---
title: "Degree of treatment, city 1200 l/s"
subject: "Sanitary Engineering"
chapter: 6
key: "semester_6/sanitary_engineering:degree-treatment-city-1200ls"
source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/degree-treatment-city-1200ls
site: BCT Engineering
---

# Degree of treatment, city 1200 l/s

Sanitary Engineering · Chapter 6: Wastewater Disposal

- Asked in: 2075 Baisakh, 8 marks
- Video: [oxygen deficit problem | DO deficit solved problem | oxygen deficit in downstream side of river](https://www.youtube.com/watch?v=4wst5Ot1YIg) — Civil Engineering Exam
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/degree-treatment-city-1200ls
- Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-6

## Question

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.

## 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.

## Answer

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%.**

## Check yourself

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

1. What is the allowable critical deficit?
   A. 4 mg/L
   B. 9.17 - 4.0 = 5.17 mg/L
   C. 9.17 mg/L
   D. 2.51 mg/L
   Answer: B. Dc(allowed) = saturation DO - minimum DO to be maintained. (Key point: Allowable deficit)

2. Why is the permissible mixture BOD found by trial?
   A. The data are missing
   B. K is unknown
   C. tc and Dc both depend on L_m, so the equations cannot be solved directly
   D. D0 changes with trial
   Answer: C. Setting Dc = 5.17 mg/L with D0 = 2.51 fixed gives L_m = 23.87 mg/L by iteration. (Key point: Trial solution)

3. How is the permissible effluent BOD found?
   A. Ls = L_m
   B. Ls = L_m x Qs/Qr
   C. Ls = 160 - L_m
   D. Ls = (L_m (Qs+Qr) - Qr Lr)/Qs
   Answer: D. The mixture load minus the river load must come from the sewage. (Key point: Effluent BOD)

4. What efficiency is required?
   A. 52.5%
   B. 23.9%
   C. 75.98%
   D. 9.17%
   Answer: A. (234.00 - 111.12)/234.00 x 100 = 52.5%, bringing BOD5 from 160 to about 75.98 mg/L. (Key point: Degree of treatment)

## Related in Sanitary Engineering

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