McKay, Donald. "Front matter" Multimedia Environmental Models ...

©2001 CRC Press LLC
N = ABDC/
Dy
mol/h
Combining B and Dy in one term k M, equal to B/Dy, with dimensions of velocity
thus appears to decrease our ignorance, since we now do not know one quantity
instead of two. Hence we write
N = Ak MDC mol/h
Term k M is termed a mass transfer coefficient, has units of velocity (m/h), and is
widely used in environmental transport equations. It can be viewed as the net
diffusion velocity. The flux N in one direction is then the product of the velocity,
area, and concentration.
For example, if, as in the lower section of Figure 7.2, diffusion is occurring in
an area of 1 m 2 from point 1 to 2, C 1 is 10 mol/m 3 , C 2 is 8 mol/m 3 , and k M is 2.0
m/h, we may have diffusion from 1 to 2 at a velocity of 2.0 m/h, giving a flux of
k MAC 1 of 20 mol/h. There is an opposing flux from 2 to 1 of k MAC 2 or 16 mol/h.
The net flux is thus the difference or 4 mol/h from 1 to 2, which of course equals
k MA(C 1 – C 2). The group k MA is an effective volumetric flowrate and is equivalent
to the term G m 3 /h, introduced for advective flow in Chapter 6.
7.3.4 Fugacity Format, D Values for Diffusion
The concentration approach is to calculate diffusion fluxes N as ABdC/dy or
ABDC/Dy or k MADC. In fugacity format, we substitute Zf for C and define D values
as BAZ/Dy or k MAZ, and the flux is then DDf, since DC is ZDf. Note that the units
of D are mol/Pa h, identical to those used for advection and reaction D values.
Worked Example 7.1
D = BAZ/Dy or D = k MAZ
N = Df 1 – Df 2 = D(f 1 – f 2)
A chemical is diffusing through a layer of still water 1 mm thick, with an area
of 200 m 2 and with concentrations on either side of 15 and 5 mol/m 3 . If the diffusivity
is 10 –5 cm 2 /s, what is the flux and the mass transfer coefficient?
Thus,
The flux N is thus
y = 10 –3 m, B = 10 –5 cm 2 /s ¥ 10 –4 m 2 /cm 2 = 10 –9 m 2 /s
k M is B/Dy = 10 –6 m/s
k MA(C 1–C 2) = 10 –6 (200(15 – 5)) = 0.002 mol/s