# Design of Sewers and Sewerage System

 Question 1
A system of seven river segments is shown in the schematic diagram. The $\mathrm{R}_{1}{ }^{\prime} \mathrm{s}, \mathrm{Q}_{1}{ }^{\prime} \mathrm{s}$, and $\mathrm{C}_{\mathrm{i}}$ 's ( $\mathrm{i}=1$ to 7 ) are the river segments, their corresponding flow rates, and concentrations of a conservative pollutant, respectively. Assume complete mixing at the intersections, no additional water loss or gain in the system and steady state condition. Given : $\mathrm{Q}_{1}=5 \mathrm{~m}^{3} / \mathrm{s} ; \mathrm{Q}_{2}=15 \mathrm{~m}^{3} / \mathrm{s} ; \mathrm{Q}_{4}=3 \mathrm{~m}^{3} / \mathrm{s} ; \mathrm{Q}_{6}=8 \mathrm{~m}^{3} / \mathrm{s} ; C_{1}=8 \mathrm{~kg} / \mathrm{m}^{3} ; C_{2}=12 \mathrm{~kg} / \mathrm{m}^{3} ; C_{6}=10 \mathrm{~kg} / \mathrm{m}^{3}$. What is the steady state concentration in $\mathrm{kg} / \mathrm{m}^{3}$, rounded off to two decimal place) of the pollutant in the river segment 7 ? ____ A 5.6 B 8.8 C 10.6 D 14.8
GATE CE 2023 SET-2   Environmental Engineering
Question 1 Explanation: At 1-1, for $\mathrm{R}_{3}$
\begin{aligned} \mathrm{Q}_{3} & =\mathrm{Q}_{1}+\mathrm{Q}_{2} \\ & =5+15=20 \mathrm{~m}^{2} / \mathrm{sec} \\ \mathrm{C}_{3} & =\frac{\mathrm{Q}_{1} \mathrm{C}_{1}+\mathrm{Q}_{2} \mathrm{C}_{2}}{\mathrm{Q}_{1}+\mathrm{Q}_{2}} \\ &=\frac{5 \times 8+15 \times 12}{5+15} \\ & =11 \mathrm{Kg} / \mathrm{m}^{3} \end{aligned}

At 2-2, For $\mathrm{R}_{5}$
$\mathrm{Q}_{5}=\mathrm{Q}_{3} \mathrm{Q}_{4}=10-3=17 \mathrm{~m}^{3} / \mathrm{sec}$
$\mathrm{C}_{5}=\mathrm{C}_{3}=11 \mathrm{Kg} / \mathrm{m}^{3}$

At 3-3 for $\mathrm{R}_{7}$
\begin{aligned} \mathrm{Q}_{7} & =\mathrm{Q}_{6}+\mathrm{Q}_{5}=17+8=25 \mathrm{~m}^{3} / \mathrm{sec} \\ \mathrm{C}_{7} & =\frac{\mathrm{Q}_{6} \mathrm{C}_{6}+\mathrm{Q}_{6} \mathrm{C}_{5}}{\mathrm{Q}_{6}+\mathrm{Q}_{5}} \\ & =\frac{8 \times 10+17 \times 11}{8+17}=10.68 \mathrm{Kg} / \mathrm{m}^{3} \end{aligned}
Answer : $10.6 \mathrm{~kg} / \mathrm{m}^{3}$.
 Question 2
Crown corrosion in a reinforced concrete sewer is caused by:
 A $H_{2}S$ B $CO_{2}$ C $CH_{4}$ D $NH_{3}$
GATE CE 2016 SET-1   Environmental Engineering
Question 2 Explanation:
Bacteria in the slime under flowing sewage convert sulphate in the sewage in sulphides. Sulphidesnin the liquid make their way to the surface of the sewage and reduce into the sewer atmosphere or Hydrogen Sulphide $(H_{2}S)$ gas. Bacterial action converts atmospheric $H_{2}S$ gas to Sulphuric Acid which causes corrosion in the crown of the pipe and this corrosion is called crown corrosion.

 Question 3
Assertion [a]: At a manhole, the crown of the outgoing sewer should not be higher than the crown of the incoming sewer.

Reason [r]: Transition from a larger diameter incoming sewer to a smaller diameter outgoing sewer at a manhole should not be made.

The CORRECT option evaluating the above statements is :
 A Both [a] and [r] are true and [r] is the correct reason for [a] B Both [a] and [r] are true but [r] is not the correct reason for [a] C Both [a] and [r] are false D [a] is true but [r] is false
GATE CE 2012   Environmental Engineering
Question 3 Explanation:
To control the backing up of sewage in the incoming sewer, the crown of the outgoing sewer should not be higher than the crown of incoming sewer.
There may be more than one incoming sewer in a manhole that's why
 Question 4
Determine the correctness or otherwise of the following Assertion [A] and the Reason [R] :

Assertion :The crown of the outgoing larger diameter sewer is always matched with the crown of incoming smaller diameter

Reason : It eliminates backing up of sewage in the incoming smaller diameter sewer.
 A Both [A] and [R] are true and [R] is the correct reason for [A] B Both [A] and [R] are true but [R] is not the correct reason for [A] C Both [A] and [R] are false D [A] is true but [R] is false
GATE CE 2008   Environmental Engineering
 Question 5
An existing 300 mm diameter circular sewer is laid at a slope of 1:28 and carries a peak discharge of 1728 $m^{3}$/d. Use the partial flow diagram shown in the given figure and assume Manning's n = 0.015 At the peak discharge, the depth of flow and the velocity are, respectively
 A 45 mm and 0.28 m/s B 120 mm and 0.50 m/s C 150 mm and 0.57 m/s D 300 mm and 0.71 m/s
GATE CE 2004   Environmental Engineering
Question 5 Explanation:
\begin{aligned} Q &=\left ( \frac{\pi }{4}D^{2} \right ).\frac{R^{\frac{2}{3}}S^{\frac{1}{2}}}{n}\; \; \; \; \; \left [ R=D/4 \right ] \\ &=\frac{\pi }{4}\times 0.3^{2}\times \left ( \frac{0.3}{4} \right )^{2/3}\times \frac{\sqrt{1/280}}{0.015} \\ &=0.05\: m^{3}/d \\ Q&=4327.3\: m^{3}/d \\ q&=1728\: m^{3}/d \\ \therefore \frac{q}{Q} &= 0.4 \end{aligned}
From graph, $\frac{d}{D}=0.4\; \; \; \; \; \; \; \; \; \therefore \; \; d=120\; mm$
From graph, $\frac{v}{V}=0.4$
$\therefore \; \; \; \; v=0.7\times \frac{1}{n}R^{2/3}\: S^{1/2}$
$=0.496\: m/s\approx 0.5\: m/s$

There are 5 questions to complete.