A completely mixed high rate-activated sludge plant with recycletreats 17,500 m3/day of industrial wastewater. The influent to theactivated sludge reactor has a BOD5 of 1000 mg/L. It is desired toreduce the influent BOD5 to 120 mg/L, prior to discharge to a municipalsewer. Pilot plant analysis gave the following results: mean cellresidence time = 6 d, MLSS concentration in reactor = 5500 mg/L,Y = 0.6 kg/kg, kd = 0.03 d–1. Determine the following:________.
a. The hydraulic retention time and volume of the activated sludge
b. The wolumetric loading rate in kg BOD,/m- day to the reactor.
c. The F/M ratio in the reactor
The mass and valume of solids wasted each day, at an underflow solids concentration, Xu = 10,000 mg/L.
e. The sludge recirculation ratio

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nmclaughlin · Accepted Answer

Answer:a) 0.488 days 11.72 hour , 8542.38 m^3b) 2.05 kg/m^3.dayc) 0.37 day^-1d) i) volume of wasted solid = 783 m^3/dayii) mass of wasted solid = 7830 kg/daye) 1.12Explanation:Given data :Flow ( Q ) = 17500 m^3/dayBOD 5 ( influent ) = 1000 mg/LBOD 5 ( efficient ) = 120 mg/Lmean cell residence time ( c ) = 6-dayMLSS concentration ( X ) = 5500 mg/La) Determine the hydraulic retention time and volume of the activated sludgei) Hydraulic retention time ( OH ) we will use the relation below X = [tex]\frac{OcY(sin-s)}{OH(1 + Kdc)}[/tex] 5500 = [tex]\frac{6*0.6(1000-120)}{OH(1 + 0.03*6)}[/tex] . therefore OH = 0.488 days 11.72 hourii) volume of the activated sludge using the relation below OH = V / Q V = OH * Q = 0.488 * 17500 = 8542.38 m^3b) Determine the volumetric loading rate in kg BOD,/m- day to the reactor loading rate = ( Q * Sin ) / V = ( 17500 * 1000 ) / 8542.38 = 2.05 kg/m^3.dayc) Determine The F/M ratio in the reactorF/M ratio : F/M = ( Q .sin ) / VX = ( 17500 * 1000 ) / ( 8542.37 * 5500) = 0.37 day^-1d) Determine The mass and volume of solids wasted each day, at an underflow solids concentrationi) volume of wasted solid = 783 m^3/dayii) mass of wasted solid = 7830 kg/dayattached below is the remaining part of the solution e) Sludge recirculation ratio = 1.12 attached below is the remaining part of the solution

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