Heat Engines and the Carnot Cycle

The Carnot Cycle

The Carnot cycle represents an idealized thermodynamic cycle that establishes the maximum theoretical efficiency for any heat engine operating between two temperature reservoirs. The cycle operates between a high-temperature reservoir, $T_h$, which provides the energy for the working substance to work on the surroundings, and a low-temperature reservoir, $T_c$, to which waste heat is released. Overall, the heat engine transfers heat from the high-temperature to the low-temperature reservoir and converting a portion of that heat into work.

The Carnot cycle consists of four reversible stages:

  1. Isothermal Expansion at $T_h$: The gas is in thermal contact with the hot reservoir at temperature $T_h$. Heat $Q_h$ is absorbed by the gas from the hot reservoir, allowing it to expand isothermally. During this stage, the gas performs work on the surroundings as it expands, and because temperature remains constant, the internal energy change, $\Delta U$, is zero. All the heat added goes into doing work.
  2. Adiabatic Expansion: In this phase, the gas is thermally isolated (no heat exchange, $\Delta Q = 0$). The gas continues to expand, but as it does so without gaining or losing heat, its temperature decreases from $T_h$ to $T_c$. The expansion is adiabatic, and the temperature decrease is due to the work done by the gas on its surroundings, which reduces its internal energy.
  3. Isothermal Compression at $T_c$: Now in contact with the cold reservoir at temperature $T_c$, the gas is compressed isothermally. During this stage, heat $Q_c$ is released to the cold reservoir as the gas is compressed. The energy lost as heat comes from the work done on the gas, keeping the internal energy constant since the temperature remains at $T_c$.
  4. Adiabatic Compression: In this final stage, the gas is again thermally isolated (no heat exchange). The gas is compressed, and the work done on the gas raises its temperature from $T_c$ back up to $T_h$, completing the cycle. Since no heat is exchanged, the temperature rise is purely due to the work done on the gas, restoring it to its initial state.

The Carnot cycle illustrates that the maximum theoretical efficiency depends solely on the temperatures of the hot and cold reservoirs, independent of the specific properties of the gas. This efficiency is given by:

Efficiency = 1 - \frac{T_c}{T_h}.

This relationship shows that as the temperature difference between the reservoirs increases, the efficiency approaches 100%. However, achieving 100% efficiency would require T_c to be absolute zero, which is unattainable, hence real engines always have efficiencies less than that of the Carnot cycle.

 

Entropy and Efficiency in the Carnot Cycle

The Carnot cycle is an idealized model that defines the upper limit of efficiency for any heat engine operating between two temperature reservoirs. Entropy plays a crucial role in the Carnot cycle, as it quantifies the energy dispersal at each stage of the cycle. Since entropy is a state function, the total entropy change $∆S_{\text{rev}}$ over one complete Carnot cycle is zero:

\[∆S_{\text{rev}} = \oint dS = \frac{∆Q_h}{T_h} + \frac{∆Q_c}{T_c} = 0\]

 

Efficiency of the Carnot Cycle

To determine the efficiency of the Carnot cycle, we need to evaluate the heat transferred during isothermal processes at high and low temperatures.

  • During the isothermal expansion at $T_h$: The heat absorbed is given by:

\[∆Q_h = n R T_h \ln \frac{V_2}{V_1}\]

  • During the isothermal compression at $T_c$: The heat released is:

\[∆Q_c = n R T_c \ln \frac{V_4}{V_3}\]

Since the cycle also includes adiabatic processes (where no heat is exchanged, $∆Q = 0$), the temperature changes during these steps relate volumes to temperatures (see 'Adiabatic processes' section):

\[\frac{T_f}{T_i} = \left( \frac{V_f}{V_i} \right)^{1 - \gamma}\]

We can rewrite this relationship to link the temperatures and volumes between stages:

\[\left( \frac{T_i}{T_f} \right)^{\frac{1}{1 - \gamma}} = \frac{V_f}{V_i}\]

Let us set a constant $a = \frac{1}{1 - \gamma}$ for simplicity. Using this constant, we get:

\[T_h^a V_1 = T_c^a V_4\]
and
\[T_c^a V_3 = T_h^a V_2\]

From these relationships, we derive:

\[\frac{T_h^a}{T_c^a} = \frac{V_3}{V_2} = \frac{V_4}{V_1} \implies \frac{V_1}{V_2} = \frac{V_4}{V_3}\]

Thus, we can express the ratio in terms of natural logarithms:

\[\ln \frac{V_4}{V_3} = - \ln \frac{V_2}{V_1}\]

Therefore:

\[∆Q_c = n R T_c \ln \frac{V_4}{V_3} \implies ∆Q_c = -n R T_c \ln \frac{V_2}{V_1}\]
 

which we can now combine with:

\[∆Q_h = n R T_h \ln \frac{V_2}{V_1}\]

to get a relatitionship between the heat exchanged in the cyclic process and the temperatures of the isothermal expansion and compression:

\[ \frac{\Delta Q_h}{\Delta Q_c} = -\frac{T_h}{T_c} \]

 

Efficiency of the Cycle

The efficiency \( \eta \) of a Carnot engine is defined as the ratio of the work done \( |W| \) to the heat absorbed \( |\Delta Q_h| \) from the hot reservoir:

\[ \eta = \frac{|W|}{|\Delta Q_h|} = \frac{|\Delta Q_h| - |\Delta Q_c|}{|\Delta Q_h|} = 1 - \frac{|\Delta Q_c|}{|\Delta Q_h|} = 1 - \frac{T_c}{T_h} \]

This formula represents the theoretical maximum efficiency of a heat engine operating between two temperatures, \( T_h \) and \( T_c \). The efficiency implies that the work is done at the expense of the heat input into the system.

 

Key Takeaways

The efficiency of the Carnot cycle depends only on the temperatures of the hot and cold reservoirs, not on the specific working substance of the engine. The entropy changes are balanced over the complete cycle, highlighting the reversible nature of the Carnot cycle. This ideal cycle sets the theoretical upper limit for efficiency, emphasizing the importance of temperature differences in achieving higher efficiencies.