Question

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.

Answer 1

A)
`COP = (16.97)/(9.8) = 1.731`

B)
`P_(IN) = 0.4763`

C) Second law efficiency 4.85%

exergy destruction for the cycle = 9.3237 kW

Step-by-step explanation:

Given data:

`T_1 = 28` degree celcius

`T_2 = 12` degree celcius

`\dot m = 1140 kg/h`

Power to refrigerator = 9.8 kW

Cp = 3.35 kJ/kg degree C

`A) Q = \dot m Cp \Delta T`

`= 1140 * 3.35* (28-12) = 61,104 kJ/h`

`Q_(abs) = 61,104 kJ/h = 16.97 kJ/sec`

`COP = (16.97)/(9.8) = 1.731`

b)

`COP ∝ (1)/(P_(in))`

`P_(in)` wil be max when COP maximum

taking surrounding temperature T_H = 20 degree celcius

`COP_(max) = (T_L)/(T_H- T_L) = (285)/(293 - 285) = 35.625`

we know that

`COP = (heat\ obsorbed)/(P_(in))`

`P_(IN) = (16.97)/(35.62) = 0.4763`

c) second law efficiency

`\eta_(11) = (COP_R)/((COP)_max) = (1.731)/(35.625) = 4.85\%`

exergy destruction os given as
`X = W_(IN) - X_(Q2)`

= 9.8 - 0.473 = 9.3237 kW