Actual efficiency of real heat engines (eta = W_net / Q_in), heat rate, air-standard cycle comparison, and a Carnot benchmark.
η = W_net / Q_in | Heat rate: HR = 3600 / η (kJ/kWh)Efficiency can never reach 100% for a heat engine; work output cannot exceed heat input.
Lower heat rate = better efficiency. A modern combined-cycle plant runs near 6,000-7,000 kJ/kWh.
Air-standard cycles are idealized (reversible, air as ideal gas). Real engines run 15-25 points below these numbers.
| Case | Result |
|---|---|
| W_net = 250 kJ, Q_in = 800 kJ | η = 31.25%, HR = 11,520.00 kJ/kWh, 10,918.85 BTU/kWh |
| Otto cycle, r = 10, γ = 1.4 | η = 60.19% |
| Diesel cycle, r = 18, ρ = 2, γ = 1.4 | η = 63.16% |
| Ideal Rankine, 3 MPa / 350 °C to 10 kPa | η = 33.45% |
| η = 36% | HR = 10,000.00 kJ/kWh, 9,478.17 BTU/kWh |
| Actual 30% vs Carnot 800 K / 300 K | Carnot = 62.50%, engine reaches 48.00% of the limit |
Thermal efficiency is the ratio of net work output to heat input for a heat engine: eta = W_net / Q_in. Example: an engine that produces 250 kJ of work from 800 kJ of heat has eta = 250/800 = 31.25%. No heat engine can reach 100%.
Heat rate is the inverse of efficiency: HR = 3600 / eta, in kJ/kWh. It tells you how much heat energy you must burn per kWh of electricity. An engine at 36% efficiency has a heat rate of 10,000.00 kJ/kWh. Lower heat rate means a more efficient plant.
Multiply kJ/kWh by 0.947817 to get BTU/kWh (1 kWh = 3412.14 BTU). Example: 10,000.00 kJ/kWh = 9,478.17 BTU/kWh. Use Mode 2 of this calculator for the conversion.
The Carnot efficiency is the theoretical maximum for perfectly reversible operation between two temperatures. Real engines suffer irreversibilities - friction, heat transfer across finite temperature differences, incomplete combustion, and heat leaks. Example: a real engine at 30% operating between 800 K and 300 K (Carnot limit 62.50%) reaches only 48.00% of the Carnot limit.
For the air-standard Otto cycle, eta = 1 - r^(1-gamma), where r is the compression ratio and gamma is 1.4 for air. With r = 10, eta = 1 - 10^(-0.4) = 60.19%. Raising the compression ratio raises efficiency, which is why high-performance engines use r = 12 or more.
For the air-standard Diesel cycle, eta = 1 - (1/gamma) * r^(1-gamma) * (rho^gamma - 1)/(rho - 1), where r is the compression ratio and rho the cutoff ratio. With r = 18, rho = 2, gamma = 1.4, eta = 63.16%. Diesels beat Otto engines mainly because they can run much higher compression ratios.
Use the steam enthalpies: eta = ((h3-h4) - (h2-h1)) / (h3-h2), where h3-h4 is turbine work, h2-h1 is pump work, and h3-h2 is heat added. For superheated steam at 3 MPa / 350 C condensing at 10 kPa, the ideal Rankine efficiency is 33.45%. Real steam plants run lower due to turbine and pump irreversibilities.
A typical coal or nuclear steam plant runs 33-40%. Simple-cycle gas turbines reach 35-40%, while combined-cycle gas turbine plants - which stack a gas turbine on top of a steam (Rankine) bottoming cycle - reach 55-60%, the highest of any commercial heat engine.
Thermal efficiency is what a real engine actually achieves (eta = W_net/Q_in, measured on a test stand). Carnot efficiency is the theoretical upper bound for the same reservoir temperatures - a ceiling no heat engine can exceed. This calculator handles actual efficiencies and uses Carnot only as a benchmark; for a dedicated theoretical-limit calculator (in Indonesian), see our Kalkulator Efisiensi Carnot.
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