Page 656 - Bird R.B. Transport phenomena

636 Chapter 20 Concentration Distributions with More Than One Independent Variable

with the boundary conditions: /(0) = 1 and /(°°) = 0. A solution of the form/(17) is possi-
ble only if the factor in parentheses is a constant. Setting the constant equal to 3 reduces
Eq. 20.3-21 to Eq. 18.6-6, for which the solution is known. Therefore we now get the
boundary layer thickness by requiring that

1 - s ' (20.3-22)
2
1 AB \h x
or

(20.3-23)
P
3
The solution of this first-order, linear equation for 8 is
A
(20.3-24)
^ A =
Hence the solution to the problem in this subsection is
f °° _ _
exp (-17) drj
J 7J (20.3-25)

which reduces to Eq. 18.6-10 for the system considered there.
Finally, we get the expression for the molar flux at the interface, which is

N
A

(20.3-26)
г
Сз) $93> АВ ft Vhjihji dx
z
For a plane surface, with h x = h z = 1 and /3 = constant, Eq. 20.3-26 reduces to Eq. 18.6-11.

EXAMPLE 20.3-1 A liquid В is flowing very slowly around a spherical bubble of gas A of radius R. Find the rate
of mass transfer of A into the surrounding fluid, if the solubility of gas A in liquid В is c .
A0
Mass Transfer for (a) Show how to use Eq. 20.3-14 to get the mass flux at the gas-liquid interface for this system.
Creeping Flow Around (b) Then get the average mass flux over the entire spherical surface.
a Gas Bubble
SOLUTION
(a) Select as the origin of coordinates the upstream stagnation point, and define the coordi-
nates x and z as follows: x = R0 and z = K(sin 0)ф, in which 6 and ф are the usual spherical
coordinates. The у direction is then the same as the r direction of spherical coordinates. The
interfacial velocity is obtained from Eq. 4B.3-3 as v s = \v^ sin в, where v x is the approach
velocity.
When these quantities are inserted into Eq. 20.3-14 we get

(20.3-27)
3
27rR Vcos в - 3 cos в + 2