A complete-mix activated sludge process treats 5 MGD of wastewater containing 22

Question

A complete-mix activated sludge process treats 5 MGD of wastewater containing 225 mgL soluble BOD_5 to 20 mg/L soluble BOD_5. The volume of the aeration basin is 111, 400 ft^3 and the MLVSS concentration is 2500 mg/L. The waste activated sludge stream has a flow rate of 0.055 MGD and a VSS concentration of 9200 mg/L. Assume negligible suspended solids leave in the final clarifier effluent. Determine the following: SVI % wasting rate for the WAS % recycle rate for the RAS HRT SRT % treatment efficiency for BOD_5 removal from the wastewater F/M U Y_obs Based on the information provided, can values be determined for mu_m, K_5, Y and k_d?

Explanation / Answer

Given , Volume = 111,400f3 = 111400*7.48= 0.83MG , BOD influent = 225mg/l ' BOD efrfluent = 20mg/l

MLVSS = 2500mg/l

flow rate = 0.055MGD

vss = 9200mg/l

c)RAS % = (Q * MLVSS ) / (VSS - MLVSS )

= Q = flow rate = 0.055MGD

MLVSS = 2500mg/l

VSS = 9200mg/l

= (0.055 * 2500) / 9200 - 2500)

= 0.020

d) HRT = (24 * Vat )/ Q

where Vat = volume of aeration tank in mg

Q = flow rate of waaste water influent to the aeration tank in MGD

= (24* 0.83) / 0.055

362hrs.

f) % treatment efficiency for BOD or removal from the waste water =

[[BOD(in) - BOD(out)] *100 ] / BOD(in)

= [(225-20) *100 ] /225

= 91.11

g) F/M ratio = (Q * S) / (MLVSS * Vat )

Q = influent flow MGD = 0.055

S = influent BOD concentration in mg/l = 225mg/l

Vat = volume of the aeration tank in MG = 0.83MG

(0.055*225)/(2500*0.83)

=0.005

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