The channel in Fig. 12–23 has a triangular cross section and is made of unfinished concrete. If the flow is 1.5 m³/s, determine the slope that produces critical flow. Take n = 0.014.

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Fluid Description. We assume steady uniform flow of an incompressible fluid.
Analysis. The critical slope is determined using Eq. 12–22, but first we need to determine the critical depth. Since

[latex]A_c = 2[\frac{1}{2} (\frac{1}{2} y_c)y_c] = \frac{1}{2} y_c^2[/latex]

[latex]b_{top} = 2(\frac{1}{2} y_c) = y_c[/latex]

Then applying Eq. 12–11,

[latex]\frac{Q^2}{g} = \frac{A_c^3}{b_{top}}[/latex]

[latex]\frac{(1.5 m^3/s)^2}{9.81 m/s^2} = \frac{(\frac{1}{2} y_c^2)^3}{y_c}[/latex]

[latex]y_c = 1.129 m[/latex]

Using this value,
[latex]b_{top}[/latex] = 1.129 m

[latex]A_c = \frac{1}{2} (1.129 m)^2 = 0.6374 m^2[/latex]

[latex]P_c = 2\sqrt{(1.129 m)^2 + (1.129 m/2)^2} = 2.525 m[/latex]

[latex]R_{hc} = \frac{A_c}{P_c} = \frac{0.6374 m^2}{2.525 m} = 0.2525 m[/latex]

Thus,

[latex]S_c = \frac{n^2gA_c}{k^2b_{top}R_{hc}^{4/3}} = \frac{(0.014)^2 (9.81 m/s^2)(0.6374 m^2)}{(1)^2(1.129 m)(0.2525 m)^{4/3}} = 0.00680[/latex]

Therefore, when the channel has a flow of Q = 1.5 m³/s, then any slope less than [latex]S_c[/latex] will produce subcritical (tranquil) flow, and any slope greater than [latex]S_c[/latex] will produce supercritical (rapid) flow.

Question 12.11: The channel in Fig. 12–23 has a triangular cross section and......

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