Carnot Engine
Definition
A machine that performs the following four processes in order is called a Carnot engine:
Step 1. Isothermal Expansion Process $A \to B$:
The volume increases from $V_{A}$ to $V_{B}$ by absorbing thermal energy $Q_{h}$ while maintaining the temperature at $T_{h}$.Step 2. Adiabatic Expansion Process $B \to C$:
The volume increases from $V_{B}$ to $V_{C}$ causing the temperature to decrease from $T_{h}$ to $T_{l}$ while the heat remains constant.Step 3. Isothermal Contraction Process $C \to D$:
The volume decreases from $V_{C}$ to $V_{D}$ by releasing thermal energy $Q_{l}$ while maintaining the temperature at $T_{l}$.Step 4. Adiabatic Contraction Process $D \to A$:
The volume decreases from $V_{D}$ to $V_{A}$ causing the temperature to increase from $T_{l}$ to $T_{h}$ while the heat remains constant.
Theorem
The thermal efficiency of a Carnot engine is as follows.
$$ \eta = 1 - \dfrac{Q_{l}}{Q_{h}} = 1 - \dfrac{T_{l}}{T_{h}} $$
Description
Although it only deals with two isothermal processes and two adiabatic processes, it is more efficient than any other complex machine. In fact, this design is only theoretically significant since the actual efficiency of heat engines is much lower than that of the Carnot engine. The reason why this seemingly useless Carnot engine is important is that its efficiency is not just good, it is the best.
Proof
Here, it is assumed that all processes are for an ideal gas.
$$ d U = \delta Q + \delta W $$
When all processes are performed once, the change in internal energy is $0$, thus following the first law of thermodynamics, the following is true.
$$ -\delta W = \delta Q $$
Therefore, the work done by the Carnot engine in one cycle is as follows.
$$ W = Q_{h} - Q_{l} $$
The Carnot engine depicted schematically is as follows.
$p V^{\gamma}$ is constant.
From the ideal gas law, since $p \propto \dfrac{T}{V}$, $pV^{\gamma} \propto T V^{\gamma - 1}$ is also constant. **Step 2. $B \to C$ is an adiabatic process, thus the following is true.
$$ { { T_{h} V_{B}^{\gamma - 1} } \over {T_{l} V_{C}^{\gamma - 1}} } = 1 $$
To summarize, one obtains the following.
$$ {{T_{h} } \over {T_{l} }} = \left( \dfrac{V_{C}}{V_{B}} \right)^{\gamma - 1} $$
**Step 4. $D \to A$ is also an adiabatic process, thus the following is true.
$$ { { T_{l} V_{D}^{\gamma - 1} } \over {T_{h} V_{A}^{\gamma - 1}} } = 1 $$
To summarize, one obtains the following.
$$ \left( \dfrac{V_{D}}{V_{A}} \right)^{\gamma - 1} = {{T_{h} } \over {T_{l} }} $$
Therefore, $\dfrac{V_{D}}{V_{A}} = \dfrac{V_{C}}{V_{B}}$ is true, and after summarizing, one obtains the following.
$$ \begin{equation} \dfrac{V_{B}}{V_{A}} = \dfrac{V_{C}}{V_{D}} \label{eq1} \end{equation} $$
$$ \Delta Q = RT \ln \dfrac{V_{2}}{V_{1}} $$
Step 1. $A \to B$ is an isothermal process, thus one obtains the following.
$$ \begin{equation} Q_{h} = RT_{h} \ln \dfrac{V_{B}}{V_{A}} \label{eq2} \end{equation} $$
Step 3. $C \to D$ is also an isothermal process, thus one obtains the following.
$$ Q_{l} = RT_{l} \ln \dfrac{V_{C}}{V_{D}} $$
Since $\eqref{eq1}$ was equal to $\dfrac{V_{B}}{V_{A}} = \dfrac{V_{C}}{V_{D}}$, substituting this into $\eqref{eq2}$ yields the following.
$$ \dfrac{Q_{h}}{Q_{l}} = \dfrac{ RT_{h} \ln \dfrac{V_{B}}{V_{A}} }{ RT_{l} \ln \dfrac{V_{C}}{V_{D}} } = \dfrac{ T_{h} \ln \dfrac{V_{C}}{V_{D}} }{ T_{l} \ln \dfrac{V_{C}}{V_{D}} } = \dfrac{T_{h}}{T_{l}} $$ Therefore, the efficiency of the Carnot engine is as follows.
$$ \eta = {{W} \over {Q_{h}}} = {{Q_{h} - Q_{l} } \over {Q_{h}}} = 1 - \dfrac{Q_{l}}{Q_{h}} = 1 - \dfrac{T_{l}}{T_{h}} $$
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