A MBR treatment plant has a flow of 0.3 mgd with 220 mg/l BOD (the soluble portion is 120 mg/l) and 230 mg/l suspended solids (184 mg/l is volatile). Effluent soluble BOD is 2 mg/l. The design calls for 5000 mg/l MLSS and sludge age of 20 days with a Y = 0.8, kd = 0.4. Calculate the aeration basin volume, detention time, BOD loading, ratio, and waste solids, both VSS and TSS.

attached below is the detailed solution

A) 8288.77 cu.ft

B) 4.96 hours

C) Vss = 131.21 IbVss/day

   Tss = 164 IbTss/day

D) attached below

E ) 0.2

F) 287.23 Ib/day

Explanation:

A) Determine the aeration basin volume

Given

∅c = 20 days

Y = 0.8Ib VSS/Ib BOD

Q = 0.3 mgd

So = 120 mg/l

Se = 2 mg/l

X = 5000 mg/l

Kd = 0.04 per day

attached below is the detailed solution

B) Determine the detention time using this relation

t = ( V / Q )* 24

  = ( 0.062 / 0.3 ) * 24  = 4.96 hours

C ) Determine Vss and Tss

we first calculate the excess biomass Px then assuming Vss ratio to be 80% for sludge age of 20 days

Vss = 131.21 IbVss/day

Tss = 164 IbTss/day

D ) determine BOD loading

  Q = 0.3 mgd , BOD = 220mg/l , V = 8288.77 cu.ft

solution attached below

e) food to microorganism ratio

F/M  = 0.2

solution attached below

f) determine the waste solid

waste solid = Q * SS * % removal of suspended solids

where : Q = 0.3 , SS = 220mgl , % = 50 %

waste solids = 0.3 * 230 * 0.5 * 8.34  = 287.23 Ib/day