An activated sludge treatment process uses a rectangular tank that can be modeled as two completely mixed reactors in series. The volume of the tank is 300 cubic meters (for modeling purposes 150 cubic meters each) and the flow rate into the system is 900 m3/day. The influent BOD concentration is 180 mg/L and the first order reaction rate coefficient is 0.5 hr-1. What is the effluent concentration of BOD? If the Biomass Yield, Y is equal to 0.6 mg SS/mg of BOD metabolized, what is the rate of SS (biomass) production in the aeration tanks (mg/L-day)? If the post-secondary clarifier effluent SS concentration is 10 mg/L, the sludge recycling rate is 33%, and the waste sludge flow rate is 5 m3/day, what is the SS concentration in the waste sludge (hint: assume the system is at steady state and complete a mass balance)? What is the sludge age?

Given Data: The volume of the tank is, V=300 m3 The influent flow rate is, Q=900 m3/day The influent BOD concentration is, L=180 mg/L The rate coefficient is, k=0.5/hr Part (1): The expression for the effluent concentration of BOD is given as, BODeff=L1-e-k×20..............(1) Here, k is the rate coefficient. Substitute the values in equation (1) BODeff=180 mg/L×1-e-0.5×20BODeff=179.99 mg/L Therefore, the effluent concentration of BOD is 179.99 mg/L. Part (2): The rate of biomass production in the aeration tanks is calculated as, u=QY0Vu=900 m3/day×0.6×180.008300 m3u=324.01 mg/L-day Therefore, the rate of biomass production in the aeration tanks is 324.01 mg/L-day. Part (3): The post-secondary clarifier effluent concentration is 10 mg/L, the waste sludge flow rate is 5 m3/day and the recycling rate is 33%. The recycling flow rate is calculated as, Qr=33×9022100Qr=2977.26 m3/day The effluent concentration is equal to ye=27 mg/L. Hence, Total concentration=5 m3/day×10 mg/L+2977.26 m3/day×27 mg/L57297 m3/dayTotal concentration=1.40 mg/L Therefore, the SS concentration in the waste sludge is 1.40 mg/L.Part (4): The expression for the sludge age is given as, θc=Mass of solid in aeration tankMass of solid wasted per dayθc=VxQwxu+Qexe.................(2) Assume xu=xe QrQ=xxu-x0.33=xxu-x0.33xu-0.33x=xxu=4.03x Assume, xe=50 mg/Lx=2500 mg/Lxu=4.03×2500 mg/L=10075 mg/LQw=23 m3/dayQe=580 m3/day Now, substitute the values in equation (1) θc=300 m3×2500 mg/L23 m3/day×10075 mg/L+580 m3/day×50 mg/Lθc=2.87 days Therefore, the sludge age is 2.87 days.

Answered: An activated sludge treatment process…

Structural Analysis

6th Edition

ISBN: 9781337630931

Author: KASSIMALI, Aslam.

Publisher: Cengage,

See similar textbooks

Concept explainers

Biological Processes

Biological solid waste includes bacterial and contaminated waste produced by living individuals. Its treatment is essential to prevent the spread of any air-borne or water-borne diseases or diseases transmitted from one organism to another in a food chai…

Question

An activated sludge treatment process uses a rectangular tank that can be modeled as two completely mixed reactors in series. The volume of the tank is 300 cubic meters (for modeling purposes 150 cubic meters each) and the flow rate into the system is 900 m³/day. The influent BOD concentration is 180 mg/L and the first order reaction rate coefficient is 0.5 hr^-1.

What is the effluent concentration of BOD?
If the Biomass Yield, Y is equal to 0.6 mg SS/mg of BOD metabolized, what is the rate of SS (biomass) production in the aeration tanks (mg/L-day)?
If the post-secondary clarifier effluent SS concentration is 10 mg/L, the sludge recycling rate is 33%, and the waste sludge flow rate is 5 m³/day, what is the SS concentration in the waste sludge (hint: assume the system is at steady state and complete a mass balance)?
What is the sludge age?

Expert Solution