Question

In: Civil Engineering

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 air per pound of BOD. Use rectangular tanks with a depth of 12 ft, a length to width ratio of 6 to 1, and a maximum length of 110 ft. Determine the following:

a) The hydraulic retention time (hr)

b) The steady-state effluent BOD concentration (mg/L)

c) The number of tanks and their lengths (ft) and widths (ft)

d) The air requirement (ft3/min)

Solutions

Expert Solution


Related Solutions

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...
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...
An activated sludge aeration blower is delivering air to diffusers in an aeration tank. You want...
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...
what is the rate limiting element in activated sludge design?
what is the rate limiting element in activated sludge design?
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: Pumping more air than the required amount
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...
Two types of flow regime are normally used in the design of an activated sludge process,...
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
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...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT