Question

A circular sewer 2 m diameter has to carry a discharge of 2 $m^3/s$ when flowing nearly full. What is the minimum required slope to initiate the flow ?

Assume manning's n = 0.015.

• A. 0.00023
• B. 0.000036
• C. 0.000091
• D. 0.000014

Answer 1

The correct option is A 0.00023

$V=\frac{Q}{A}=\frac{2}{\frac{\pi }{4}(2{)}^2}=0.637m/sec$

For circular section, flowing full, the HMD is

$R=\frac{d}{4}=\frac{2}{4}=0.5$

From Manning's formula

$V=\frac{1}{n}{R}^{2/3}{S}^{1/2}$

$\sqrt{S}=\frac{0.637\times 0.015}{(0.5{)}^{2/3}}=0.015$
$S=0.00023$