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

©2001 CRC Press LLC
N = VdC W/dt = –k OWA(C W – C A/K AW)
where C A and C W are concentrations in air and water and k OW the overall MTC.
Assuming C A to be zero, integrating gives
or
C W = C WO exp(–k OWAt/V) = C WO exp (–k OWt/Y)
f W = f WO exp(–D Vt/VZ W)
Plotting C W on semilog paper vs. linear time gives a measurable slope of –k OW/Y,
hence k OW can be estimated. A system of this type has been described by Mackay
and Yuen (1983) and is illustrated in Figure 7.8.
A very useful quantity is the evaporation half-life, which is 0.693Y/k OW and
0.693 VZ W/D V. Often, an order of magnitude estimate of this time is sufficient to
show that volatilization is unimportant or that it dominates other processes, such as
reaction.
As noted earlier, measurement of the individual contributing air and water D
values or MTCs is impossible, because the interfacial concentrations cannot be
measured. If, however, the evaporation rates of a series of chemicals of different
K AW are measured, it is possible to deduce k W and k A or D W and D A.
The relationship 1/k OW = 1/k W + 1/k AK AW suggests plotting, as in Figure 7.8,
1/k OW versus 1/K AW for a series of chemicals. The intercept will be 1/k W and the
slope 1/k A.
A correction may be necessary for molecular diffusivity differences. k W or D W
is measured by selecting chemicals of high K AW for which the term 1/k AK AW or 1/D A
is negligible. Alkanes, oxygen, or inert gases are convenient. k A or D A is measured
by choosing chemicals of low K AW such that 1/k AK AW or 1/D A is large compared to
1/k W or 1/D W. Alcohols are convenient for this purpose.
Worked Example 7.4
A tank contains 2 m 3 of water at 25°C, 50 cm deep, with dissolved benzene
(K AW = 0.22) and naphthalene (K AW = 0.017), each at a concentration of 0.1 mol/m 3 .
After 2 hours, these concentrations have dropped to 47.1 and 63.9% of their initial
value, respectively. What are the overall and individual MTCs and D values?
In each case, C W = C WO exp(–k OWt/Y), Y being 0.5 m. Thus, k OW = –(Y/t) ln
(C W/C WO). Substituting gives
Benzene k OW = 0.188 m/h
Naphthalene k OW = 0.112 m/h
Assuming each k OW to be made up of identical k W and k A values, i.e.,
1/k OW = 1/k W + 1/(K AWk A)