Question

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 = 0.60 mg VSS/mg BOD, Kd= 0.06 day-1, Ks= 60 mg/Lof BOD, (K) or μmax= 5.0 day-1. Determine the BOD removal efficiency, F/M ratio, volume of the aeration tank, aeration time, concentration of substrate (BOD) in the effluent, and the sludge production rate.

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 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...

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...

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...

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 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 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...

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...

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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