Part load conditions

During part- load operation, the evaporator temperature rises and the condenser temperature
falls, effectively increasing the COP. But at the same time, deviation from the design
operation point and the fact that mechanical losses form a greater proportion of the total
power negate the effect of improved COP, resulting in lower part- load efficiency.

iread the above ^^^ in a book,the thing is i do not understand this.

How would operating at part load increase the COP( change the evaporator and condensor temperature respectively)
from what i noe of refrigeration:
COP=te/tc-te
where te= is the temperature of the refrigerant in the evaporator
tc= the condensing temperature of the refrigerant

By part load they mean when the refrigerator is not full( say with food in a domestic one)
please explain ?
thnks in advance

Are you referring to Co-efficient of Performance (COP) ?


COP can be expressed as

COP = hh / hw (1)

where
COP = Coefficient of Performance
hh = heat produced (Btu/h)
hw = equivalent electric energy input (Btu/h) = 3413 Pw

where
Pw = electrical input energy (W)

If a heat pump delivers 3 units of heat for every unit of energy input - the COP is 3.

Example of COP on a Heat Pump

Cooling Cycle: A heat pump delivering 60000 Btu/h with a total input of 9 kW:

COP = (60,000 Btu/h / 3413) / 9 (kW)
= 1.95

Heating Cycle : A heat pump delivering 50000 Btu/h with a total input of 7 kW:

COP = (50,000 Btu/h / 3413) 7 (kW)
= 2.1

Attachment 9003Balance Point