Question

**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**: K_{s} = 20
mg/L COD, k_{d} = 0.1 per day, mu_{max} = 3
gVSS/gVSS-day,

Y = 0.4 gVSS/gCOD

**Primary settling Tank Effluent
Characteristics:**

Q= 20MGD, biodegradable COD, bCOD = 300 mg/L, SS = 100 mg/L of which 20 mg/L are inert solids, (i.e. non-biodegradable).

**Aeration Tank**: SVI = 125 ml/gram MLSS,
MLVSS:MLSS ratio = 0.80, T = 20^{o}C

**Final Settling Tanks**: Operating at an overflow
rate = 600 gpd/sq.ft, Solids Loading Rate = 10 lb SS/ day-sq.ft.,
Final Settling Tank Effluent, (FSTE), SS = 20 mg/L

Calculate/ Estimate: (Questions 1-9, 5-pts. each, Questions 10 & 11, 10pts. each)

- The soluble bCOD in the final settling tank, (FST), effluent in mg/L ____, estimate the total COD in the effluent _____________
- The Q
_{RAS}in MGD ___________ - The concentration of MLVSS in mg/L ____________ and the MLSS in mg/L ___________. Hydraulic retention time, HRT, in the aeration tank in hrs ______
- The net growth rate in lb VSS/ day _______________
- The flow rate of the waste activated sludge, WAS, taken from the RAS line in gpd _____
- The oxygen demand in lb/day ______________________
- The specific oxygen uptake rate in: g oxygen / g MLVSS-day ______________
- The value of
**K**_{L}**a**that will enable the aeration tank to be operated at a DO concentration of 2 mg/L, in hours^{-1}_______ - If the aeration rate being applied equals 14,835 standard cubic feet/ minute, scfm, Calculate the oxygen transfer efficiency in % ________________
- Estimate the number of diffusers required for aeration_______________________

(Use **Sanitaire** Silver Series II diffusers- disk
shape)

- Sketch to scale the process flow block diagram of the activated
sludge process that should include
**two**common wall, four-pass aeration tanks followed by**four**common wall rectangular aeration tanks. In a separate sketch show a plan view of the bottom of**one**aeration tanks and the actual number of diffusers that were selected. Aeration tank dimensions: Depth=20 feet and width= 40 feet, Length = ?

Answer #1

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 ?

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%...

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 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 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 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...

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...

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...

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?

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 5 minutes ago

asked 29 minutes ago

asked 32 minutes ago

asked 45 minutes ago

asked 45 minutes ago

asked 51 minutes ago

asked 59 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago