REFRIGERATION CYCLES 723

dharm

\M-therm\Th14-1.pm5

In this system, work gained from expander is employed for compression of air, consequently
less external work is needed for operation of the system. In practice it may or may not be done e.g.,
in some aircraft refrigeration systems which employ air refrigeration cycle the expansion work
may be used for driving other devices.
This system uses reversed Brayton cycle which is described below :
Figs. 14.5 (a) and (b) shows p-V and T-s diagrams for a reversed Brayton cycle. Here it is
assumed that (i) absorption and rejection of heat are constant pressure processes and (ii) Com-
pression and expansion are isentropic processes.
p (Pressure)

V (Volume)

(a)

14

2 3

Expansion

Compression

T (Temp.)

s (Entropy)

(b)

1

4

3

2

Expansion

Constant pressure lines

Compression

Fig. 14.5. (a) p-V diagram. Fig. 14.5. (b) T-s diagram.
Considering m kg of air :
Heat absorbed in refrigerator, Qadded = m × cp × (T 3 – T 2 )
Heat rejected is cooler, Qrejected = m × cp × (T 4 – T 1 )
If the process is considered to be polytropic, the steady flow work of compression is given by,
Wcomp =
n
n− 1 (p^4 V^4 – p^3 V^3 ) ...(14.6)
Similarly work of expansion is given by,
Wexp. =
n
n− 1 (p^1 V^1 – p^2 V^2 ) ...(14.7)
Equations (14.6) and (14.7) may easily be reduced to the theoretical isentropic process shown
in Fig. 14.5 (b) by substituting γ = n and the known relationship.

R = cp γ

γ

F −

HG

I

KJ

(^1) J
The net external work required for operation of the cycle
= Steady flow work of compression – Steady flow work of expansion
= Wcomp. – Wexp.

−
F
HG
I
KJ
−−+

−
F
HG
I
KJ
−−+

−
F
HG
I
KJ
−−+
n
n
pV pV pV pV
n
n
mR T T T T
n
n
mR
J
TTTT
1
1
1
44 33 11 22
4312
4312
()
()
()
Q pV mRT
pV mRT
pV mRT
pV mRT
11 1
22 2
33 3
44 4

=

=
L
N
M M M M M M
O
Q
P P P P P P
in heat units.