Answer:
h1 = 1.4545h1 - 0.2273
h1 = 0.2727 m
Step-by-step explanation:
To solve this problem, we can use the concept of continuity equation, which states that the flow rate of water is constant along the channel. The flow rate can be calculated as the product of the cross-sectional area of flow and the velocity of flow.
Let's first calculate the cross-sectional area of flow in the trapezoidal channel. The side slopes of 1.5h:1v mean that the ratio of the height to the top width of the trapezoid is 1.5:1. Let h be the depth of flow, and b be the bottom width of the trapezoid. Then we have:
h/b = 1.5/1
h = 1.5b
The area of the trapezoid can be calculated as:
A = (1/2)(b1 + b2)h
= (1/2)(4.3 + 4.3 + 1.5h)(h)
= (1/2)(8.6 + 1.5h)(h)
= 4.3h + 0.75h^2
Substituting the given values, we get:
A = 4.3(2.2) + 0.75(2.2)^2
= 11.99 m^2
The flow rate can be calculated as:
Q = Av
= 11.990.17
= 2.0383 m^3/s
Now, let's consider the bridge placed in the middle of the channel. The cross-sectional area of flow upstream and downstream of the bridge should be equal to maintain continuity of flow. Let's assume that the depth of flow adjacent to the pier is h1, and the width of the pier is w. Then the cross-sectional area of flow upstream of the pier can be calculated as:
A1 = (1/2)(4.3 + 1.5h1)(h1)
= 2.15h1 + 0.75h1^2
The cross-sectional area of flow downstream of the pier can be calculated as:
A2 = (1/2)(4.3 + 1.5h2)(h2)
= 2.15h2 + 0.75h2^2
Since the flow rate is constant, we have:
A1v1 = A2v2
The velocity of flow can be assumed to be the same upstream and downstream of the pier, since the water level is expected to be constant. Therefore, we have:
v1 = v2 = 0.17 m/s
Substituting the given values and solving for h2, we get:
h2 = (4.3 + 1.5h1 - w)(h1)/(4.3 + 1.5h1 + w)
Now, we can use this equation to calculate the depth of flow adjacent to the pier for different values of w. For example, if we assume w = 0.5 m, we get:
h2 = (4.3 + 1.5h1 - 0.5)(h1)/(4.3 + 1.5h1 + 0.5)
= 1.4545h1 - 0.2273
To find the maximum value of w that will not cause a rise in the water surface upstream of the pier, we need to find the value of w that makes h2 = h1. This gives: