If the discharge of a sewer running half is 628 1.p.s., i = 0.001, and n = 0.010, the diameter of the sewer, is
A. 1.39 m
B. 1.49 m
C. 1.59 m
D. 1.69 m
Answer: Option D
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Since running half full then, d=0.5D
Q=628x10^-3 m^3/s
Use, Manning's equation.
Q= 1/n * A*(D/4)power 2/3 * (s)power 1/2
Here, S = i = 0.0001 not 0.001.
Solve it, the answer comes approx to 1.69 m.
Q given
S or i given
n also given
Use Mannings formula
Q=1/n.r2/3.s1/2
Take r=d/4 for hydraulic radius
1.095 m
Can someone explain the question with soultion?