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How to Use the System Capacity Calculator
Select the refrigerant from the nameplate, then choose the pressure unit and temperature unit. Enter the site elevation if you read gauge pressures.
Type the suction and discharge pressures from the manifold, the suction line temperature near the compressor, the liquid line temperature at the expansion device, and the discharge line temperature.
Enter the compressor electrical power and the discharge temperature for a power balance, and the displacement and speed for the displacement method. Add the expected capacity from the manufacturer data to compare.
Check the capacity in kW and tons, the refrigerating effect and the comparison with the expected value. Add the project details, then use PDF, CSV, Copy or Print.
Estimating refrigeration capacity from field readings
Capacity is the mass of refrigerant moving through the evaporator each second multiplied by the heat each kilogram picks up. Neither number can be read from a gauge, but both can be worked out from the pressures, the line temperatures and the compressor power. The calculator turns your readings into enthalpies from CoolProp 8 data, finds the mass flow by whichever of two methods your inputs allow, and compares the answer with the capacity you expect at those conditions.
The refrigerating effect from pressures and temperatures
The suction pressure gives the evaporating temperature at the dew point, and the line temperature above it gives the superheat. The discharge pressure gives the condensing temperature, and the liquid line temperature gives the subcooling. The refrigerating effect is the suction enthalpy minus the enthalpy of the liquid at the expansion device, which the page takes as saturated liquid at the liquid line temperature. In the first example, 120 psig suction and 360 psig discharge on R-410A give 40.7°F evaporating and 108.9°F condensing. A 54°F suction line and a 100°F liquid line give 431.1 kJ/kg at the compressor inlet and 262.3 kJ/kg ahead of the valve, so each kilogram carries 168.8 kJ of cooling.
Mass flow from the power balance
A suction-cooled compressor puts nearly all of its electrical input into the refrigerant, so the power equals the mass flow times the rise in enthalpy from suction to discharge, less any heat lost to the room. With a discharge temperature, the page finds the discharge enthalpy at the discharge pressure and divides the power by the rise. At 3.5 kW and a discharge enthalpy 43.7 kJ/kg above the suction value, the flow is 0.08 kg/s and the capacity is 13.5 kW, which is 3.84 tons. Pacific Northwest National Laboratory tested virtual mass-flow sensors of this kind and reported errors under 5% RMS in fault-free operation. It also noted that an energy-balance estimate stays accurate when a leaking compressor valve cuts the real flow, and that differences between estimates can flag a fault. The page lets you enter the heat lost to the room as a percentage of input power, and the capacity falls by that same percentage, so a 5% loss takes the example from 13.5 kW to 12.8 kW.
Mass flow from displacement
The other route multiplies the swept volume per second by the volumetric efficiency and divides by the specific volume of the suction gas. The efficiency is the weak point. A textbook relation for a compressor with a clearance volume fraction C is one plus C minus C times the pressure ratio raised to the power one over the isentropic exponent, as in Purdue's ME 418 lecture notes, which use 5% clearance in their example. The page uses that figure unless you enter an efficiency, giving 96.6% at a pressure ratio of 2, 93.6% at 3, 90.9% at 4 and 85.9% at 6 for R-410A. Real compressors lose more to leakage and suction heating, so an entered value of 85% instead of the 94% estimate lowers the example capacity by 10%. Use the manufacturer's volumetric efficiency or isentropic data when you have it.
Superheat, subcooling and where the heat comes from
The refrigerating effect is measured from the compressor inlet, so any heat the suction line picks up counts as capacity even if it came from a hot attic and not from the space. In the displacement method, raising the superheat from 7 K to 17 K moves the result from 13.54 kW to 13.53 kW, because the larger enthalpy gain is cancelled by the thinner gas. The liquid side matters more: a liquid line 5 K cooler adds 5.4% to the refrigerating effect. Because the line gain is not useful cooling, an air-side check is worth making. The coil load calculator finds the load from the airflow and the entering and leaving air, and it should land near the refrigerant-side figure when the readings are good.
Comparing with the expected capacity
A manufacturer rates capacity at one evaporating and condensing temperature, so compare the result with the performance table read at your own conditions. Weather alone can move the answer a long way: at a 5°C evaporating temperature, dropping the condensing temperature from 43°C to 30°C lifts the displacement-method capacity from 14.2 kW to 16.5 kW, a gain of 16%. The page also reports the cooling COP, which is 3.86 in the first example. Multiplying that by 3.412 gives an EER of about 13.2 Btu/h per watt, and the EER calculator carries the same relationship further. The tolerance field starts at 10% so that normal measurement scatter does not raise a false alarm. It is your choice and not a published limit.
Gauge and probe error
Small reading errors move the two methods in different directions. A suction gauge that reads 3 psi low changes the example displacement-method capacity by -2.4% and the power-balance capacity by 1.6%, so a large gap between them points to a bad reading or a faulty compressor. Zero the gauges, enter the site elevation when the site is high, and clamp the probes on clean copper under insulation. The result is a field estimate, and it will not replace a calorimeter test.
Frequently Asked Questions
Capacity is the refrigerant mass flow multiplied by the refrigerating effect. The refrigerating effect is the enthalpy of the suction gas minus the enthalpy of the liquid ahead of the expansion device. In the R-410A example, 120 psig suction and 360 psig discharge give a 40.7°F evaporating temperature and a 108.9°F condensing temperature, and the readings give a refrigerating effect of 168.8 kJ/kg. A compressor input of 3.5 kW and a discharge temperature of 172°F give a mass flow of 0.08 kg/s from the energy balance, so the capacity is 13.5 kW, or 3.84 tons.
The power balance is usually the better check, because it uses only measured values. The displacement method needs a volumetric efficiency, and the page estimates one from a 5% clearance volume unless you enter the manufacturer value. PNNL found that an energy-balance estimate of mass flow stays accurate when the compressor has internal faults that cut flow, while a compressor-map estimate does not. In the example the displacement method gives 14.1 kW against 13.5 kW from the power balance, a difference of 4.6%, and the calculator shows both when you enter both sets of inputs.
A nameplate figure belongs to a single evaporating and condensing temperature. At the same 5°C evaporating temperature, 7 K of superheat and 5 K of subcooling, lowering the condensing temperature from 43°C to 30°C raises the displacement-method capacity of the example compressor from 14.2 kW to 16.5 kW, a gain of 16%. Cool weather is the usual reason. Compare the result with the manufacturer's data at the same evaporating and condensing temperatures, and check the gauges and probes if the conditions match.
In the example, a suction gauge that reads 3 psi low changes the displacement-method capacity by -2.4% and the power-balance capacity by 1.6%, so the two methods move in opposite directions. A liquid line probe that reads 5 K too warm cuts the refrigerating effect by about 5.1%. Zero the gauges, insulate the probes on clean pipe, and treat a result within a few percent of the expected figure as inside the measurement error.