Modern Engineering Thermodynamics

778 CHAPTER 19: Introduction to Coupled Phenomena
and, when J M = 0, we get a situation of pure thermal conduction, which is described by Fourier’s law,
Eq. (19.18), as
dT
J Q = _q F = _Q F /A = − k t
dx
where k t is the thermal conductivity of the fluid. Setting J M = 0 in Eq. (19.47) and using Eq. (19.18) to solve for
L QQ gives
L QQ = k t T 2 + L 2 QM /L MM
Using the results from Eqs. (19.38) and (19.42) for L QM and L MM in this equation produces
L QQ = k t T 2 + Tμk 2 o /k p (19.48)
Substituting the formula for L QQ , L QM = L MQ , and L MM back into Eqs. (19.34) and (19.35) gives the final formula
for the coupled fluxes as
J Q = − k t + μk2 o dT dp
− k o
(19.49)
Tk p dx dx
and
J M = − ρk o
T
dT
dx
− ρk p
μ
dp
dx
Substituting these results into the total EPRD formula of Eq. (19.14) gives
σ thermomechanical = k t + μk
o 1 dT 2
2k + o dT dp
Tk p T dx T 2 dx dx
+ k p
μT
(19.50)
2 dp
(19.51)
dx
When the system has reached a steady state condition with J M = 0, then Eq. (19.50) shows us that the pressures
and temperatures are coupled, so that
dp/dx JM
= dp JM
= − μk o
(19.52)
dT/dx
=0
dT
=0
Tk p
Here, dp is the pressure change “induced” by the temperature change dT under the condition of zero mass flow
rate. It is called the thermomolecular pressure difference.
Similarly, when the system has reached a steady state condition with J Q = 0, Eq. (19.49) gives
dp JQ
= − k t
− μk o
= − k
t
+
dp
dT
=0
k o Tk p k o dT (19.53)
J M =0
It is conceivable that these equations may some day be found to provide a basic understanding of various
phenomena of industrial value, such as the vortex tube temperature separation effect discussed earlier.
EXAMPLE 19.4
A membrane with a permeability of 1.00 × 10 −6 m 2 separates two chambers filled with carbon dioxide gas. The gas has a
temperature of 300. K on one side of the membrane and 305 K on the other side. The osmotic heat conductivity (k o ) of the
membrane with CO 2 is 2.00 × 10 4 m 2 /s and the viscosity of the CO 2 is 1.50 × 10 −5 kg/(m · s). Determine the steady state
thermomolecular pressure difference across the membrane.
Solution
Since this is a steady state problem, we can use the result given in Eq. (19.52),
dp = −
k p
μk o
dT
T
and, assuming μ, k o , and k p are all constants over the small temperature range of 300. to 305 K, we can integrate this equation
to find
p 2 − p 1 = − μk
o
k p
ln T
2
T 1
= − ð1:50 × 10−5 kg/ðm.sÞÞð2:00 × 10 4 m 2
/sÞ
1:00 × 10 −6 m 2
ln 305
= −4960 N/m 2
300: