Heat-engine efficiency
Define and apply thermal efficiency and COP, with clear sign conventions and physical bounds.
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ContinueA performance ratio is meaningful only when its useful output, energy input and system boundary are clear. A heat engine converts part of its absorbed heat into work; a heat pump uses work to transfer heat. Their numerators describe different benefits, so thermal efficiency and coefficient of performance are different measures.
Use the cycle energy balance. Here Qₕ,Qc and work are positive magnitudes, all per cycle or all rates over the same steady interval. The working device returns to its starting energy after a complete cycle. A startup interval with accumulating energy needs a storage term.
Define engine efficiency from net cycle work
For an engine, Qₕ is absorbed from the hot source, Qc is rejected, and net cycle work is Wₒᵤₜ = Qₕ-Qc. Thus η = Wₒᵤₜ/Qₕ = 1-Qc/Qₕ. Thermal efficiency η is also written as e in the short examples and checks below. “Net cycle work” already subtracts mechanical work the working substance needs on compression steps; it is not the sum of expansion work alone. A plant can also consume external auxiliary power in pumps, fans or controls. State whether the reported output is gross device output or useful output after those loads.
Rates obey the same steady balance, W dot ₒᵤₜ = Q dot ₕ-Q dot c, when storage is negligible. Do not divide energy for one interval by power or heat for another. A percentage efficiency is 100 times the fractional ratio. Being below 100% is necessary for the stated heat-only engine, but the Carnot bound can be much stricter.
Choose the heat-pump benefit
For a pump or refrigerator, supplied work gives Qₕ = Qc + Wᵢₙ. Heating benefits from Qₕ delivered to the hot space; cooling benefits from Qc extracted from the cold space: Kₕ = Qₕ/Wᵢₙ, Kc = Qc/Wᵢₙ, Kₕ-Kc = 1. COP may exceed one because delivered or removed heat is not all created from the work input. The difference identity applies to the same cycle and boundary with every energy channel counted. If the reported useful heating excludes some delivered heat, or the input includes auxiliaries whose heat leaves elsewhere, reconstruct that product boundary before using the identity.
For two fixed reservoirs, the reversible COP bounds are K_(h,C) = Tₕ/(Tₕ-Tc) and K_(c,C) = Tc/(Tₕ-Tc), using kelvin. A first-law COP is not automatically feasible just because it is positive.
Short worked examples
Engine efficiency and output
Worked example 1
Engine efficiency
Problem
Show full solution
Find the fraction converted to work
Method
Use e = 1-Qc/Qₕ.Reason
The rejected fraction cannot contribute to work output; cycle energy storage is zero.Working
Answer
e = 1-Qc/Qₕ = 1-650/1000 = 0.35.State efficiency and work
Working
Answer
The efficiency is 35%, and Wₒᵤₜ = eQₕ = 0.35(1000) = 350 J. The output plus rejected heat equals the 1000 J input.
Heat-pump coefficients of performance
Worked example 2
Heating and cooling COP
Problem
Show full solution
Choose the heating benefit
Method
Divide delivered heat Qₕ by work input.Reason
In heating mode the useful output is heat delivered to the hot side, not heat extracted from the cold side.Working
Answer
Kₕ = Qₕ/Wᵢₙ = 4.8/1.2 = 4.0.Find the cooling COP
Working
Answer
Kₕ = 4.0 and Kc = Kₕ-1 = 3.0. Equivalently, Qc = 4.8-1.2 = 3.6 kJ and Kc = 3.6/1.2 = 3.0. Both ratios are dimensionless and exceed one because the pump moves heat as well as receiving work.
Account for auxiliary loads and recovered heat
Guided practice 3
Separate work efficiency from combined heat use
Problem
In steady operation, a heat-engine device absorbs 20 kW of heat and rejects 12 kW, with no energy storage. Its gross work output supplies a separate 2.0 kW auxiliary load before useful export. An installation also supplies 5.0 kW of useful recovered heat from the rejected heat. Find useful exported work power, gross work efficiency and useful exported work efficiency. Would counting the recovered heat describe the same efficiency? Enter efficiencies as percentages.
