A complete-mix activated sludge process with secondary clarification and sludge

Question

A complete-mix activated sludge process with secondary clarification and sludge recycle is used to treat a dairy wastewater at a flow rate of 1000m3/day with a influent biodegradable COD concentration of 3000mg/L. There is no inert non-biodegradable VSS in the influent. The aeration tank has a VSS concentration of 3000mg/L. The effluent VSS concentration is 20mg/L, hydraulic retention time Tao is 24hrs, returned VSS concentration is 10,000mg/L, and waste activated sludge flow rate from the recycle line 80m3/day a) Calculate the daily sludge production in kg/day as VSS b) Determine the system SRT in days c) Determine the biomass yield Y in g VSS/g COD, assuming that k(sub d)=0.1/day and f(sub d)=0.15 gVSS/gVSS

Explanation / Answer

a)

VSS -> sludge

if the reacction is the 1 orden we have than ra = KCa = dCa/dt =) Kt = -Ln(Caf/Cao) where t = 24h and Caf = 20mg/l and Cao = 3000mg/l ra = (Cao-Caf)/t =) ra =3000mg/l-20mg/l )/24h = 124.2mg/hl

ra t = Ca =) 124.2mg/hl x 24h = 2981mg/l of VSS are consumed therefore are produced 2981mg/l in a day = 3x10-3Kg/l

if we see all reaction whit a one unity V/vo = (Cao-Caf)/ra where vo = 1000m3/day and V = volume of unitiy

vo = 1000m3/day x day/24h + 80m3/day x day/24h = 41.66m3/h + 3.33m3/h = 45m3/h =  45x103l/h

V/vo = (Cao-Caf)/ra =  V = 45x103l/h/h(3000mg/l-20mg/l )/(24h )  = 5587500l

3x10-3Kg/l x 5587500l = 16762,5kg of sludge per day

b) V/vo = t for the sludge vo = 80m3/day V = 5587500x10-3m3 t = 5587500x10-3m3/80m3/day = 69.8 days

c) 15500g / 3000g/l = 5

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