REFRIGERATION CYCLES 725

dharm

\M-therm\Th14-2.pm5

Solution. Fig. 14.6 shows the working cycle of the refrigerator.

Given : p 2 = 1.0 bar ;

p 1 = 8.0 bar ;

T 3 = 9 + 273 = 282 K ;

T 4 = 29 + 273 = 302 K.

p (bar)

V(m )^3

14

2 3

pV1.35= C

p=8 1

p = 1.0 2

Fig. 14.6

Considering polytropic compression 3-4, we have

T

T

4

3

= p

p

n

1 n

2

(^1) 135 1
8 135
1
F
HG
I
KJ
=F
HG
I
KJ
−. −
.
= (8)0.259 = 1.71
or T 4 = T 3 × 1.71 = 282 × 1.71 = 482.2 K
Again, considering polytropic expansion 1-2, we have
T
T
p
p
n
1 n
2
1
2
(^1) 135 1
8 135
1

F
HG
I
KJ
=F
HG
I
KJ
−. −
.
= 1.71
T 2 =
T 1
71
302
1.711.
= = 176.6 K
Heat extracted from cold chamber per kg of air
= cp (T 3 – T 2 ) = 1.003 (282 – 176.6) = 105.7 kJ/kg.
Heat rejected in the cooling chamber per kg of air
= cp (T 4 – T 1 ) = 1.003 (482.2 – 302) = 180.7 kJ/kg.
Since the compression and expansion are not isentropic, difference between heat rejected
and heat absorbed is not equal to the work done because there are heat transfers to the sur-
roundings and from the surroundings during compression and expansion.
To find the work done, the area of the diagram ‘1-2-3-4’ is to be considered :
Work done =
n
n− 1
(p 4 V 4 – p 3 V 3 ) –
n
n− 1
(p 1 V 1 – p 2 V 2 )