Sanitary Sewer Design Example # 1

By: Syed Ahmad Amin Shah / On: Feb 05, 2019 / Design of, Sewers

Problem Statement

Design a sanitary sewer to serve a population of 5,000 people, if the average consumption is 400 liters per capita per day (lpcd). How many extra persons can be served if the slope is doubled? Using "n" value of 0.013 in the Manning's formula & the return flow as 70%. Check the minimum self-cleaning velocity. Neglect infiltration & inflow?

*Given Data:*
Population (P)	= 5000 Persons
Average water consumption (q)	= 400 lpcd
Manning Coefficient (n)	= 0.013
Return Flow	= 70 %
Assume Slop (s)	= 0.005

Required:

1. Find the Velocity (V) =? Also check minimum self cleaning velocity

2. When the slope is doubled find the extra population to be served =?

*Solution:*
PART 1
Average waste water flow (q_w)	= P * Return flow (%) * q
	= 5000 * 0.7 * 400
	= 1,400,000 lpcd
	= 0.0162 m³/sec
Let take peaking factor (P.F)	= 3
Peak Hourly Waste Water Flow	= 3 * 0.0162
	= 0.0486 m³/sec
Now finding the diameter of sewer pipe line
Using discharge formula;
Q = AV
0.0162 = ( $\frac{\Pi }{4}$ D²) ( $\frac{1}{n}$ R^2/3 S^1/2)
D = 0.264 m (264mm or 10.40”)
Use 12” Dia pipe
Checking the minimum self cleaning velocity
V = ( $\frac{1}{n}$ R^2/3 S^1/2) Where R = (D/4)^2/3 -------- for circular pipe
V = 0.96 m/sec
V_min = 0.6 m/sec < V = 0.96 m/sec < V_max = 2.5 m/sec
PART 2
Doubling the slope i.e. 2 S = 0.01
Q = ( $\frac{\Pi }{4}$ D²) ( $\frac{1}{n}$ R^2/3 S^1/2)
Q = 0.0331 (m³/s)
Q = P * q_w
P = Q/q_wWhere q_w = Return flow (%) * q (= 280 lpcd )
P = $\frac{2,860,429.78}{280}$ $\frac{(Liters/Day)}{(Liters/Capita/Day)}$
P = 10,216 Persons