Question
In: Civil Engineering
A wastewater treatment plant with activated-sludge process treats wastewater for 120,000 m3/d. The effluent standard for...
A wastewater treatment plant with activated-sludge process
treats wastewater for 120,000 m3/d. The effluent standard for both
BOD and TSS is 20 mg/L. Influent TSS concentration is 450 mg/L and
influent BOD concentration is 220 mg/L. MLSS concentration is 3500
mg/L (Assume no nitrification and temperature is 20°C). Determine:
(20 pts) (a) Influent BOD and TSS concentration to the aeration
tank if the treatment efficiencies in primary clarifier are 30% and
70% for BOD and TSS, respectively. (b) Determine sludge production
as kg TSS/d (c) What is the design volume and retention time of the
aeration basin (m3)? (d) Determine the F/M ratio in units of kg
BOD5/kg MLVSS-day.
(According to your design you can assume some parameters i.e SRT
but do not forget to check both your assumptions and your results
are in the desirable design conditions)
Solutions
Expert Solution
Hope this helps
Please feel free to comment on any further queries
All the best :)
Ally Wells answered 7 months agoRelated Solutions
Design an aeration tank for activated sludge wastewater treatment designed to meet an effluent standard of...
Design an aeration tank for activated sludge wastewater
treatment designed to meet an effluent standard of 10 mg/L
BOD5. The flow from the primary effluent is 10
m3/min with a primary effluent BOD5 of 250
mg/L. The design calls for the suspended solids to equal 3500 mg/L
for an SRT of 5 days. The decay rate constant is 0.03
day-1 and the yield is 0.75 mg/mg. Determine:
the tank volume required to meet these specifications,
the resulting mass of sludge...
A conventional activated-sludge plant treats 10.0 MGD of municipal wastewater with a BOD concentration of 240...
A conventional activated-sludge plant treats 10.0 MGD of
municipal wastewater with a BOD concentration of 240 mg/l and
suspended solids of 200 mg/l. The sludge flow pattern is shown in
Figure 11-1 of the textbook with a gravity belt thickener to
concentrate the excess activated sludge. Primary and thickened
activated sludge are pumped separately to the anaerobic digester.
The primary clarifier removes 50% of the suspended solids and 35%
of the BOD. The primary sludge solids content is 4.0%. The...
A conventional activated-sludge plant treats 10.0 MGD of municipal wastewater with a BOD concentration of 240...
A conventional activated-sludge plant treats 10.0 MGD of
municipal wastewater with a BOD concentration of 240 mg/l and
suspended solids of 200 mg/l. The sludge flow pattern is shown in
Figure 11-1 of the textbook with a gravity belt thickener to
concentrate the excess activated sludge. Primary and thickened
activated sludge are pumped separately to the anaerobic digester.
The primary clarifier removes 50% of the suspended solids and 35%
of the BOD. The primary sludge solids content is 4.0%. The...
A wastewater plant with the activated sludge process received 5 × 106 L/day of wastewater with...
A wastewater plant with the activated sludge process received 5
× 106 L/day of wastewater with a BOD5 of 500
mg/L. The primary clarifier removes 30% of the BOD and primary
clarify effluent has no biomass (Xi=0 mg/L). Recycle
(Qr) and waste sludge (Qw) flows are 25% and
2% of Qin, respectively. Effluent BOD5 from secondary
clarifier and waste sludge is 20 mg/L. Microorganism concentration
in the waste sludge (Xw) is 6,000 mg/L. Soluble
BOD5 is biologically converted into CO2...
A wastewater treatment plant that treats wastewater to secondary treatment standards uses a primary sedimentation process...
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 municipal wastewater treatment plant discharges 20,000 m3/d of treated wastewater to a stream. The wastewater...
A municipal wastewater treatment plant discharges 20,000 m3/d of
treated wastewater to a stream. The wastewater has a BOD5 of 30
mg/L with a k1 of 0.23 d-1. The temperature of the wastewater is 20
oC, and the dissolved oxygen is 2.0 mg/L. The stream just above the
point of wastewater discharge flows at 0.65 m3/s, has a BOD5 of 5.0
mg/L, and is 90% saturated with oxygen. The temperature of the
stream is also 20 oC. After mixing, the...
The following are average operating data from a secondary activated sludge: Wastewater flow = 29160 m3/d...
The following are average operating data from a secondary
activated sludge:
Wastewater flow = 29160 m3/d
Vol. of aeration tanks = 8400 m3
Influent SS = 120 mg/L Influent BOD5 = 170 mg/L
Effluent SS = 22 mg/L Effluent BOD5 = 20 mg/L
Volatile fraction of mixed liquor and effluent suspended solids
is 0.65.
Mixed liquor suspended solids = 2500 mg/L
Waste sludge flow = 200 m3/d SS in Waste sludge = 9800
mg/L
Determine the following:
Aeration period, Food...
A 100-mL water sample is collected from the activated sludge process of municipal wastewater treatment. The...
A 100-mL water sample is collected from the activated sludge
process of municipal wastewater treatment. The sample is placed in
a drying dish (weight = 0.5000 g before the sample is added), and
then placed in an oven at 104°C until all the moisture is
evaporated. The weight of the dried dish is recorded as 0.5625 g. A
similar 100-mL sample is filtered and the 100-mL liquid sample that
passes through the filter is collected and placed in another drying...
An activated sludge process with an aeration basin volume of 0.3 MG is treating wastewater with...
An activated sludge process with an aeration basin volume of 0.3
MG is treating wastewater with the average daily flow of 2 MGD. The
raw sewage entering the treatment plant has an average BOD5 of 400
mg/L. The primary treatment removes 25% of BOD5 and the subsequent
activated sludge process is designed to remove 90% of BOD5.
Given:
Plant effluent BOD5 concentration = 30 mg/L
Biomass concentration in the aeration tank = 2,000 mg/L
Biomass concentration in the plant effluent...
An activated sludge plant receive 5.0 MGD of wastewater with a BOD of 220 mg/L. The...
An activated sludge plant receive 5.0 MGD of wastewater with a
BOD of 220 mg/L. The primary clarifier removes 35% of the BOD. The
sludge is aerated for 6 hr. The food-to-microorganism ratio is
0.30. The recirculation ratio is 0.2. The surface loading rate of
the secondary clarifier is 800 gal/day-ft2. The final effluent has
a BOD of 15 mg/L. What are the (a) BOD removal efficiency of the
activated sludge treatment processes (secondary BOD removal), (b)
aeration tank volume,...
ADVERTISEMENT
ADVERTISEMENT
Latest Questions
- In java based on the classes provided finish the program per the requirements: ----------------------------------------------Class1 package cardgame;...
- Identify two cultures and analyze the nurturing process within each culture and discuss the impact the...
- Digital security is an increasing concern is the Internet age. In order to protect sensitive information...
- the challenges experienced by companies considering full application of customer retention strategy in the face of...
- Respond to the following in a minimum of 175 words: The flow of a program is...
- Write a version of the binary search algorithm that can be used to search a string...
- Using the following reduction potentials: I2(s) + 2e- <---> 2I-(aq) E0 = 0.535 V I2 (aq)...
ADVERTISEMENT