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

Given:

Discharge, Q = 628L/s = 0.628m^3/s
Hydraulic gradient, S = 0.001
Manning's coefficient, n = 0.010
Sewer is running half full
For a circular sewer flowing half full:
A=(pi D^2/8)
R=D/4
Using Manning's formula:
Q=1/n R^2/3 S^0.5
0.628=(1/0.010)(D/4)^2/3 (0.001)^0.5
D=1.31m

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?