The document describes how to determine the diameter of a circular channel that carries the same flow rate as a rectangular channel of width b that is one-third full. It shows that the discharge equations for both channels set equal to each other can be solved to find that the diameter of the circular channel should be 1.12 times the width of the rectangular channel.

1. Water flows in a rectangular channel of width b at a depth of b/3. Determine the diameter of a
circular channel (in terms of b) that carries the same flow rate when it is half-full. Both channels
have the same Manning coefficient and slope.
Solution
For rectangular section
Area = b xd = b x b/3 = 0.33b 2
wetted perimeter = P = b + 2d = b+2b/3 = 1.67b
hydrallic mean depth = m = A/P = 0.33b 2
/ 1.67b = 0.2b
velocity = c sqrt(mi) = c sqrt(0.2b x i)
i = bed slope
Q = AV = b x d x c sqrt(0.2b x i) = 0.33b 2
x c sqrt(0.2b x i)
For circular section
depth = D/2
Area = 3.14 x D 2
/ 8
wetted perimeter = P = 3.14D/2
m = A/P = D/4
Q = (3.14 x D 2
/ 8) x c x sqrt(Di/4)
discharge for rectangular and circular flow are equal
(3.14 x D 2
/ 8) x c x sqrt(Di/4) = 0.33b 2
x c sqrt(0.2b x i)

2. 0.39D 2
x sqrt(Di/4) = 0.33b 2
x sqrt(0.2b x i)
SOBS
0.152D 4
x 0.25D = 0.109B 4
x 0.2B
1.74 D 5
= B 5
B = (1.74 D 5
) (1/5)
B = 1.12D