Vol. 05, Issue 03, March 2020, Available Online: www.ajeee.co.in/index.php/AJEEE THE JOULE–THOMSON (KELVIN) EFFECT AND COEFFICIENT: AN ANALYSIS
Dr. Jeet Singh
Associate Professor, Department of Physics, LSSSS Government Degree College, Mant, Mathura (U.P.)
Abstract - In thermodynamics, the Joule–Thomson effect (also called the Joule–Kelvin effect or Kelvin–Joule effect) describes the activity of a true gas or liquid (as differentiated from a perfect gas) when it's forced through a valve or porous plug while keeping it insulated so no heat is exchanged with the environment.[1][2][3] This procedure is named a throttling process or Joule–Thomson process.[4] At temperature, all gases except hydrogen, helium, and neon cool upon expansion by the Joule–Thomson process when being throttled through an orifice; these three gases experience the identical effect but only at lower temperatures.[5][6] Most liquids like hydraulic oils are going to be warmed by the Joule–
Thomson throttling process.
The gas-cooling throttling process is often exploited in refrigeration processes like liquefiers.[7][8] In hydraulics, the warming effect from Joule–Thomson throttling is wont to find internally leaking valves as these will produce heat which might be detected by thermocouple or thermal-imaging camera. Throttling may be a fundamentally process. The throttling thanks to the flow resistance in supply lines, heat exchangers, regenerators, and other components of (thermal) machines may be a source of losses that limits their performance.
For most real gases at around ambient conditions, μ is positive—i.e., the temperature falls because it passes through the constriction. For hydrogen and helium, it's negative and also the temperature increases. At higher temperatures, for many gases, μ falls and will even become negative, μ may also become negative through application of pressure, even at ambient temperature, but pressures in more than 200 bar are normally necessary to realize this.
Keywords: Thermodynamics, Joule–Thomson Effect, Coefficient, Temperature, Real gas, Liquid.
1 INTRODUCTION
In thermodynamics, the Joule–Thomson effect describes the natural action of a true gas or liquid (as differentiated from a perfect gas) when it's forced through a valve or porous plug while keeping it insulated so no heat is exchanged with the environment.[1][2][3] This procedure is termed a throttling process or Joule–
Thomson process.[4] At temperature, all gases except hydrogen, helium, and neon cool upon expansion by the Joule–
Thomson process when being throttled through an orifice; these three gases experience the identical effect but only at lower temperatures.[5][6] Most liquids like hydraulic oils are going to be warmed by the Joule–Thomson throttling process.
The gas-cooling throttling process is often exploited in refrigeration processes like liquefiers.[7][8] In hydraulics, the warming effect from Joule–Thomson throttling is wont to find internally leaking valves as these will produce heat which may be detected by thermocouple or thermal-imaging camera. Throttling could be a fundamentally process. The throttling because of the flow resistance
in supply lines, heat exchangers, regenerators, and other components of (thermal) machines may be a source of losses that limits their performance.
2 JOULE-THOMSON COEFFICIENT When a gas in steady flow passes through a constriction, e.g., in an orifice or valve, it normally experiences a change in temperature. This can be partially thanks to changes in mechanical energy, but there's another part contributed by the non-ideality of the gas. If the upstream and downstream ducts are sufficiently large for mechanical energy to be negligible at these stations, upstream and downstream temperatures are measured far enough off from the disturbance created by the constriction and therefore the system is adiabatic; the measured effect is thanks to the non-ideality alone.
The rate of change of temperature T with reference to pressure P in an exceedingly Joule–Thomson process (that is, at constant enthalpy H) is the Joule–
Thomson (Kelvin) coefficient
JT. ThisVol. 05, Issue 03, March 2020, Available Online: www.ajeee.co.in/index.php/AJEEE coefficient can be expressed in terms of
the gas's volume
V
, its heat capacity at constant pressureC
p, and its coefficient of thermal expansion
as:[1][3][17]JT
p
( 1)
H
T V
P C T
See the Derivation of the Joule–Thomson coefficient below for the proof of this relation. The value of
JTis typically expressed in °C/bar (SI units: K/Pa) and depends on the type of gas and on the temperature and pressure of the gas before expansion. Its pressure dependence is usually only a few percent for pressures up to 100 bar.Fig. 1 – Joule–Thomson coefficients for various gases at atmospheric pressure All real gases have an inversion point at which the value of
JTchanges sign. The temperature of this point, the Joule–Thomson inversion temperature, depends on the pressure of the gas before expansion.
