The table provides suspended solids concentrations in several different wastestr
ID: 1921076 • Letter: T
Question
The table provides suspended solids concentrations in several different wastestreams at a municipal wastewater treatment plant. The BOD5 measured in the sewer just before the treatment plant is 250 mg/L, after primary treatment is 150 mg/L, and after secondary treatment is 15 mg/L. Total nitrogen levels in the plant are approximately 30 mg N/L.
Process stream
Suspended Solids Concentration (mg SS/L)
Plant influent
200
Primary sludge
5,000
Secondary sludge
15,000
Aeration basin effluent
3,000
If the design hydraulic retention time of each of four aeration basins operated in parallel equals 6 hours and the total plant flow is 5 million gallons per day, what is the F/M ratio in units of lbs BOD5/lbs MLVSS-day?
Process stream
Suspended Solids Concentration (mg SS/L)
Plant influent
200
Primary sludge
5,000
Secondary sludge
15,000
Aeration basin effluent
3,000
Explanation / Answer
F/M ratio = ?
F = mass of BOD removed
F = Q.Yo
Q = waste water flow = 5 million gallons per day = 18927 m3/day
Yo = influent BOD = 200 gm/day
F = (18927 x 200)/ 1000
= 3785.4 kg
M = mass of MLSS
= V x Xt
V = volume of aeration tank, assume volume of aeration tank 10900 m3
Xt = Mixed liquor suspended solids (MLSS) 3000 mg/l
M = (10900 x 3000)/1000
= 32700 kg
F/M = 3785.4/32700
= 0.11 kg
= 0.24 lbs BOD per day/lb of MLSS
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