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Ideal Refrigeration Cycle COP Calculator

The best COP a vapor-compression cycle can reach between two temperatures, with real refrigerant properties — a ceiling to compare real equipment against.

Method reviewed October 26, 2026. Version 1.0.0. How it works

Your inputs

Unit system

Only refrigerants whose properties agree with independent published tables are offered. R-454B is not offered yet.

The saturation (dew-point) temperature at the suction pressure. Get it from a gauge reading with the refrigerant PT calculator.

The saturation (dew-point) temperature at the discharge pressure.

Result

Ideal cycle COP (cooling)
6.52
Ideal compressor power per ton
0.539kW/ton
Isentropic discharge temperature
125.5°F

Ideal COP 6.52 (0.539 kW/ton) for R-410A at 40 °F / 100 °F.

  • Real systems reach well below this ideal COP: compressors are not isentropic, and pressure drops, motor losses and fans all add power.
Refrigeration effect
69.06BTU/lb
Compressor work (isentropic)
10.59BTU/lb
Heat rejected per pound
79.65BTU/lb
Equivalent EER (3.412 × COP)
22.3
Mass flow per ton
2.896lb/min per ton
Pressure ratio (absolute)
2.5
Carnot COP between the saturation temperatures
8.33
Show the working
  1. 1 → 2: saturated vapour compressed at constant entropy to the condensing pressure
    h₁ = 181.7 BTU/lb, work = 10.587 BTU/lb, T₂ = 125.5 °F
  2. 3: liquid leaves the condenser
    99.78 °F, h₃ = 112.6 BTU/lb
  3. 3 → 4: expansion at constant enthalpy; 4 → 1: evaporation
    refrigeration effect = 181.7 − 112.6 = 69.06 BTU/lb
  4. COP = refrigeration effect ÷ work
    69.06 ÷ 10.587 = 6.523
  5. Carnot COP = T_evap ÷ (T_cond − T_evap), absolute
    499.7 ÷ 60 = 8.33

The ideal cycle

Saturated vapour leaves the evaporator and is compressed at constant entropy to the condensing pressure. It condenses, leaves as liquid (with any subcooling you enter), and expands through the metering device at constant enthalpy back to the evaporator.

COP = refrigeration effect ÷ compressor work. The calculator also shows the isentropic discharge temperature, the heat rejected per pound, the pressure ratio and the Carnot COP between the same saturation temperatures.

Why real systems do worse

Real compressors are not isentropic, motors and drives have losses, lines and coils have pressure drops, and fans use power. Real equipment therefore runs at a fraction of this COP. The ideal figure is still useful for seeing how much raising the evaporating temperature or lowering the condensing temperature is worth.

The property tables were checked against the NIST Chemistry WebBook: for R-134a between 40 °F and 100 °F the hand cycle from NIST data gives a COP of 6.95, the same as this calculator.

Worked example

These are the example values the calculator opens with, and the result the engine calculates for them.

Inputs

Refrigerant
R-410A
Saturated evaporating temperature
40 °F
Saturated condensing temperature
100 °F
Liquid subcooling
0 °F

Result

Ideal cycle COP (cooling)
6.52
Ideal compressor power per ton
0.539 kW/ton
Isentropic discharge temperature
125.5 °F
Refrigeration effect
69.06 BTU/lb
Compressor work (isentropic)
10.59 BTU/lb
Heat rejected per pound
79.65 BTU/lb

Ideal COP 6.52 (0.539 kW/ton) for R-410A at 40 °F / 100 °F.

Assumptions

  • Ideal cycle: saturated vapour enters the compressor, compression is isentropic, the expansion keeps enthalpy constant and there are no pressure drops.
  • Evaporating and condensing temperatures are dew points; the liquid leaves at the bubble point minus the subcooling.
  • Properties generated with CoolProp 8.0.0 and checked against NIST WebBook (R-134a, R-22, R-32) and Arkema’s REFPROP-based R-410A table.

Limitations

  • This is an upper bound for comparison, not a prediction of a real system’s efficiency or ratings.
  • Superheat at the compressor inlet is not modelled.

Frequently asked questions

What is the ideal COP for R-410A at 40 °F / 100 °F?

About 6.5 with saturated liquid leaving the condenser.

Why does subcooling raise the COP?

It increases the refrigeration effect per pound without adding compressor work in this cycle.

Is this the same as SEER or EER?

No. Ratings are measured on real equipment at test conditions. The ideal EER shown here (3.412 × COP) is an upper bound, not a rating.

Sources

  1. CoolProp — open-source thermophysical property library (version 8.0.0) — CoolProp developers (MIT licence). Equations of state for pure fluids and mixtures, incompressible fluids and brines, and IAPWS-IF97 water/steam; free for commercial use under its licence. Used here to generate the refrigerant saturation tables.
  2. Ideal-cycle property tables: generator and validation (packages/calc/reference/b7-g4/generate_g4_data.py) — FreeHVACCalculators.com. CoolProp 8.0.0 tables of liquid and vapor enthalpy, isentropic work and discharge temperature; agree with NIST WebBook (R-134a, R-22, R-32) to 0.001 % and with Arkema/REFPROP R-410A to 0.1 %; interpolation error under 0.15 %.
  3. NIST Chemistry WebBook, SRD 69 — saturation tables and isobars for R-134a, R-22 and R-32 — National Institute of Standards and Technology. Ideal cycle check, 40 °F evaporating / 100 °F condensing: refrigeration effect 63.702, 68.810 and 105.419 Btu/lb and isentropic work 9.169, 9.858 and 15.653 Btu/lb (thermochemical Btu) for R-134a, R-22 and R-32; discharge 106.53, 126.65 and 144.82 °F.
  4. Forane 410A Thermodynamic Properties (Saturation), English units — Arkema (generated with NIST REFPROP 9.0). R-410A saturated enthalpies: liquid 88.97 / vapor 181.80 Btu/lb at 40 °F; liquid 112.82 / vapor 183.13 Btu/lb at 100 °F (used only for comparison).
  5. University Physics Volume 2, §4.3 Refrigerators and Heat Pumps — OpenStax, Rice University (CC BY 4.0). Energy conservation W = Q_h − Q_c; refrigerator coefficient of performance K_R = Q_c ÷ W; heat pump K_P = Q_h ÷ W.

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