Question

A wastewater treatment plant that treats wastewater to secondary treatment standards uses a primary sedimentation process followed by conventional activated sludge process composed of covered aeration tanks and secondary clarifiers. You are asked to calculate the different operation parameters to make sure that the system is operated within normal process ranges listed in the reference tables below. Table 5-20 Typical design information for primary sedimentation tanks U.S. customary units Item Unit Range Typical Primary sedimentation tanks followed by secondary treatment Detention time 1.5-2.5 20 Overflow rate Average flow gal/142- 800-1200 1000 Peak hourly flow gal/l-d 2000-3000 2500 Weir loading gal/Itd 10,000-40,000 20,000 h TABLE 12.3 General Loading and Operational Parameters for Activated-Sludge Processes BOD Loading Ib BODI 1000 ft day Ib BODI day/lb of MLSS Average Return Sludge Rates (5) Aeration Period (hr) Sludge Age (days) 5-15 5-15 Process 30-50 30-40 0.2-0.5 0.2-0.5 5.0-7.0 6.0-75 50 30 Step aeration Conventional (tapered aeration) Contact stabilization Extended aeration High-purity oxygen 30-50 10-30 120+ 0.2-0.5 0.05-0.2 0.6-1.5 5-15 20+ 5-10 6.0-9.0 20-30 1.0-3.0 100 100 30 1.0 lb/1000 ft/day - 16 g/ m.d.

Answer 1

Let's Proceed

16, 17

Page 16 GivenamG Detention time is given by De= Volume of one clasifier one Clarifies # total flowetz GOMGD Flow over #flow over ne clasifies a total flow no- of tanks GOMGD 6 &ZLOMGA => De= २ Dt- o.lday = (4.868 17 Are the process operated under normal Condition? for this cue need to Compare the obtained results with Standards - If they are in the Limits then the unit can be opel ated under normal Conditions need to increase the volun we If not Either op Detention time =) As ped Question - 16 result 4-8ho Detention time is in the limit so, it can be operated But need to Consider all the results

18,19

18 Flm ratio is Food/micoo organisms It is amount of sludge recieing into the aeration Colo activated sludge unit. = when there is more primary Clasificos the sludge Inflow will Increase activated sludge unit Parmazy Sludge is not treated so flusatio will also Increas Pinto ases If P. C not operational new sludge activated slodge into Asitank = 1 Primary Sludge so (Flmn Q As there is Do palmary clasified in that case the Boo removal is not achived so; total 8 o influent Puto aeration tank is will be a boong! (Inflow of P. () instead of yomg/L (Effluent of PCC) » So for this Extra amount of Bod, more forggen is needed to onidize the BOD = Oxygen - demand will be increased

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