Convert between cooling load units like kW/ton, COP and EER.
Chiller efficiency depends on heat removed and energy consumed.
- kW/ton = 12 / EER
- kW/ton = 12 / (COP x 3.412)
- COP = EER / 3.412
- COP = 12 / (kW/ton) / 3.412
- EER = 12 / (kW/ton)
- EER = COP x 3.412
If the efficiency of a chiller is rated to 1 kW/ton, then
- COP = 3.52
- EER = 12
Cooling Load in - kW/ton
The term kW/ton is commonly used for larger commercial and industrial air-conditioning, heat pump and refrigeration systems.
The term is defined as the ratio of energy consumption in kW to the rate of heat removal in tons at the rated condition. The lower the kW/ton - the more efficient the system.
kW/ton = P c / Q r (1)
P c = electrical power consumption (kW)
Q r = heat removed (ton)
Coefficient of Performance - COP
The Coefficient of Performance - COP - is the basic parameter used to report efficiency of refrigerant based systems. CO P is the ratio between useful cooling or heating output and power input and can be expressed as
COP = P c / P (2)
COP = Coefficient of Performance
P c = useful cooling or heating power output (Btu/h, W)
P = power input (Btu/h, W)
The COP is an instantaneous measurement in that the units are power which can be measured at one point in time. COP can be used to define the cooling efficiency for a cooling system - or the heating efficiency for a heat pump system.
- Cooling - COP is defined as the ratio of of the heat removal to the power input to the compressor
- Heating - COP is defined as the ratio of the heat delivered to the power input to the compressor
- higher COP - more efficient system
COP can be treated as an efficiency where COP of 2.0 = 200% efficiency. For unitary heat pumps, ratings at two standard outdoor temperatures of 47 o F and 17 o F ( 8.3 o C and -8.3 o C ) are typically used.
Example - COP for an Air Conditioner Unit
At an instantaneous moment an air conditioner units cools air from 30 o C and 70% moisture to 20 o C and 100% moisture. The air flow through the unit is 0.1 m 3 /s and the electrical power consumption of the unit is 600 W.
From the Mollier diagram we can see that the enthalpy of the input air is aprox. 78 kJ/kg and the enthalpy of the output air is aprox. 57 kJ/kg.
The heat removed from the air can be calculated as
P c = ((78 kJ/kg) - (57 kJ/kg)) (0.1 m 3 /s) (1.2 kg/m 3 )
= 2.5 kW
COP for the unit can be calculated as
COP = (2.5 kW) / (0.6 kW)
Example - COP for a Heat Pump
A heat pump delivers 4.8 kW (16378 Btu/h) of heat with electric power consumption 1.2 kW (4094 Btu/h) . COP for the heat pump at the actual conditions can be calculated as
COP = (4.8 kW) / (1.2 kW)
Energy Efficiency Ratio - EER
The Energy Efficiency Ratio - EER - is a term generally used to define cooling energy efficiency of unitary air-conditioning and heat pump system.
The efficiency is determined at a single rated condition specified by an appropriate equipment standard and is defined as the ratio of net cooling capacity - or heat removed in Btu - to the total input rate of electric energy applied - in Wh . The units of EER are Btu/Wh .
EER = Q c / E (3)
EER = energy efficient ratio (Btu/Wh)
Q c = net cooling energy (Btu)
E = applied electrical energy (Wh)
This efficiency term typically includes the energy requirement of auxiliary systems such as the indoor and outdoor fans.
- higher EER - more efficient system
Example - EER for an Air Conditioner Unit
The heat removed and electrical power consumed for the air conditioner unit can be measured and calculated in different ways. One simple alternative is to calculate mean values from some instantaneous samples.
|Instantaneous Heat Removed |
|Instantaneous Electrical Power |
The heat removed for 3 hours can be estimated to
Q c = ((8500 Btu/h + 10000 Btu/h + 7000 Btu/h) / 3) (3 h)
= 25500 Btu
The electrical consumption for 3 hours can be estimated to
E = ((600 W + 700 W + 550 W) / 3) (3 h)
= 1850 Wh
EER for the air conditioner unit can be estimated to
EER = (25500 Btu) / (1850 Wh)