Heat Engines — Carnot and Otto Cycles

Heat engine schematic: a heat engine operates cyclically between a hot reservoir at temperature $T_H$ (supplying heat $Q_H$) and a cold reservoir at $T_C$ (absorbing heat $Q_C$). The net work output is $W = Q_H - Q_C$.

Efficiency: $\eta = W/Q_H = 1 - Q_C/Q_H$. Since $Q_C > 0$ always (Second Law), we have $\eta < 1$: no heat engine can convert all absorbed heat into work.

Fundamental limit: efficiency increases with higher $T_H$ and lower $T_C$. That is why modern thermal power plants operate at very high temperatures and pressures (superheated steam at ~600°C) and use cooling water at low temperature.

Inverse machines:

• Refrigerator/heat pump: absorbs work $W$ to transfer heat $Q_C$ from the cold reservoir to the hot one. Coefficient of performance $COP_{ref} = Q_C/W = Q_C/(Q_H - Q_C)$.
• Heat pump for heating: $COP_{hp} = Q_H/W = T_H/(T_H - T_C)$ in the ideal case.

Real-world examples: steam engine, gas turbine, internal combustion engine (Otto), Diesel engine, refrigerator.

The ideal cycle: the Carnot cycle consists of 4 reversible processes: two isothermals (at $T_H$ and $T_C$) alternating with two adiabatics. It is the cycle with the maximum possible efficiency between two given temperatures.

• 1→2: isothermal expansion at $T_H$. The gas absorbs $Q_H$ from the hot reservoir and expands, doing work.
• 2→3: adiabatic expansion. The gas cools from $T_H$ to $T_C$ without exchanging heat, continuing to do work.
• 3→4: isothermal compression at $T_C$. The gas releases $Q_C$ to the cold reservoir. Work is done on the gas.
• 4→1: adiabatic compression. The gas warms from $T_C$ to $T_H$ without exchanging heat.

Carnot efficiency: $\eta_{Carnot} = 1 - T_C/T_H$. It depends only on the absolute temperatures of the reservoirs. To increase $\eta$, one must increase $T_H$ or decrease $T_C$.

Carnot theorem: no heat engine operating between two reservoirs at $T_H$ and $T_C$ can have efficiency greater than $\eta_{Carnot}$. $\eta_{real} \leq \eta_{Carnot}$, with equality only for reversible processes.

Fundamental relation: for the Carnot cycle: $Q_H/T_H = Q_C/T_C$. This relation is the basis for the definition of entropy as a state function.

Numerical example: a thermal power plant with $T_H = 800$ K (527°C) and $T_C = 300$ K (27°C) has $\eta_{Carnot} = 1 - 300/800 = 62.5\%$. Real efficiencies are ~35-45% due to irreversibilities, friction, and thermal losses.

The 4-stroke cycle (Nikolaus Otto, 1876): the Otto cycle describes the operation of a petrol engine. It consists of 4 processes: two adiabatics and two isochorics.

1. 1→2 — Adiabatic compression: the piston compresses the air-fuel mixture. $T$ and $P$ increase. Work is negative (done on the system).
2. 2→3 — Isochoric combustion: the spark plug ignites the mixture. Heat $Q_H$ is released at constant volume. $P$ and $T$ rise sharply. Work is zero ($V$ const).
3. 3→4 — Adiabatic expansion (power stroke): the hot gases expand, pushing the piston. The gas does positive work. $T$ and $P$ decrease.
4. 4→1 — Isochoric exhaust: the exhaust valve opens, burnt gases are expelled at constant volume. The system releases heat $Q_C$ to the environment.

Otto efficiency: $\eta_{Otto} = 1 - 1/r^{\gamma-1}$, where $r = V_{max}/V_{min}$ is the compression ratio.

• Typical $r$ for petrol engines: $8 \div 12$, giving $\eta \approx 40\div 55\%$.
• Higher $r$ means higher efficiency, but also higher risk of knocking (spontaneous combustion before the spark).
• Diesel engines have $r \approx 15\div 22$ and $\eta \approx 50\div 65\%$.

Comparison with Carnot: the Otto cycle has lower efficiency than Carnot because heat is added at constant volume (not isothermally), and heat is rejected at variable temperature during expansion.

Equivalence of the two statements: the Second Law of thermodynamics can be formulated in two equivalent ways:

Kelvin–Planck: it is impossible to perform a thermodynamic process whose sole result is the complete conversion of heat absorbed from a single reservoir into work. In other words, $\eta < 1$ — a heat engine cannot have unit efficiency.

• Consequence: it is impossible to build a perpetual motion machine of the second kind (which would violate only the Second Law, not the First).

Clausius: it is impossible to perform a thermodynamic process whose sole result is the transfer of heat from a colder body to a hotter body. External work is always required for cooling.

• Consequence: a refrigerator needs electricity to operate. An air conditioner consumes energy to cool rooms.

Why they are equivalent: if we violated Kelvin–Planck, we could use the work produced to run a heat pump that transfers heat from cold to hot, violating Clausius. And vice versa.

Maximum theoretical COP:

• Refrigerator: $COP_{ref,max} = T_C/(T_H - T_C)$ — cooling is more efficient when the temperature difference is small.
• Heat pump: $COP_{hp,max} = T_H/(T_H - T_C)$ — always $> 1$, so it is always more convenient than direct electric heating.