Refrigeration and heat pump calculator
Calculate the vapour-compression cycle of a refrigeration plant or a heat pump: from the four enthalpies you read off the log p-h chart of your refrigerant you get mass flow, refrigerating capacity, compressor power and condenser duty. In both operating modes the calculator gives the COP for refrigeration and for heat pump operation, compares them with the Carnot cycle between the same temperature levels and forms the second-law efficiency from it, live with every input.
Calculation
Refrigeration: the benefit is the capacity of the evaporator.
The calculation uses the vapour-compression cycle built from four enthalpies that you read off the log p-h chart of your refrigerant; there is therefore no selection field for the refrigerant and the calculator is tied to no property database. Both coefficients of performance are shown in either mode, because the same plant delivers both benefits. The second-law efficiency is classified against empirical values: real vapour-compression plants reach about 0.4 to 0.6 of the Carnot value, above that you are in the idealised range. This is a classification, not a verification against a standard.
Model: steady vapour-compression cycle from four enthalpies read off the chart, isenthalpic expansion, piping outside the suction line assumed free of heat pickup and pressure drop. The output is the internal compressor power, not the coupling or motor power; drive and fan power are not contained in the coefficients of performance. The curve over the evaporating temperature assumes the second-law efficiency of the operating point to be constant. A design and comparison tool, not a replacement for the compressor manufacturer's performance data.
Results
Calculating …
Calculation in your browser, inputs go to our server only when you export or save.
Formulas and fundamentals
The four states in the log p-h chart
The vapour-compression cycle is not calculated from a property database but from four enthalpies read off the chart. In flow direction these are: h₁ at the evaporator outlet (suction gas, possibly superheated), h₁* at the compressor inlet, h₂ at the compressor outlet (hot gas) and h₃ for the liquid ahead of the expansion valve. If the suction gas is not heated in the suction line, h₁* = h₁. The expansion is isenthalpic, so h₃ is at the same time the enthalpy at the evaporator inlet and no fifth state is needed:
Specific quantities from the enthalpies
Every component is balanced across the enthalpy difference at its own ports. This is the technical core: the refrigerating capacity arises only in the evaporator, the compressor work only downstream of the suction line, and the condenser takes the hot gas all the way back to the liquid enthalpy, so it covers desuperheating, condensation and subcooling:
Mass flow and duties
The specific refrigerating capacity q₀ links the size of the plant to the refrigerant flow. If the refrigerating capacity follows from the cooling load, the mass flow is the result; if the compressor is already fixed, the mass flow is given and the refrigerating capacity is the result. The suction volume flow follows from the specific volume v₁* at the compressor inlet and governs the compressor frame size:
Coefficients of performance and energy balance
The coefficient of performance relates benefit to effort. For a refrigeration plant the benefit is the capacity of the evaporator, for a heat pump the heating duty of the condenser; the effort is the compressor power in both cases:
If the suction gas picks up additional heat between evaporator outlet and compressor inlet, that heat ends up in the condenser without loading the compressor. The balance Q̇_c = Q̇₀ + P then no longer holds, and ε_W exceeds ε_K by more than one:
Carnot comparison and second-law efficiency
The Carnot cycle between the same two temperature levels is the upper bound for any plant. Evaporating temperature T₀ and condensing temperature T_c have to be entered in kelvin. The second-law efficiency η_G places the achieved COP within that frame:
For the Carnot cycle ε_C,W = ε_C,K + 1 holds exactly, because no heat enters at a suction line there.
Worked example
A vapour-compression plant with R134a is to provide 8 kW of refrigerating capacity. The evaporating temperature is t₀ = −10 °C, the condensing temperature t_c = +25 °C. From the log p-h chart one reads: h₃ = 235 kJ/kg for the saturated liquid ahead of the expansion valve, h₁ = 392 kJ/kg for the dry saturated vapour at the evaporator outlet and h₂ = 417 kJ/kg for the hot gas after isentropic compression, plus v₁ = 0.100 m³/kg at the compressor inlet.
This gives q₀ = 392 − 235 = 157 kJ/kg and a refrigerant flow of ṁ = 8/157 = 0.051 kg/s. The compressor work is w = 417 − 392 = 25 kJ/kg, so P = 1.27 kW, and the condenser rejects q_c = 417 − 235 = 182 kJ/kg, so Q̇_c = 9.3 kW. The compressor draws V̇ = 0.051 · 0.100 = 0.0051 m³/s, that is 18.3 m³/h. The coefficients of performance are ε_K = 157/25 = 6.3 and ε_W = 182/25 = 7.3.
