Joule Thomson Coefficient

1. Joule Thomson Coefficient

The application of the Joule Thomson Coefficient is illustrated through the following example (that is schematically described in Figure 21 ). Assume that you can plug a thick ce- ramic pipe with a porous sponge and flow a high–pressure gas through it. Significant pressure losses will occur without a consequential change in the kinetic or potential energies. From en- ergy balance analysis for an open system we can show that the sum of the enthalpies and the

kinetic and potential energies is constant, i.e., h i + ke i + pe i =h e + ke e + pe e = constant. Here, the subscripts i and e, respectively, denote the inlet and exit. When the inlet and exit areas are equal, i.e., A i =A e , then ρ i v i = ρ e v e , where v denotes the flow velocity. For a negligible change

in the potential energy, h + ke

= h e + ke i ( ρ i / ρ e ) . Consequently, the temperature across the sponge changes as the pressure is decreased, as illustrated in Figure 21 . In general, changes in

the kinetic energy are not large and we can assume that h i = h e . (Although, strictly speaking,

for compressible fluids ρ e ≠ρ i and, hence, h e ≠ h i .)

Throttling is essentially an isenthalpic expansion process. The Joule Thomson coeffi- cient ( µ JT ) is defined as,

µ JT =( ∂T/∂P) h (154) Since h = h(T,P),

dh = ( ∂h/∂T) P dT + ( ∂h/∂P) T dP = c p dT + ( ∂h/∂P) T dP. For an isenthalpic process, dh = 0, i.e.,

µ JT =( ∂T/∂P) h = –( ∂h/∂P) T /c P .

For an ideal gas, ∂h o / ∂P = 0, so that the above equation can be re-written as µ JT =– ∂((h – h o )/c p )/ ∂P.

(155) where h- h 0 is known for any specified equation of state (Eq. (66)) or from enthalpy correction

charts ( Figure B-3 ) As illustrated by curve “iIABe” in Figure 21(b) , for inlet pressures P i < P I , the temperature decreases with a decrease in pressure and µ JT > 0. On the other hand, for a

flow with the same enthalpy, if P i > P I , initially, µ JT < 0 and temperature increases with de- crease in pressure. Later, when P < P I , the temperature will fall. The point I Figure 21(b) is

called the inversion point where µ JT changes sign. If inlet temperature is changed at same Pi enthalpy changes and as such inversion points change as shown by inversion curve KIL.

jj. Example 36 The experimentally determined value of µ JT for N is 0.31 K bar –1 2 at–70 °C and 5

MPa. If the gas is throttled from 6 MPa and –67ºC to 4 MPa, what is the final exit temperature?

Solution µ JT = ( ∆h/c p )/ ∆P = ∆T/∆P = (T e –203)/(40 – 60) = 0.31 K bar –1 . Therefore, T e =

–73.2ºC.

a. Evaluation of µ JT Recall that

dh = c p dT + (v – T( ∂v/∂T) P )dP. Since dh ≈ 0 during throttling, we obtain the relation µ 2

JT = (dT/dP) h = –(v – T( ∂v/∂T) P )/c P = (T /c P ) ∂(v/T)/∂T, or (156)

µ JT = (dT/dP) h = –(1 – T β P )/(c P /v).

b. Remarks For ideal gases, v = RT/P so that T( ∂v/∂T) = v and, hence, µ JT = 0. There is no tem-

perature change due to throttling for ideal gases. For incompressible fluids ∂v/∂T = β P = 0, and µ JT = –v/c P has a negative value. There- fore, liquids generally heat up upon throttling. If v < T( ∂v/∂T) P or T β P < 1, µ JT > 0, and vice versa.

At the inversion point, µ JT = 0. Inversion occurs when T inv β P = (T/v)( ∂v/∂T) P = 1. For

a real gas Pv = ZRT, i.e., ∂v/∂T = ZR/P + (RT/P)∂Z/∂T. If (T/v)(∂v/∂T) P = 1, this im- plies that ZRT/Pv + (RT 2 /Pv) ∂Z/∂T = 1, or (∂Z/∂T)

P = 0.

For cooling to occur, the inlet pressure must be lower than the inversion pressure. Cooling of a gas can also be accomplished using isentropic expansion. We define µ s =

( ∂T/∂P) s that is related to the temperature decrease due to the work delivered during an isentropic process. Recall that

ds = c p dT/T – ( ∂v/∂T) P dP. For an isentropic process, ds = 0, and µ s = T( ∂v/∂T) P /c P = Tv β P /c P .

(158) Dividing Eq.(157) by Eq.(158), µ JT / µ s = 1 – (1/(T β P ))

(159) Equation (159) provides a relation to indicate relative degree of heating of the isen-

tropic to isenthalpic throttling process. For substances that expand upon heating β P > 0 and, hence, µ s > 0. If β P T > 1, then

µ JT < µ s , i.e., the isentropic expansion results in greater cooling than isenthalpic ex- pansion for the same pressure ratio. The values of T inv , β , β T , and P sat P can be directly obtained from the state equations,

while µ JT , h, u, c v , and other such properties depend both on the equations of state and