A lake with constant volume of 10 times 10^6 m^3 is fed by a pollution-free stre
ID: 980907 • Letter: A
Question
A lake with constant volume of 10 times 10^6 m^3 is fed by a pollution-free stream with a flow rate 40 m^3/s. A factory dumps 5 m^3/s of a nonconservative waste with a concentration of 100 mg/L into the lake. The pollutant has a reaction rate coefficient K of 0.18/day. Assuming the lake is completely mixed, find the steady-state concentration of pollutant in the lake. If the factory is forced to install treatment to reduce the concentration of its waste by 80%, what will be the pollutant concentration in the lake 14 days after the treatment process begins operating?Explanation / Answer
a) the characteritic equation for a CSTR is T= CAOXA/-rA
for first order KT = XA/(1-XA) where XA= conversion K= rate constant =0.18/day
where T= space time= V/VO , V= volume of reactor and Vo= volumetric flow rate =40+5= 45 m3/s
T= 10*106/ (40+5)=222222.2sec= 222222.2/60 min =3703.74 min =3703.74/60 hr =61.7284 hr= 61.7284/60 days= 2.57 days
once the stream gets mixed , inlet concentration, CAO = 0+5*100mg/L /45= 11.11 mg/L
KT= XA/(1-XA)
0.18*2.57= XA/(1-XA)
0.4626= XA/(1-XA)
XA =0.4626- 0.4626XA
1.4626XA =0.4626 XA=0.4626/1.4626=0.32
1-CA/CAO =XA where CA= concentratino at any time t and CAO= initial concentration
1-CA/CAO= 0.32
CA/CAO= 0.68
CA=0.68*11.11 mg/L=7.5548 mg/L
b) give T= 14 days
KT= XA/(1-XA)
0.18*14= 2.52= XA/(1-XA)
2.52- 2.52XA= XA and 3.52XA= 2.52 and XA= 2.52/3.52= 71.5%
CA= CAO*(1-XA)= 11.11*(1-0.715)= 3.16636 mg/L
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