Bananas are to be cooled from 28°C to 12°C at a rate of 1140 kg/h by a refrigerator that operates on a vapor-compression refrigeration cycle. The power input to the refrigerator is 9.8 kW. Determine (a) the rate of heat absorbed from the bananas, in kJ/h, and the COP, (b) the minimum power input to the refrigerator, and (c) the second-law efficiency and the exergy destruction for the cycle. The specific heat of bananas above freezing is 3.35 kJ/kg·°C.

Respuesta :

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

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