760 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th14-3.pm5

Fig. 14.35

= 0.6854 + 0.733 loge

T 3

303

F

HG

I

KJ

∴ loge T^3

303

F

HG

I

KJ

= 0.0488

i.e., T 3 = 318 K
Now, enthalpy at ‘3’, h 3 = h 3 ′ + cp (T 3 – T 3 ′)
= 199.6 + 0.733 (318 – 303) = 210.6 kJ/kg.
Also, enthalpy at 4′, hf 4 ′ = hf 4 – (cp)liquid (T 4 – T 4 ′) = 64.6 – 1.235 × 6 = 57.19 kJ/kg
For the process 4′-1,
Enthalpy at 4′ = enthalpy at 1 = 57.19 kJ/kg
For specific volume at 2,
v
T

v

T

2

2

2

2

′

′

=

∴ v 2 =

v

T

2

2

′

′ × T^2 = 0.0767 ×

270

263 = 0.07874 m

(^3) /kg
(i)Refrigerating effect per kg
= h 2 – h 1 = 188.33 – 57.19 = 131.14 kJ/kg. (Ans.)
(ii)Mass of refrigerant to be circulated per minute for producing effect of 2400 kJ/
min.

2400
131.14
= 18.3 kg/min. (Ans.)