Why aeration is important for activated sludge process? What
would be the effect on the DO...
Why aeration is important for activated sludge process? What
would be the effect on the DO profile with time (provide a sketch)
and BOD removal and sludge production if:
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...
An activated sludge aeration blower is delivering air to
diffusers in an aeration tank. You want to estimate the isentropic
efficiency and the work input (by the motor). At the blower inlet,
the volumetric flow rate is measured at 85 m3/min, the pressure is
0.98 atm and the temperature is 20⁰C. If the pressure is increased
to 1.61 atm and the blower outlet temperature is 350 K: a) what is
the blower isentropic efficiency, and b) what is the work...
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.
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 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...
Data from a field study on a step-aeration activated-sludge
secondary are as follows:aeration tank volume = 120,000 ft3= 0.898
milliongalloninfluent wastewater BOD = 128 mg/Leffluent SS = 26
mg/Leffluent wastewater BOD = 22 mg/Leffluentwastewater flow = 3.67
mgdSS in waste sludge = 11,000 mg/LMLSS in aeration tank = 2350
mg/Lwaste sludge flow = 18,900 gpd = 0.0189 mgd return sludge flow
= 1.27 mgd Using these data, calculate BOD loading and MLSS
inaeration tank. Also, compute the sludgeage (solids retention...
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...
What is an activated sludge system and what are the processes
that occur in the system? What is the role of "mixed liquor
suspended solids"? (hint: draw the system diagram and think about
what can be happening)