Question

A wastewater treatment plant discharges 1 m3/s of effluent with an ultimate BOD of 40 mg/L into a stream flowing at 10 m3/s. Just upstream of the discharge point, the stream has an ultimate BOD of 3 mg/L. The deoxygenation constant (kd) is estimated to be 0.22 1/d. (a) Assuming complete and instantaneous mixing, find the ultimate BOD of the mixture of the waste and the river just downstream of the outfall. (b) Assuming a constant cross-sectional area for the stream equal to 55 m2, what ultimate BOD would you expect to find at a point 10,000 m downstream

Answer 1

(a) 6.36 mg/L

(b) 5.60 mg/L

Step-by-step explanation:

(a)

Using the formula below to find the required ultimate BOD of the mixture.

`L_o = (Q_wL_w+ Q_rL_r)/(Q_W+Q_r)`

where;

Q_w = volumetric flow rate wastewater

Q_r = volumetric flow rate of the river just upstream of the discharge point

L_w = ultimate BOD of wastewater

Replacing the given values:

`L_o = ((1 \ m^3/L ) (40 \ mg/L) + (10 \ m^3/L) (3 \mg/L))/((1m^3/L) +(10 \ m^3/L)) \\ \\ L_o = 6.36 \ mg/L`

(b)

The Ultimate BOD is estimated as follows:

Recall that:

`time(t) = (distance )/(speed)`

replacing;

distance with 10000 m and speed with
`(11 \ m^3/s)/(55 \ m^2)`

`time =(10000 \ m)/(\Bigg((11 \ m^3/s)/(55 \ m^2)\Bigg))\Bigg((1 \ hr)/(3600 \ s)\Bigg) \Bigg((1 \ day)/(24 hr)\Bigg)`

time (t) = 0.578 days

Finally;
`L_t = L_oe^(-kt)`

here;

k = rate of coefficient reaction

`L_ t= (6.36) * e^(-(0.22/day)(0.5758 \ days))\\ \\ \mathbf{L_t =5.60 \ mg/L}`

Thus, the ultimate BOD = 5.60 mg/L