TITLE.PM5

REFRIGERATION CYCLES 753

dharm

\M-therm\Th14-3.pm5

Example 14.19. A Freon-12 refrigerator producing a cooling effect of 20 kJ/s operates on
a simple cycle with pressure limits of 1.509 bar and 9.607 bar. The vapour leaves the evaporator
dry saturated and there is no undercooling. Determine the power required by the machine.
If the compressor operaters at 300 r.p.m. and has a clearance volume of 3% of stroke
volume, determine the piston displacement of the compressor. For compressor assume that the
expansion following the law pv1.13 = constant.
Given :

Temperature ps vg Enthalpy kJ/kg Entropy kJ/kg K Specific

°C bar m^3 /kg hf hg sf sg heat

kJ/kg K

• 20 1.509 0.1088 17.8 178.61 0.073 0.7082 —
  40 9.607 — 74.53 203.05 0.2716 0.682 0.747
  (U.P.S.C. 1996)
  Solution. Given : (From the table above) :
  h 2 = 178.61 kJ/kg ; h 3 ′ = 203.05 kJ/kg ; hf 4 = 74.53 kJ/kg = h 1

40°C

3 ′

3

Compression

• 20°C^2

4

s

Evaporation

Condensation

Throttling

253

313

T(K)

1

Fig. 14.30

Now, cooling effect = m&(h 2 – h 1 )

20 = m&(178.61 – 74.53)

∴ m& =

20

(178.61 74.53)− = 0.192 kg/s

Also, s 3 = s 2

s 3 ′ + cp ln T

T

3

3 ′

F

HG

I

KJ

= 0.7082

0.682 + 0.747 ln T^3

313

F

HG

I

KJ

= 0.7082