Question

② Wastewater is being-treated in a mplete mixe d achvated Sludge rocess witn recycle. The falla ata are availabl e e ollo ng So Dsratutne Influentこ250mgIL(35% removal in Primary Cla er luent Bon the Permit 4mG/L Moss-42000rnalL (75% volatile ge(ra): 2 days Linethc Constants t 12C aWithis Sustem be able fo meet the bWhat wil be the HRT this C) What will be the excess Solids RoduCHon uent BoDs on the permit? equure stem Chrs? f.. 'the Waste Solids are 75%

This is a wastewater problem. Please can I get a correct answer for part b and part c for this problem. Thank you

Answer 1

Answers:

75% is volatile

so VSS / SS = 0.75

X = 0.75* MLSS

= 0.75*2000

= 1500 mg/L

In aeration tank,

V*X = Y*Q*$\theta$c * (S0 - S) / (1 +kd*$\theta$c)

(V/Q)*X = Y*$\theta$c * (S0 - S) / (1 +kd*$\theta$c)

V/Q = hydraulic retention time

HRT *1500 = 0.65*2 * (250 - 4) / (1 +0.04*2)

HRT = 0.1974 days

= 4.737 hrs

= 5 hrs

net VSS produced = V*X /$\theta$c

V = Q * HRT

= 4 *0.1974

= 0.7896 MG

= 0.7896* 10^6 * 3.785 liters

VSS = 0.7896* 10^6 * 3.785 * 1500 / 2

= 0.224 * 10^10 mg/day

= 2241.477 kg/day

So SS produced = 2241.477 / 0.75 = 2988.636 kg/day