Environmental Engineering FOURTH EDITION

194 ENVIRONMENTAL ENGINEERING

case that X = 4OOOmgL The hydraulic retention is then obtained by rearranging
Eq. (9.30):

• t=

0.5(300 - 30)(200 + 30)

= 0.129day = 3.1 h.

2(30) (4000)

The volume of the tank is then

V = fQ = 4000(0.129) = 516m3.

The sludge age is

= 3.8 days.

(4OOOmg/L)(0.129 day)

(0.5kgSS/kgBOD)(300 - 30) mgL

0, =

Since

kg sludge was Wday

kg sludge in aeration tank '

• 1
  0,

_-

from Eq. (9.28)

= 543 kglday.

XV

0, 3.8

(4000) (5 16) ( lo3 Urn3) (1 / lo6 kg/mg)

XrQw = - - -

EXAMPLE 9.7. Using the same data as in the preceding example, what mixed liquor

solids concentration would be necessary to attain 95% BOD removal, or S = 15 ma?

The substrate removal velocity is

(2 day-') 15 mg/L

= 0.28 day-'

' = (0.5)(200 + 15) mg/L

and

300 - 15

X= = 7890mg/L.

(0.129)(0.28)

The sludge age would be

= 7.1 days.

1

(0.28)(0.5)

0, =

Higher removal efficiency requires more microorganisms in the aeration tank.

A simple batch settling test can be used to estimate the return sludge pumping

rates. After 30 min of settling, the solids in the cylinder are at a 55% concentration that

would be equal to the expected return sludge solids, or

HX

x, = -

h

(9.33)