Work through the model, then compare your reasoning
Hints
Hint 1
Check the model
Keep the working-device heat balance separate from the larger installation’s useful outputs.Show solution step by step
Close the working-device balance
Method
Use W dot _gross = Q dot ₕ-Q dot c.Reason
The device has no steady storage and its two heat rates determine its cycle work output.Working
Answer
Gross work output is 20-12 = 8.0 kW, giving η_gross = 8/20 = 0.40 = 40%. After the 2.0 kW auxiliary demand, useful work export is 6.0 kW.
Name the useful-output measure
Method
Divide useful work by the same absorbed heat rate.Reason
Adding useful recovered heat changes the numerator and the question being answered.Working
Answer
ηₑₓₚₒᵣₜ = 6/20 = 0.30 = 30%. A combined useful-energy ratio would be (6 + 5)/20 = 0.55 = 55% under the supplied boundary, but it is not the work conversion efficiency. The recovered 5 kW is part of the rejected heat, not extra work. Reservoir data would still be needed to test the second-law feasibility of the stated heat engine.
Combine cycles with different heat inputs
Check your understanding 1
An engine runs one cycle absorbing 100 J and rejecting 50 J, then one cycle absorbing 900 J and rejecting 720 J. Each cycle returns the device to its starting energy, and there are no other transfers. Find the work output and thermal efficiency over both cycles. A colleague averages the individual efficiencies: is that the correct aggregate? What further information would be needed for a second-law check?
Show answer
The outputs are 50 J and 180 J, giving 230 J of work from 1000 J of absorbed heat. Aggregate efficiency is 230/1000 = 0.23 = 23%. The arithmetic mean of 50% and 20% is 35%, which weights unequal heat inputs equally. The correct ratio is heat-input weighted: (100 × 0.50 + 900 × 0.20)/(100 + 900). The energy balance alone cannot prove feasibility; the relevant source and sink temperatures and transfer conditions are still needed.
Keep COPs on the same installation boundary
Check your understanding 2
A cyclic heat-pump device delivers 4.8 kJ indoors using 1.2 kJ of work. A separate control unit uses another 0.30 kJ of electrical work during the same interval and rejects all of that energy as heat to the same outdoor reservoir from which the pump extracts heat. The devices have no net energy storage and no other transfers. Define the whole installation’s cooling benefit as net heat removed from that outdoor reservoir. Find the installation’s heating COP, net outdoor heat removal and cooling COP. Compare these with the working heat-pump device’s COPs, and explain which pair obeys the difference identity.
Show answer
The working pump extracts Qc = 4.8-1.2 = 3.6 kJ, so its own COPs are Kₕ = 4 and Kc = 3. For the whole installation, total work input is 1.2 + 0.30 = 1.50 kJ, while indoor heat remains 4.8 kJ. Thus K_(h,installation) = 4.8/1.50 = 3.2.
Net outdoor heat removal is 3.6-0.30 = 3.3 kJ, giving K_(c,installation) = 3.3/1.50 = 2.2. The complete installation balance is 4.8 = 3.3 + 1.5 kJ, so the installation COP difference is 3.2-2.2 = 1. The working pump’s pair also differs by one, 4-3 = 1.
Using gross extraction with total installation work would give 3.6/1.5 = 2.4, which differs from 3.2 by only 0.8 because it omits the outdoor control heat. That would mix useful-effect definitions. Check that your explanation identifies the same boundary, interval and complete heat/work account for each COP pair; these energy ratios alone do not prove second-law feasibility.
Practice set
Check your understanding 3
An engine absorbs 1500 J and rejects 900 J per cycle. Find its efficiency.
Show hint
Find the fraction of input left after heat rejection.
Show answer
e = 1-900/1500 = 0.40 = 40%. Its work output is the remaining 600 J per cycle.
Check your understanding 4
A refrigerator extracts 1800 J using 600 J of work per cycle. Find its cooling COP and heat rejected per cycle.
Show hint
Show answer
Check your understanding 5
A heat pump has heating COP Kₕ = 3.6. Find the cooling COP for the same cycle.
Show hint
Show answer
Check your understanding 6
Can a cyclic heat engine, with heat as its only energy input, have e = 1.05? Explain.
Show hint
Show answer
What to keep
State the useful effect and boundary before forming a performance ratio. Close the cycle or steady-rate balance, distinguish gross and exported work, and weight combined performance by energy inputs. COP above one is compatible with conservation, but every ratio still needs the applicable second-law check.
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Syllabus and review details
No official syllabus alignment is listed for this lesson.