Answer:
a) t = 165 days
b) 9.9 kg/day
Explanation:
Given data:
final lelvel of BOD is 20 mg/l
Ks = 100 mg BOD/L,
kd = 0.10 day-1 ,
μm = 1.6 day-1 ,
Y = 0.60 mg SS/mg BOD
a) we know that
[tex]c_{out} = \frac{c_{in}}{1 + kt}[/tex]
[tex]20 = \frac{350}{1 + 0.1 t}[/tex]
solving for t
t = 165 days
b) mass of microbes [tex]= Q(mlD) × C(mg/l)[/tex]
[tex]= 30\times 10^{-3} (350-20) = 9.9 kg/day[/tex]