Question 12.20: A food storage locker requires a refrigeration capacity of 5...
A food storage locker requires a refrigeration capacity of 50 kW. It works between a condenser temperature of 35°C and an evaporator temperature of – 10°C. The refrigerant is ammonia. It is sub-cooled by 5°C before entering the expansion valve by the dry saturated vapour leaving the evaporator. Assuming a single cylinder, single-acting compressor operating at 1000 r.p.m. with stroke equal to 1.2 times the bore.
Determine : (i) The power required, and
(ii) The cylinder dimensions.
Properties of ammonia are :
| Saturation temperature, °C |
Pressure bar |
Enthalpy (kJ/kg) | Entropy (kJ/kg K) | Specific volume (m³/kg) |
Specific heat (kJ/kg K) |
||||
| Liquid | Vapour | Liquid | Vapour | Liquid | Vapour | Liquid | Vapour | ||
| – 10 35 |
0.9157 13.522 |
154.056 366.072 |
1450.22 1488.57 |
0.82965 1.56605 |
5.7550 5.2086 |
— 1.7023 |
0.417477 0.095629 |
— 4.556 |
2.492 2.903 |
(UPSC)
Given : From the table above:
h_{2} = 1450.22 kJ/kg ; h_{3} ^{ ′} = 1488.57 kJ/kg ; h_{f_4} = 366.072 kJ/kg ;
h_{f_4} ^{ ′} = h_{1} = h_{f_4} – 4.556 (308 – 303)
= 366.07 – 4.556 (308 – 303) = 343.29 kJ/kg
Also s_{3} = s_{2}
or s_{3} ^{′} + c_{p} \ln (\frac{T_{3}}{T_{3} ^{′}}) = 5.755
5.2086 + 2.903 \ln (\frac{T_{3}}{308}) = 5.755
or \ln (\frac{T_{3}}{308}) = \frac{5.755 – 5.2086}{2.903} = 0.1882
\frac{T_{3}}{308} = e^{0.1882} = 371.8 K
Now, h_{3} = h_{3} ^{′} + c_{p} (T_{3} – T_{3} ^{′})
= 1488.57 + 2.903 (371.8 – 308) = 1673.8 kJ/kg
Mass of refrigerant, \dot{m} = \frac{50}{ h_{2} – h_{1}} = \frac{50}{1450.22 – 343.29}
= 0.04517 kJ/s
(i) Power required :
Power required = \dot{m} (h_{3} – h_{2})
= 0.04517 (1673.8 – 1450.22) = 10.1 kW.
(ii) Cylinder dimensions :
\dot{m} = \frac{π}{4} D² × L × \frac{N}{60} × 0.417477 = 0.04517 (calculated above)
or \frac{π}{4} D² × 1.2 D × \frac{1000}{60} × 0.417477 = 0.04517
or D³ = \frac{ 0.04517 × 4 × 60}{π × 1.2 × 1000 × 0.417477 } = 0.006888
∴ Diameter of cylinder, D = (0.006888)^{1/3} = 0.19 m.
and, Length of the cylinder, L = 1.2D = 1.2 × 0.19 = 0.228 m.
