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Percentage purification, town of 50000

Short answer

Required BOD removal is 91.9% (effluent BOD5 ≈ 40.6 mg/l). Mixed flow 0.211 m³/s, initial deficit 3.52 mg/l, permissible mixture ultimate BOD 25.5 mg/l. Raw sewage ultimate BOD 731 mg/l, permissible effluent ultimate BOD 59.4 mg/l.

The treated domestic sewage of a town is to be discharged in a natural stream. Calculate the percentage purification required in the treatment plant with the following data: Population = 50000; BOD contributed per capita = 0.07 kg/day; BOD of stream on U/S side = 3 mg/lit; DO to be maintained in D/S = 5 mg/lit; Domestic sewage = 140 lpcd; Lean period discharge of stream = 0.13 m³/sec. Assume rate constants K1 = 0.1 per day, K2 = 0.5 per day and stream with a saturation DO as 9.17 mg/lit.

Data and assumptions

  • Sewage flow =50000×140=7000 m3/day=0.0810 m3/s=50000\times 140 = 7000\ \text{m}^3/\text{day} = 0.0810\ \text{m}^3/\text{s}; river lean flow Qr=0.13Q_r=0.13 m³/s.
  • Raw sewage BOD: 0.07×50000×106 mg/day7000×103 l/day=500.0\dfrac{0.07\times 50000\times 10^6\ \text{mg/day}}{7000\times 10^3\ \text{l/day}} = 500.0 mg/l (taken as 5-day BOD).
  • Assumed: upstream river is saturated (DOr=9.17DO_r=9.17 mg/l, 20°C) and the treated sewage carries zero DO (conservative). K1=0.1K_1=0.1, K2=0.5K_2=0.5 (base 10), f=5f=5. The deficit must not exceed 9.17−5=4.179.17-5=4.17 mg/l.

Step 1: Mixing

DOmix=0.13×9.17+0.0810×00.2110=5.65 mg/lD0=9.17−5.65=3.52 mg/l\begin{aligned} DO_{mix} &= \frac{0.13\times 9.17+0.0810\times 0}{0.2110} = 5.65\ \text{mg/l}\\ D_0 &= 9.17-5.65 = 3.52\ \text{mg/l} \end{aligned}

Step 2: Permissible BOD of the mixture

Using tc=1K(f−1)log⁡10[f{1−(f−1)D0Lm}]t_c=\dfrac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\dfrac{D_0}{L_m}\right\}\right] and Dc=Lmf10−Ktc=4.17D_c=\dfrac{L_m}{f}10^{-Kt_c}=4.17 mg/l, trial gives Lm=25.51L_m=25.51 mg/l (tc=0.875t_c=0.875 days).

Step 3: Permissible BOD of the effluent

Lr=30.6838=4.39 mg/lLs=Lm(Qs+Qr)−QrLrQs=25.51×0.2110−0.13×4.390.0810=59.39 mg/lLraw=500.00.6838=731.24 mg/l\begin{aligned} L_{r} &= \frac{3}{0.6838} = 4.39\ \text{mg/l}\\ L_s &= \frac{L_m(Q_s+Q_r)-Q_rL_r}{Q_s} = \frac{25.51\times 0.2110 - 0.13\times 4.39}{0.0810} = 59.39\ \text{mg/l}\\ L_{raw} &= \frac{500.0}{0.6838} = 731.24\ \text{mg/l} \end{aligned} Percentage purification=731.24−59.39731.24×100=91.9%\text{Percentage purification} = \frac{731.24-59.39}{731.24}\times 100 = 91.9\%

Answer: about 91.9% BOD removal is required (effluent BOD5 about 40.61 mg/l).

✓ Answer checked by Claude Opus 5.5 (high effort)

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