In a gas expansion the pressure decreases, so the sign of
P
is negative by definition. With that in mind, the following table explains when the Joule–Thomson effect cools or warms a real gas:If the gas temperature is then
JTis since P
is thus T
must be so the gas below the inversion temperature positive always negative negative cools above the inversion temperature negative always negative positive warms Helium and hydrogen are two gaseswhose Joule–Thomson inversion temperatures at a pressure of 1 atmosphere are very low (e.g., about 45 K (−228 °C) for helium). Thus, helium and hydrogen warm when expanded at constant enthalpy at typical room temperatures. On the opposite hand, nitrogen and oxygen, the 2 most abundant gases in air, have inversion temperatures of 621 K (348 °C) and 764 K (491 °C) respectively: these gases is cooled from temperature by the Joule–Thomson effect.[1]
For an ideal gas,
JTis always equal to zero: ideal gases neither warm nor cool upon being expanded at constant enthalpy.2.1 Proof that the precise enthalpy remains constant
In thermodynamics so-called "specific"
quantities are quantities per unit mass (kg) and are denoted by lower-case characters. So h, u, and v are the particular enthalpy, specific internal energy, and specific volume (volume per unit mass, or reciprocal density), respectively. In a very Joule–Thomson process the precise enthalpy h remains
constant.[20] To prove this, the primary step is to compute the online work done when a mass m of the gas moves through the plug. This amount of gas contains a volume of V1 = m v1 within the region at pressure P1 (region 1) and a volume V2 = m v2 when within the region at pressure P2 (region 2). Then in region 1, the "flow work" done on the number of gas by the remainder of the gas is: W1 = m P1v1. In region 2, the work done by the number of gas on the remainder of the gas is: W2 = m P2v2. So, the overall work done on the mass m of gas is
1 1 2 2
.
W mPv mP v
The change in internal energy minus the total work done on the amount of gas is, by the first law of thermodynamics, the total heat supplied to the amount of gas.
U W Q
In the Joule–Thomson process, the gas is insulated, so no heat is absorbed. This means that
2 1 1 1 2 2
( mu mu ) ( mPv mP v ) 0
1 1 1 2 2 2
mu mPv mu mP v
1 1 1 2 2 2
u Pv u P v
where u1 and u2 denote the specific internal energies of the gas in regions 1
Vol. 05, Issue 03, March 2020, Available Online: www.ajeee.co.in/index.php/AJEEE and 2, respectively. Using the definition of
the specific enthalpy h = u + Pv, the above equation implies that
1 2
h h
where h1 and h2 denote the specific enthalpies of the amount of gas in regions 1 and 2, respectively.
3 THROTTLING IN THE T-S DIAGRAM
Fig. 2 – T-s diagram of nitrogen.
The red dome represents the two-phase region with the low-entropy side (the saturated liquid) and the high-entropy side (the saturated gas). The black curves give the T-s relation along isobars. The pressures are indicated in bar. The blue curves are isenthalps (curves of constant specific enthalpy). The specific enthalpies are indicated in kJ/kg. The specific points a, b, etc., are treated in the main text.
A very convenient way to get a quantitative understanding of the throttling process is by using diagrams such as h-T diagrams, h-P diagrams, and others. Commonly used are the so-called T-s diagrams. Figure 2 shows the T-s diagram of nitrogen as an example.[21]
Various points are indicated as follows:
a) T = 300 K, p = 200 bar, s = 5.16 kJ/(kgK), h = 430 kJ/kg;
b) T = 270 K, p = 1 bar, s = 6.79 kJ/(kgK), h = 430 kJ/kg;
c) T = 133 K, p = 200 bar, s = 3.75 kJ/(kgK), h = 150 kJ/kg;
d) T = 77.2 K, p = 1 bar, s = 4.40 kJ/(kgK), h = 150 kJ/kg;
e) T = 77.2 K, p = 1 bar, s = 2.83 kJ/(kgK), h = 28 kJ/kg (saturated liquid at 1 bar);
f) T = 77.2 K, p = 1 bar, s = 5.41 kJ/(kgK), h =230 kJ/kg (saturated gas at 1 bar).
As shown before, throttling keeps h constant. E.g. throttling from 200 bar and 300 K (point a in fig. 2) follows the
isenthalpic (line of constant specific enthalpy) of 430 kJ/kg. At 1 bar it leads to point b which includes a temperature of 270 K. So throttling from 200 bar to 1 bar gives a cooling from temperature to below the melting point of water.
