Derive the expression for Carnot engine efficiency.

Explanation

Head engine: Heat engine is a device which takes heat as input and converts this heat in to work by undergoing a cyclic process.
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.
`eta = ("work done")/("heat extracted ") = (W)/(Q_H)` ......(1)
From the first law of thermodynamics, `W = Q_H - Q_L`
` eta = (Q_H - Q_L)/(Q_H) = 1 - (Q_L)/(Q_H)` ......(2)
Applying isothermal conditions, we get,
` Q_H = mu RT_H ln (V_2/V_1) `
`Q_L = mu 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
` therefore (T_L)/(T_H) = (Q_L ln (V_3/V_4) )/(Q_H ln (V_2/V_1) )` .......(4)
By applying adiabatic conditions, we get,
` T_H V_(2)^(gamma -1) = T_L V_(3)^(gamma -1) `
` T_h V_(1)^(gamma - 1) = T_L V_(4)^(gamma -1)`
By dividing the above two equations, we get
`(V_2/V_1)^(gamma - 1) = (V_3/V_4)^(gamma - 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)
` therefore ` The efficiency `eta = 1 - T_L/T_H` ....(7)
Note : ` T_L` and `T_H` should be expressed in kelvin scale .
Impotant result :
`eta ` is always less than 1 because `T_L` is less than `T_H` · This implies the efficiency cannot bel00%.
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, higher the efficiency.
3. When `T_H = T_L` the efficiency `eta` = 0. No engine can work having source · and sink at the same temperature.