Question 12.9: A trapezoidal channel has a bottom width of 6.1 m and side s...
A trapezoidal channel has a bottom width of 6.1 m and side slope of 2 horizontal to 1 vertical. When the depth of water is 1.07 m, the flow is 10.47 \mathrm{~m}^{3} / \mathrm{s} . (i) What is the specific energy of flow? (ii) Is the flow tranquil or rapid?
The cross sectional area of flow,
A=6.1(1.07)+2\left(\frac{1}{2}\right)(1.07)(2.14)=8.82 \mathrm{~m}^{2}From Eq. (12.31), we can write
E_{s}=h+\left(\frac{q^{2}}{2 g}\right) \frac{1}{h^{2}} (12.31)
E_{s}=h+\frac{1}{2 g}(Q / A)^{2}where Q is the volumetric flow rate.
Hence, E_{S}=1.07+\frac{1}{2 \times 9.81}\left(\frac{10.47}{8.82}\right)^{2}=1.14 \mathrm{~m}
To determine the critical depth, we have to first find out a similar relation as given in Eq. (12.32) for a channel whose width varies with the depth. For this purpose, we start with Eq. (12.31) as
h_{c}=\left(q^{2} / g\right)^{1 / 3} (12.32)
E_{S}=h+\frac{1}{2 g}(Q / A)^{2}where Q is the flow rate and A is the cross-sectional area. At critical condition, i.e., for minimum specific energy.
\frac{\mathrm{d} E_{S}}{\mathrm{~d} h}=1+\frac{Q^{2}}{2 g}\left(-\frac{2}{A^{3}} \frac{\mathrm{d} A}{\mathrm{~d} h}\right)=0substituting \mathrm{d} A=B^{\prime} \mathrm{d} h ( B^{\prime} is the width at the water surface), we get
\left(Q^{2} B^{\prime}\right) /\left(g A_{c}^{3}\right)=1Or Q^{2} / g=A_{c}^{3} / B^{\prime}
From the geometry of the channel, A_{c}=6.1 h_{c}+2 h_{c}^{2}
And B^{\prime}=6.1+4 h_{c}
(where \mid h_{c} is the critical depth and \mid A_{c} is the corresponding cross-sectional area of flow)
Therefore,
Solving by trial, h_{c}=0.625 \mathrm{~m}
Since the actual depth exceeds the critical one, the flow is tranquil.