Question

Question 12
? In a vapour compression
refrigeration plant, the enthalpy
values at different points are:
1.Enthalpy at exit of the
evaporator = 350 kJ/kg 2.
Enthalpy at exit of the
compressor = 375 kJ/kg 3
Enthalpy at exit of the condenser
= 225 kJ/kg The refrigerating
efficiency of the plant is 0.8. What
is the power required per kW of
cooling to be produced?
Options (Marks: 4)​

Answer 1

Sorry the question is way too big

Answer 2

Answer:

The power required per kW of cooling to be produced is 0.25kW.

Explanation:

Given:

Enthalpy at exit of the evaporator (h1) = 350 kJ/kg

Enthalpy at exit of the compressor (h2) = 375 kJ/kg

Enthalpy at exit of the condenser (h3 = h4) = 225 kJ/kg

Efficiency (η) = 0.8

To find:

Power required (P)

Step 1:

We first find the ideal COP.

$(COP)_{ideal} = \frac{h1 - h4}{h2-h1}$

where, h1 - h4 = refrigerating effect, and h2 - h1 = work input

Substituting the given values, we get

$(COP)_{ideal} = \frac{350- 225}{375-350}$

$(COP)_{ideal} = \frac{125}{25}$

$(COP)_{ideal} = 5$

Step 2:

We know that the actual COP is given by multiplying the ideal COP by the efficiency of the plant. Therefore,

$(COP)_{actual} =$ η × $(COP)_{ideal}$

Substituting the values in the above formula, we get

$(COP)_{actual} = 0.8 * 5$

$(COP)_{actual} = 4$

Step 3:

The power required per kW of cooling produced is

$P = \frac{1kW}{(COP)_{actual}}$

$P = \frac{1kW}{4}$

P = 0.25kW

Therefore, the power required is 0.25kW.

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