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In: Physics
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 operating F/M ratio for the activated-sludge process is 0.3 lb BOD per day per pound of MLSS. The solids content of the waste sludge removed from the secondary clarifier is 16,000 mg/l; the gravity belt thickening captures 95% of the solids in the sludge and increases the solids content of the thickened sludge to 7%. Calculate: a) Flow rate of primary sludge, in gals/day b) Flow rate of thickened secondary sludge, gals/day. c) Flow rate and solids concentration of the blended sludge
Solutions
genius_generous answered 1 year agoRelated Solutions
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...
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...
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 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...
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...
. The average flow of wastewater into the Metropolitan Wastewater Treatment Plant is 180 MGD. What...
. The average flow of wastewater into the Metropolitan
Wastewater Treatment Plant is 180 MGD. What is the loading of total
solids in this wastewater (lbs/day)? What is the loading of
volatile solids in this wastewater (lbs/day)?
Design a set of activated sludge aeration tanks. The flow rate to treat is 5.6 MGD...
Design a set of activated sludge aeration tanks. The flow rate
to treat is 5.6 MGD and the BOD concentration is 150 mg/L. The
design solids concentration (X) at steady-state is 2,000 mg/l. The
design MCRT is 7 days. The kinetic coefficients are as follows:
K = 2 g BODg cells⋅d,
Ks = 25 BOD/L, Kd
= 0.06 day-1, Y = 0.5 g
BODg cells⋅d. The influent ammonia concentration is
40 mg/l and nitrification is needed. It takes 1400
ft3of...
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.
You have wastewater plant discharging into a clean river. The BOD of the Wastewater flow is...
You have wastewater plant discharging into a clean river. The
BOD of the Wastewater flow is 300 mg/L, the wastewater flow is 15
m3/sec, the river flow upstream of the mixing point is 120 m3/sec
and BOD upstream of river is 0. The DO of the river at the mixed
point is 8 mg/L and the saturation DO in the river would be 10
mg/L, the deoxygenation rate coefficient is 0.30/day and the
reaeration constant is 0.90/day. The river is...
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