Audited 31 Jul 2026·Last updated 15 Sept 2026·3 citations·Tier 2·0 uses

Thermal Efficiency Calculator

Thermal efficiency calculator: η = useful output ÷ heat input × 100%. Compare engines and power cycles against typical values and the Carnot ceiling.

Thermal Efficiency Calculator

J
J
Thermal efficiency
35
Result of η = W_useful / Q_in × 100% using the entered coherent-SI magnitudes.
Model scope
Boundary-consistent reported ratio only; it does not infer losses or a thermodynamic cycle, and a result above 100% signals inconsistent inputs rather than a physical heat engine.

Background.

Thermal efficiency answers the bluntest question you can ask of an engine, boiler, or power cycle: of the heat you paid for, what fraction came back as the output you wanted? The definition is a plain ratio, η = W_useful / Q_in, and this page turns two measured energies into that percentage.

The measurements must share one boundary and one time interval — that discipline is where real efficiency figures are won or lost. Count the fuel's full heating value in but only shaft work out, and a petrol engine lands near 30%; quietly switch the output to “heat delivered to the cabin plus shaft work” and the number inflates past any fair comparison. The calculator treats the inputs as your declared boundary and does the arithmetic; the declaring is your job, and the model-scope note beside the result says so.

Typical magnitudes are worth memorising as sanity anchors: spark-ignition car engines 25–35%, modern diesels around 40%, simple-cycle gas turbines 35–40%, combined-cycle plants up to about 64%, home gas boilers 85–95% (a boiler's output is heat, so it escapes the work-conversion penalty entirely). A figure that beats these classes by a wide margin usually signals a boundary error, not a breakthrough.

The deep ceiling is Carnot's: no heat engine operating between a hot source at T_h and a cold sink at T_c can exceed η = 1 − T_c/T_h. A gas turbine breathing 1,500 K combustion gas against a 300 K ambient is capped at 80% before a single irreversibility is counted — real machines then lose their share to friction, finite-rate heat transfer, and exhaust enthalpy.

What is thermal efficiency calculator?

Thermal efficiency is the dimensionless ratio of useful energy delivered to heat energy supplied, over the same boundary and the same interval, expressed here as a percentage. For a heat engine the useful term is net work; for a furnace or boiler it is delivered heat. It is a first-law bookkeeping figure — it tells you what fraction of the input survived as intended output, and by subtraction how much left as waste heat, but it does not by itself say where those losses occur.

How to use this calculator.

  1. Fix the boundary first: decide what counts as the machine, over which interval you are accounting, and stick to it for both numbers.
  2. Enter the useful energy out — shaft work for an engine, delivered heat for a boiler — in joules (kWh work too, as long as both inputs use the same unit).
  3. Enter the heat input over the same interval; for fuel, that is mass burned times heating value, and stating which heating value (higher or lower) matters at the few-percent level.
  4. Read the percentage and place it against the typical range for the machine's class — 30% is respectable for a petrol engine and disastrous for a boiler.
  5. If the result tops 100%, do not celebrate: some input energy crossed your boundary uncounted, or the output includes energy that was never part of Q_in.

The formula.

η = W_useful / Q_in × 100%

The first law guarantees the energy budget balances: Q_in = W_useful + Q_rejected. Efficiency takes the ratio of the term you want to the term you paid for, η = W_useful/Q_in, and multiplying by 100 states it as a percentage — the calculator's only two operations, done in Decimal arithmetic with a single rounding at the end. The second law then bounds what the ratio can be for any cyclic heat engine: η ≤ 1 − T_c/T_h (temperatures in kelvin), the Carnot limit, reached only by a reversible cycle operating infinitely slowly. The gap between a machine's measured η and its Carnot ceiling is the engineer's loss budget: exhaust enthalpy, incomplete combustion, friction, and heat leaking around the working fluid. Note what the ratio deliberately excludes — it says nothing about power (a slow engine can be efficient and useless), emissions, or cost per joule.

A worked example.

Example

A bench test burns fuel releasing 1,000 J of heat into a small engine, and the dynamometer logs 350 J of shaft work over the same run. The efficiency is the ratio of what you got to what you paid: η = 350 / 1,000 = 0.35, or 35%. The unglamorous remainder, 1,000 − 350 = 650 J, left through the exhaust and the cooling system — the first law insists it went somewhere. Is 35% credible? It sits exactly in the 25–35% band of a decent spark-ignition engine, so yes. And the second law shows the headroom: if this engine's combustion gas peaks near 900 K against a 300 K ambient, Carnot allows at most 1 − 300/900 = 66.7%. The 31.7-point gap between ceiling and measurement is not failure — it is the ordinary price of finite-speed, irreversible machinery.

useful Energy J350
heat Input J1,000

Frequently asked questions.

Can thermal efficiency exceed 100%?
For a heat engine, never — that would create energy. A reading above 100% means the accounting boundary leaked: some input (a second fuel stream, electrical preheating, ambient heat) was not counted in Q_in, or the “useful” figure includes recycled energy. The one legitimate-looking exception is a heat pump's COP, which is a different ratio measured against work, not heat input.
Is 35% a good thermal efficiency?
Entirely depends on the machine class. It is solid for a petrol car engine (25–35% typical), mediocre for a large marine diesel (up to ≈50%), poor for a combined-cycle power plant (≈60%+), and catastrophic for a condensing gas boiler, which should deliver 90%+ because its product is heat rather than work.
What is the Carnot limit and why does it matter here?
It is the second-law ceiling for any cyclic heat engine between a hot source T_h and cold sink T_c: η_max = 1 − T_c/T_h in kelvin. It converts your measured efficiency into a meaningful grade — 350 J from 1,000 J looks modest until you know the temperature pair only permits 66.7%, at which point the machine is running at over half its theoretical best.
Do my two inputs have to be in joules?
They must share a unit; which unit is irrelevant because the ratio is dimensionless. kWh over kWh, BTU over BTU, and J over J all give the same percentage. The classic mistake is mixing — fuel energy in kWh against work in MJ — which silently scales the answer by 3.6.
Which heating value of the fuel should I count as heat input?
State it and stay consistent. Using the lower heating value (LHV) flatters efficiency by a few percent relative to the higher heating value (HHV), because LHV writes off the latent heat in exhaust water vapour. Condensing boilers advertised near 98% are HHV figures made possible by recovering exactly that latent heat; the same hardware quoted on LHV can exceed 100%, which is a definitional artefact, not free energy.

References& sources.

  1. [1]OpenStax, University Physics Volume 2, section 4.5, The Second Law of Thermodynamics.
  2. [2]Çengel and Boles, Thermodynamics: An Engineering Approach, 10th ed. (PRINT).
  3. [3]BIPM, The International System of Units (SI Brochure), 9th ed., version 3.01, coherent derived units and quantity equations.

How this page was produced

Published by
Quanta Calculator
Primary sources
3 cited below
Method
η = W_useful / Q_in × 100%
Published
Last verified

Built with AI assistance and verified by automated tests against the cited sources — every worked example on this page is computed by the same code that runs the calculator. How we build and check calculators.

In this category

Embed

Quanta Pro

Paid features are coming later.

  • All 1560 calculators remain free
  • No billing is enabled
Coming soon