Treated waste water is discharged at the rate of 1.5$m^3$/sec into a river of minimum flow of 5$m^3$/sec. The temperature of river flow and waste water flow may be assumed at 25°C.The BOD removal rate constant $k_1$ is 0.12/d, water upstream of the waste water is 200 mg/s and that of the river water upstream of the waste water outfall is 1 mg/l. The efficiency of waste water treatment is 80%. Evaluate

1. $BOD_5$ at 20°C . if river water receives untreated waste water.
2. $BOD_5$ at 25°C , if river water receives treated waste water
3. ultimate BOD of the river water after it receives treated waste water.

Given:-

Discharge of waste water=$Q_w=1.5m^3/s$

Discharge of river=$Q_r=5m^3/s$

Temperature=T=25°C

$K_d(25°)=k_1=0.12/d$

$C_w=conc of BOD_5 $ for untreated waste water

=200 mg/l

$C_r=conc. of BOD_5 $ for river water=1mg/l

1. conc. of$BOD_5$ of the mixture if untreated waste water is discharged into the river

$C=\frac{C_w*Q_w+C_R*Q_R}{Q_w+Q_R}$

$C=\frac{200*1.5+1*5}{1.5+5}$

C=46.92mg/l

1. $BOD_5 $ of the treated wastwater is given by,

$C_TW$=20% of the $BOD_5$ of untreated wastwater

∵efficiency of wastwater treatment is 80%

$C_TW = 20 \% * C_w$

= 20%*200 mg/l

$C_TW=40 mg/l$

$BOD_5$ of mixture if treated wastwater is discharged into the river

$C^1=\frac{C_TW*Q_w+C_R*Q_R}{Q_w+Q_R}$

$=\frac{40*1.5+1*5}{1.5+5}$

$C^1=10mg/l$

1. $BOD_5$ of river water after it receives treated wastewater = 10 mg/l

ultimate BOD of this mixture=$Y_u$=L=?

$Y_t(day)=L[1-10^-K_Dt]$

$Y_s=L[1-10^-0.12*5]$

$10=L[1-10^-0.6]$

L=13.35 mg/l