Physics > First Law of Thermodynamics > 7.0 Heat engine
First Law of Thermodynamics
1.0 Introduction
2.0 Three important terms in first law of thermodynamics.
3.0 First law of thermodynamics
4.0 Different thermodynamic processes
4.1 Isobaric Process:
4.2 Isochoric process
4.3 Isothermal process
4.5 Adiabatic process
4.6 Polytropic process
5.0 Graphs
6.0 Efficiency of cyclic process
7.0 Heat engine
8.0 Refrigerator
7.2 Carnot Engine
4.2 Isochoric process
4.3 Isothermal process
4.5 Adiabatic process
4.6 Polytropic process
Carnot cycle consists of the following four stages:
- Isothermal expansion ( process $AB$ )
- Adiabatic expansion ( process $BC$ )
- Isothermal compression ( process $CD$ )
- Adiabatic compression ( process $DA$ )
The $P$-$V$ diagram of the cycle is shown in figure.
In process $AB$ heat ${{Q_1}}$ is taken by the working substance at constant temperature ${T_1}$ and in process $CD$ heat ${{Q_2}}$ is liberated from the working substance at constant temperature ${{T_2}}$. The net work done is area of graph ABCD. After doing the calculations for different processes we can show that:$$\frac{{{Q_2}}}{{{Q_1}}} = \frac{{{T_2}}}{{{T_1}}}$$Therefore, efficiency of the cycle is,$$\eta = 1 - \frac{{{T_2}}}{{{T_1}}}$$
Note:
That efficiency of carnot engine is maximum( not 100%) for given temperatures ${{T_1}}$ and ${{T_2}}$. But still carnot engine is not a practical engine because many ideal situations have been assumed while designing this engine which can practically not be obtained.
Question18. Carnot's engine takes in a thousand kilo calories of heat from a reservoir at ${827^o}C$ and exhausts it to a sink at ${27^o}C$. How much work does it perform ? What is the efficiency of the engine ?
Solution: Given, $$\begin{equation} \begin{aligned} {Q_1} = {10^6}\,calories \\ {T_1} = \left( {827 + 273} \right) = 1100K \\ and\;{T_2} = \left( {27 + 273} \right) = 300K \\ as,\;\frac{{{Q_2}}}{{{Q_1}}} = \frac{{{T_2}}}{{{T_1}}} \\ {Q_2} = \frac{{{T_2}}}{{{T_1}}}.{Q_1} = \left( {\frac{{300}}{{1100}}} \right)\left( {{{10}^6}} \right) \\ = 2.72 \times {10^5}\;cal \\\end{aligned} \end{equation} $$Efficiency of the cycle,$$\begin{equation} \begin{aligned} \eta = \left( {1 - \frac{{{T_2}}}{{{T_1}}}} \right) \times 100 \\ \eta = \left( {1 - \frac{{300}}{{1100}}} \right) \times 100 \\ = 72.72\% \\\end{aligned} \end{equation} $$
Question 19. Two carnot engines $A$ and $B$ are operated in series. A receives heat at $800 K$ and rejects heat at $T K$. The engine $B$ receives heat a $T K$ and rejects it at $300 K$. If their efficiencies are equal, what is $T$?
Solution: $$\begin{equation} \begin{aligned} \eta = 1 - \frac{{{T_2}}}{{{T_1}}} \\ \therefore \;1 - \frac{T}{{800}} = 1 - \frac{{300}}{T} \\ \frac{T}{{800}} = \frac{{300}}{T} \\ {T^2} = 240000 \\ T = 100\sqrt {24} = 489.9K \\\end{aligned} \end{equation} $$
