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

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©2001 CRC Press LLC<br />

f W = f WO exp(–D Vt/VZ W)<br />

D V for benzene = 1.38 ¥ 10 –3 (note that this is k OWAZ W)<br />

D V for naphthalene = 10.6 ¥ 10 –3<br />

Now, 1/D V equals (1/D A + 1/D W), the D A value being common to both chemicals.<br />

D W contains the variable Z W; therefore, D W for naphthalene is D W for benzene times<br />

23.7 ¥ 10 –3 /1.836 ¥ 10 –3 or 12.9.<br />

Benzene D A = 0.0242 D W = 0.00146<br />

Naphthalene D A = 0.0242 D W = 0.0189<br />

In practice, it is unwise to rely on only two chemicals, it being better to use at<br />

least five, covering a wide range of K AW values. The air phase resistance, when<br />

viewed as 1/D A, is 41.3 units in both cases, but the water phase resistance for benzene<br />

is 685, while for naphthalene it is 52.9. Benzene experiences 5.7% of the transfer<br />

resistance in the air, while naphthalene experiences 44% resistance in the air, because<br />

it has a much lower K AW.<br />

Example 7.5<br />

Ten kilograms each of benzene, 1,4 dichlorobenzene, and p cresol are spilled<br />

into a pond 5 m deep with an area of 1 km 2 . If k W is 0.1 m/h, and k A is 10 m/h,<br />

what will be the times necessary for half of each chemical to be evaporated? Use<br />

the property data from Chapter 3, and ignore other loss processes.<br />

Answer<br />

Benzene, 36 h, dichlorobenzene 38 h, p cresol 12400 h.<br />

Some of the earliest environmental modeling was of oxygen transfer to oxygendepleted<br />

rivers in which a “reaeration constant,” k 2, was introduced (with units of<br />

reciprocal time) using the equation<br />

dC W/dt = k 2(C E – C W) mol/m 3 h<br />

where C E is the equilibrium solubility of oxygen in water. Another term is usually<br />

included for oxygen consumption, but we ignore it here. Now, if the volume in<br />

question is 1 m 2 in horizontal area and Y m deep, it will have a volume of Y m 3 ,<br />

and the flux N will be YdC w/dt mol/h. But<br />

N = k M(C E – C W) = YdC W/dt = Yk 2(C E – C W)<br />

k M is thus equivalent to Yk 2. A typical k 2 of 1 day –1 in a river of depth 2.4 m<br />

corresponds to a mass transfer coefficient of 2.4 m/day or 0.1 m/h. Oxygen reaeration

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