At steady state, a refrigeration cycle operating between hot and cold reservoirs at 300 k and 275 k, respectively, removes energy by heat transfer from the cold reservoir at a rate of 600 kw.(a) if the cycle’s coefficient of performance is 4, determine the power input required, in kw.(b) determine the minimum theoretical power required, in kw, for any such cycle.

a) Wcycle=150KW

b) Wcycle=54.55KW

Explanation:

Given data:

Hot reservoir Temp Th=300 K

Cold reservoir Temp Tc=275 K

Heat transfer at cold reservoir Qc=600KW

To find out

a) Coefficient of performance \beta actual=4

    Cycle power in KW

b)  The minimum theoretical power required Wcycle in KW

Assumption:

Constant average values is used.

Solution:

a) Actual cycle formula

               Wcycle=Qc÷\beta actual

                           =600/4

                           =150KW

b) maximum cycle coefficient formula

                B_{max}=Tc÷(Th-Tc )

                              =275÷(300-275 )

                               =11

 For minimal theoretical power required

     \beta actual=B_{max}=11

  Actual cycle coefficient of performance can be determined by

     \beta actual=Qc÷Wcycle

             Wcycle=Qc÷\beta actual

                         =600/11

                         =54.55KW

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explanation: