Question

Sooner City Reclamation Plant is a completely mixed activated
sludge process designed to treat 20,000 m^{3}/d of domestic
wastewater having an influent BOD_{5} concentration of 350
mg/L (assume a 28.6% reduction in during primary treatment). Use
typical activated sludge kinetic constants except for k_{d}
= 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 BOD_{5} and TSS
concentrations do not exceed 20 mg/L on an annual basis.

Part 1: What is the effluent soluble BOD_{5}
concentration, in mg/L, necessary to meet the total BOD_{5}
treatment criteria?

Part 2: What is the mean cell residence time, in days, necessary to meet the treatment criteria?

Part 3: What is the volume of the aeration basin in cubic meters?

Part 4: What is the quantity of daily sludge wasted, in kg/d, at the mean cell residence time determined in Part 2?

Answer #1

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 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...

A complete mixed activated sludge process aeration tank treats 5
MLD sewage having the influent and effluent soluble BOD
concentrations of 220 and 20 mg/L respectively. The volume of the
aeration tank is 5060 m3 and the mean cell residence time is 10
days. MLVSS in aeration tank is 3,400 mg/L, MLVSS/MLSS = 0.8 and
sludge wastage rate is 1 m3/hour. Calculate the concentration of
return sludge suspended solids ?

Design a completely mixed activated sludge process using three
different design approaches (i.e. calculate the volume of the
aeration basin). Base your design for the first approach on
detention time, for the second one on BOD loading, and the third
one on kinetics. The influent design flow rate to the activated
sludge process is 10,000 m3 /day with a BOD5 concentration of
200mg/L. Completely mixed activated sludge processes typically have
a detention time ranging from 3 to 6 hours and...

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...

Estimate the volume of air to be supplied (m3/d) for an
activated sludge system with a
flow of 0.25 m3/sec, net waste activated sludge production of
340 kg/day VSS, and an
influent soluble BOD5 of 74 mg/L. Assume that soluble BOD5 is
58% of BODL, BOD5 in
the aeration tank is 14 mg/L, and that the oxygen transfer
efficiency is 9%. Density of air
is 1.185 kg/m3, and air has 23.2% oxygen.

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...

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...

An activated sludge tank has the following
characteristics: Q = 4.18 m3/sec; Influent
BOD5 = 225 mg/L; Effluent BOD5 = 4.5 mg/L;
Ks = 100 mg/L; μm = 2.5 day-1;
kd = 0.05 day-1. The mean cell residence time
(MCRT) in days is most nearly:
a.12.4
b.7.0
c.2.5
d.17.4
QUESTION 38
The hydraulic detention time, θc, of the
activated sludge tank above (in Question #37) in hours is most
nearly: (Note: Cells in the activated sludge tank, X = 3,500...

A rectangular, well-mixed aeration lagoon is 75 m long, 5 m
wide, and 2 m deep. It receives 400 m3/d of wastewater
with a BOD5of 367mg BOD5/L for treatment. The
biodegradation rate constants for the wastewater are the following:
yield coefficient = 0.8 mg VSS/mg BOD5, endogenous decay
constant = 0.08 d-1, maximum specific microbial growth
rate = 1 d-1, and half velocity constant = 76.0 mg
BOD5/L.
What must the hydraulic detention time be in the aeration pond?
(days)...

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