Question

Two identical sedimentation tanks are to be designed for an activated sludge plant with a flow of 2 MGD (flow is split evenly between the two clarifiers). The MLSS will be 1950 mg/L. The underflow (return sludge) concentration will be 12,000 mg/L. A settling curve was developed (using a column with Ho = 2.13 ft) and is shown below. For clarification area use a safety factor of 2.0 and for thickening area use a safety factor of 1.5. Tanks are available in 5-ft increments. What is the design diameter (ft)?

Answer #1

ANSWER:----

Two identical sedimentation tanks are to be designed for an
activated sludge plant with a flow of 2 MGD (flow is split evenly
between the two clarifiers). The MLSS will be 1950 mg/L. The
underflow (return sludge) concentration will be 12,000 mg/L. A
settling curve was developed (using a column with Ho = 2.13 ft) and
is shown below. For clarification area use a safety factor of 2.0
and for thickening area use a safety factor of 1.5. Tanks are...

Two identical sedimentation tanks are to be designed for an
activated sludge plant with a flow of 2 MGD (flow is split evenly
between the two clarifiers). The MLSS will be 1950 mg/L. The
underflow (return sludge) concentration will be 12,000 mg/L. A
settling curve was developed (using a column with Ho =
2.13 ft) and is shown below. For clarification area use a safety
factor of 2.0 and for thickening area use a safety factor of 5.
Tanks are...

Mass of solids Generated from the activated sludge tanks.
13. Assume: The total design flow is 3.75 MGD (15,000
m3/day). The NPDES limit is 25/30. Assume that the waste
strength is 170 mg/L BOD after primary
clarification. Y = 0.55 kg/kg.
What is the mass of solids generated each day (Kg/day or
lbs/day) in the activated sludge tankage?

An activated sludge system has three square reactor tanks in
series (CFSTRs) that are each 60 ft wide, 60 ft long, and 18 ft
deep. The engineers have decided to add a new reactor in parallel.
It will be a plug flow reactor (PFR) with a depth of 18 feet and a
width of 18 feet. The length is to be calculated. The flowrate of
wastewater is 2 MGD and the BOD is 165 mg/L. This will be split in...

ANSWER as much as possible:
Design an activated sludge process consisting of a complete mix
flow through aeration tank followed by final settling tanks. The
sizing of the reactor and all appurtenances should be based on a
solids retention time, (SRT), of 4 days and the particulates given
below.
Biological characteristics: Ks = 20
mg/L COD, kd = 0.1 per day, mumax = 3
gVSS/gVSS-day,
Y = 0.4 gVSS/gCOD
Primary settling Tank Effluent
Characteristics:
Q= 20MGD, biodegradable COD, bCOD...

A wastewater treatment plant that treats wastewater to secondary
treatment standards uses a primary sedimentation process followed
by an activated sludge process composed of aeration tanks and a
secondary clarifiers.
This problem focuses on the activated sludge process.
The activated sludge process receives a flow of 15 million
gallons per day (MGD).
The primary sedimentation process effluent has a BOD5
concentration of 180 mgBOD5/L (this concentration is
often referred as S). Before flowing into the activated sludge
process, the primary...

Dechlorination contact tank (Wastewater Treatment)
An activated sludge plant has a design flow (average daily flow)
of 11 MGD. Determine dimensions of dechlorination tank. The design
parameters: use 1 tank. detention time = 30 sec, depth = 10 ft,
rectangular tank: L = 4 W. use maximum hourly flow (3 * Average
design flow)

A completely mixed activated-sludge process is being designed
for a wastewater flow of (2.64 mgd) using the kinetics equations.
The influent BOD of 120 mg/Lis essentially all soluble and the
design effluent soluble BOD is 7 mg/L, which is based on a total
effluent BOD of 20 mg/L. For sizing the aeration tank, the mean
cell residence time is selected to be10 days and the MLVSS 2000
mg/L. The kinetic constants from a bench-scale treatability study
are as follows: Y...

Problem #3: Biochemical Oxygen Demand (BOD) and Stream
Quality
A wastewater
treatment plant designed for a community of 45,000 people has a
flow of 100 gal/person/day and a BOD5 loading of 0.2
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
mg/L in this stretch of the river.
The minimum BOD
removal efficiency that the treatment...

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