Page 444 - Advanced thermodynamics for engineers

434 CHAPTER 18 LIQUEFACTION OF GASES

!
r ffiffiffiffiffiffiffiffiffiffi! r ffiffiffiffiffiffiffiffiffiffi
a <bT i <bT i
p ¼ 2 1 3 1 (18.24)
b 2a 2a
The maximum and minimum inversion temperatures are achieved when the pressure, p, is zero.
This gives
2a ^ 2a T i
b
T i ¼ ; and T i ¼ ¼ (18.25)
b
<b 9<b 9
The value of the critical temperature for a van der Waals gas is
8a
T c ¼ ; (18.26)
27<b
and hence

T i Maximum inversion temperature
b
¼ ¼ 6:75
T c Critical temperature
(18.27(a, b))
^
T i Minimum inversion temperature
¼ ¼ 0:75
T c Critical temperature
These equations will now be applied to air: van der Waals equation for air was given as Eqn (18.15).
Substituting these values of coefficients into Eqn (18.24), which can be solved as a quadratic equation
in T i , gives the diagram shown in Fig. 18.9. This shows that it is not possible to liquefy air, if it is above
a pressure of about 340 bar.
The maximum pressure for which inversion can occur is when dp/dT ¼ 0. This can be related to
Van der Waals equation in the following way, using the expression for x introduced in Eqn (18.23).
dp dp dx dx dp dp
¼ ; and since is a simple single term expression; ¼ 0 when ¼ 0:
dT dx dT dT dT dx
dp a
¼ 3ð1 xÞ ð3x 1Þ (18.28)
dx b 2
Equation (18.28) is zero when 3(1 x) (3x 1) ¼ 0, i.e. when x ¼ 2/3.
Hence the maximum pressure at which inversion can occur is

a 2 2 a
p ¼ 2 1 3 1 ¼ 2 ¼ 9p c (18.29)
b 3 3 3b
Thus the maximum pressure at which a Van der Waals gas can be liquefied using the Joule–
Thomson effect is nine times the critical pressure. Substituting for a and b in Eqn (18.29) gives the
maximum pressure as 341.6 bar.
Figure 18.9 indicates that it is not possible to cool air using the Joule–Thomson effect, if it is at a
temperature of greater than 900 K, or at one less than about 100 K. Similar calculations for hydrogen