Why is the permissible mixture BOD found by trial?

tc and Dc both depend on L_m, so the equations cannot be solved directly

2075 Baisakh · 8 marks

Degree of treatment, city 1200 l/s

Short answer

Mixing gives initial deficit 2.51 mg/L. Permissible critical deficit is 5.17 mg/L (to maintain DO ≥4 mg/L). Trial solution yields maximum permissible mixture ultimate BOD 23.87 mg/L. Sewage ultimate BOD must be reduced from 234 mg/L to 111.12 mg/L, requiring 52.5% treatment efficiency.

A city discharges sewage at the rate of 1200 l/s, into a stream whose minimum flow is 5000 liters/sec, the temperature of both being 20°C. The 5 day BOD at 20°C for sewage is 160 mg/l and that of river water is 2 mg/l. The DO content of sewage is zero while that of stream is 90% of the saturation DO. Find out the degree of treatment required if the minimum DO to be maintained in the stream is 4 mg/lit. Assume deoxygenation coefficient as 0.10 (base 10) and re-oxygenation coefficient as 0.30 (base 10). Given saturation DO at 20°C as 9.17 mg/lit.

Flows are in m³/s ( $Q_s=1.2$ , $Q_r=5.0$ ). River DO $=0.9\times 9.17=8.253$ mg/l; sewage DO $=0$ ; the temperature is 20°C so $DO_{sat}=9.17$ mg/l.

Step 1: Mixing and permissible deficit

\begin{aligned} DO_{mix} &= \frac{1.2\times 0 + 5.0\times 8.253}{6.200} = 6.66\ \text{mg/l}\\ D_0 &= DO_{sat}-DO_{mix} = 9.17-6.66 = 2.51\ \text{mg/l}\\ D_{c(allowed)} &= DO_{sat}-DO_{min} = 9.17-4.0 = 5.17\ \text{mg/l} \end{aligned}

Step 2: Permissible BOD of the mixture

With $K=0.1$ and $R=0.3$ (base 10), $f=R/K=3.00$ . The critical deficit is

t_c=\frac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\frac{D_0}{L_m}\right\}\right],\qquad D_c = \frac{L_m}{f}10^{-Kt_c}

and $D_c$ must not exceed 5.17 mg/l. Solving these two equations for $L_m$ by trial (with $D_0=2.51$ fixed) gives $L_m=23.87$ mg/l, with $t_c=1.872$ days (check: $D_c=5.17$ mg/l).

Step 3: Permissible BOD of the effluent

\begin{aligned} L_{raw} &= \frac{BOD_5}{1-10^{-5K}} = \frac{160}{0.6838} = 234.00\ \text{mg/l}\\ L_r &= 2.92\ \text{mg/l (river)}\\ L_s &= \frac{L_m(Q_s+Q_r)-Q_rL_r}{Q_s} = \frac{23.87\times 6.200 - 5.0\times 2.92}{1.2} = 111.12\ \text{mg/l} \end{aligned}

So the treated effluent can have $BOD_5 \approx 75.98$ mg/l.

\text{Degree of treatment} = \frac{234.00-111.12}{234.00}\times 100 = 52.5\%

Answer: the sewage BOD5 must be reduced from 160 mg/l to about 75.98 mg/l, i.e. a treatment efficiency of about 52.5%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Degree of treatment, city 1200 l/s

Step 1: Mixing and permissible deficit

Step 2: Permissible BOD of the mixture

Step 3: Permissible BOD of the effluent

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