "T"_"L""V"_4^(γ - )`

By dividing the above two equations, we get

`("V"_2/"V"_1)^(γ - 1) = ("V"_3/"V"_4)^(γ - 1)`

Which implies that `"V"_2/"V"_1 = "V"_3/"V"_4` .........(5)

Substituting equation (5) in (4), we get

`"Q"_"L"/"Q"_"H" = "T"_"L"/"T"_"H"` .........(6)

∴ The efficiency η = `1 - "T"_"L"/"T"_"H"` .........(7)

Note: T_L and T_H should be expressed in the Kelvin scale.

Important results:

1. η is always less than 1 because T_L is less

than T_H. This implies the efficiency cannot be 100%.

2. The efficiency of the Carnot’s engine is independent of the working substance. It depends only on the temperatures of the source and the sink. The greater the difference between the two temperatures, the higher the efficiency.

3. When T_H = T_L the efficiency η = 0. No engine can work having source and sink at the same temperature.