Heat Engine and Carnot’s Cycle

Heat Engine:

It is a device which continuously converts heat energy into the mechanical energy in a cyclic process.

Efficiency of heat engine:

$\eta$ = $\frac{workoutput}{heatinput}$ = $\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1}=1-\frac{Q_2}{Q_1}$

Where Q₁ is the heat supplied by the source and Q₂ is the heat rejected to the sink.

Carnot Engine:

It is an ideal heat engine which is based on Carnot's reversible cycle. It works in four steps viz. Isothermal expansion, adiabatic expansion, isothermal compression and adiabatic compression. The efficiency of a Carnot engine is given by

$\eta$= 1 - $\frac{Q_2}{Q_1}$ = 1 - $\frac{T_2}{T_1}$

where T₁ and T₂ are the temperatures of source and sink respectively.

Example 1: The efficiency of a Carnot cycle is 1/6. If on reducing the temperature of the sink by 65⁰C, the efficiency becomes 1/3, find the initial and final temperatures between which the cycle is working.

Solution: Given $\eta$₁ = $\frac{1}{6}$, $\eta$₂ = $\frac{1}{3}$

If the temperatures of the source and the sink between which the cycle is working are T₁ and T₂, then the efficiency in the first case will be

$\eta$₁ = 1 - $\frac{T_2}{T_1}$ = $\frac{1}{6}$

In the second case $\eta$₂ = 1 - $\frac{T_2-65}{T_1}$ =$\frac{1}{3}$

Solving T_­1 = 390 K and T₂ = 325 K.