Question

A completely mixed activated-sludge process is being designed for a wastewater flow of 10,000 m3/d (2.64 mgd) using the kinetics equations. The influent BOD5 of 180 mg/l is essentially all soluble, and the design effluent BOD5 is 20 mg/L. The effluent biosolids concentration is 15 mg/l, of which 80% is volatile and 65% is biodegradable. For sizing the aeration tank, the SRT is selected to be 8 d and the MLVSS 3000 mg/L. The kinetic constants are as follows:
Y=0.6 mg VSS/mg BOD, ka = 0.06 d-1, Ks = 60 mg/L of BOD, and k = 5.0 day-1.
Determine the efficiency of soluble BOD removal, using the formula
E=S0-Se/S0 X 100
Here, S0 is influent soluble BOD concentration and Se is influent soluble BOD concentration.

Answer 1

Answer:

The efficiency of soluble BOD removal is 88.8889%

Explanation:

For this question not all the given parameters will be used, only the following:

BODi = influent soluble concentration = 180 mg/L

BODe = effluent soluble concentration = 20 mg/L

Question: Determine the efficiency of soluble BOD removal, η = ?

The efficiency of soluble BOD removal can be calculated using the following formula:

$\eta =\frac{BOD_{i}-BOD_{e}}{BOD_{i} } *100=\frac{180-20}{180} *100=88.8889$%