# Cooling Load - Convert between *kW/ton* to *COP* or *EER*

Chiller efficiency depends on heat removed and energy consumed.

Absorption chillers are often rated in fuel consumption per ton cooling. Electric motor driven chillers are rated in kilowatts per ton cooling.

*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*

*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)

where

P_{c}= electrical power consumption (kW)

Q_{r}= heat removed (ton)

**Coefficient of Performance - ***COP*

*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)

where

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*and

^{o}C*-8.3*) are typically used.

^{o}C### 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)*

* = 4.2*

### 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)*

* = 4 *

**Energy Efficiency Ratio - ***EER*

*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)

where

EER= energy efficient ratio (Btu/Wh)

Q_{c}= net cooling energy (Btu)

E= appliedelectrical 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.

Time (h) | Instantaneous Heat RemovedP_{c}(Btu/h) | Instantaneous Electrical PowerP(W) |
---|---|---|

0 | 8500 | 600 |

1 | 10000 | 700 |

2 | 7000 | 550 |

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) *

* = 13.8*

## Related Topics

### • Air Conditioning

Air Conditioning systems - heating, cooling and dehumidification of indoor air for thermal comfort.