Discussion
A regime channel carrying a discharge of 25 m3 s−1 is designed using Lacey’s regime theory. The side slope of the channel is ½ H : 1 V, and Lacey’s silt factor is unity. The bottom width and depth of flow in the channel, in m, respectively are
Answer: B
The keyed pair is 20.26 m bottom width and 1.56 m depth.
Lacey's regime equations with silt factor f = 1 give the velocity, area and wetted perimeter:
\[ \begin{aligned} V &= \left(\frac{Qf^{2}}{140}\right)^{1/6} = 0.750\ \text{m/s} \\ A &= Q/V = 33.3\ \text{m}^{2} \\ P &= 4.75\sqrt{Q} = 23.75\ \text{m} \end{aligned} \]
For a side slope of ½ H : 1 V, \(P = b + \sqrt{5}\,d\) and \(A = bd + d^{2}/2\). Solving these two equations gives a depth of about 1.6 m and a bottom width of about 20.2 m, which matches the keyed pair to within the rounding of the regime constants.
The pair with 1.38 m depth gives an area of only about 29 m2, too small for the regime velocity. 23.75 m is the whole wetted perimeter mistaken for the bottom width, and 32.78 m is wider still.
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