Applications of Chemical Kinetics in Environmental Systems - Elements of Environmental Engineering: Thermodynamics and Kinetics - page 395

org increases in an approximate first-order form to approach the equilibrium
value K BW [A] w .
When we move up the food chain (zooplankton

Thus, [ A ]

top predator), we have
to consider not only the rate of direct uptake from water by sorption/desorption, but
also by ingestion of the lower species and the portion actually assimilated into the
tissues of the animal.Additionally, we also have to take into account the mass excreted
by the animal (Figure 6.71). Thus, we have three rate processes included in the overall
uptake rate: (i) the rate of uptake by sorption, (ii) the rate of uptake by ingestion, and
(iii) the rate of loss via desorption and excretion.

→ fish →

E XAMPLE 6.36 A PPROACH TO S TEADY -S TATE U PTAKE IN THE F OOD C HAIN

The earliest attempts at modeling bioaccumulation involved exposing the first level of
aqueous species to organic chemicals and obtaining the rate of uptake. We will now
ascertain the approach to steady state in this system. Figure 6.72 treats the organism
and water as distinct compartments (CSTRs) between which an organic chemical is
exchanged. Neely (1980) described such a model; the following discussion is based on
his approach.
As explained earlier, let [A] org represent the moles of chemical A per weight of
organism, and [A] w be the concentration of chemical A in water (mol/dm 3 or mol/kg).
Expanding the left-hand side of the equation for the rate of uptake

d t (W org [ A ] org ) = W org d [ A ] org

+[ A ] org d W org
d t

d

. (6.269)

d t

If the weight of the organism is constant, d W org /d t = 0. However, since the organism
grows as it consumes the chemical, its concentration changes in relation to the growth.
The overall equation is then

W org d [ A ] org
d t

= k s W org [ A ] w − k d W org [ A ] org −[ A ] org d W org
d t

. (6.270)

To proceed further, we denote the growth rate of the organisms as

1
W org

d W org
d t

μ = (6.271)

Hence,

d [ A ] org
d t

= k s [ A ] w − (k d + μ ) [ A ] org . (6.272)

Iftheconcentrationinwaterisassumedtochangeonlyslightlysuchthat[A] w isconstant
during a time interval, τ , we can integrate the above equation to yield

k s
k d + μ

( 1 − e − (k d +μ )t ) .

[ A ] org

] w =

(6.273)

[ A

If μ = 0 and τ is very large, we have [ A ] org / [ A ] w = k s /k d = K BW . The approach to
equilibrium will be delayed if the organism grows as it feeds on the pollutant A.
continued