2. (Learning objective #3, #7) In an activated sludge system, 1 MGD is leaving t

Question

2. (Learning objective #3, #7) In an activated sludge system, 1 MGD is leaving the primary clarifier with a BOD concentration of 130 ppm and a VSS of 200 ppm. The effluent flow is 0.95 MGD with a VSS of 20 ppm and a BOD concentration of 5 ppm. Qr/Q = 1,MLVSS = 2000 ppm. If the residence time in the primary clarifier was 4.5 hours, calculate the BOD and SS coming into the primary clarifier. What is the VSS in the recycle stream?(3791 ppm) c. How many lbs/day of sludge exit the system?(1581 lb/day) What is the efficiency of the activated sludge system for BOD removal?

Explanation / Answer

Given:

Effluent from the primary clarifiier (Qe) = 1 MGD

BOD concentration of the effluent = 130 ppm

VSS of effluent = 200 ppm

The effluent flow from the ASP process = 0.95 MGD

BOD concentration of the effluent from the ASP process = 5 ppm

VSS of the effluent from the ASP process = 20 ppm

Return activated sludge = (Qr/Q) * 100

R = (1) * 100 = 100 %

MLVSS = 2000 ppm

Solution:

As we know the hydraulic retention time we can find the volume of the the primary clarifier

HRT = 24 * V / Qe

4.5 = 24 * V / 0.95

Volume V = 0.178 MG

Volumetric BOD Loading (VBL)

VBL = 8.34 (Q * S) / V

Where Q = effluent flow

S = BOD concentratin

V = Volume of the primary clarifier

VBL = 8.34 (1 * 130) / 178000

BOD Loading = 0.006 lb BOD/1000ft3 * day

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