An ideal vapor-compression refrigeration cycle that uses refrigerant-134a as its working fluid maintains a condenser at 800 kPa and the evaporator at −12°C. Determine this system's COP and the amount of power required to service a 150 kW cooling load.

COP = 4.846

Explanation:

From the table A-11 i attached, we can find the entropy for the state 1 at -12°C.

h1 = 243.3 KJ/Kg

s1 = 0.93911 KJ/Kg.K

From table A-12 attached we can do the same for states 3 and 4 but just enthalpy at 800 KPa.

h3 = h4 = hf = 95.47 KJ/Kg

For state 2, we can calculate the enthalpy from table A-13 attached using interpolation at 800 KPa and the condition s2 = s1. We have;

h2 = 273.81 KJ/Kg

The power would be determined from the energy balance in state 1-2 where the mass flow rate will be expressed through the energy balance in state 4-1.

W' = m'(h2 - h1)

W' = Q'_L((h2 - h1)/(h1 - h4))

Where Q'_L = 150 kW

Plugging in the relevant values, we have;

W' = 150((273.81 - 243.3)/(243.3 - 95.46))

W' = 30.956 Kw

Formula foe COP is;

COP = Q'_L/W'

COP = 150/30.956

COP = 4.846

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