Carnot Engine - Practice Questions & MCQ

• 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
$W_1=Q_1=\int_{V_1}^{V_2} P d V=R T_1 \log _e\left(\frac{V_2}{V_1}\right)=$ 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 $T_1$ to $T_2$

$
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 $T_2$.
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

$
\begin{aligned}
& W=W_1+W_2+\left(-W_3\right)+\left(-W_4\right) \\
& \text { 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}
\end{aligned}
$

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)}
$

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}
$

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}
$

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)
$

Put equation (4) in equation (1) we get
The efficiency of the Carnot engine as

$
\eta=1-\frac{T_2}{T_1}
$
 

   

$
\text { So }^\eta=\frac{W}{Q_1}=1-\frac{T_2}{T_1}
$

where

$
T_1=\text { Source temperature }, T_2=\operatorname{Sink} \text { Temperature and }\left(T_1>T_2\right)
$

and $\quad T_1$ and $T_2$ are in kelvin
From the above formula, we can conclude that
1. The efficiency of a heat engine depends only on temperatures of source and sink and is independent of all other factors.
2. As a Carnot engine is the ideal engine, So no heat engine can be more efficient than Carnot engine.
3. All reversible heat engines working between same temperatures are equally efficient
4. Since $T_1>T_2$ So the efficiency of a Carnot engine is always lesser than unity.
- Carnot Theorem (The relation between Heat and Temperature in the Carnot cycle)
i.e $\frac{Q_1}{T_1}=\frac{Q_2}{T_2}$
where
$Q_1=$ Heat absorbed
$Q_2=$ Heat released
$T_2=$ Low temperature
$T_1=$ Higher Temperature