Thermodynamic Cycles and Machines §

The four steps (heat engine mode, working fluid is an ideal gas):

1. Isothermal expansion at $T_H$: absorb $Q_H$, do work.
2. Adiabatic expansion: temperature drops from $T_H$ to $T_C$.
3. Isothermal compression at $T_C$: reject $Q_C$, work is done on the gas.
4. Adiabatic compression: temperature rises from $T_C$ to $T_H$.

Why it is the maximum. The Carnot cycle is fully reversible. Any irreversibility (friction, finite-$\Delta T$ heat transfer, turbulence) reduces efficiency. The Second Law and Entropy sets the ceiling.

Practical limitation. Even at $T_H = 600$ K, $T_C = 300$ K, $\eta_{\text{Carnot}} = 50\%$. Real steam plants achieve 55–80% of the Carnot limit, depending on technology generation (subcritical ≈ 55–65%, supercritical/ultra-supercritical ≈ 70–80%).

Steps: (1) isentropic compression, (2) constant-volume heat addition (combustion), (3) isentropic expansion (power stroke), (4) constant-volume heat rejection (exhaust).

Limitation. High compression ratio increases efficiency but also peak temperature, risking auto-ignition (knock). Practical gasoline engines: $r \approx 8$–$12$, $\eta \approx 30$–$40\%$.

Energy accounting. The First Law and Internal Energy applied to each step: $Q_{\text{in}} = m\,c_V(T_3 - T_2)$, $Q_{\text{out}} = m\,c_V(T_4 - T_1)$, $W_{\text{net}} = Q_{\text{in}} - Q_{\text{out}}$.

Steps: (1) isentropic compression, (2) constant-pressure heat addition (fuel injection), (3) isentropic expansion, (4) constant-volume heat rejection.

Difference from Otto. Heat addition at constant pressure (not volume) allows higher compression ratios without knock, since peak temperature is lower for a given heat input. Diesel engines: $r \approx 15$–$25$, $\eta \approx 35$–$50\%$.

Steps: (1) isentropic compression (compressor), (2) constant-pressure heat addition (combustor), (3) isentropic expansion (turbine), (4) constant-pressure heat rejection (exhaust).

All four steps are at constant pressure or isentropic — no volume constraints, making it well-suited to continuous-flow (open) machines like gas turbines and jet engines.

Energy accounting. Per the First Law and Internal Energy: $w_{\text{comp}} = c_P(T_2 - T_1)$, $w_{\text{turb}} = c_P(T_3 - T_4)$, $q_{\text{in}} = c_P(T_3 - T_2)$.

Regeneration. A heat exchanger (regenerator) preheats the compressor outlet using the turbine exhaust, reducing fuel consumption. With perfect regeneration, the regenerative Brayton cycle efficiency is $\eta = 1 - T_2/T_3$. Unlike the Ericsson or Stirling cycles (which achieve Carnot efficiency with perfect regeneration because all external heat transfer occurs isothermally), the regenerative Brayton cycle still falls below Carnot because its heat addition and rejection occur over a range of temperatures at constant pressure.

Steps: (1) isentropic compression (pump, liquid water), (2) constant-pressure heat addition (boiler: liquid → superheated vapor), (3) isentropic expansion (turbine), (4) constant-pressure heat rejection (condenser: vapor → liquid).

Why not Brayton for steam? Water’s large latent heat and the ability to condense at low temperature make the Rankine cycle more efficient for high-temperature heat sources (fossil, nuclear, solar thermal).

Energy accounting. Per the First Law and Internal Energy: $w_{\text{pump}} = v_f(P_2 - P_1)$ (small), $w_{\text{turb}} = h_3 - h_4$, $q_{\text{boiler}} = h_3 - h_2$.

Improvements. Reheat (expand in two turbine stages with intermediate reheating), regenerative feedwater heating (extract steam to preheat feedwater), and supercritical operation ($P > 22.1\text{ MPa}$) push efficiency toward 60%.

The four steps:

1. Isothermal expansion at $T_H$: absorb $Q_H$, do work.
2. Isochoric cooling (constant volume): temperature drops from $T_H$ to $T_C$. In a real engine, the regenerator absorbs this heat.
3. Isothermal compression at $T_C$: reject $Q_C$, work is done on the gas.
4. Isochoric heating (constant volume): temperature rises from $T_C$ to $T_H$. The regenerator returns the stored heat.

The regenerator. A thermal mass (wire mesh, ceramic) placed between the hot and cold ends. During step 2 it absorbs heat from the working fluid; during step 4 it returns that heat. With perfect regeneration, the only external heat transfers are the two isothermal steps, so the cycle is thermodynamically equivalent to Carnot: $\eta = 1 - T_C/T_H$.

Comparison with Ericsson cycle. The Ericsson cycle replaces the two isochoric legs with two isobaric (constant-pressure) legs, also with a regenerator. Like Stirling, it achieves Carnot efficiency with perfect regeneration because all external heat transfer is isothermal. The key distinction from the regenerative Brayton cycle: in Brayton, heat addition and rejection occur at constant pressure over a range of temperatures, so even perfect regeneration cannot make it Carnot-equivalent.

Practical use. Stirling engines are used in solar thermal concentrators, cogeneration, and niche automotive applications. Their low speed and quiet operation are advantages; the regenerator adds complexity and limits power density.

Refrigerator vs. heat pump. Same cycle, different purpose. A refrigerator removes heat from a cold space ($Q_C$ is the desired output). A heat pump delivers heat to a warm space ($Q_H$ is the desired output). Since $Q_H = Q_C + W$, the heat-pump COP is always 1 greater than the refrigerator COP.

Vapor-compression cycle. The standard refrigeration cycle: (1) isentropic compression of vapor, (2) constant-pressure condensation (heat rejection), (3) isenthalpic throttling (expansion valve), (4) constant-pressure evaporation (heat absorption). The throttling step is irreversible, so real COP < Carnot COP.

Typical values. Household refrigerator: COP ≈ 2–4. Air-source heat pump: COP ≈ 2–4. Carnot COP at $T_H = 293$ K, $T_C = 273$ K: $\mathrm{COP}_{\text{ref}} = 13.7$ — real machines achieve 25–35% of the Carnot COP.

Carnot heat-pump COP:

Power input:

Heat extracted outdoors:

The heat pump moves 9.17 kW from outside and adds 0.83 kW of work to deliver 10 kW indoors.

Thermal efficiency:

Peak pressure ratio. From the isentropic compression (1→2):

Constant-volume heat addition (2→3): $P_3/P_2 = T_3/T_2$:

Isentropic relation for pressure: $P_2/P_1 = r^{\gamma} = 10^{1.4} = 25.12$. Therefore:

The peak pressure is about 73 times the intake pressure. This high ratio is why practical engines limit $r$ to 8–12: beyond that, knock and mechanical stress become prohibitive.