Applications of Equilibrium Thermodynamics - Elements of Environmental Engineering: Thermodynamics and Kinetics - page 150

(i) For NaCl, v = 2, m s = 0.1, M s = 58. B 1 = 0.66 / 298 = 2.2 × 10 − 3 μ m, B 2 =

3.44 × 10 13 (2) (5 × 10 − 13 )/ 58 = 0.55 ( μ m) 3 . ln( P w /P w ) = ( 2.2 × 10 − 3 ) −
( 0.55 / 1 3 ) =− 0.55.Hence P w /P w = 0.57.Since P w = 223.75, P w = 13.7 mm Hg.
(ii) σ a / w = 72 erg/cm 2 , V w = 18 cm 3 /mol, r = 1 × 10 − 4 cm, R = 8.314 × 10 7 erg/
K mol, T = 298 K. Hence ln( P w /P w ) = ( 2 )( 72 )( 18 )/( 1 × 10 − 4 × 8.314 ×

× 298 ) = 1.04 × 10 − 3 . Hence P w /P w = 1.001. Therefore P w = P w =
23.75 mm Hg.

10 7

4.2.6 A IR /W ATER E QUILIBRIUM IN W ASTE T REATMENT S YSTEMS
The equations and methods of calculation of flux in the atmosphere, as described in
Section 4.2.4, also apply to the estimation of air emissions from wastewater storage
impoundments, settling ponds, and treatment basins (Springer et al., 1986). Waste-
watersareplacedtemporarilyinopenconfinedabovegroundfacilitiesmadeofearthen
material. In some cases, the storage of wastewater in such impoundments is for the
purpose of settling of solids by gravity. In other cases, wastewater treatment using
mechanical agitation or submerged aerators is conducted in these impoundments
for the purposes of neutralization, chemical reaction, biochemical oxidation, and
evaporation. All these activities cause a significant fraction of the species from water
to volatilize to the atmosphere. Natural forces in the environment can dominate the
transfer process. Examples of these are wind, temperature, humidity, solar radiation,
ice cover, and so on at the particular site. All of these effects have to be properly
accounted for if accurate air emissions of VOCs are to be made.
Consider a surface impoundment as shown in Figure 4.7. The natural variable that
has the most influence on air emissions is wind over the water body. Both k w and k a
depend on wind velocity.An average overall mass transfer coefficient K w over a 24-h
period can be estimated. The flux of material from water will be given by

C i w − .

C i a
K aw

N i = K w · (4.56)

C i a /K aw , and the net overall flux is
zero. In other words, at equilibrium the rates of both volatilization and absorption
are the same, but opposite in direction. The air movement over the water body due to
wind-induced friction on the surface is transferred to deeper water layers, thereby set-
tinguplocalcirculationsinbulkwaterandenhancingexchangebetweenairandwater.
Thus the wind-induced turbulence tends to delay approach to equilibrium between
the phases. The effects of wind are far greater on k a than on k w , and hence depend-
ing on which one of these two controls mass transfer the overall effect on K w will
be different.

If the two phases are at equilibrium, C i w =