---
title: "Percentage purification, town of 50000"
subject: "Sanitary Engineering"
chapter: 6
key: "semester_6/sanitary_engineering:percent-purification-town-50000"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/percent-purification-town-50000
site: BCT Engineering
---

# Percentage purification, town of 50000

Sanitary Engineering · Chapter 6: Wastewater Disposal

- Asked in: 2073 Magh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/percent-purification-town-50000
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-6

## Question

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.

## 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.

## Answer

#### Data and assumptions

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

#### Step 1: Mixing

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

#### Step 2: Permissible BOD of the mixture

Using $t_c=\dfrac{1}{K(f-1)}\log_{10}\left[f\left\{1-(f-1)\dfrac{D_0}{L_m}\right\}\right]$ and $D_c=\dfrac{L_m}{f}10^{-Kt_c}=4.17$ mg/l, trial gives $L_m=25.51$ mg/l ($t_c=0.875$ days).

#### Step 3: Permissible BOD of the effluent

$$
\begin{aligned}
L_{r} &= \frac{3}{0.6838} = 4.39\ \text{mg/l}\\
L_s &= \frac{L_m(Q_s+Q_r)-Q_rL_r}{Q_s} = \frac{25.51\times 0.2110 - 0.13\times 4.39}{0.0810} = 59.39\ \text{mg/l}\\
L_{raw} &= \frac{500.0}{0.6838} = 731.24\ \text{mg/l}
\end{aligned}
$$

$$
\text{Percentage purification} = \frac{731.24-59.39}{731.24}\times 100 = 91.9\%
$$

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the raw BOD of 500 mg/L obtained?
   A. 0.07 x 140
   B. BOD load 0.07 x 50000 kg/day divided by sewage flow 7000 m3/day
   C. 50000/140
   D. 7000/0.07
   Answer: B. 3500 kg/day of BOD in 7000 m3/day equals 500 mg/L. (Key point: Raw sewage BOD)

2. What is the maximum allowable deficit?
   A. 5 mg/L
   B. 9.17 mg/L
   C. 9.17 - 5 = 4.17 mg/L
   D. 3.52 mg/L
   Answer: C. DO must stay at least 5 mg/L, so Dc max = 4.17 mg/L. (Key point: Allowable deficit)

3. What was assumed for DO in mixing?
   A. Both zero
   B. Both saturated
   C. River zero, sewage saturated
   D. Upstream river saturated (9.17) and treated sewage DO zero
   Answer: D. This gives DO_mix = 5.65 and D0 = 3.52 mg/L, a conservative assumption. (Key point: Assumption on DO)

4. What percentage purification is required?
   A. 91.9%
   B. 40.6%
   C. 25.5%
   D. 59.4%
   Answer: A. (731.24 - 59.39)/731.24 x 100 = 91.9%. (Key point: Result)

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