Question

1. An industry produces a degradable waste at flow rate of 0.72 m/day. Concentration of pollutant in the raw waste is 850 mg/L with first-order reaction rate constant of 2.5/day. The industry splits the flow into three parallel continuous stirred tank reactors at 0.25 m/day to reactor one, 0.3 m/day to reactor two and the remaining 0.17 m®/day to reactor three. If each reactor must achieve removal efficiency of 90%, what is the required detention time and the volume for each of the reactors? 15 pts

Answer 1

- Soli raw waste pollutant concentration = econgle reaction rate comtant = 26/day 907. efficienay of removal. effluent concentration = 850 *(1-0.9)=85 mgic. 2 we know that, . 4 = Lo ciekt) ->) 85 = 850 Cicekt) - lekt = 85 -) ekt = 1-011209 =) e 25xt = 0.9 850

) é 2.6*+=0.9 =) -2.5xt = ln (0.9) t = 0.042144 days. ... detention time required = 0.042144 days. volume of reactor with Qc = 0.25 m3/day V = Qot = 0.25X0.042144 = 0.010536m3 Volume of reactor with Qo = 0.3 ms Iday V= Qot = 0.340.042144 = 0.0126432 mm). reactor with Qo=o.lam volume of y = Qot = 0.17X0.042144= of. 16448X163 m3