Carnot Engine - Practice Questions & MCQ

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Carnot Engine is considered one of the most asked concept.
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Carnot Engine

Carnot Engine-

As shown in the below figure, It consists of the following parts

1.A cylinder with perfectly non-conducting walls and a perfectly conducting base containing an ideal gas as working
substance and fitted with a non-conducting frictionless piston.

2. A source of infinite thermal capacity maintained at a constant higher temperature T₁

3. A sink of infinite thermal capacity maintained at a constant lower temperature T₂

4. A perfectly non-conducting stand for the cylinder.

Carnot cycle-

As shown in the below figure, It consists of the following 4 processes.

1. Isothermal expansion (curve AB)

The cylinder containing ideal gas as working substance allowed to expand slowly at this constant temperature T₁

So Work done = Heat absorbed by the system

$W_{1}=Q_{1}=\int_{V_{1}}^{V_{2}} P d V=R T_{1} \log _{e}\left(\frac{V_{2}}{V_{1}}\right)=\text { Area of ABGE }$

2. Adiabatic expansion (curve BC)

The cylinder is then placed on the non-conducting stand and the gas is allowed to expand adiabatically till the
temperature falls from T₁ to T₂

$W_{2}=\int_{V_{2}}^{V_{3}} P d V=\frac{R}{(\gamma-1)}\left[T_{1}-T_{2}\right]=\text { Area of } B C H G$

3. Isothermal compression (curve CD)

The cylinder is placed on the sink and the gas is compressed at constant temperature T₂.
Work done = Heat released by the system

$W_{3}=Q_{2}=-\int_{V_{3}}^{V_{4}} P d V=-R T_{2} \log _{e} \frac{V_{4}}{V_{3}}=R T_{2} \log _{e} \frac{V_{3}}{V_{4}}=\text { Area of } C D F H$

4. Adiabatic compression (curve DA)

Finally, the cylinder is again placed on a non-conducting stand and the compression is continued so that gas returns to its initial stage.

$W_{4}=-\int_{V_{4}}^{V_{1}} P d V=-\frac{R}{\gamma-1}\left(T_{2}-T_{1}\right)=\frac{R}{\gamma-1}\left(T_{1}-T_{2}\right)=\text { Area of } A D F E$

The efficiency of the Carnot cycle ( $\eta$ )

The efficiency of the engine is defined as the ratio of work done to the heat supplied.

$\eta=\frac{\text { work done }}{\text { Heat input }}=\frac{W}{Q_{1}}$

Net work done during the complete cycle

$\\ W=W_{1}+W_{2}+\left(-W_{3}\right)+\left(-W_{4}\right)\\ \ As \ W_{2}=W_{4}\\ \Rightarrow W=W_{1}-W_{3}=\text { Area of } A B C D$

$\eta=\frac{W}{Q_{1}}=\frac{W_{1}-W_{3}}{W_{1}}=\frac{Q_{1}-Q_{2}}{Q_{1}}=1-\frac{W_{3}}{W_{1}}=1-\frac{Q_{2}}{Q_{1}}$

Putting the values we get

$\eta=1-\frac{R T_{2} \log _{e}\left(V_{3} / V_{4}\right)}{R T_{1} \log _{e}\left(V_{2} / V_{1}\right)}$ ..... (1)

Since points, B and C lie on the same adiabatic curve

$\therefore T_{1} V_{2}^{\gamma-1}=T_{2} V_{3}^{\gamma-1} \text { or } \frac{T_{1}}{T_{2}}=\left(\frac{V_{3}}{V_{2}}\right)^{\gamma-1}$ ......(2)

Also, point D and A lie on the same adiabatic curve

$\therefore T_{1} V_{1}^{\gamma-1}=T_{2} V_{4}^{\gamma-1} \text { or } \frac{T_{1}}{T_{2}}=\left(\frac{V_{4}}{V_{1}}\right)^{\gamma-1}$ ....(3)

From the equation (2) and (3) we get

$\frac{V_{3}}{V_{2}}=\frac{V_{4}}{V_{1}} \text { or } \frac{V_{3}}{V_{4}}=\frac{V_{2}}{V_{1}} \Rightarrow \log _{e}\left(\frac{V_{3}}{V_{4}}\right)=\log _{e}\left(\frac{V_{2}}{V_{1}}\right)$ .....(4)