Carnot efficiency calculator
The best a heat engine can do between two temperatures, and how a real one compares.
What is the Carnot efficiency formula?
The most efficient heat engine that can run between a hot reservoir at Th and a cold reservoir at Tc has the efficiency e = 1 − Tc ÷ Th, with both temperatures in kelvins (OpenStax University Physics, Volume 2, section 4.5). Degrees Celsius become kelvins by adding 273.15, and degrees Fahrenheit by (°F − 32) × 5/9 + 273.15; the calculator does this for you. OpenStax asks about a Carnot engine working between 400°C and 30°C: e = 1 − 303.15 ÷ 673.15 = 55%. The efficiency grows with a hotter source or a colder sink, and it reaches 100% only if the sink were at absolute zero.
How do I find the real efficiency, the work and the heat rejected?
A real engine takes in the heat Qh, turns part of it into work W and rejects the rest, Qc, to the cold reservoir: W = Qh − Qc and efficiency e = W ÷ Qh = 1 − Qc ÷ Qh (section 4.2). Give the calculator any two of Qh, Qc, W and the efficiency and it finds the others. OpenStax Example 4.1: a lawn mower rated 25.0% efficient and 3.00 kW does 180 kJ of work in a minute, so it must take in 720 kJ and reject at least 540 kJ. Energies and powers both work, as long as every number uses the same unit.
How can I tell whether my engine breaks the second law?
Compare it with the Carnot limit for the same two temperatures. OpenStax states that all real engines between two reservoirs are less efficient than a reversible one (Carnot’s principle), so the real efficiency must be 100% or less of the Carnot value; that ratio is the second-law efficiency shown in the table. If the calculator says “impossible”, the usual causes are temperatures typed in the wrong unit (the formula needs the absolute scale, which the unit menu handles) or a mix-up between heat in and heat out.
How do I calculate the COP of a refrigerator or a heat pump?
A refrigerator’s coefficient of performance is the heat removed from the cold side per unit of work, K_R = Qc ÷ W; a heat pump’s is the heat delivered per unit of work, K_P = Qh ÷ W = K_R + 1 (section 4.3). The ideal Carnot values are K_R = Tc ÷ (Th − Tc) and K_P = Th ÷ (Th − Tc) (section 4.5). OpenStax Example 4.3: a Carnot heat pump between 0°C outside and 20.0°C inside needs only about 2 kJ of work (2.05 kJ with 273.15 K) to deliver 30.0 kJ. A Carnot refrigerator between 0°C and 25°C has K_R = 10.9. Real air conditioners quote the same idea as EER = COP × 3.412; the SEER calculator converts between them.
Why does a real machine do worse than the Carnot figure?
The Carnot cycle is a reversible idealization with no friction, no heat leaking through the walls and no temperature drop between the working substance and the reservoirs. Real engines and heat pumps have all of those, so they reach only a fraction of the Carnot limit. The limit still tells you which way to push: the Carnot COP of a heat pump or a cooler is higher the smaller the temperature difference between the two sides, and the Carnot efficiency of an engine is higher the larger it is. This page gives the upper bound, not a prediction for a particular machine.
Sources: OpenStax University Physics, Volume 2, sections 4.2 (heat engines, Example 4.1), 4.3 (refrigerators and heat pumps) and 4.5 (the Carnot cycle, Examples 4.2 and 4.3, and the problems); EER = COP × 3.412 as on our SEER calculator. Related: SEER calculator (COP and EER of an air conditioner), blackbody radiation calculator and heat conduction calculator.