Question

A completely mixed activated-sludge plant is being designed for a wastewater flow of 5.0 MGD. The soluble BOD concentration of the influent to the aeration basin is 200 mg/l. The effluent soluble BOD requirement is 10 mg/l. The mean cell residence time is to be 8.0 days at an MLSS concentration of 2500 mg/l. From pilot plant data, Y = 0.6 lb VSS/lb BOD, kd = 0.06 day–1 , and 80% of the MLSS are VSS. Calculate:
a) The volume of the aeration basin, in gals.
b) The dry mass of excess solids in the waste sludge in lb/day.
c) The aeration period, q in hours.
d) Food-to-microorganism ratio, F/M.
e) BOD loading, lb/day-1000 ft3 .

Answer 1

Answer:

a) 154054 gals

b) 20850 lb/day

c) 0.7392hr

d) 2.466day^-1

e) 7.23 lb/day-1000 ft^3

Explanation:

a)

$Vx = \frac{Qy(So - Se)}{1+kdQc}= \frac{5x10^{6}(0.6)(200-10)}{1+0.06(8)}=154054 gals$

b)

Dry mass = 5 MGD * (0.20) * (2500 mg/l) * 8.34 = 20850 lb/day

c)

Aeration period Q = $\frac{V}{Q} =\frac{154054}{5x10^{6} } =0.0308day=0.7392hr$

d)

$\frac{F}{M} =\frac{Q(So-Se)}{Vx}=\frac{5x10^{6}(200-10) }{154054(2500)}=2.466day^{-1}$

e)

BOD loading = $\frac{8.34(5)(200)}{154.054(7.48)} = 7.23 lb/day-1000 ft^{3}$

Hope this helps!