Question

Question 1 A refrigerator has working temperatures in the evaporator and condenser coils of 234 K and 305 K respectively. 1.1 What is the maximum COP possible? (2) 1.2 If the actual refrigerator has a COP of 0.75 calculate the required power input for a refrigerating effect of 5 kW. [10 Marks] (8)

Answer 1

Theory

The Refrigerator works on reverse carnot cycle. A refrigerator working on reverse carnot cycle will have the maximum possible coefficient of performance (COP). The coefficient of performance of a refregirator working on a reversce carnot cycle is given by,

$COP = \frac{Heat\,Absorbed}{Work\,Done}=\frac{Refrigerating\,effect}{Power\,in}$

Below is a typical Reverse Carnot Cycle which is represented in Temperature-Entropy graph.

2 3 72 T 4 Entropy

where

Hence the heat absorbed can be written as,

Therefore,

$COP_{max} =\frac{T_1*(s_2-s_3)}{(T_2-T_1)*(s_2-s_3)}= \frac{T_1}{T_2-T_1}$

Solution

Part 1: The maximum possible COP

Given,

Temperature of Evaporator =

Temperature of Condensor =

Hence

$COP_{max} = \frac{T_1}{T_2-T_1} = \frac{234(K)}{305(K)-234(K)} = 3.296$

Hence the maximum coefficient of performance is 3.296.

Part 2: Required Power Input

IMPORTANT NOTE: In question it has been stated that COP is 0.75. This is an error because coefficient of performance of a refrigerator is never less than 1. It is always greater than 1 and at the maximum for 100% efficieny can be equal to one. This is because coeficient of performance of refrigerator is inverse of its efficiency. Efficiency can never be greater than 100%.

Hence below solution is for

Hence the coefficient of performance,

Hence

$COP = \frac{Refrigerating\,effect}{Power\,in}$

$Power\,in = \frac{5(kW)}{2.472} = 2.022(kW)$

The required power input is 2.022(kW).

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