Subject: Physics

It is defined as the ratio of external work done, W by the engine to the amount of heat energy. Q_{1} absorbed from the heat source i.e.

\begin{align*} \eta = \frac {W}{Q_1} = \frac{Q_1 –Q_2}{Q_1} \\ &= 1 - \frac {Q_2}{Q_1} \\ &= 1 - \frac {RT_2 \log _e V_3/ V_4}{RT_1\log _e V_2/V_1} \\ &= 1 - \frac {T_2 \log _e V_3/ V_4}{T_1\log _e V_2/V_1} \dots (vi) \\ \end{align*}

As shown in figure, as A (P_{1}, V_{1}) and B (P_{2} , V_{2}) lie on the same isothermal.

$$ \therefore \: P_1V_1 = P_2V_2 \dots (vii) $$

Also the points B (P_{2}, V_{2}) and C (P_{3}, V_{3}) lie on the same adiabatic,

$$\therefore \: P_2V_2^{\gamma } = P_3V_3^{\gamma } \dots (viii) $$

Further the points C (P_{3}, V_{3}) and D (P_{4}, V_{4}) lie on the same isothermal,

$$ \therefore \: P_3V_3 = P_4V_4\dots (ix) $$

Finally, the points D (P_{4}, V_{4}) and A (P_{1}, V_{1}) lie on the same adiabatic

$$\therefore \: P_1V_1^{\gamma } = P_4V_4^{\gamma } \dots (x) $$

\begin{align*} \text {Dividing Equation} \: (vii) \: \text {by equation} \: (x), \text {we get} \\ \frac {P_2V_2^\gamma }{P_1V_1^\gamma } = \frac {P_3V_3^\gamma }{P_4V_4^\gamma } \dots (xi)\\ \text {From equation} \: (vi) \: P_1V_1 = P_2V_2 \\ \text {or,} \: \frac {P_2}{P_1} = \frac {V_1}{V_2} \\\text {From equation} \: (vii)\\ P_4V_4 &= P_3V_3 \\ \text {or,} \: \frac {P_3}{P_4} = \frac {V_4}{V_3} \\ \end{align*}

Substituting these ratios of pressure in equation (xi), we get

\begin{align*} \frac {P_2V_2^{\gamma}} {P_1V_1^{\gamma}} = \frac {P_3V_3^{\gamma}} {P_4V_4^{\gamma}} \\ \text {or,} \: \frac {V_2}{V_1} \times \frac {V_2^{\gamma }}{V_1^{\gamma }} = \frac {V_4}{V_3} \times \frac {V_3^{\gamma }}{V_4^{\gamma }} \\ \text {or,} \: \left (\frac {V_2}{V_1} \right )^{\gamma -1} &= \left (\frac {V_4}{V_3} \right )^{\gamma -1} \\ \text {or,} \: \frac {V_2}{V_2} = \frac {V_3}{V_4} \\ \end{align*}

Therefore, equation (vi) can be written as

\begin{align*} \eta &= 1 - \frac {T_2 \log _e V_3/ V_4}{T_1\log _e V_2/V_1} \\ &=1 - \frac {T_2 \log _e V_3/ V_4}{T_1\log _e V_3/V_4} \\ &= 1 - \frac {T_2}{T_1} \\ \end{align*}

So, efficiency of carnot engine, \(\eta = \left ( 1- \frac {T_2}{T_1} \right ) \).

This expression shows that the efficiency depends on the temperature of the source T_{1} and that of the sink T_{2} but does not depend on upon the nature of the working substance. As T_{2} is always less than T_{1}, so the efficiency of a heat engine is always less than one or efficiency cannot be 100%.

The Carnot cycle is perfectly reversible because

- There is no friction between the cylinder and piston.
- The operations on the gas should be performed very slowly.
- The loss of heat due to conduction is prevented by using an insulating piston and the insulating piston and the insulating walls of the cylinder.
- During an isothermal change, the temperature remains constant, because the Carnot engine uses a conducting base for the cylinder and, the sink and source have large heat capacity.

A refrigerator is a machine that operates in a manner opposite to that of a heat engine.

In a refrigerator, the working substance absorbs a certain quantity of heat Q_{2} from the sink at low-temperature T_{2}. If W is the external work done on the working substance, then the quantity of heat delivered to the source at higher temperature T_{1} is Q_{1} = Q_{2} + W, which is shown as in the figure. Therefore, a refrigerator transfers heat from a cold body to a hot body when external work is done on the machine. Thus cold body becomes colder. The significance of this principle is that heat can be made to flow a cold body to a hot body only if work is done on the system.

The effectiveness of a refrigerator is measured by its coefficient of performance. It is defined as the ratio of the amount of heat absorbed from the cold body to the work done in running machine.

\begin{align*} \text {Coefficient of performance,} \: \beta = \frac {Q_2}{W} \\ &= \frac {Q_2}{Q_1 – Q_2} \\ \text {where} \: W &= Q_1 – Q_2 \\ \text {or,} \: \beta &= \frac {Q_2/Q_1}{Q_1/Q_1 – Q_2/Q_1} = \frac {Q_2/Q_1}{1- Q_2/Q_1} \\ \text {In a Carnot cycle} \\ \frac {Q_2}{Q_1} &= \frac {T_2}{T_1} \\ \therefore \beta &= \frac {T_2/T_1}{1- T_2/T_1} \\ \beta &= \frac {T_2}{T_1 – T_2} \\ \end{align*}

Smaller the value of T_{1} – T_{2}, larger will be the value of coefficient of performance. Figure shows the schematic diagram of a refrigerator. In the diagram P is the compressor in which the working substance is compressed to a high pressure. The hot high-pressure gas coming out through the value V_{2} is cooled by passing it through a spiral pipe S in the condenser. Cold water is circulated in the condenser.

The compressed vapor gets liquefied. Through the valve V_{3}, the high-pressure liquid enters the spiral tube in the evaporator. The necessary heat is taken from the materials kept inside the evaporator. The vapor coming out of the evaporator is sucked back into the compressor through the valve V_{1} and the process is repeated. Therefore, continuous cooling produced.

1,efficiency of carnot engine, \(\eta = \left ( 1- \frac {T_2}{T_1} \right ) \).

2,The Carnot cycle is perfectly reversible ,

3,A refrigerator is a machine that operates in a manner opposite to that of a heat engine.

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