Question 11.2: A discharge of 12 m³/s takes place in a rectangular channel ......

From the given data,             v_{1}={\cfrac{Q}{A}}={\cfrac{12}{4\times1.5}}=2~\mathrm{m/s}

Froude number,             F_{1}={\frac{v_{1}}{\sqrt{g y_{1}}}}={\frac{2}{\sqrt{9.81\times1.5}}}=0.52\lt 1

Thus, the flow in the channel is subcritical.

{\frac{v_{1}^{2}}{2g}}={\frac{2^{2}}{2\times9.81}}=0.204\;\mathrm{m} \\ E_{1}=y_{1}+{\frac{v_{1}^{2}}{2g}}=1.5+0.204=1.704~{\mathrm{m}}

The discharge intensity at the downstream section, q_{2}={\frac{12}{3}}=4\;{\mathrm{m}}^{2}/{\mathrm{s}}

Critical depth,       y_{c2}={\dot{\sqrt[3]{(q_{2}^{2}/g)}}}={\dot{\sqrt[3]{(4^{2}/9.81)}}}=1.177\,{\mathrm{m}} \\ E_{c2}={\frac{3}{2}}y_{c2}={\frac{3\times1.177}{2}}=1.765\,{\mathrm{m}}

The available specific energy at Section 2, E_2 = E_1 – Δz = 1.70 – 0.30 = 1.40 m < E_{c2}

As the available energy is less than the minimum value required to create a critical condition in the contraction, the upstream water level must rise and the flow will take place under choked condition. The new specific energy at section 1, E_{12} = E_{c2} + Δz = 1.765 + 0.30 = 2.065 m. If y_{1n} is the new depth of the approaching flow, one can write

or             2.065=y_{1n}+\frac{3^{2}}{2\times9.81\times y_{1n}^{2}}=y_{1n}+\frac{0.459}{y_{1n}^{2}}

By trial,          y_{1n}=1.95\,{\mathrm{m}}

The change in water surface elevation is, y_{1n}-y_{1}=1.95-1.5=0.45~{\mathrm{m}}