Physics › Engineering physics › Heat engines and heat pumps
Heat engines and heat pumps
Why no engine, however perfect its engineering, can turn all its heat into work, and how the same machine run backwards heats your house with more energy than you feed it. The loop on the p-V diagram, and the honest arithmetic of real engines.
Builds on The first law of thermodynamics and Torque, angular momentum and rotational power.
IN THIS TOPIC
- Describe the engine cycle and calculate efficiency from W, Q_H and Q_C, and its theoretical ceiling from the kelvin temperatures.
- Explain why no heat engine can be 100% efficient, and why the ceiling rises with a hotter source or colder sink.
- Audit a real engine with input, indicated, brake and friction power.
- Treat refrigerators and heat pumps as reversed engines and calculate both coefficients of performance.
WHAT YOU PROBABLY THINK
A perfect engine would turn all the fuel's heat into useful work.
The engine cycle
Every heat engine, petrol, diesel, steam turbine, is the same machine in outline. A working substance, usually a gas, is carried round a repeating cycle. Each cycle it absorbs heat QH from a hot source, converts part of that heat to work W, and must dump the remainder QC into a cold sink, the surroundings. Energy conservation fixes W = QH − QC, and the fraction of the input that became work is the efficiency:
Why must QC exist at all? Because the working gas has to be returned to its starting state to go round again, and resetting it means compressing it, which demands rejecting heat somewhere colder. An engine that dumped nothing would need its exhaust as hot as its source, and no net work would be left. That is the second-law heart of the topic, and it caps the efficiency below one for any engine whatever. The cap has a clean form. The best possible cycle running between source temperature TH and sink temperature TC achieves
with both temperatures in kelvin. The formula is a ratio of absolute temperatures, so Celsius numbers wreck it; only the kelvin scale starts at the true zero the ratio needs. Real engines fall short of even this ceiling, through friction, turbulence and heat leaking where it should not.
WORKED EXAMPLE
A power station's ceiling
A turbine takes steam at 850 K and rejects heat to cooling towers at 300 K. Find its maximum theoretical efficiency, and comment on its measured efficiency of 40%.
Maximum efficiency = (TH − TC)/TH = (850 − 300)/850 = 0.65.
The real 40% sits well below the 65% ceiling, and the gap is friction, imperfect insulation and the compromises of running fast. Engineers chase the ceiling from both ends, hotter steam and colder cooling water, because both raise the theoretical limit itself.
Auditing a real engine
On the p-V diagram a full cycle is a closed loop, traversed clockwise for an engine. Expansion happens along the high-pressure top, compression along the low-pressure bottom, so the work out exceeds the work back in, and the difference, the net work per cycle, is the area enclosed by the loop.
That area feeds a chain of powers, and AQA asks for the whole audit. The input power is chemical energy arriving as fuel, the fuel's calorific value (joules per kilogram) times the rate it is burned. The indicated power is what the gas develops inside the cylinders, the loop area per cycle times the cycles per second, times the number of cylinders. The brake power is what actually reaches the output shaft, measured as P = Tω. What friction and moving parts eat between cylinder and shaft is the friction power:
WORKED EXAMPLE
The full audit
A four-cylinder engine has a p-V loop of area 380 J per cylinder and runs at 25 cycles per second. Its crankshaft turns at 314 rad s−1 delivering a torque of 102 N m, burning fuel of calorific value 45 MJ kg−1 at 2.0 × 10−3 kg s−1. Audit it.
Input power = 45 × 106 × 2.0 × 10−3 = 90 kW.
Indicated power = 380 × 25 × 4 = 38 kW. The other 52 kW left as exhaust heat and cooling, the QC of the cycle.
Brake power = Tω = 102 × 314 = 32 kW, so friction power = 38 − 32 = 6 kW.
Overall efficiency = 32/90 = 0.36. A modern engine wastes most of its fuel's energy, and this audit shows precisely where each share went.
