Thermodynamic Cycles

Carnot, Otto, Diesel, and Rankine - the engine cycles that power civilization.

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Thermal Efficiency Calculator

Formula
η = 1 − (T_cold / T_hot) [Carnot maximum]
T_H Hot Temperature
°C
T_C Cold Temperature
°C
Maximum Efficiency
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1 The Carnot Cycle - The Theoretical Maximum

The Carnot cycle is the most efficient heat engine possible operating between two temperatures. No real engine can exceed its efficiency - it's a fundamental limit set by the second law of thermodynamics.

η_Carnot = 1 − T_C/T_H
T must be in Kelvin (add 273.15 to °C)η depends ONLY on temperature ratioHigher T_H or lower T_C → higher efficiency
Why 100% is impossible: For η = 100%, you'd need T_C = 0 K (absolute zero), which is physically unreachable. Real engines achieve 30-60% of Carnot efficiency due to friction, irreversibilities, and practical constraints.
Worked Example 1

Carnot Efficiency - Coal Power Plant

Problem: A coal plant operates with steam at 540°C and condenser at 30°C. What is the maximum theoretical efficiency? If the actual efficiency is 38%, what fraction of the Carnot limit is achieved?
Carnot efficiency
T_H = 540 + 273 = 813 K, T_C = 30 + 273 = 303 K
η = 1 − 303/813 = 1 − 0.373
η_Carnot = 62.7%
Actual/Carnot = 38/62.7
= 60.6% of theoretical maximum
Answer: Carnot: 62.7%, actual: 38% (61% of Carnot). The remaining 39% of Carnot efficiency is lost to friction, heat leaks, pumping losses, and incomplete combustion.

2 The Otto Cycle - Gasoline Engines

The cycle behind every gasoline car engine. Four strokes: intake, compression, power (combustion), exhaust. Efficiency depends on the compression ratio r.

η_Otto = 1 − 1/r^(γ−1)
r = V_max/V_min (compression ratio, typically 8-12)γ = 1.4 for air (ratio of specific heats)r = 10 → η = 60.2% (theoretical)
Worked Example 2

Otto Cycle - Car Engine Efficiency

Problem: A gasoline engine has compression ratio r = 9.5. Calculate the theoretical Otto efficiency. If the engine produces 150 kW, how much fuel energy is consumed per second?
Otto efficiency
η = 1 − 1/9.5^(0.4) = 1 − 1/9.5^0.4
9.5^0.4 = e^(0.4×ln9.5) = e^(0.4×2.251) = e^0.900 = 2.460
η = 1 − 1/2.460 = 59.3%
In practice ~30% (friction, heat loss). Fuel power = 150/0.30
= 500 kW input (350 kW wasted as heat)
Answer: Theoretical: 59.3%, practical: ~30%. The engine consumes 500 kW of fuel energy to produce 150 kW of useful work - the rest heats the engine block, exhaust, and radiator.

3 The Diesel Cycle

Similar to Otto but fuel is injected into pre-compressed hot air (no spark plug needed). Higher compression ratios (14-25) give better efficiency than Otto, but the cycle itself is less thermodynamically ideal.

η_Diesel = 1 − (1/r^(γ−1)) × (r_c^γ − 1)/(γ(r_c − 1))
r = compression ratio (14-25)r_c = cutoff ratio (volume after/before combustion)Higher r → higher η, but also heavier engine

4 The Rankine Cycle - Steam Power Plants

The cycle used in 80% of the world's electricity generation (coal, nuclear, solar thermal, geothermal). Water is boiled to steam, expanded through a turbine, condensed, and pumped back.

1 → 2
Pump: liquid water pressurized (small work in)
2 → 3
Boiler: water → superheated steam (heat in)
3 → 4
Turbine: steam expands, drives generator (work out)
4 → 1
Condenser: steam → liquid water (heat rejected)

5 Refrigeration & Heat Pumps

Refrigerators and heat pumps run heat engine cycles in reverse - they use work input to move heat from cold to hot. Performance is measured by COP (Coefficient of Performance) rather than efficiency.

COP_cooling = T_C / (T_H − T_C)
COP_heating = T_H / (T_H − T_C) = COP_cooling + 1Typical fridge: COP ≈ 3-5 (moves 3-5× more heat than work input)Heat pump: COP ≈ 3-4 (300-400% "efficient")
Worked Example 3

Heat Pump vs. Electric Heater

Problem: Compare heating a house with a 5 kW electric heater vs. a heat pump (COP = 3.5) using the same 5 kW of electricity. Outside: 5°C, inside: 20°C.
Compare heat delivered
Electric heater: Q = W = 5 kW (100% conversion)
Heat pump: Q = COP × W = 3.5 × 5
Q_HP = 17.5 kW of heating (3.5× more!)

The heat pump delivers 17.5 kW of heat using only 5 kW of electricity by extracting 12.5 kW from the cold outside air.

Answer: Heat pump delivers 17.5 kW vs electric heater's 5 kW - 3.5× more heating per watt. This is why heat pumps are replacing gas boilers globally. They appear to be ">100% efficient" because they move heat rather than creating it.

6 PV Diagrams & Real-World Efficiencies

Pressure-Volume (PV) diagrams show thermodynamic cycles visually. The enclosed area equals the net work done. Real engines always fall short of theoretical due to friction, irreversibilities, and heat losses.

Car engine (Otto)
25-35% actual efficiency
Diesel truck
35-45%
Coal power plant
33-40%
Gas turbine (combined)
55-62% (most efficient heat engine)
Nuclear plant
33-37%
Large ship diesel
45-55% (largest, slowest = most efficient)
Combined cycle gas turbines (CCGT): Burn gas in a turbine (~40%), then use exhaust heat to make steam for a second turbine (~20% more). Total: ~60% - the highest thermal efficiency of any heat engine. This is why natural gas is displacing coal in electricity generation.
Otto Cycle - PV Diagram V P 1 2 3 4 Adiabatic compression (1→2) Q_in (2→3) Adiabatic expansion (3→4) Q_out

Otto Cycle Efficiency Calculator

Formula
η = 1 − 1/r^(γ−1)
r Compression ratio
γ Heat capacity ratio (1.4 for air)
Thermal Efficiency
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Refrigerator COP Calculator

Formula
COP = T_C / (T_H − T_C)
T_C - Cold reservoir (°C)
T_H - Hot reservoir (°C)
Coefficient of Performance
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