Answer:
A) [tex]COP = \frac{16.97}{9.8} = 1.731[/tex]
B) [tex]P_{IN} = 0.4763[/tex]
C) Second law efficiency 4.85%
exergy destruction for the cycle = 9.3237 kW
Explanation:
Given data:
[tex]T_1 = 28[/tex] degree celcius
[tex]T_2 = 12[/tex] degree celcius
[tex]\dot m = 1140 kg/h[/tex]
Power to refrigerator = 9.8 kW
Cp = 3.35 kJ/kg degree C
[tex]A) Q = \dot m Cp \Delta T[/tex]
[tex]= 1140 \times 3.35\times (28-12) = 61,104 kJ/h[/tex]
[tex]Q_{abs} = 61,104 kJ/h = 16.97 kJ/sec[/tex]
[tex]COP = \frac{16.97}{9.8} = 1.731[/tex]
b)
[tex]COP ∝ \frac{1}{P_{in}}[/tex]
[tex]P_{in}[/tex] wil be max when COP maximum
taking surrounding temperature T_H = 20 degree celcius
[tex]COP_{max} = \frac{T_L}{T_H- T_L} = \frac{285}{293 - 285} = 35.625[/tex]
we know that
[tex]COP = \frac{heat\ obsorbed}{P_{in}}[/tex]
[tex]P_{IN} = \frac{16.97}{35.62} = 0.4763[/tex]
c) second law efficiency
[tex]\eta_{11} = \frac{COP_R}{(COP)_max} = \frac{1.731}{35.625} = 4.85\%[/tex]
exergy destruction os given as [tex]X = W_{IN} - X_{Q2}[/tex]
= 9.8 - 0.473 = 9.3237 kW