Question 12.26: A refrigeration system of 10.5 tonnes capacity at an evapora...
A refrigeration system of 10.5 tonnes capacity at an evaporator temperature of – 12°C and a condenser temperature of 27°C is needed in a food storage locker. The refrigerant ammonia is sub-cooled by 6°C before entering the expansion valve. The vapour is 0.95 dry as it leaves the evaporator coil. The compression in the compressor is of adiabatic type.
Using p-h chart find :
(i) Condition of volume at outlet of the compressor
(ii) Condition of vapour at entrance to evaporator
(iii) C.O.P.
(iv) Power required, in kW.
Neglect valve throttling and clearance effect.
Refer to Fig. 36.
Using p-h chart for ammonia,
• Locate point ‘2’ where – 12°C cuts 0.95 dryness fraction line.
• From point ‘2’ move along constant entropy line and locate point ‘3’ where it cuts constant pressure line corresponding to + 27°C temperature.
• From point ‘3’ follow constant pressure line till it cuts + 21°C temperature line to get point ‘4’.
• From point ‘4’ drop a vertical line to cut constant pressure line corresponding to – 12°C and get the point ‘5’.
The values as read from the chart are :
h_{2} = 1597 kJ/kg
h_{3} = 1790 kJ/kg
h_{4} = h_{1} = 513 kJ/kg
t_{3} = 58°C
x_{1} = 0.13.
(i) Condition of the vapour at the outlet of the compressor
= 58 – 27 = 31°C superheat.
(ii) Condition of vapour at entrance to evaporator,
x_{1} = 0.13.
(iii) C.O.P. = \frac{ h_{2} – h_{1}}{h_{3} – h_{2} } = \frac{1597 – 513}{1790 – 1597} = 5.6.
(iv) Power required :
C.O.P. = \frac{Net refrigerating effect }{Work done} = \frac{R_{n}}{W}
5.6 = \frac{10.5 × 14000}{W × 60}
∴ W = \frac{10.5 × 14000}{5.6 × 60} kJ/min = 437.5 kJ/min.
= 7.29 kJ/s.
i.e., Power required = 7.29 kW.
