Page 443 - Advanced thermodynamics for engineers

18.2 LIQUEFACTION BY EXPANSION – METHOD (II) 433

vT 1 a v b 2a
and hence; ¼ p þ ;
vv p < v 2 < v 3
(18.17)
vv <
giving ¼
vT a 2a
p
p þ ðv bÞ
v 2 v 3
Thus, from Eqn (18.17), the Joule–Thomson coefficient for a van der Waals gas is not zero at all
points, being given by
2 3
<T
1 6 7
m ¼ 6 v 7 (18.18)
4 a 2a 5
c p
p þ 2 ðv bÞ 3
v v
18.2.2 MAXIMUM AND MINIMUM INVERSION TEMPERATURES
The general form of Van der Waals equation, Eqn (18.15), can be rewritten as

a ðv bÞ

T ¼ p þ : (18.16)
v 2 <
The inversion temperature is defined as
vT

T i ¼ v (18.19)
vv
p
which can be evaluated from Eqn (18.16) in the following way.

vT 1 a v b 2a
¼ p þ ;
vv < v 2 < v 3
p
(18.20)

vT 1 a 2a
giving v ¼ v p þ ðv bÞ
vv < v 2 v 2
p
It is now necessary to solve for v in terms of T i alone, and this can be achieved by multiplying
Eqn (18.16) by v and Eqn (18.20) by (v b). This gives
a vðv bÞ

vT i ¼ p þ 2 ; (18.21)
v <
2
vðv bÞ a ðv bÞ 2a
and ðv bÞT i ¼ p þ (18.22)
< v 2 < v 2
Subtracting Eqns (18.21) and (18.22) gives
r ffiffiffiffiffiffiffiffiffiffi

v b <bT i b
¼ ¼ x ¼ 1 (18.23)
v 2a v
Substituting Eqn (18.23) in Eqn (18.15) gives