Question

Two types of flow regime are normally used in the design of an
activated sludge process,

namely completely mixed and plug flow (both are continuous). Using
first order kinetic

reaction, demonstrate which flow regime provide a better
performance, given the

followings:

Q = 550 m3/d t = 6 hours Co = 225 mg/L k = 0.15 d-1

Answer #1

Design a completely mixed activated sludge process using three
different design approaches (i.e. calculate the volume of the
aeration basin). Base your design for the first approach on
detention time, for the second one on BOD loading, and the third
one on kinetics. The influent design flow rate to the activated
sludge process is 10,000 m3 /day with a BOD5 concentration of
200mg/L. Completely mixed activated sludge processes typically have
a detention time ranging from 3 to 6 hours and...

A completely mixed activated-sludge process is being designed
for a wastewater flow of (2.64 mgd) using the kinetics equations.
The influent BOD of 120 mg/Lis essentially all soluble and the
design effluent soluble BOD is 7 mg/L, which is based on a total
effluent BOD of 20 mg/L. For sizing the aeration tank, the mean
cell residence time is selected to be10 days and the MLVSS 2000
mg/L. The kinetic constants from a bench-scale treatability study
are as follows: Y...

Mass of solids Generated from the activated sludge tanks.
13. Assume: The total design flow is 3.75 MGD (15,000
m3/day). The NPDES limit is 25/30. Assume that the waste
strength is 170 mg/L BOD after primary
clarification. Y = 0.55 kg/kg.
What is the mass of solids generated each day (Kg/day or
lbs/day) in the activated sludge tankage?

Sooner City Reclamation Plant is a completely mixed activated
sludge process designed to treat 20,000 m3/d of domestic
wastewater having an influent BOD5 concentration of 350
mg/L (assume a 28.6% reduction in during primary treatment). Use
typical activated sludge kinetic constants except for kd
= 0.06 1/d. Assume that the MLVSS concentration in the aeration
basin is maintained at 3,000 mg/L and the ratio of VSS:TSS is 0.75.
Permit requires that the effluent BOD5 and TSS
concentrations do not exceed...

A completely mixed activated sludge process is designed to treat
20,000 m3 /d of domestic wastewater having a BOD5 concentration of
250 mg/L following primary treatment. The NPDES permit requirement
for effluent BOD5 and TSS is 20 mg/L for each. The biokinetic
parameters have been established as: yield = 0.6 , k = 5/d, Ks = 60
mg/L, and kd = 0.06/d. The MLVSS concentration in the aeration tank
is to be maintained at 3000 mg/L and the ratio of...

Question 1
You are assigned to carry out a preliminary design for a water
treatment process. A pollutant of concern, denoted ‘A’, is known to
degrade in water. But you do not know the rate law or reaction rate
constant. You conduct a series of experiments. In each case, you
established an initial concentration of 50 mg/L of pollutant ‘A’ in
water in a sealed vessel. After a certain period, you open the
vessel and measure the remaining concentration of...

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