Question 12.19: A Freon-12 refrigerator producing a cooling effect of 20 kJ/...
A Freon-12 refrigerator producing a cooling effect of 20 kJ/s operates on a simple cycle with pressure limits of 1.509 bar and 9.607 bar. The vapour leaves the evaporator dry saturated and there is no undercooling. Determine the power required by the machine.
If the compressor operaters at 300 r.p.m. and has a clearance volume of 3% of stroke volume, determine the piston displacement of the compressor. For compressor assume that the expansion following the law pv^{1.13} = constant.
Given :
| Temperature °C |
p_{s} bar |
v_{g} m³/kg |
Enthalpy h_{f} |
kJ/kg h_{g} |
Entropy s_{f} |
kJ/kg K s_{g} |
Specific heat kJ/kg K |
| – 20 40 |
1.509 9.607 |
0.1088 — |
17.8 74.53 |
178.61 203.05 |
0.073 0.2716 |
0.7082 0.682 |
— 0.747 |
(UPSC)
Given : From the table above :
h_{2} = 178.61 kJ/kg ; h_{3} ^{′} = 203.05 kJ/kg ; h_{f_4} = 74.53 kJ/kg = h_{1}
Now, cooling effect = \dot{m} (h_{2} – h_{1})
20 = \dot{m} (178.61 – 74.53)
∴ \dot{m} = \frac{20}{(178.61 – 74.53)} = 0.192 kg/s
Also, s_{3} = s_{2}
s_{3} ^{′} + c_{p} \ln (\frac{T_{3}}{T_{3} ^{′}}) = 0.7082
0.682 + 0.747 \ln (\frac{T_{3}}{313}) = 0.7082
or \ln (\frac{T_{3}}{313}) = \frac{0.7082 – 0.682}{ 0.747 } = 0.03507
or \frac{T_{3}}{313} = e^{0.03507} = 1.0357
∴ T_{3} = 313 × 1.0357 = 324.2 K
Now, h_{3} = h_{3} ^{′} + c_{p} (324.2 – 303)
= 203.05 + 0.747 (324.2 – 313) = 211.4 kJ/kg
Power required :
Power required by the machine = \dot{m} (h_{3} – h_{2})
= 0.192 (211.4 – 178.61) = 6.29 kW.
Piston displacement, V :
Volumetric efficiency, η_{vol.} = 1 + C – C (\frac{p_{d}}{p_{s}})^{1/n}
= 1 + 0.03 – 0.03 (\frac{9.607}{1.509})^{\frac{1}{1.13}} = 0.876 or 87.6%
The volume of refrigerant at the intake conditions is
\dot{m} × v_{g} = 0.192 × 0.1088 = 0.02089 m³/s
Hence the swept volume = \frac{0.02089}{η_{vol.}} = \frac{0.02089}{0.876} = 0.02385 m³/s
∴ V = \frac{0.02385 × 60}{300} = 0.00477 m³.
