Discussion
A trapezoidal canal, having a bottom width of 5.0 m and a side slope of 1 H : 1 V, is carrying a discharge of 20 m3 s−1. The critical depth, in m, is
Answer: A
The critical depth is about 1.09 m.
At critical flow \(Q^{2}/g = A^{3}/T\), where for this section \(A = (5 + y)y\) and the top width is \(T = 5 + 2y\). The left side is \(20^{2}/9.81 = 40.77\). Trying y = 1.09 m:
\[ \begin{aligned} A &= (6.09)(1.09) = 6.638\ \text{m}^{2} \\ T &= 7.18\ \text{m} \\ A^{3}/T &= 292.5/7.18 = 40.74 \end{aligned} \]
That matches 40.77, so y = 1.09 m. 1.18 m is the critical depth of a 5 m wide rectangular channel, \((q^{2}/g)^{1/3}\) with q = 4 m2/s, which ignores the extra area of the side slopes; at 1.18 m the trapezoidal expression is already about 52.6. 2.12 and 2.62 m are far deeper than either.
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