Heat Engines, Refrigerators, and Heat Pumps

A heat engine takes heat Q_h from a hot reservoir, converts part of it to work W and rejects the rest Q_c to a cold reservoir, with thermal efficiency η = W ÷ Q_h = 1 − Q_c ÷ Q_h; a refrigerator or heat pump does the opposite, using work to move heat from cold to hot, with coefficients of performance COP_R = Q_c ÷ W and COP_HP = Q_h ÷ W. In brief: energy is conserved: Q_h = W + Q_c for an engine, and Q_h = W + Q_c for a refrigerator (heat rejected is the heat extracted plus the work). The efficiency is always less than 1. A heat pump delivers more heat than the electrical energy it uses: COP_HP = COP_R + 1.

Car engines, power stations, refrigerators and air conditioners are heat engines or their reverse. This guide describes how they work with energy-flow diagrams, defines efficiency and coefficient of performance, discusses real engines such as the Otto, diesel and steam turbines, compares the machines in a table, works through numerical examples and corrects common mistakes.

How does a heat engine work?

A heat engine operates in a cycle with a working substance (steam, air-fuel mixture, gas). In each cycle: it absorbs heat Q_h from a hot source (burning fuel, a nuclear reactor), does work W (the area of the cycle on the PV diagram), and rejects the waste heat Q_c to a colder sink (air, a river, the sea). After a complete cycle, ΔU = 0, so by the first law W = Q_h − Q_c. The thermal efficiency is η = W ÷ Q_h = 1 − Q_c ÷ Q_h. The second law of thermodynamics prohibits an efficiency of 100 percent: some heat must always be rejected to a colder reservoir. Typical efficiencies: a petrol car engine about 25 to 30 percent, a diesel engine 35 to 45 percent, a coal power plant 33 to 40 percent, and a combined-cycle gas power plant up to about 60 percent.

How do refrigerators and heat pumps work?

A refrigerator extracts heat Q_c from the cold space (the inside of the fridge), using work W (the electric compressor), and rejects Q_h = Q_c + W to the warmer surroundings (the coils at the back). The performance is measured by the coefficient of performance COP_R = Q_c ÷ W, which can be greater than 1 (typically 2 to 5). A heat pump uses the same cycle to heat a building, moving heat from the cold outside to the warm inside; it is rated by COP_HP = Q_h ÷ W = COP_R + 1. A heat pump with a COP of 3.5 delivers 3.5 kWh of heat for 1 kWh of electricity, which makes it more efficient than direct electric heating, because most of the heat is moved, not generated. An air conditioner is a refrigerator that cools a room. The working fluid (refrigerant) evaporates in the cold coil, absorbing heat, is compressed, condenses in the hot coil, releasing heat, and expands through a valve.

What are real engines like?

• Otto cycle (petrol engine): four strokes: intake, compression, power (ignition by a spark) and exhaust. The ideal efficiency is 1 − r^1−γ, where r is the compression ratio (about 8 to 10); for r = 9 and γ = 1.4, η = 1 − 9^−0.4 = 58 percent, and in real engines it is about half this value because of friction, heat losses and incomplete combustion.
• Diesel engine: higher compression ratio (15 to 20), with ignition of the fuel by the compressed hot air.
• Steam turbine (Rankine cycle): boiler, turbine, condenser, pump; used in power stations.
• Gas turbines and jet engines (Brayton cycle).

The waste heat of the engines can be used (combined heat and power systems), which raises the total energy efficiency to above 80 percent.

Engines, refrigerators and heat pumps compared

Worked example: efficiency and COP

1. Heat engine. An engine absorbs 1,000 J of heat per cycle and rejects 700 J. W = 1,000 − 700 = 300 J. η = 300 ÷ 1,000 = 30 percent.
2. Power. The engine runs at 50 cycles per second: P = 300 × 50 = 15 kW. The fuel heat input rate is 1,000 × 50 = 50 kW, and the waste heat is 35 kW.
3. Car fuel. A car engine with an efficiency of 28 percent delivers 21 kW of mechanical power. The fuel provides 21 ÷ 0.28 = 75 kW of heat. At 34 MJ/L for petrol, this is 75,000 ÷ 34 × 10⁶ = 2.2 × 10⁻³ L/s, or 7.9 L/h.
4. Refrigerator. A refrigerator with COP = 3.0 extracts 600 J from the cold space per cycle. W = 600 ÷ 3.0 = 200 J, and Q_h = 600 + 200 = 800 J is released into the room.
5. Heat pump. A heat pump with COP = 4.0 uses 2.0 kW of electric power. It delivers 8.0 kW of heat: 2.0 kW of the electrical energy plus 6.0 kW taken from the outside air. With a direct electric heater, 8.0 kW of heat would need 8.0 kW of electricity.
6. Conversion of the cost. If electricity costs 0.25 per kWh, direct heating costs 0.25 per kWh of heat, and the heat pump (COP = 4.0) costs 0.0625 per kWh of heat.

How were they developed?

Thomas Newcomen built the first practical steam engine in 1712, and James Watt improved it in the 1760s to 1780s with a separate condenser. Sadi Carnot analyzed the efficiency of ideal heat engines in 1824. Nikolaus Otto built the four-stroke engine in 1876, Rudolf Diesel patented his engine in 1892, and Jacob Perkins made the first vapor-compression refrigerator in 1834. William Thomson (Lord Kelvin) proposed the heat pump in 1852.

Which mistakes are common?

Mistake 1: Expecting 100 percent efficiency in an engine

The second law requires some waste heat to be rejected.

Mistake 2: Thinking that COP greater than 1 breaks energy conservation

The machine moves heat, not creates it: Q_h = Q_c + W, so the energy is conserved.

Mistake 3: Using the efficiency formula for refrigerators

Use the coefficient of performance Q_c ÷ W for refrigerators and Q_h ÷ W for heat pumps.

Mistake 4: Forgetting to include the waste heat in the energy balance

Q_h = W + Q_c. All the energy has to be accounted for.