Page 442 - Advanced thermodynamics for engineers

432 CHAPTER 18 LIQUEFACTION OF GASES

1 vv
This can be written in terms of the coefficient of expansion b ¼ , and Eqn (18.10)
becomes v vT p
v
m ¼ ½bT 1Š: (18.11)
c p
At the inversion temperature

vT 1

¼ 0; giving T i ¼ : (18.12)
vp b
h
It is possible to evaluate the Joule–Thomson coefficient for various gases from their state equations.
For example the Joule–Thomson coefficient for a perfect gas can be evaluated by evaluating m from
Eqn (18.10) by differentiating the perfect gas equation

pv ¼ RT (18.13)
Now,
" #
1 vv
m ¼ T v :
c p vT p
and, from Eqn (18.13) the derivative

vv R
¼ (18.14)
vT p
p
Hence,
1 R
m ¼ T v ¼ 0:
c p p
This means that it is not possible to cool a perfect gas by the Joule–Thomson effect. This is what
would be expected, because the enthalpy of an ideal gas is not a function of pressure. However, it does
not mean that gases which obey the ideal gas law at normal atmospheric conditions (e.g. oxygen,
nitrogen, etc.) cannot be liquefied using the Joule–Thomson effect, because they cease to obey this law
close to the saturated vapour line. The possibility of liquefying a gas obeying the van der Waals
equation is considered below.
3
2
Van der Waals equation for air may be written, using a and b for air as 1.358 bar (m /kmol) and
3
0.0364 m /kmol respectively,
0:083143T 1:358
p ¼ 2 (18.15)
ðv 0:0364Þ v
To assess whether a van der Waals gas can be liquefied using the Joule–Thomson effect, it is
necessary to evaluate the Joule–Thomson coefficient, m. This is related to (vv/vT) p , which can be
evaluated in the following way.
From Eqn (18.15), in general form

a ðv bÞ

T ¼ p þ 2 (18.16)
v <