Question

A wastewater plant with the activated sludge process received 5
× 10^{6} L/day of wastewater with a BOD_{5} of 500
mg/L. The primary clarifier removes 30% of the BOD and primary
clarify effluent has no biomass (Xi=0 mg/L). Recycle
(Q_{r}) and waste sludge (Q_{w}) flows are 25% and
2% of Qin, respectively. Effluent BOD_{5} from secondary
clarifier and waste sludge is 20 mg/L. Microorganism concentration
in the waste sludge (X_{w}) is 6,000 mg/L. Soluble
BOD_{5} is biologically converted into CO_{2} and Y
(yield coefficient) is 0.5 mgVSS /mgBOD

**Show me how to solve for
Volume of tank on this. I can get every thing else after that. I
know I have to find theta and use that to find it show me
how.**

Answer #1

The volumetric flow to the ACME wastewater treatment plant is 25
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(biomass) residence time of 10 days. The yield coefficient is 0.6
mg biomass/mg substrate, and the biomass decay rate is 0.09/day.
The reactor is to operate at a biomass concentration of 2500 mg/L.
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A conventional activated sludge plant treating a domestic flow
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overflow contains no particulate material and the MLVSS of the
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A wastewater treatment plant that treats wastewater to secondary
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This problem focuses on the activated sludge process.
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A completely mixed activated sludge process is designed to treat
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Problem #3: Biochemical Oxygen Demand (BOD) and Stream
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A wastewater
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lb/person/day. The upstream characteristics of the receiving waters
are:
stream flow rate =
20 cfs and BOD5 = 2 mg/L
The water quality
standards require the in-stream BOD5 to be less than 7
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The minimum BOD
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You have been retained to design a wastewater treatment plant using
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mg/l.
Assume the following design criteria are applicable:
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A completely mixed activated-sludge process is being designed
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14. Sludge wasting rate (Qw) from the solids
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Assume that the waste strength is 170
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XA=MLSS = 2200 mg/L,
Xw = Xu =
XR = 6,600 mg/L,
qc = 8 days.
Make sure you account for the...

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