A waste water effluent of 620 lit/sec with a BOD=45 mg/lit, D.O.=3.0mg/lit and temperature of 24°C enters a river where the flow is 27 m/sec, BOD=3.0 mg/lit, DO=8.2mg/lit, and temperature of 18° C, Ki of the waste is 0.10 per day at 20°C. The velocity of water in the river downstream is 0.18m/sec and depth of 1.2m. Determine the following after mixing of waste water with the river water. i) Combined discharge ii) BOD 1) DO iv) Temperature

Particulars of sewage thrown:

Qs =620 l/s= $0.62m^3$/sec

Concentrations(Cs):

BOD= 45mg/l

Do= 3 mg/l

Temperature= $24^o$C

K1 at $20^o$C= 0.1 per day

Particulars of river:

Qr = $27 m^3$/sec

Concentrations(Cr):

BOD= 3 mg/l

Do= 8.2 mg/l

Temperature= $18^o$C

Combined discharge= =Qs+Qr = 0.62 + 27= $27.62 m^3$/sec

C= $\frac{Cs.Qs + Cr.Qr}{Qs + Qr }$

BOD of mix = $\frac{45 x 0.62 + 3 x 27}{0.62 + 27 }$ = 3.94 mg/l

DO of mix = $\frac{3 x 0.62 + 8.2 x 27}{0.62 + 27 }$ = 8.08 mg/l

Temperature of Mix = $\frac{24 x 0.62 + 18 x 27}{0.62 + 27 }$ = $18.13^o$C