A schematic flow diagram of a completely mixed biological reactor with provision for recycling of solids is shown in the figure. Q-15000 m³ d S = 100 mg L Xo-0 mg L Acration tank S = 15 mg L X = 3000 mg L Detention time=2h Q-5000 m³/d S = 15 mg L X-10000 mg L Sludge retum Secondary settling tank Q, QrQw = flow rates, Q-Q=14950 S = 15 mg L X-0 mg L Q = 50 m³/d S = 15 mg L X-10000 mg L So, S = readily biodegradable soluble BOD, mg/L Xo, X, X₁, X₁ = microorganism concentrations (mixed-liquor volatile suspended solids or MLVSS), mg/L The mean cell residence time (in days, up to one decimal place) is

A schematic flow diagram of a completely mixed biological reactor with provision for recycling of solids is shown in the figure. Q-15000 m³ d S = 100 mg L Xo-0 mg L Acration tank S = 15 mg L X = 3000 mg L Detention time=2h Q-5000 m³/d S = 15 mg L X-10000 mg L Sludge retum Secondary settling tank Q, QrQw = flow rates, Q-Q=14950 S = 15 mg L X-0 mg L Q = 50 m³/d S = 15 mg L X-10000 mg L So, S = readily biodegradable soluble BOD, mg/L Xo, X, X₁, X₁ = microorganism concentrations (mixed-liquor volatile suspended solids or MLVSS), mg/L The mean cell residence time (in days, up to one decimal place) is

Solid Waste Engineering

3rd Edition

ISBN:9781305635203

Author:Worrell, William A.

Publisher:Worrell, William A.

Chapter2: Municipal Solid Waste Characteristics And Quantities

Section: Chapter Questions

Problem 2.17P

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he raw water supply for a community contains 160 mg/L total particulate matter. It isto be treated by addition of 86 mg/L alum.Essentially all of the added alum precipitatesby this reaction:Al2(SO4)3 · 14H2O ⇒Al(OH)3(s) + 3SO2−4 + 6H+ + 8H2O. For a total flow of 5500 m3/day, compute the daily alum requirement. Answer in units of kg/day. Determine the concentration of solids in the water following alum addition (i.e., the sum of the solids initially present and those formed during treatment). Answer in units of mg/L.
A waste with a 5-day BOD (BOD)=200mg/L and k = 0.1 day¹ is discharged to a river at a rate of 1 m/s. The river has flow rate of 9 m/s and a background ultimate BOD (Lo) of 2 mg/L upstream of the discharge. (a) Calculate the ultimate BOD (Le) of the waste. (b) Furthermore, assuming instantaneous mixing after discharge, calculate Lo of the river water after it has received the waste. Ans.: (a) Ultimate BOD of wastewater = 508.3 mg/L, (b) Ultimate BOD of river water after discharge of waste-52.63 mg/L.
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As an operator of a wastewater treatment facility, you have learned that your plant runs well at an SRT (based on MLVSS (X) in the aeration basin) of 12 days. If today’s MLVSS (X) concentration is 2,700 mg/L and the RAS (Xr) concentration is 5,500 mg/L, how many L of sludge should you waste from the secondary clarifier today? Other design conditions: Average influent flow = 12,000 m3/d Influent soluble BOD = 150 mg/L Required effluent TSS = 30 mg/L Aeration tank Volume = 1,000 m3
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