TITLE.PM5

756 ENGINEERING THERMODYNAMICS

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Example 14.21. A refrigeration cylce uses Freon-12 as the working fluid. The temperature
of the refrigerant in the evaporator is – 10°C. The condensing temperature is 40°C. The cooling
load is 150 W and the volumetric efficiency of the compressor is 80%. The speed of the compressor
is 720 rpm. Calculate the mass flow rate of the refrigerant and the displacement volume of the
compressor.
Properties of Freon-12
Temperature (°C) Saturation Enthalpy (kJ/kg) Specific volume
pressure (MPa) (m^3 /kg)
Saturated vapour
Liquid Vapour

• 10 0.22 26.8 183.0 0.08
  40 0.96 74.5 203.1 0.02
  (GATE, 1995)
  Solution. Given : Cooling load = 150 W ; ηvol. = 0.8 ; N = 720 r.p.m.
  p

h

4 40°C 3

1

Cond.

Evap.

Th. Comp.

2

• 10°C

Fig. 14.32

Mass flow rate of the refrigerant m& :

Refrigerating effect = h 2 – h 1

= 183 – 74.5 = 108.5 kJ/kg

Cooling load = m& × (108.5 × 1000) = 150

or m& =
150
108 5 1000. × = 0.001382 kJ/s. (Ans.)
Displacement volume of the compressor :
Specific volume at entry to compressor,
v 2 = 0.08 m^3 /kg (From table)
∴ Displacement volume of compressor =

& ..

vol..

mv 2 0 001382 0 08

η 08

= ×

= 0.0001382 m^3 /s. (Ans.)

Example 14.22. In a simple vapour compression cycle, following are the properties of the

refrigerant R-12 at various points :