Determine the required volume of a completely mixed activated sludge aeration tank for a conventional activated sludge system treating a design flow rate of 34,560 m³/d, where the effluent standards are 30.0 mg/L for BOD5 and 30.0 mg/L for total suspended solids (TSS). Assume that the BOD5 of the effluent TSS is 70% of the TSS concentration. Assume the BOD5 concentration leaving the primary clarifier is 128 mg/L that the MLVSS concentration (Xa) is 2,500 mg/L. Assume the following values for the growth constants: • K = 100 mg/L BOD5 • μm = 2.5 d−1 • kd = 0.050 d 1 Y = 0.50 mg VSS/mg BOD5 removed Express your answer in m³ and round to the nearest integer.
Determine the required volume of a completely mixed activated sludge aeration tank for a conventional activated sludge system treating a design flow rate of 34,560 m³/d, where the effluent standards are 30.0 mg/L for BOD5 and 30.0 mg/L for total suspended solids (TSS). Assume that the BOD5 of the effluent TSS is 70% of the TSS concentration. Assume the BOD5 concentration leaving the primary clarifier is 128 mg/L that the MLVSS concentration (Xa) is 2,500 mg/L. Assume the following values for the growth constants: • K = 100 mg/L BOD5 • μm = 2.5 d−1 • kd = 0.050 d 1 Y = 0.50 mg VSS/mg BOD5 removed Express your answer in m³ and round to the nearest integer.
Solid Waste Engineering
3rd Edition
ISBN:9781305635203
Author:Worrell, William A.
Publisher:Worrell, William A.
Chapter6: Biological Processes
Section: Chapter Questions
Problem 6.1P
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Transcribed Image Text:Determine the required volume of a completely mixed activated sludge aeration
tank for a conventional activated sludge system treating a design flow rate of
34,560 m³/d, where the effluent standards are 30.0 mg/L for BOD5 and 30.0 mg/L
for total suspended solids (TSS). Assume that the BOD5 of the effluent TSS is 70%
of the TSS concentration. Assume the BOD5 concentration leaving the primary
clarifier is 128 mg/L that the MLVSS concentration (Xa) is 2,500 mg/L. Assume the
following values for the growth constants:
•
K = 100 mg/L BOD5
• μm = 2.5 d−1
•
kd = 0.050 d 1
Y = 0.50 mg VSS/mg BOD5 removed
Express your answer in m³ and round to the nearest integer.
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