TITLE.PM5

REFRIGERATION CYCLES 763

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\M-therm\Th14-4.pm5

(v)C.O.P.
(vi)Heat removed through condenser.
(vii)Theoretical bore and stroke of the compressor.
Neglect valve throttling and clearance effect.
Solution. Refer Fig. 14.37. 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 :

h 2 = 615 kJ/kg

h 3 = 664 kJ/kg

h 4 = h 1 = 446 kJ/kg

v 2 = 0.14 m^3 /kg.

(i)Refrigerating effect per kg = h 2 – h 1 = 615 – 446 = 169 kJ/kg. (Ans.)

p (Pressure)

h (Enthalpy)

446 615 664

1

2

• 28°C

Evaporation

Condensation

+ 23°C

23°C

20°C

4

Sub cooling

3

ThrottlingThrottling

CompressionCompressionConstant

entropy line

Constant

entropy line

• 20°C

23°C

20°C20°C

Fig. 14.37

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

m =

20 14000

169 60

×

× = 27.6 kg/min. (Ans.)

(iii)Theoretical piston displacement

= Specific volume at suction × Mass of refrigerant used/min

= 0.14 × 27.6 = 3.864 m^3 /min

(iv)Theoretical power

= m × (h 3 – h 2 ) = 27.6

60

(664 – 615) = 22.54 kJ/s

= 22.54 kW. (Ans.)

(v) C.O.P. =

hh

hh

21

32

615 446

664 615

−

−

= −

−

= 3.45. (Ans.)