REFRIGERATION CYCLES 755

dharm

\M-therm\Th14-3.pm5

35°C

3 ′

3

–10°C^2

Evap.

Cond. Comp.

Th.

4 ′

s

263

308

T(K)

1

303

4

Fig. 14.31

Also s 3 = s 2

or s 3 ′ + cp ln

T

T

3

3 ′

F

HG

I

KJ = 5.755

5.2086 + 2.903 ln

T 3

308

F

HG

I

KJ = 5.755

or ln
T 3
308

F

HG

I

KJ =

5.755 5.2086

2.903

−

= 0.1882

T (^3) e0 1882
308
=. = 371.8 K
Now, h 3 = h 3 ′ + cp(T 3 – T 3 ′)
= 1488.57 + 2.903 (371.8 – 308) = 1673.8 kJ/kg
Mass of refrigerant, m& =
50
hh 21 −

−
50
1450.22 343.29
= 0.04517 kJ/s
(i)Power required :
Power required = m& (h 3 – h 2 )
= 0.04517 (1673.8 – 1450.22) = 10.1 kW. (Ans.)
(ii)Cylinder dimensions :
m& = π
460
DL^2 ×× ×N 0 417477. = 0.04517 (calculated above)
or
π
4
12
1000
60
DD^2 ××. × 0.417477 = 0.04517
or D^3 =
0 04517 4 60
1 2 1000 0
.
.
××
π×× ×.417477 = 0.006888
∴ Diameter of cylinder, D = (0.006888)1/3 = 0.19 m. (Ans.)
and, Length of the cylinder, L = 1.2D = 1.2 × 0.19 = 0.228 m. (Ans.)