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

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ajeigbeibraheem ajeigbeibraheem · Answer 1 · 2021-04-14T18:33:06+02:00

Answer:

(a) 6.36 mg/L

(b) 5.60 mg/L

Explanation:

(a)

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

[tex]L_o = \dfrac{Q_wL_w+ Q_rL_r}{Q_W+Q_r}[/tex]

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:

[tex]L_o = \dfrac{(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[/tex]

(b)

The Ultimate BOD is estimated as follows:

Recall that:

[tex]time(t) = \dfrac{distance }{speed}[/tex]

replacing;

distance with 10000 m and speed with [tex]\dfrac{11 \ m^3/s}{55 \ m^2}[/tex]

[tex]time =\dfrac{10000 \ m}{\Bigg(\dfrac{11 \ m^3/s}{55 \ m^2}\Bigg)}\Bigg(\dfrac{1 \ hr}{3600 \ s}\Bigg) \Bigg(\dfrac{1 \ day}{24 hr}\Bigg)[/tex]

time (t) = 0.578 days

Finally; [tex]L_t = L_oe^{-kt}[/tex]

here;

k = rate of coefficient reaction

[tex]L_ t= (6.36) \times e^{-(0.22/day)(0.5758 \ days)}\\ \\ \mathbf{L_t =5.60 \ mg/L}[/tex]

Thus, the ultimate BOD = 5.60 mg/L