A flow of 2.6 MGD leaves a primary clarifier with a BOD of 131 mg/L. Determine the aeration period (hydraulic detention time) of an activated sludge tank that has a BOD loading of 35 lbsper 1000 ft3per day. Determine the sludge age of the facility given the following data: VSS concentration is 2100 mg/L, and the mass rate of wasted sludge is equal to 44 percent of the BOD loading.

8.5 days

Explanation:

Given data :

Flow ( Q ) = 2.6 MGD = 11819.834 m^3/day

BOD = 131 mg/L

BOD loading rate = 35 Ibs/1000 ft^3   per day = 0.5606 kg/m^3/day

Calculate the sludge age of the facility

Given  the BOD  applied to the aeration tank  = 11819.834 m^3/day * 131mg/l

                                                                            = 1548.398 kg/day

first calculate the volume of the aeration tank

V = BOD applied / BOD loading rate

V = 1548.398 / 0.5606 = 2762.03 m^3

Hydraulic Detention time = V / Q

                                          = 2762.03 / 11819.834  = 0.2336 day = 5.6 hour

next : determine the mass rate of the waste

= 44% * 0.5606 kg/m^3/day

= 0.2466 kg/m^2/day

finally determine the sludge age

= V * Xt / ∅w * R

= ( 2762.03 m^3 * 2100 * 10^-3 ) / ( 0.2466 * 2762.03 kg/day )

= 8.5 days

answer:

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explanation: