Feb 26, 2016

Feb 26, 2016

Feb 26, 2016

Sanitary Sewer Design Example # 1

/ On: Feb 05, 2019 /

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 (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 = ($\frac{\Pi&space;}{4}$ 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 = ($\frac{\Pi&space;}{4}$ 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.