1- The influent suspended solids concentration to a primary settling tank is 435
ID: 1827499 • Letter: 1
Question
1- The influent suspended solids concentration to a primary settling tank is 435mg/L.
The average flow rate is 0.050m3/s. If the suspended solids removal efficiency is 60
percent, how many kilograms of suspended solids are removed in the primary settling
tank each day?
2- The town of Camp Verde has been directed to upgrade its primary WWTP to a
secondary plant that can meet an effluent standard of 25 mg/L BOD5 and 30 mg/L of
suspended solids. They have selected a completely mixed activated sludge system for the
upgrade. The existing primary treatment plant has a flowrate of 0.029m3/s. The effluent
from the primary tank has a BOD5 of 240mg/L. Using the following assumptions,
estimate the required volume of the aeration tank:
(a). BOD5 of the effluent suspended solids is 70 percent of the allowable suspended
solids concentration.
(b). Growth constants values are estimated to be: Ks = 100mg/L BOD5; kd= 0.025 d-1;
Explanation / Answer
1)
Step1 Concentration of solids= .435g/l ;Flow rate=.050x1000 = 50l/s; efficiency= 60%
Step2 Mass of solids removed in 1 s = (60/100)x .435 x50 =13.05 g = 13.05x10^-3 kg
Step3 mass of solid removed in 1 day = 13.05x10^-3 x86400 =1127.52g
= 1.1275 kg
2)
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