Question
In: Civil Engineering
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 suspended solids is equal to 60% of the suspended solids concentration. The biokinetic coefficients were measured and determined to be:
Ks=70 mg/L BOD5
um=2.5 day-1
kd = 0.06 day-1
Y = 0.6 mg VSS/mg BOD5 removed
Find:
SRT in days
The Aeration basin volume (L)
The Aeration period (days)
The kg of secondary solids wasted each day
Solutions
Ally Wells answered 9 months agoRelated Solutions
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 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 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 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 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 completely mixed activated sludge process is to be used to treat a flow rate of...
A completely mixed activated sludge process is to be used to
treat a flow rate of 2.5 MGD. Design a rectangular activated sludge
reactor with a 4:1 ratio of length to width. (design only 1 reactor
for all flow). The effluent from the plant must not exceed 15 mg/L
of BOD5. Use the following design criteria:
a. BOD5 influent = 250.0 mg/L
b. MLVSS (X) = 3500 mg/L
c. Reactor depth = 11 ft
d. Effluent BOD5 must be less...
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...
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...
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...
ADVERTISEMENT
ADVERTISEMENT
Latest Questions
- The Coca-Cola Company A description of a strategy in the selected organization to evaluate. Analysis of...
- For each case history provide: a diagnosis; list the criteria required for the diagnosis; and provide...
- (a) Let S = {a, b, ab, aba}. How many different factorizations are there of (ab)11 ? It is not enough to...
- 1. Prove that if a set A is bounded, then A-bar is also bounded. 2. Prove...
- Using Visual Studio Code (JavaScript) For this lab you must use a reasonable faceted search example,...
- Python 3 Fix the code so if the user enter the same bar code more than...
- JavaScript each item must have at least three categorical attributes and at least one numeric attribute....
ADVERTISEMENT