Question

A wastewater treatment plant discharges 1.0 m3/s of effluent having an ultimate BOD of 40.0 mg/ L into a stream flowingat 10.0 m3/s. Just upstream from the discharge point, the stream has an ultimate BOD of 3.0 mg/L The deoxygenationconstant kd is estimated at 0.22/day.(a) Assuming complete and instantaneous mixing. find the ultimate BOD of the mixture of waste and river just downstreamfrom the outfall.(b) Assuming a constant cross-sectional area for the stream equal to 55 m2, what ultimate BOD would you expect to find ata point 10.000 m downstream?

Answer 1

Answer:

a) 6.4  mg/l

b) 5.6 mg/l

Explanation:

Given data:

effluent Discharge Q_w = 1.0 m^3.s

Ultimate BOD L_w = 40 mg/l

Discharge of stream Q_r = 10 m^3.s

Stream ultimate BOD L_r = 3  mg/l

a) Ultimate BOD of mixture$= \frac{Q_w l_w + Q_r L_r}{Q_w + Q_r}$

                                         $= \frac{1*40 + 10*3}{10 +1} = 6.4 mg/l$

b) utlimate BOD at 10,000 m downstream

$t =\frac{distance}{speed} = \frac{10,000}{\frac{Q_r +Q+w}{55}} \times \frac{hr}{3600} \times \frac{day}{24 hr}$

putting $Q_r + Q_w = 1+ 10 = 11 m^3/s$

t = 0.578  days

we know

$L_t = L_o e^{-kt}$

$L_t = 6.4 \times e^{-0.22 \times 0.578}$

$L_t = 5.6 mg/l$