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A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at 2308C by rejecting its waste heat to cooling water that enters the condenser at 188C at a rate of 0.25 kg/s and leaves at 268C. The refrigerant enters the condenser at 1.2 MPa and 658C and leaves at 428C. The inlet state of the compressor is 60 kPa and 2348C and the compressor is estimated to gain a net heat of 450 W from the surroundings. Determine (a) the quality of the refrigerant at the evaporator inlet, (b) the refrigeration load, (c) the COP of the refrigerator, and (d) the theoretical maximum refrigeration load for the same power input to the compressor.

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batolisis

Answer:

hello your question is incomplete attached below is the missing part and also attached is the solution

answer: a) 0.4801

              b) 5.398 kw

              c) 2.14

              d) 12.72

Explanation:

The quality of the refrigerant at the evaporator inlet

h4 = hf4 + x4(hfx4)

Refrigeration load

Ql = m(h1-h4)

COP of the refrigerator

Ql / m(h2-h1) - Qm

Theoretical maximum refrigeration load

( Ql )max = COPr.rev * [m(h2-h1) - Qin]

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The quality that will exist at the inlet of the refrigerant's evaporator would be:

a). [tex]0.4801[/tex]

The load of the refrigeration would be          

b). [tex]5.398 kW[/tex]

The refrigerator's COP would be:

c). [tex]2.14[/tex]

The maximum refrigeration load would be as follows:          

d) [tex]12.72[/tex]

a). The determination of the quality of the refrigerant at the inlet of the evaporator would be:

[tex]h4 = hf_{4} + x_{4}(hfx_{4})[/tex]  

As given,

[tex]hf_{4}[/tex] [tex]= 3.841[/tex]

[tex]hf_{2,4}[/tex][tex]= 223.95[/tex]

[tex]h_{4} = 111.37[/tex]

Now,

solving for [tex]x_{4}[/tex] [tex]= (111.37 - 3.841)/223.95[/tex]

[tex]= 0.4801[/tex]

b). Refrigeration load

[tex]Q_{l}[/tex] [tex]= m(h_{1} - h_{4})[/tex]

[tex]= 0.0455(230.01-111.37)[/tex]

[tex]= 5.398 kW[/tex]

c). COP of the refrigerator

[tex]Q_{l}[/tex]/[tex]m(h_{2} - h_{1}) - Q_{m}[/tex]

by putting the values, we get

∵ COP [tex]= 2.14[/tex]

d). Theoretical maximum refrigeration load

[tex](Q_{l})[/tex]max [tex]= COPr.rev[/tex] × [tex]{m(h_{2} - h_{1}) - Q_{in}[/tex]

by putting the values, we get

∵ [tex]Q_{L}max = 12.72[/tex]

Learn more about "Compressor" here:

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