The Carnot Cycle and Heat Engine Efficiency
Carnot cycle
Carnot cycle
The Carnot cycle represents an idealized model of a heat engine that achieves the maximum possible efficiency for given reservoir temperatures.
The Carnot cycle consists of four reversible processes:
- Isothermal Expansion (Process 1-2)
- Gas expands at constant temperature $T_h$
- Absorbs heat $Q_h$ from hot reservoir
- Work is done by the gas
- Adiabatic Expansion (Process 2-3)
- No heat exchange with surroundings
- Temperature drops from $T_h$ to $T_c$
- Work is done by the gas
- Isothermal Compression (Process 3-4)
- Gas compressed at constant temperature $T_c$
- Releases heat $Q_c$ to cold reservoir
- Work is done on the gas
- Adiabatic Compression (Process 4-1)
- No heat exchange with surroundings
- Temperature increases from $T_c$ to $T_h$
- Work is done on the gas
The Adiabatic Equation for a Monatomic Ideal Gas
- During an adiabatic process, the relationship between pressure and volume for a monatomic ideal gas is given by: $$PV^{\frac{5}{3}} = \text{constant}$$
- This equation arises from combining the ideal gas law with the first law of thermodynamics under adiabatic conditions.
- Suppose a gas undergoes adiabatic expansion from an initial state $P_1$, $V_1$ to a final state $P_2$, $V_2$.
- The adiabatic equation ensures: $$P_1V_1^{\frac{5}{3}} = P_2V_2^{\frac{5}{3}}$$
Heat Engine Cycles
A heat engine is a device that converts thermal energy into mechanical work by operating in a cyclic process.
Efficiency of a Heat Engine
Efficiency
The efficiency ($\eta$) of a heat engine measures how effectively it converts input energy into useful work.
It is defined as:
$$\eta = \frac{\text{Useful work}}{\text{Input energy}} = \frac{W}{Q_{\text{in}}}$$
If a heat engine receives 500 J of energy from a hot reservoir and performs 200 J of work, its efficiency is:
$$\eta = \frac{200 \, \text{J}}{500 \, \text{J}} = 0.4 \, \text{or} \, 40\%$$
Carnot Efficiency
Carnot efficiency
The Carnot efficiency represents the maximum possible efficiency of a heat engine operating between two temperatures, $T_h$ (hot reservoir) and $T_c$ (cold reservoir).
It is given by:
$$\eta_{\text{Carnot}} = 1 - \frac{T_c}{T_h}$$
For an engine operating between a hot reservoir at 600 K and a cold reservoir at 300 K, the Carnot efficiency is:
$$\eta_{\text{Carnot}} = 1 - \frac{300}{600} = 0.5 \, \text{or} \, 50\%$$
- The Carnot efficiency is an ideal limit.
- No real engine can exceed this efficiency due to practical limitations such as friction and heat loss.
- This is because it assumes perfectly reversible processes with no energy losses due to friction, turbulence, or heat dissipation.
- In reality, all real engines experience irreversibilities, making the Carnot cycle an upper limit on efficiency rather than an achievable process.
Real Heat Engines and Their Efficiency
- In real heat engines, such as the Otto and Diesel engines, the cycle approximates but does not replicate the Carnot cycle.
- These engines have lower efficiencies due to:
- Mechanical energy losses
- Non-reversible processes
- Non-ideal working conditions, such as imperfect insulation
Comparing Thermodynamic Processes
| Process | Constant Quantity | Key Equation | Work Done |
|---|---|---|---|
| Isovolumetric | Volume ($V$) | $Q = \Delta U$ | $W = 0$ |
| Isobaric | Pressure ($P$) | $Q = \Delta U + P \Delta V$ | $W = P \Delta V$ |
| Isothermal | Temperature ($T$) | $Q = W$ | $W = nRT \ln \frac{V_2}{V_1}$ |
| Adiabatic | No heat exchange ($Q = 0$) | $PV^{\gamma} = \text{constant}$ | $W = \frac{P_1V_1 - P_2V_2}{\gamma - 1}$ |
- What is the main difference between an isothermal and an adiabatic process?
- How is the efficiency of a heat engine calculated?
- Why can no real engine exceed the Carnot efficiency?