Throttling from 200 bar and an initial temperature of 133 K (point c in fig. 2) to 1 bar leads to point d, which is within the two-phase region of nitrogen at a temperature of 77.2 K. Since the enthalpy is an in depth parameter the enthalpy in d (hd) is adequate the enthalpy in e (he) multiplied with the mass fraction of the liquid in d (xd) plus the enthalpy in f (hf) multiplied with the mass fraction of the gas in d (1 − xd). So
(1 ) .
d d e d f
h x h x h
With numbers: 150 = xd 28 + (1 − xd) 230 so xd is about 0.40. This means that the mass fraction of the liquid in the liquid–
gas mixture leaving the throttling valve is 40%.
3.1 Derivation of the Joule–Thomson coefficient
It is difficult to think physically about what the Joule–Thomson coefficient,
JT, represents. Also, modern determinations of
JTdo not use the original method used by Joule and Thomson, but instead measure a different, closely related quantity.[22] Thus, it is useful to derive relationships between
JTand other, more conveniently measured quantities, as described below.The first step in obtaining these results is to note that the Joule–Thomson coefficient involves the three variables T, P, and H. A useful result is immediately obtained by applying the cyclic rule; in terms of these three variables that rule may be written
1.
H P T
T H P
P T H
Each of the three partial derivatives in this expression has a specific meaning.
The first is
JT, the second is the constant pressure heat capacity,C
p, defined byp
C H
T
Vol. 05, Issue 03, March 2020, Available Online: www.ajeee.co.in/index.php/AJEEE and the third is the inverse of the
isothermal Joule–Thomson coefficient,
T, defined byT
T
H
P
.This last quantity is more easily measured than
JT.[23][24] Thus, the expression from the cyclic rule becomesT
JT
.
C
p
This equation can be used to obtain Joule–Thomson coefficients from the more easily measured isothermal Joule–
Thomson coefficient. It is used in the following to obtain a mathematical expression for the Joule–Thomson coefficient in terms of the volumetric properties of a fluid.
To proceed further, the starting point is the fundamental equation of thermodynamics in terms of enthalpy;
this is
d H T S V P d d .
Now "dividing through" by dP, while holding temperature constant, yields
T T
H S
T V
P P
The partial derivative on the left is the isothermal Joule–Thomson coefficient,
T, and the one on the right can be expressed in terms of the coefficient of thermal expansion via a Maxwell relation.The appropriate relation is
T P
S V
P T V
where α is the cubic coefficient of thermal expansion. Replacing these two partial derivatives yields
T
TV V .
This expression can now replace
Tin the earlier equation for
JTto obtain:JT
p
( 1).
H
T V
P C T
This provides an expression for the Joule–Thomson coefficient in terms of the commonly available properties heat capacity, molar volume, and thermal
expansion coefficient. It shows that the Joule–Thomson inversion temperature, at which
JTis zero, occurs when the coefficient of thermal expansion is equal to the inverse of the temperature. Since this is true at all temperatures for ideal gases, the Joule–Thomson coefficient of an ideal gas is zero at all temperatures.[25]3.2 Joule's second law
It is easy to verify that for an ideal gas defined by suitable microscopic postulates that αT = 1, so the temperature change of such an ideal gas at a Joule–Thomson expansion is zero. For such an ideal gas, this theoretical result implies that:
The internal energy of a fixed mass of an ideal gas depends only on its temperature (not pressure or volume).
This rule was originally found by Joule experimentally for real gases and is known as Joule's second law. More refined experiments found important deviations from it.
4 CONCLUSION
The Joule–Thomson expansion occurs at constant enthalpy through a valve or throttling device
dH=CpdT+[V−T(∂V∂T)P]dP=0
And the Joule–Thomson coefficient μ is defined as the ratio of the temperature change to the pressure drop, and is expressed in terms of the thermal expansion coefficient and the heat capacity
μJT=(∂T∂P)H=−[V−T(∂V/∂T)P]Cp=−V(1−βT) Cp
The Joule–Thomson coefficient are going to be zero at a degree called inversion point (T = 1/β) for all real gases.
Expansion of most real gases causes cooling when the Joule–Thomson coefficient is positive and also the gas temperature is below the inversion temperature. However, at air pressure, because the inversion temperature for hydrogen is low (202K) and hence hydrogen will warm during a Joule–
Thomson expansion at temperature. Since there's no change of temperature when a perfect gas expands through a throttling device, a nonzero Joule–Thomson coefficient refers to a true gas.
For most real gases at around ambient conditions, μ is positive—i.e., the temperature falls because it passes
Vol. 05, Issue 03, March 2020, Available Online: www.ajeee.co.in/index.php/AJEEE through the constriction. For hydrogen
and helium, it's negative and therefore the temperature increases. At higher temperatures, for many gases, μ falls and should even become negative, μ can even become negative through application of pressure, even at ambient temperature, but pressures in way over 200 bar are normally necessary to attain this.
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