Question 11.1: A 2.0 m wide rectangular channel carries a discharge of 4 m³......
A 2.0 m wide rectangular channel carries a discharge of 4 m³/s and the depth of flow is 1.50 m. A smooth hump is constructed in the channel to raise the bed level of the channel by 0.2 m. Calculate the elevation of the water surface on the elevated surface.
Given, q = 4/2 = 2 m²/s
Therefore y_{c}=\sqrt[3]{\frac{{2^{2}}}{9.81}}=0.741\,{\mathrm{m}} \\ E_{c}=y_{c}+\left\lgroup{\frac{v_{c}^{2}}{2g}}\right\rgroup={\frac{3}{2}}y_{c}=1.112\;\mathrm{m} \\ v_{1}=\frac{q}{y_{1}}=\frac{2}{1.5}=1.33\,\frac{\mathrm{m}}{\mathrm{s}} \\ {\frac{v_{1}^{2}}{2g}}=0.091\,{\mathrm{m}} \\ \begin{array}{l}{{E_{1}=y_{1}+\frac{v_{1}^{2}}{2g}=1.50+0.091=1.591\,\mathrm{m}}}\\ {{E_{2}=E_{1}-\Delta z=1.591-0.2=1.391\,\mathrm{m}\gt E_{c}}}\end{array}
Therefore, the flow on top of the hump will be subcritical.
From the definition of the specific energy,
E_{2}=y_{2}+{\frac{v_{2}^{2}}{2g}}=y_{2}+{\frac{q^{2}}{2g y_{2}^{2}}}or 1.31=y_{2}+{\frac{2^{2}}{2\times9.81\times y_{2}^{2}}}=y_{2}+{\frac{0.209}{y_{2}^{2}}}
By trial, y_2 = 1.26 m.
With reference to the datum, the elevation of the water surface on top of elevated channel bottom = 0.2 + 1.26 = 1.46 m < 1.5 m (upstream depth).
The drop in the free surface level = 1.50 – 1.46 = 0.04 m = 4 cm.
An important observation in Example 11.1 is that the presence of the hump did not affect the upstream water level. The hump height can be increased further without affecting the upstream level, till y_2 becomes the critical depth, y_c. For such a case, E_{c}={\frac{3}{2}}y_{c} and the maximum height of the hump, Δz_{max} \ = \ (E_1 – E_c). In this example, Δz_{max} = 1.591 – 1.112 = 0.479 m ≈ 48 cm. However, if one provides a hump height Δz exceeding Δz_{max}, the upstream water level will rise (i.e., the flow is choked) and the position of E_1 will shift to the right. The change in E_1 will be such as to produce a critical depth y_c = 0.741 on top of the hump. Thus, the new value of E_1 = Δz + E_c.
In the given example, suppose the height of the hump provided = 0.6 m
The upstream specific energy, E_1 = E_c + Δz = 1.112 + 0.60 = 1.712 m
From the relation, E_{1}=y_{1}+{\frac{v_{1}^{2}}{2g}}=y_{1}+{\frac{q^{2}}{2gy_{1}^{2}}}, we have
1.712=y_{1}+{\frac{2^{2}}{2\times9.81\times y_{1}^{2}}}=y_{1}+{\frac{0.209}{y_{1}^{2}}}By trial, y_1 = 1.63 m. Therefore, the upstream water level rises by 0.13 m because of the hump.