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

ANS:D - 1.69 m

The Manning's equation for flow rate in an open channel is: Q=1.486/n​⋅A⋅R^2/3⋅S^1/2 Given:

• Q=628L/s=0.628m³/s
• n=0.010
• S=0.001

First, we need to find the cross-sectional area A and the hydraulic radius R of the sewer. For a circular sewer, the cross-sectional area A is given by: A=π⋅D^2/4​ And the wetted perimeter P is given by: P=π⋅D The hydraulic radius R is defined as the ratio of the cross-sectional area A to the wetted perimeter P: R=A/P ​=π⋅D^2/4/π⋅D ​​=D/4​ Now, substituting these values into the Manning's equation: 0.628=1.486​ /0.010 ⋅π⋅D^2 / 0.0101​⋅(D/4​)^2/3⋅0.001​ Solving for D: D^5/3=​0.628⋅0.010⋅4^2/3​ / 1.486⋅π⋅0.001 D5^/3= 0.628⋅0.010⋅1.5874​ /1.486⋅π⋅0.0316 D^5/3≈1.688 D≈(1.688)^3/5 D≈1.687m Rounding to two decimal places, the diameter of the sewer is approximately 1.69 m1.69m. Hence, the correct answer is 1.69 m.