Environmental Engineering FOURTH EDITION

Wastewater Treatment 193

Using a mass balance in terms of microorganisms

Net rate of change Rate of Rate of Rate of

in microorganism = microorganism - microorganism + microorganism

substrate inflow ou~ow production

Again assuming the steady state and Xo = Xe = 0,

Xt Qw

xv

p= -* (9.27)

Note that p is the reciprocal of the sludge age, or mean cell residence time so that

(9.28)

Substituting the expression for p from Eq. (9.14), the substrate removal velocity is

(9.29)

Setting Eq. (9.29) equal to Eq. (9.23) and solving for So - S, the reduction in BOD,

bsxt

so-s=

Y(K, + S)’

Then from Eq. (9.29), the mean cell residence time (sludge age) is

and the concentration of microorganisms in the reactor (MLSS) is

so - s

X=-.

*4

(9.30)

(9.31)

(9.32)

EXAMPLE 9.6. An activated sludge system operates at a flow rate (0) of 4000m3/day,

with an incoming BOD(&) of 300 mg/L. A pilot plant showed the kinetic constants to

be Y = 0.5 kg SSikg BOD, K, = 200 ma, p = 2/day. We need to design a treatment

system that will produce an effluent BOD of 30mgL (90% removal). Determine (a)

the volume of the aeration tank, (b) the MLSS, and (c) the sludge age. How much

sludge will be wasted daily?

The MLSS concentration is usually limited by the ability to keep an aeration

tank mixed and to transfer sufficient oxygen to the microorganisms. Assume in this