Modern Engineering Thermodynamics

780 CHAPTER 19: Introduction to Coupled Phenomena
Then, these equations give
dp JM
= − ρ _Q i
dT
=0
T _m = − ρ _S i
_m
(19.56)
where _S i = _Q i /T is the isothermal entropy transport rate “induced” by the thermomechanical mass flow rate _m :
19.5.3 Zero Pressure Gradient
When both vessels are at different temperatures but at equal pressures, p = constant and dp/dx = 0. Then,
Eqs. (19.49) and (19.50) give
and
Then,
J p=constant Q = − k t + μk2 o dT
Tk p dx
J p=constant M = − ρk o
T
J Q
J M
dT
dx
p=constant
= Tk t
ρk o
+ μk o
ρk p
(19.57)
This time, we define the isobaric mass flow rate _m “induced” by the thermomechanical heat transfer rate _Q i as
J Q p=constant Q
= _ _m = Tk t
+ μk o
(19.58)
ρk o ρk p
J M
Combining Eqs. (19.57) and (19.58) gives the isobaric mass flow rate induced by the thermomechanical heat
transfer in the thermomechanical system as
_m i =
ρ _Q
Tk t /k o + μk o /k p
(19.59)
EXAMPLE 19.5
Both of the large vessels shown in Figure 19.10 are filled with saturated liquid water at 30.0°C. They are maintained isothermal
but have a pressure difference of 10.0 kPa. The interconnecting tube has an inside diameter of 0.0100 m and is 0.100 m
in length. It is filled with a porous material having a permeability of 1.00 × 10 −12 m 2 . Careful measurements reveal that the
isothermal energy transport rate in this system is 15.0 J/s. Determine
a. The thermomechanical mass flow rate between the vessels.
b. The osmotic heat conductivity coefficient.
c. The isothermal entropy transport rate induced by the thermomechanical mass flow rate.
Saturated
liquid
water
Pressure p 1
Temperature T 1
0.100 m
Pressure p 2
p 2 − p 1 = 10.0 kPa
Temperature T 2
T 2 = T 1 = 30.0°C
Tube filled with porous material
FIGURE 19.10
Example 19.5.