Carnot between 263.15 K and 298.15 K gives ε_C,K = 263.15/35 = 7.52. The second-law efficiency of the example is therefore η_G = 6.3/7.52 = 0.84. That is above the empirical range of 0.4 to 0.6 for real plants, and rightly so: the calculation used the isentropic compression taken from the chart, that is without compressor losses, without pressure drops and without heat pickup in the suction line. Replacing h₂ = 417 kJ/kg with the actual discharge enthalpy of a compressor with an internal efficiency of about 66 %, that is h₂ = 430 kJ/kg, lowers ε_K to 4.13 and the second-law efficiency to 0.55. This is the default setting of the calculator.
Extending the same case by 5 K of subcooling and 10 K of suction superheat, with the suction gas heated further to +15 °C in the suction line, gives h₃ = 228, h₁ = 401, h₁* = 414 and h₂ = 443 kJ/kg. The specific refrigerating capacity rises to 173 kJ/kg, the mass flow drops to 0.046 kg/s. The compressor work is w = 443 − 414 = 29 kJ/kg and not 443 − 401 = 42 kJ/kg; the wrong difference would yield P = 1.94 kW instead of the correct 1.34 kW. The suction line picks up Q̇_Saug = 0.60 kW, so the condenser rejects 9.94 kW: 8 + 1.34 + 0.60. Source of the figures: Langeheinecke/Jany/Thieleke, Thermodynamik für Ingenieure, 6th edition, example 9.2, pp. 200 f.
Frequently asked questions
Why is there no selection field for the refrigerant?
Because the calculator deliberately works without a property database. In practice the design goes through the log p-h chart or the vapour table of the refrigerant in question: you enter the two pressures, draw the cycle and read off the four enthalpies. Exactly these four values are what the calculator takes. It is therefore equally valid for every refrigerant, for R134a as much as for R290, R744 or ammonia, and it does not claim property data it could not substantiate. The price is one working step at the chart, which you take anyway if you want to understand the cycle.
Why is the heat pump COP not always exactly one above the refrigeration COP?
Because the relation ε_W = ε_K + 1 only holds as long as all the heat rejected in the condenser comes from the evaporator and the compressor: Q̇_c = Q̇₀ + P, divided by P, gives ε_W = ε_K + 1. If the suction gas is heated in the suction line between evaporator outlet and compressor inlet, a third heat flow is added. That heat leaves through the condenser again without the compressor having to do work for it, and the balance reads Q̇_c = Q̇₀ + P + Q̇_Saug. In the example with suction superheat ε_K = 5.97 while ε_W is 7.41 and not 6.97. The calculator therefore always forms ε_W from the actual duties and never as ε_K + 1.
What second-law efficiency do real plants reach?
As an empirical value, executed vapour-compression plants reach about 0.4 to 0.6 of the Carnot COP, measured between evaporating and condensing temperature. Small plants and those with a large pressure ratio tend to sit at the lower end, large and well controlled plants at the upper end. Values above 0.6 as a rule mean that the idealised cycle was calculated, that is with isentropic compression and without losses; the ideal vapour-compression cycle itself reaches roughly 0.8 to 0.85 and stays below Carnot because of the throttling and the superheat alone. Clearly below 0.4 points to a poor compressor efficiency, excessive temperature differences in the heat exchangers or pressure drops. These are empirical values for classification, not a standard and not a verification.
Which enthalpy belongs at the compressor outlet: the isentropic or the real one?
That depends on what you want to know. Enter the isentropic enthalpy h₂s, that is the point on the isentrope through the compressor inlet, and you calculate the ideal reference cycle and get the smallest possible compressor power. For a real machine enter the actual discharge enthalpy: h₂ = h₁* + (h₂s − h₁*)/η_i with the internal efficiency η_i, which for refrigerant compressors is about 0.6 to 0.8 depending on design and pressure ratio. Only then is the reported second-law efficiency comparable with the empirical range of real plants.
What do subcooling of the liquid and superheating of the suction gas achieve?
Subcooling lowers h₃ and thus directly increases the specific refrigerating capacity q₀ = h₁ − h₃ without changing the compressor work. It is therefore almost always a gain and only costs area in the condenser. Suction superheat raises h₁ and increases q₀ as well, but at the same time increases the specific volume at the compressor inlet and hence the required suction volume flow, and the discharge temperature rises. A few kelvin of superheat are necessary anyway to protect the compressor against liquid slugging. In the example q₀ rises from 157 to 173 kJ/kg and the mass flow drops from 0.051 to 0.046 kg/s.
Why does the COP fall so sharply when the evaporating temperature drops?
Because the COP essentially lives off the temperature difference the cycle has to bridge. Carnot gives ε_C,K = T₀/(T_c − T₀): the numerator gets smaller and the denominator larger as soon as T₀ falls, so the COP drops twice over. Going from t₀ = −10 °C to −25 °C at t_c = +25 °C, ε_C,K falls from 7.52 to 4.97, a loss of about one third. In practice this means that every kelvin the evaporator is allowed to run warmer, and every kelvin the condenser runs cooler, pays off directly in the electrical demand. The chart in the calculator shows this relationship for your operating point.
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