Running it backwards
Drive the cycle anticlockwise, feeding work in, and heat is pumped the wrong way, from cold to hot. One machine, two names, chosen by which end you care about. A refrigerator wants the heat QC extracted from the cold space. A heat pump wants the heat QH = QC + W delivered to the hot space, a house in winter. Each is scored by a coefficient of performance, the benefit over the work paid:
Unlike efficiency, a COP happily exceeds one, and no law is broken. The work does not become most of the delivered heat; it drives the transfer of heat that was already there, outdoors. A domestic heat pump with a COP of 4 delivers four joules of heat per joule of electricity, against exactly one for the best possible electric fire, and both COPs climb as the temperature gap being pumped across narrows.
YOUR TURN
Heating a house
A heat pump delivers 6.0 kW of heat into a house using 1.5 kW of electrical power. Find its coefficient of performance and the rate it extracts heat from the winter air outside.
Show the working
COPhp = QH/W = 6.0/1.5 = 4.0.
QC = QH − W = 6.0 − 1.5 = 4.5 kW drawn from the cold outdoors. The house gets the outdoor heat plus the electrical work, and a plain electric heater on the same 1.5 kW would deliver only a quarter as much.
TRY IT UNSEEN
Scoring a refrigerator
In each cycle a refrigerator extracts 120 J from its cold compartment and rejects 168 J into the kitchen. Find the work input per cycle and the coefficient of performance.
Show the working
W = QH − QC = 168 − 120 = 48 J per cycle.
COPref = QC/W = 120/48 = 2.5. Note the kitchen receives more heat than the food lost; a running fridge with its door open warms the room.
THE EXAM BIT
- Kelvin, both temperatures, every time. The maximum efficiency formula is a ratio of absolute temperatures, and one Celsius value voids the whole calculation.
- Why below 100%? Answer with the cycle. The gas must be recompressed to repeat, so heat QC must be rejected to a colder sink, so W is always less than QH. Friction is a second, separate reason, so name both when asked about a real engine.
- Keep the power ladder in order, input, indicated, brake, and check each rung is smaller. Friction power is the indicated-minus-brake gap, and an answer with brake above indicated has slipped somewhere.
- Loop questions: net work per cycle is the enclosed area, clockwise for an engine, and indicated power multiplies it by cycles per second and by cylinders. Forgetting the cylinder count is the standard slip.
- COP definitions are benefit over work. QC/W for a refrigerator, QH/W for a heat pump, and a COP above 1 is expected, not an error, because the machine moves heat rather than creating it.
CHECK YOURSELF
In each cycle an engine absorbs 2000 J from a source at 750 K and rejects 1440 J to a sink at 310 K. Find the work per cycle, the efficiency, and the maximum theoretical efficiency, and comment on the comparison.
Show a hint
W from conservation, then two efficiency ratios, one from heats, one from kelvin temperatures.
Show the answer
W = QH − QC = 2000 − 1440 = 560 J per cycle.
Efficiency = W/QH = 560/2000 = 0.28.
Maximum = (750 − 310)/750 = 0.59. The engine achieves under half its theoretical ceiling, which is typical; the ceiling assumes a perfect, friction-free, ideally slow cycle no real engine can run.
An engine taps the flow from hot to cold: W = Q_H − Q_C, and rejecting Q_C is compulsory.
The ceiling is (T_H − T_C)/T_H, in kelvin, and only a colder sink or hotter source raises it.
Power audit: input from fuel, indicated in the cylinders, brake at the shaft, friction the gap.
Reversed, the cycle pumps heat uphill, and its COP is the benefit over the work paid.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Describe the engine cycle and calculate efficiency from W, Q_H and Q_C, and its theoretical ceiling from the kelvin temperatures.
- Explain why no heat engine can be 100% efficient, and why the ceiling rises with a hotter source or colder sink.
- Audit a real engine with input, indicated, brake and friction power.
- Treat refrigerators and heat pumps as reversed engines and calculate both coefficients of performance.
Open the full revision checklist to track your progress across the whole unit.