Question

For the following activated sludge process unit, find the correct SRT(hr) and F/M ratio, respectively, as: HRT: 1.5 hr Biological reactor Influent Secondary clarifier Effluent Q: 12000 m3/d X= 3500 mg/L BOD: 10 mg/L Setting BOD: 100 mg/L Derby matter XR=8000 mg/L Retum activated sludge Qw/Q=5% Surplus sludge

Answer 1

Se xe Q, SO Xo Secondary daritter Aeration tank tu QR XR Qw, YR 3 mg Influent BOD = 100 SO E ma Influent flowrate = Q = Q = 12,000 m a MLSS Conch 3500 mg Se 10 mg ma effluent BOD = 0,0625 day HRT = 1.5hr - Hydraulic Retention time Return sludge XR 8000 mg mass concentration step 1: Calculate volume of aeration tank Biological reactor using HRT, Q 3 V = (Q) CHRT) 12,000 m 0.0625 day = 750 m² day Х / -- .

→ and no .: solids beaning per day. Step-2 writing Biomass balance along the boundary. Accumulation inflow - outflow & net grouth. v dx Q Ko - [Q– Q..)X + Q. Xe] + Yav. dt 1 at steadystate vdx becomes zero. biomass conch in influent water i Xo=0 rg V (Q- Qw) Xe + Q W NR. -Qw) Xe't QuikR both sides divided by X (a-Qw) Xe + Q w XR This term's inversion is :SRT Crolid retention times meaning ixx solids present in reactor : rg is rg A VX -- > . .! 11 VX SRT = 1 l-Qw) Xe t Q w XR if conch of biomass leaving (Xe) is very low. Xe no

1, 1 VX SRT Qw XR Step 3: with the known valures V.Qw = 0.05), X, XR 750 x 3500 mx mail SRY (0.05)(12000) * 8000 sy} ng 2625000 0.546 day 600 X 8000 SRT E 13. Ihr where QSC X M L XV 12000 x 100 m3 mg Roo न 3500 x 750 me NLUSS 0.457 mg BOD mg Murss.day