(a) A 3 m wide rectangular channel discharges $5.1 \mathrm{~m}^{3} / \mathrm{s}$, the depth of flow being 0.75 m . If due to a sudden release of flow, the upstream discharge rate is doubled, find the velocity of the consequent surge. Find also the new depth of flow.
Explanation
The velocity of the consequent surge is calculated using the equation: v = (2 * Q1 / A) where Q1 is the initial discharge and A is the cross-sectional area of the channel. The new depth of flow is given by the equation: h = (Q1 / (A * v)) where v is the velocity of the surge.
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