TITLE.PM5

746 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th14-3.pm5

Temp. Enthalpy (kJ/kg) Entropy of liquid Entropy of vapour

°C (kJ/kg K) kJ/kg K

Liquid Vapour

25 100.04 1319.22 0.3473 4.4852

• 15 – 54.56 1304.99 – 2.1338 5.0585
  Take latent heat of ice = 335 kJ/kg.

Solution. Theoretical C.O.P. =

hh

hh

21

32

−

−

Here, h 3 = 1319.22 kJ/kg ;

h 1 = h 4 (i.e., hf 4 ) = 100.04 kJ/kg ...From the table above.

T (K)

s (kJ/kg K)

44 33

CondensationCondensation

ThrottlingThrottling

WW

CompressionCompression

EvaporationEvaporation

RRnn

11 22

(25 + 273)

= 298

(–15 + 273)

= 258

Fig. 14.24

To find h 2 , let us first find dryness at point 2.

Entropy at ‘2’ = Entropy at ‘3’ (Process 2-3 being isentropic)

sf 2 + x 2 sfg 2 = sg 3

• 2.1338 + x 2 × [5.0585 – (– 2.1338)] = 4.4852

∴ x 2 =

4.4852 2.1338

5.0585 2.1338

+

+ = 0.92

∴ h 2 = hf 2 + x 2 hfg 2 = – 54.56 + 0.92 × [1304.99 – (– 54.56)]

= 1196.23 kJ/kg.

∴ C.O.P.(theoretical) =

1196.23 100.04

1319.22 1196.23

−

−

= 8.91.

∴ C.O.P.(actual) = 0.62 × C.O.P.(theoretical) ... Given
i.e., C.O.P.(actual) = 0.62 × 8.91 = 5.52