Question
In: Civil Engineering
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 and Y (yield coefficient) is 0.5 mgVSS /mgBOD
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- Calculate the value of X and the volume of the tank
Solutions
Ally Wells answered 10 months agoRelated Solutions
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,...
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...
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...
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...
The Lolla Hart hospital has a small activated sludge plant to treat its wastewater. The average...
The Lolla Hart hospital has a small activated sludge plant to
treat its wastewater. The average daily hospital discharge is 1,500
L per day per bed, and the average soluble BOD5 after
primary settling is 500 mg/L. Their permit dictates that the
effluent BOD5 and TSS cannot exceed 30 and 20 mg/L,
respectively, on an annual basis. Assume the MLVSS concentration in
the aeration basin will be maintained at 3,000 mg/L. Also, assume
that the concentration of BOD5 of the...
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...
1-S0 in an activated sludge process represents _________________________ concentration. 2-X in an activated sludge process represents...
1-S0 in an activated sludge process represents
_________________________ concentration.
2-X in an activated sludge process represents
___________________________ concentration.
3-Biomass present in the influent is a product of biomass
concentration in the influent multiplied by ______________
___________.
4-Hydraulic detention time is a ratio of ______________ and
_________________.
5- ___________________ represents the portion of microorganisms
discharged from the process.
9. The following data are given for an activated sludge process: BOD5 (influent) = 250 mg/L...
9. The following data are given for an activated sludge process:
BOD5 (influent) = 250 mg/L (soluble) BOD5 (effluent) = 20 mg/L
(soluble) MCRT = 10 days MLSS in the aeration tank = 2500 mg/L
Concentration of SS in the recycle line = 10,000 mg/L Concentration
of SS in the effluent = 30 mg/L (70% biodegradable) BOD5/BODu =
0.75, No nutrient limitations. Yield co-efficient, Y = 0.65
Endogenouse decay coefficient, kd = 0.06/day , Q=5MGD
Determine:
(a) Find soluble BOD5...
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...
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