Question

Que 3.23. The depth and velocity of flow in a rectangular channel are 1 m and 1.5 m/sec respectively. If the rate of inflow at the upstream end is suddenly doubled, what will be the height and absolute velocity of the resulting surge and the celerity of the wave ? ​

Answer 1

To calculate the height and absolute velocity of the resulting surge, as well as the celerity of the wave, we can use the principles of open channel flow. Let's assume the original flow conditions are denoted by subscript "1" and the new conditions after the inflow rate is doubled are denoted by subscript "2".

• Given:
• Depth of flow (h1) = 1 m
• Velocity of flow (V1) = 1.5 m/sec
• First, let's calculate the new depth of flow (h2) using the principle of continuity:
• Q1 = A1 * V1 = A2 * V2
• Where:
• Q1 = Inflow rate at the upstream end
• A1 = Cross-sectional area of flow at original conditions
• A2 = Cross-sectional area of flow at new conditions
• V2 = Velocity of flow at new conditions
• Since the inflow rate is suddenly doubled, we have:
• Q2 = 2 * Q1
• We know that the cross-sectional area (A) of a rectangular channel is given by:
• A = b * h
• Where:
• b = Width of the channel (assumed to be constant)
• Since the width of the channel is constant, we can write:
• A1 * h1 = A2 * h2
• To simplifying, we have:
• h2 = (A1 * h1) / A2
• Now, let's calculate the new velocity of flow (V2) using the principle of continuity:
• V2 = (Q2 / A2)
• Next, let's calculate the celerity of the wave (C) using the formula:
• C = √(g * h1)
• Where:
• g = Acceleration due to gravity (approximately 9.81 m/s²)
• Finally, let's calculate the absolute velocity of the resulting surge (Va) using the equation:
• Va = √(V2² + C²)
• Substituting the given values into the formulas, we can find the numerical values for the height, absolute velocity, and celerity.

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