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

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significant resistance to transfer at the interface. It appears that if it does exist, it is<br />

small and unmeasurable. In any event, we do not know how to estimate it, so it is<br />

convenient to ignore it and assume that equilibrium applies. We thus argue that there<br />

is no interfacial resistance, and C WI and C AI are in equilibrium.<br />

and<br />

©2001 CRC Press LLC<br />

C AI/C WI = K AW = Z A/Z W = H/RT<br />

C AI/K AW = C WI<br />

The solute then diffuses in the air from C AI to C A in the bulk air with a mass<br />

transfer coefficient k A. We can write the flux equations for each phase, noting that<br />

the fluxes N must be equal, otherwise there would be net accumulation or loss at<br />

the interface.<br />

or more conveniently,<br />

or<br />

which is<br />

N = k WA(C W – C WI) = k AA(C AI – C A) mol/h<br />

C W – C WI = N/k WA<br />

C AI – C A = N/k AA<br />

C AI/K AW – C A/K AW = N/(k AAK AW)<br />

C WI – C A/K AW = N/(k AAK AW)<br />

Adding the first and last equations to eliminate C WI gives<br />

or<br />

where<br />

C W – C A/K AW = N(1/k WA + 1/k AAK AW) = N/k OWA<br />

N = k OWA(C W – C A/K AW) = k OWA(C W – P/H)<br />

1/k OW = 1/k W + 1/k AK AW = 1/k W + RT/Hk A<br />

The term k OW is an “overall” mass transfer coefficient that contains the individual<br />

k W and k A terms and K AW. It should not be confused with K OW, the octanol-water

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