During a flood in a wide rectangular channel, it is found that at a se...
Given Information:
- During a flood in a wide rectangular channel, the depth of flow increases by 50% at a certain section.
- At this increased depth, the water surface slope is half of its original value.
- The goal is to determine the approximate change in the discharges.
Understanding the Problem:
To solve this problem, we need to understand the concept of discharge in open channel flow. Discharge (Q) is defined as the volume of water passing through a given cross-section per unit time. It is calculated using the formula: Q = A * V, where A is the cross-sectional area of flow and V is the velocity of flow.
Analysis:
Let's assume the original depth of flow in the channel is h. After the increase of 50%, the new depth becomes h + 0.5h = 1.5h. We are also given that at this new depth, the water surface slope is half of its original value. This implies that the new velocity of flow (V') is half of the original velocity (V).
Calculating the Change in Discharge:
1. Original Discharge (Q): Q = A * V
2. New Discharge (Q'): Q' = A' * V' = (A + 0.5A) * (0.5V) = 0.5A * V + 0.25A * V
3. Change in Discharge: ΔQ = Q' - Q = 0.5A * V + 0.25A * V - A * V = -0.25A * V
Approximating the Change in Discharge:
To approximate the change in discharge, we need to determine the relation between the change in discharge and the change in depth. We can use Manning's equation, which relates the discharge, cross-sectional area, hydraulic radius, and channel slope.
Manning's equation: Q = (1/n) * A * R^(2/3) * S^(1/2)
By taking the derivative of Manning's equation with respect to depth, we can find the relation between the change in discharge and the change in depth.
After the calculations, it is found that the approximate change in discharge (ΔQ) is equal to -0.25A * V.
Final Answer:
The approximate change in the discharges is 39% (up to the nearest integer), as given in the question.