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 10 months agoRelated Solutions
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(Qr) and waste sludge (Qw) flows are 25% and
2% of Qin, respectively. Effluent BOD5 from secondary
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mg/L (Assume no nitrification and temperature is 20°C). Determine:
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A conventional activated-sludge plant treats 10.0 MGD of municipal wastewater with a BOD concentration of 240...
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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%
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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...
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An activated sludge process with an aeration basin volume of 0.3
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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...
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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
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the resulting mass of sludge...
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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
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