The evaporator and condenser temperatures of 20 tonnes capacity freezer are – 28°C and 23°C respectively. The refrigerant – 22 is sub-cooled by 3°C before it enters the expansion valve and is superheated to 8°C before leaving the evaporator. The compression is isentropic. A sixcylinder

Question

The evaporator and condenser temperatures of 20 tonnes capacity freezer are – 28°C and 23°C respectively. The refrigerant – 22 is sub-cooled by 3°C before it enters the expansion valve and is superheated to 8°C before leaving the evaporator. The compression is isentropic. A sixcylinder single-acting compressor with stroke equal to bore running at 250 r.p.m. is used. Determine :
(i) Refrigerating effect/kg.
(ii) Mass of refrigerant to be circulated per minute.
(iii) Theoretical piston displacement per minute.
(iv) Theoretical power.
(v) C.O.P.
(vi) Heat removed through condenser.
(vii) Theoretical bore and stroke of the compressor.
Neglect valve throttling and clearance effect.

Accepted Answer

Refer to Fig. 37. Following the procedure as given in the previous example plot the points 1, 2, 3 and 4 on p-h chart for freon-22. The following values are obtained :
[latex] h_{2} [/latex] = 615 kJ/kg
[latex] h_{3} [/latex] = 664 kJ/kg
[latex] h_{4} = h_{1} [/latex] = 446 kJ/kg
[latex] v_{2} [/latex] = 0.14 m³/kg.

(i) Refrigerating effect per kg = [latex] h_{2} - h_{1} [/latex] = 615 – 446 = 169 kJ/kg.

(ii) Mass of refrigerant to be circulated per minute,

m = [latex] \frac{20 × 14000}{169 × 60} [/latex] = 27.6 kg/min.

(iii) Theoretical piston displacement

= Specific volume at suction × Mass of refrigerant used/min
= 0.14 × 27.6 = 3.864 m³/min.

(iv) Theoretical power

= m × [latex] (h_{3} - h_{2}) = \frac{27.6}{60} [/latex] (664 – 615) = 22.54 kJ/s

= 22.54 kW.

(v) C.O.P. = [latex] \frac{h_{2} - h_{1}}{h_{3} - h_{2}} = \frac{615 - 446}{664 - 615} [/latex] = 3.45.

(vi) Heat removed through the condenser

= m [latex] (h_{3} - h_{4}) [/latex] = 27.6 (664 – 446) = 6016.8 kJ/min.

(vii) Theoretical displacement per minute per cylinder

= [latex] \frac{Total displacement/ min.}{Number of cylinders} = \frac{3.864}{6} [/latex] = 0.644 m³/min

Let diameter of the cylinder = d
Then, stroke length, l = d

Now, [latex] \frac{π}{4} d² × l = \frac{0.644}{950} [/latex]

or [latex] \frac{π}{4} d² × d = \frac{0.644}{950} [/latex]

i.e., d = 0.0952 m or 95.2 mm.

and l = 95.2 mm.

Question 12.27: The evaporator and condenser temperatures of 20 tonnes capac...

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