Joule-Thomson Effect - Wiley Online Library

Joule-Thomson Effect
CHRISTOPH WINDMEIER, Linde AG, Pullach (M€unchen), Germany
RANDALL F. BARRON, Louisiana Tech University, Ruston, United States
A fluid flow encountering an obstruction like,
e.g., an expansion valve, an orifice plate, a
porous plug, or a capillary tube, will experience
a pressure drop. Since outer heat transfer and
changes in kinetic energy are usually negligible, and because there is no transfer of work
from and to the fluid flow, the resulting expansion can be considered to be isenthalpic. A
non-ideal fluid will hence experience a change
in temperature. This fundamental phenomenon
is called the Joule–Thomson (J–T) effect and it
enables the production of cold temperature
levels by isenthalpic expansion of highpressure fluids.
The origin of the J–T effect lies within the
occurrence of intermolecular forces in nonideal fluids: A pressure reduction is accompanied by an increase of specific volume. Consequently the average intermolecular distances
also increase. The internal work done on
attractive or repulsive interactions will result
to a respective change of system temperature.
The constant enthalpy lines on the temperature–
pressure plane are shown in Figure 1. The effect
of temperature change for an isenthalpic
change in pressure is represented by the
Joule–Thomson coefficient, mJT, defined by
mJT ¼
@T
@p
h
The Joule–Thomson coefficient is equal to
the slope of the isenthalpic lines in Figure 1. In
the region where mJT < 0, isenthalpic expansion results into an increase in temperature,
whereas in the region where mJT > 0, respective expansion results into a temperature
decrease. The curve that separates the two
regions is called the inversion curve. The
J–T coefficient is zero along the inversion
# 2015 Wiley-VCH Verlag GmbH & Co. KGaA, Weinheim
10.1002/14356007
curve, because the slope of the isenthalpic
line is zero.
Thermodynamics show that the J–T
coefficient may be expressed in terms of volumetric and caloric properties as follows:
mJT ¼
" #
1
@v
T
v
cp
@T p
ð1Þ
where v is the specific volume and cp the
specific heat at constant pressure of the
material.
Since an ideal gas does not possess any
intermolecular forces one can expect a value
of zero for its J–T coefficient. Substituting the
equation of state of an ideal gas v ¼ (RT)/p into
Equation 1 gives:
@v
@T
p
¼
R v
¼
p T
where R is the gas constant and p is the pressure.
Substitution into Equation 1 gives mJT ¼ 0 for
an ideal gas. For a real gas, the J–T coefficient
may be negative, zero, or positive, depending
on the initial temperature and pressure of the
material.
The simplest equation of state for real gases
is the van der Waals equation of state:
pþ
a
ðv bÞ ¼ RT
v2
where the constant a gives a measure of the
intermolecular forces between the molecules,
and the constant b provides a measure of the
finite size of the gas molecules. Constants a
and b are thus dependent upon the species and
can be determined using the first and second
derivative of the isothermal line at the critical
point.
2
Joule-Thomson Effect
The inversion curve is formed by all points at
which the J–T coefficient is zero. The inversion
temperature Ti from Equation 2 is given by:
Ti ¼
2a
ð1 b=vÞ2
bR
The maximum inversion temperature for a
van der Waals fluid is the temperature on the
inversion curve where p ¼ 0 (or b/v ¼ 0) or Ti,
max ¼ 2a/bR. The maximum inversion temperature (in K) for several gases is as follows:
Figure 1. General temperature–pressure diagram for a
real gas
The expression for the J–T coefficient for a
van der Waals gas is given by Equation 2:
mJT ¼
ð2a=RT Þð1 b=vÞ2 b
h
i
cp 1 ð2=vRT Þð1 b=vÞ2
ð2Þ
For large values of the specific volume, as
typically occurring in gas phases, Equation 2
can be approximated by:
mJT ¼
1 2a
b
cp RT
This expression shows that the J–T
coefficient is positive when T < 2a/bR and
negative when T > 2a/bR.
Helium [4He]
Carbon monoxide
Hydrogen
Neon
Argon
Air
Oxygen
Nitrogen
Methane
45
652
205
250
794
603
761
621
939
The maximum inversion temperatures for
helium, hydrogen, and neon are below ambient
temperature. When starting out from ambient
temperature, liquefaction systems that use J–T
expansion alone to produce low temperature
cannot be used to liquefy these gases: Additional means such as external sources of refrigeration or expansion engines must be used in
order to liquefy these gases.