The width of a rectangular sewer is twice its depth while discharging 1.5 m/sec. The width of the sewer is

ANS:C - 1.36 m

Let's denote the width of the rectangular sewer as W and its depth as D. Given:

• The width of the sewer is twice its depth, which means W=2D.
• The velocity of discharge is 1.5 m/s.

We know that for an open channel flow, the flow velocity (v) can be calculated using the Manning's equation: v=n1​⋅R2/3⋅S1/2 Where:

• n is the Manning's roughness coefficient,
• R is the hydraulic radius,
• S is the slope of the channel.

For a rectangular channel, the hydraulic radius R is calculated as the ratio of the cross-sectional area A to the wetted perimeter P: R=A/P​=WD/2W+2D   ​ Given that W=2D, we can substitute �W with 2D in the equation above: R=(2D)+2D2D⋅D​=6D2D2​=3D​ Now, let's substitute R into the Manning's equation: 1.5=1(3)2/3⋅1/21.5=n1​⋅(3D​)2/3⋅S1/2 We need to know the Manning's roughness coefficient (n) and the slope (S) of the sewer to solve the equation. Without these values, we cannot determine the specific width of the sewer. Please provide the Manning's roughness coefficient (n) and the slope (S) of the sewer so we can proceed with the calculations.