Question

The town of Camp Verde has been directed to upgrade its primary Waste Water Treatment Plant (WWTP) to a secondary plant that can meet an effluent standard of 25 mg/L BOD5 and 30 mg/L suspended solids. They have selected a completely mixed activated sludge system for the upgrade. The existing primary treatment plant has a flow rate of 0.029 m3/s. The effluent from the primary tank has a BOD5 of 240 mg/L. Using the following assumptions, estimate the required volume of the aeration tank

- BOD5 of the effluent suspended solids is 70% of the allowable suspended solids concentration.

- Growth constants values are estimated to be: Ks=100 mg/L BOD5; kd = 0.025 d-1; μm= 10 d-1; Y=0.8 mg VSS/mg BOD5 removed

. - The design MLVSS is 3000 mg/L

Answer #1

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...

A completely mixed activated sludge process is designed to treat
20,000 m3 /d of domestic wastewater having a BOD5 concentration of
250 mg/L following primary treatment. The NPDES permit requirement
for effluent BOD5 and TSS is 20 mg/L for each. The biokinetic
parameters have been established as: yield = 0.6 , k = 5/d, Ks = 60
mg/L, and kd = 0.06/d. The MLVSS concentration in the aeration tank
is to be maintained at 3000 mg/L and the ratio of...

A conventional activated sludge plant treating a domestic flow
of 150 ML/d is operated at a SRT of 10 d with a MLVSS of 3500 ppm.
The activated sludge plant is required to reduce the influent BOD
from 200 ppm to less than 10 ppm prior to discharge. The discharge
from the aeration basin is clarified, such that the clarifier
overflow contains no particulate material and the MLVSS of the
recycle stream is 8000 ppm. You are required to determine...

Sooner City Reclamation Plant is a completely mixed activated
sludge process designed to treat 20,000 m3/d of domestic
wastewater having an influent BOD5 concentration of 350
mg/L (assume a 28.6% reduction in during primary treatment). Use
typical activated sludge kinetic constants except for kd
= 0.06 1/d. Assume that the MLVSS concentration in the aeration
basin is maintained at 3,000 mg/L and the ratio of VSS:TSS is 0.75.
Permit requires that the effluent BOD5 and TSS
concentrations do not exceed...

A 14 mgd wastewater treatment plant has a primary clarifier that
treats an influent with 800 mg/L of solids and a removal efficiency
of 55%. Sludge at the bottom of the tank is pumped at a rate of 0.1
mgd.
a. What is the effluent solids concentration from the clarifier
(in mg/L)?
b. What is the solids concentration in the sludge at the bottom
of the tank (in mg/Land %)?
c. The primary sludge is further processed in a digester...

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...

If the F/M of a 0.44 m3/s completely mixed activated
sludge plant is 0.20 mg/mg.d, the influent BOD5 after
primary settling is 180 mg/L, and the MLVSS is 3,200 mg/L,
Calculate the following:(you can assume any missing data)
The volume of the aeration tank?
The HRT
The cell residence time
The recirculated sludge
The waste activated sludge produced each day from the plant in
kg/d
The mass of oxygen to be supplied (kg/d).

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...

You have been retained to design a wastewater treatment plant using
a conventional activated sludge process for a town of 40,000
people. The industrial flow contribution to the sewer system is
300,000 gallons per day. Effluent discharge standards require
effluent BOD5 and total suspended solids concentration of <10
mg/l.
Assume the following design criteria are applicable:
Domestic sewage flow = 100 gal/cap/day (including I/I)
BOD loading = 0.18 lbs/cap/day
TSS loading = 0.20 lbs/cap/day
BOD removal in primary = 33%...

The volumetric flow to the ACME wastewater treatment plant is 25
MGD. The influent soluble BOD5 is 200 mg/L and the effluent soluble
BOD5 is 8 mg/L. The plant is designed to operate with a mean cell
(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.
The thickened underflow biomass concentration from the secondary
clarifier is...

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