Question

An activated sludge system with recycle (CMFR) uses to treat primary effluent after sedimentation.Use the appropriate kinetic coefficients for the influent water characteristics to CMFR. Influent Characteristics: Flow = 1000 m3/day; bsCOD= 192 g/m3; nbVSS= 30 g/m3; inert inorganics= 10 g/m3

Aeration Tank Data: MLVSS = 2500; g/m3; SRT or θc = 6-day

Calculate:

1.- Effluent bsCOD

2.- Hydraulic retention time

3.- Daily sludge production in Kg/day as

a) VSS

b)TSS

Answer #1

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...

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...

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 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
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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 ?

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 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.

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...

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...

Design a completely mixed activated sludge process using three
different design approaches (i.e. calculate the volume of the
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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...

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