Why is treatment judged necessary here?

The untreated critical deficit 8.95 mg/L exceeds the permissible 4.77 mg/L

The sewage BOD is above 200 mg/L

The river flow is less than the sewage flow

The initial deficit is 2.16 mg/L

How is the permissible mixture L_m found?

By trial using the tc and Dc equations so that Dc = 4.77 mg/L with D0 fixed

How is the percentage reduction computed?

(raw Ls - permissible Ls)/raw Ls x 100 = 49.9%

(permissible - raw)/permissible

2076 Bhadra · 8 marks

Is treatment necessary? Industry near river

Short answer

Initial deficit 2.16 mg/L, ultimate BOD 64.59 mg/L gives critical deficit 8.95 mg/L, exceeding permissible 4.77 mg/L, so treatment is needed. Maximum permissible mixture ultimate BOD is 33.06 mg/L; required BOD reduction of sewage is 49.9% (to 112.72 mg/L BOD5).

An industry is proposed to be established near the river. The characteristics of raw sewage and the characteristics of river in which the sewage is discharged are as follows: a) Sewage discharge of a town = 1.72 m³/s; b) BOD5 at 20°C of sewage = 225 mg/l; c) Minimum river discharge = 7.24 m³/s; d) BOD5 at 20° of river water = 1.2 mg/l; e) Temperature of the river after mixing = 20°C; f) De-oxygenation constant = 0.1/day at 20°C; g) Re-oxygenation constant = 0.5/day at 20°C; h) Initial DO deficit just at mixing = 2.16 mg/l; i) Critical DO deficit = 4.77 mg/l. Is treatment necessary? If so, what will be the percentage reduction of BOD of sewage?

Treatment is necessary if the critical deficit of the mixture exceeds the permissible value, 4.77 mg/l (the limit given). Rates are base 10: $K=0.1$ , $R=0.5$ , so $f=R/K=5$ .

Step 1: Deficit with no treatment

\begin{aligned} BOD_5^{mix} &= \frac{1.72\times 225+7.24\times 1.2}{8.96} = 44.16\ \text{mg/l}\\ L_0 &= \frac{44.16}{1-10^{-0.5}} = 64.59\ \text{mg/l}\\ t_c &= \frac{1}{0.1(5-1)}\log_{10}\left[5\left\{1-4\times\frac{2.16}{64.59}\right\}\right] = 1.591\ \text{days}\\ D_c &= \frac{64.59}{5}\,10^{-0.1\times 1.591} = 8.95\ \text{mg/l} \end{aligned}

Since $D_c = 8.95 > 4.77$ mg/l, treatment is necessary.

Step 2: Maximum permissible BOD of the mixture

The ultimate BOD $L_m$ of the mixture that just gives $D_c = 4.77$ mg/l (with $D_0=2.16$ ) is found from $D_c=\dfrac{L_m}{f}10^{-Kt_c}$ and the $t_c$ equation by trial. Solving gives $L_m = 33.06$ mg/l (check: $t_c=1.419$ days, $D_c=4.77$ mg/l).

Step 3: Permissible BOD of the sewage

River $L_r = 1.2/0.6838 = 1.75$ mg/l. From $Q_sL_s + Q_rL_r = L_m(Q_s+Q_r)$ :

L_s = \frac{33.06\times 8.96 - 7.24\times 1.75}{1.72} = 164.85\ \text{mg/l}\quad (BOD_5 = 112.72\ \text{mg/l})

Raw sewage $L_s = 225/0.6838 = 329.06$ mg/l.

\text{BOD reduction} = \frac{329.06-164.85}{329.06}\times 100 = 49.9\%

Answer: treatment is necessary; BOD of the sewage must be reduced by about 49.9% (to about 112.72 mg/l BOD5).

✓ Answer checked by Claude Opus 5.5 (high effort)

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Is treatment necessary? Industry near river

Step 1: Deficit with no treatment

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Step 3: Permissible BOD of the sewage

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