Question
Describe the working of Carnot engine. Obtain an expression for its efficiency.
✓
Answer
Carnot used a set of four devices—a source at a high temperature (say $T_1$), a sink at a low temperature (say $T_2$), a non-conducting base and a cylinder with a working substance, frictionless piston made of conducting base and non-conducting walls and piston. He carried the system through a step of four processes to complete a cycle as shown here. Process I-HW is an isothermal expansion. $Q_1$ energy flows in and temperature is maintained at $T_1$ by placing the cylinder over the source.
Process II-WN is an adiabatic processes at a temperature flow from $T_1$ to $T_2$ conducted with cylinder over (NCB = Non conducting base). Process III-NF is an isothermal process at $T_2$ which gives out an energy $Q_2$ with cylinder over sink Process IV-FH is an adiabatic process which changes the temperature from $T_2$ to $T_1$ conducted with the cylinder placed over (NCB = Non conducting base).
In the process, a net work equalling the area HWNFH is done with the net heat intake, $Q_1 - Q_2$ So, Efficiency $=\frac{\text{Work done}}{\text{heat supplied}}$ $=\frac{\text{Q}_1-\text{Q}_2}{\text{Q}_1}=1-\frac{\text{Q}_2}{\text{Q}_1}=1-\frac{\text{T}_2}{\text{T}_1}$
Process II-WN is an adiabatic processes at a temperature flow from $T_1$ to $T_2$ conducted with cylinder over (NCB = Non conducting base). Process III-NF is an isothermal process at $T_2$ which gives out an energy $Q_2$ with cylinder over sink Process IV-FH is an adiabatic process which changes the temperature from $T_2$ to $T_1$ conducted with the cylinder placed over (NCB = Non conducting base).
In the process, a net work equalling the area HWNFH is done with the net heat intake, $Q_1 - Q_2$ So, Efficiency $=\frac{\text{Work done}}{\text{heat supplied}}$ $=\frac{\text{Q}_1-\text{Q}_2}{\text{Q}_1}=1-\frac{\text{Q}_2}{\text{Q}_1}=1-\frac{\text{T}_2}{\text{T}_1}$
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