Sanitary Sewer Design Example # 1
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 m^{3}/sec | ||
Let take peaking factor (P.F) |
= 3 | |
Peak Hourly Waste Water Flow |
= 3 * 0.0162 | |
= 0.0486 m^{3}/sec | ||
Now finding the diameter of sewer pipe line |
||
Using discharge formula; |
||
Q = AV |
||
0.0162 = ( D^{2}) ( 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 = ( 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 = ( D^{2}) ( R^{2/3} S^{1/2}) |
||
Q = 0.0331 (m^{3}/s) |
||
Q = P * q_{w} |
||
P = Q/q_{w }Where q_{w} = Return flow (%) * q (= 280 lpcd ) |
||
P = |
||
P = 10,216 Persons |
So, if the slop is doubled then total of 10,216 extra persons can be served.
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