Use this balanced equation:   Note that this equation is m…

Use this balanced equation:   Note that this equation is made up just for this Question and is not correct stoichiometrically. Molecular weight of acetate (CH3COOH) is 60 g/mole. Molecular weight of cell material (C5H7O3N) is 129 g/mole. Molecular weight of oxygen (O2) is 32 g/mole. COD of acetate is 1.07 g COD/g substrate. 1 g of cells can be assumed to be 1 g VSS.

This illustration was provided during the lecture to help yo…

This illustration was provided during the lecture to help you understand type of solid (A, B, or C in red color) generated during secondary treatment process. Questions from 4 to 10 describes type of solid (in a descriptive way or equation). Choose right solid type (A, B, or C in red color) for the given solid in the following questions. 

Substrate utilization rate curve is provided as below. The s…

Substrate utilization rate curve is provided as below. The substrate utilization rate is approaching to its maximum, 1000 g/d/m3 as substrate concentration increases. Using the information provided, determine Ks in mg bsCOD/L. X = 100 mg VSS/L k = 10 mg bsCOD/mg VSS/d