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Modern Engineering Thermodynamics

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566 CHAPTER 14: Vapor and Gas Refrigeration Cycles<br />

EXAMPLE 14.9 (Continued )<br />

4h<br />

3<br />

Condenser<br />

Q refrig. = 422 kJ/h<br />

Refrigeration evaporator<br />

5<br />

2s<br />

Compressor<br />

1<br />

Freezer evaporator<br />

6h<br />

Q freezer = 422 kJ/h<br />

2s<br />

T<br />

3<br />

4h<br />

5<br />

6h<br />

1<br />

s<br />

FIGURE 14.24<br />

Example 14.9.<br />

Solution<br />

The properties of the refrigerant at the states shown in Figure 14.24 are as follows (see Tables C.7e, C.7f, and C.8d for the<br />

numerical values):<br />

Station 1—Compressor inlet<br />

Station 2s—Compressor outlet<br />

x 1 = 1:00<br />

s 2s = s 1 = 0:9315 kJ/kg.K<br />

T 1 = −18:0°C<br />

p 2s = p 3 = p sat ð30:0°CÞ = 0:770 MPa<br />

h 1 = 236:53 kJ/kg<br />

h 2s = 271:0 kJ/kg<br />

s 1 = 0:9315 kJ/kg.K<br />

ðinterpolation in Table C:7fÞ<br />

Station 3—Condenser outlet<br />

Station 4h—Refrigerator evaporator inlet<br />

x 3 = 0:00 h 4h = h 3<br />

T 3 = 30:0°C<br />

T 4h = 4:00°C<br />

p 3 = p sat ð30:0°CÞ = 0:770 MPa<br />

No further information is needed:<br />

h 3 = h f ð30:0°CÞ = 91:49 kJ/kg<br />

Station 5—Refrigerator evaporator outlet Station 6h—Freezer evaporator inlet<br />

x 5 = ? h 6h = h 5 = ?<br />

T 5 = T 4h = 4:00°C<br />

T 6h = −18:0°C<br />

h 5 = ?<br />

No further information is needed:<br />

The isentropic efficiency of the compressor is 80.0%, or (η s ) comp = 0.800.<br />

a. The COP for this system is the ratio of the total heat rate removed from the refrigeration plus freezer compartments<br />

divided by the power input to the compressor, or<br />

COP =<br />

but since h 5 = h 6h here, this equation reduces to<br />

COP =<br />

_ Q R + _Q F<br />

_W C<br />

= ðh 5 − h 4h Þ + ðh 1 − h 6h Þ<br />

ðh 2s − h 1 Þ/ðη s Þ comp<br />

h 1 − h 4h<br />

=<br />

236:53 − 91:49<br />

ðh 2s − h 1 Þ/ðη s Þ comp<br />

ð271:0 − 236:53Þ/0:800 = 3:37<br />

so that the system COP does not depend on states 5 or 6h.<br />

b. Since h 5 = h 6h , we can write the refrigeration and freezer compartment cooling rates as<br />

_Q R = _m ref ðh 5 − h 4h Þ and _Q F = _m ref ðh 1 − h 6h Þ = _m ref ðh 1 − h 5 Þ. Solving these two equations for _m ref gives<br />

_m ref =<br />

Q _ R + _Q F<br />

h 1 − h 4h<br />

=<br />

ð422 + 422 kJ/hÞð1 h/60minÞ<br />

236:53 − 91:49 kJ/kg<br />

= 0:0970 kg/min

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