B4-4. Heat Engines, Efficiency, and The Carnot CycleHL Only

1. Macro-Operations of a Heat Engine

A heat engine is a practical device designed to continuously transform thermal energy into useful mechanical work by operating in a repeating thermodynamic cycle. The engine absorbs heat energy ($Q_H$) from a high-temperature reservoir at temperature $T_H$, converts a portion into work output ($W$), and dumps the unavoidable leftover thermal waste ($Q_C$) into a low-temperature sink at temperature $T_C$.

Because the engine operates in a complete closed loop, the final state matches the initial state precisely, meaning the net change in internal energy over a full cycle is zero ($\Delta U_{\text{cycle}} = 0$).

Net Work Output Formula:

$$W = Q_H - Q_C$$

Thermal Efficiency ($\eta$):

The efficiency of a heat engine is the ratio of useful work output to total thermal energy input:

$$\eta = \dfrac{W}{Q_H} = \dfrac{Q_H - Q_C}{Q_H} = 1 - \dfrac{Q_C}{Q_H}$$

2. The Ideal Reversible Carnot Cycle

The Carnot Cycle establishes the absolute maximum theoretical limit for the efficiency of any heat engine operating between two specific temperatures. It operates through four completely reversible stages, ensuring no generation of unnecessary global entropy:

  1. Isothermal Expansion ($A \rightarrow B$): The gas expands at a constant high temperature $T_H$, absorbing thermal energy $Q_H$ from the hot reservoir.
  2. Adiabatic Expansion ($B \rightarrow C$): The gas expands without heat transfer ($Q = 0$). Its temperature drops from $T_H$ to $T_C$.
  3. Isothermal Compression ($C \rightarrow D$): The surroundings compress the gas at a constant low temperature $T_C$, forcing it to reject thermal waste $Q_C$ into the cold reservoir.
  4. Adiabatic Compression ($D \rightarrow A$): The gas is compressed without heat transfer ($Q = 0$). Its temperature rises from $T_C$ back to the starting point $T_H$.

For a perfectly reversible Carnot cycle, the entropy changes cancel out perfectly across the system loop, resulting in a direct proportionality between heat transfers and absolute temperatures: $$\frac{Q_C}{Q_H} = \frac{T_C}{T_H}$$.

Carnot Efficiency Limit Equation:

$$\eta_C = 1 - \dfrac{T_C}{T_H}$$

Note: Temperatures $T_C$ and $T_H$ must absolutely be in Kelvin (K). No real-world heat engine can ever equal or exceed this efficiency because friction and unconstrained heat paths always generate additional irreversible entropy.

Mathematically Exact Carnot Cycle TH (Hot Isotherm) TC (Cold Isotherm) A B C D QH (In) QC (Out) Wnet (Area = ∮p dV) p V Cyclic Energy Balance Hot Reservoir (TH) Cold Reservoir (TC) Working Fluid ΔUcycle = 0 QH (Input) QC (Waste) Wout = QH - QC First Law: Q = ΔU + W Because ΔU = 0, Net Heat equals Net Work

Example 1: Evaluating Thermodynamic Possibility

Problem: An engineer proposes a heat engine design that operates between an energy supply at $227^\circ\text{C}$ and an exhaust environment at $27^\circ\text{C}$. The design specifications claim that for every $200$ kJ of thermal energy absorbed from the hot source, the engine performs $90$ kJ of useful mechanical work. Determine if this engine is thermodynamically possible according to the laws of physics.

Solution:

  • Step 1: Convert reservoir temperatures to absolute Kelvin units.
    $T_H = 227 + 273.15 = 500.15$ K
    $T_C = 27 + 273.15 = 300.15$ K
  • Step 2: Calculate the maximum possible theoretical efficiency ($\eta_C$).
    $$\eta_C = 1 - \dfrac{T_C}{T_H} = 1 - \dfrac{300.15}{500.15} = 1 - 0.600 = 0.400 = \mathbf{40.0\%}$$
  • Step 3: Calculate the proposed engine's efficiency ($\eta$).
    The input energy $Q_H = 200$ kJ and useful work $W = 90$ kJ.
    $$\eta = \dfrac{W}{Q_H} = \dfrac{90 \text{ kJ}}{200 \text{ kJ}} = 0.450 = \mathbf{45.0\%}$$
  • Step 4: Draw a definitive thermodynamic conclusion.
    The proposed design claims an efficiency of $45\%$, which directly exceeds the maximum theoretical Carnot limit of $40\%$. This violates the Second Law of Thermodynamics. Therefore, this engine is completely impossible to manufacture.