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Sanitation Engineering

Sanitary Sewer Design Example 3

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

Example 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 (qw)

 = 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} R2/3 S1/2)

D = 0.264 m (264mm or 10.40”)

Use 12” Dia pipe

Checking the minimum self cleaning velocity

V = ( \frac{1}{n} R2/3 S1/2 ) Where R = (D/4)2/3 for circular pipe

V = 0.96 m/sec

Vmin = 0.6 m/sec < V = 0.96 m/sec < Vmax = 2.5 m/sec
PART 2

Doubling the slope i.e. 2 S = 0.01

Q = ( \frac{\Pi }{4} D2) ( \frac{1}{n} R2/3 S1/2)

Q = 0.0331 (m3/s)

Q = P * qw

P = Q/qw Where qw = Return flow (%) * q (= 280 lpcd )

P = \frac{2,860,429.78}{280} \frac{(Liters/Day)}{(Liters/Capita/Day)}

P = 10,216 Persons

So, if the slop is doubled then total of 10,216 extra persons can be served.

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