Question

Consider a population with an initial biomass (X0) of 5 mg VSS/L, a maximum specific growth rate (µ_max) of 0.9/day, and a decay coefficient of 0.15/day. Determine the biomass concentration (mg VSS/L) at the end of 5 days and 20 days.

Answer 1

mg VSS The initial concentration of biomass, X is 5 0.9 The maximum specific growth rate coefficient, μ max is 0.15 The respiration (or) decay coefficient, kj is The expression for combined (maximum specific growth rate coefficient and respiration (or) decay coefficient) growth rate is as follows: dX dt dX dt The expression for the concentration at time t (X.) is as follows Rearrange equation (1) as follows: dX dX Integrate the above equation with the following limits: X (Concentration): Initial concentration (XoConcentration at time t (X,) t (time ):0>t

dt 6 =In X+C) 贡 Since, ( Hmax k Since, In-= Ina-in b k,)t

Calculate the concentration of biomass (mg VSS/L) at the end of 5 days as shown below: mg VSS day X, = 5 mgvssx ea. 0.15) days IX) days day day 0.75 5 days dary X,-5 mg VSS mg VSS X. 212.605 X 213 mg VSS mg VSS Therefore the concentration of biomass (mg VSS/L) at the end of 5 days is 213

Calculate the concentration of biomass (mg VSS/L) at the end of 20 days as shown below 0.9 0.15 dav k20 days mg VSS 0.75 20 dasís 15 X. 16345086.86 Multiply and divide with 10' X, = 16345086.86 mg VSS 107 10 X = 1.634508686×107 ng VSS X, ~ 1.6×107 Therefore, the concentration of biomass (mg VSS/L) at the end of 20 days is 1.6×107-mgVSS