Water runs down a \(50 \mathrm{~m}\) wide spillway at \(280 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) on to a long concrete apron ( \(n=0.015\) ) having a uniform downward slope of 1 in 2500 . At the foot of the spillway the depth of the flow is \(600 \mathrm{~mm}\). How far from the spillway will a hydraulic jump occur? (For this very wide channel taking \(m=h\) gives acceptable accuracy.)

In fluid mechanics, flow rate is a crucial concept that describes the volume of fluid flowing past a point in a given time. It's usually denoted by the symbol Q and is measured in cubic meters per second (\text{m}^3/\text{s}) for water flow. When solving hydraulic problems, the formula for flow rate is used as follows: \ \[ Q = A \times V \] \ Where \(A\) is the cross-sectional area of the flow and \(V\) is the velocity of the water. To find the area, we multiply the width of the channel or spillway by the depth of the water. In the given exercise, the flow rate of water running down the spillway is \(280 \text{ m}^3/\text{s}\). The spillway is 50 meters wide, and the depth is 0.6 meters. So, the area of flow \(A\) is: \ \[ A = \text{Width} \times \text{Depth} = 50 \times 0.6 = 30 \text{ m}^2 \] \ Given the flow rate (\(Q\)), we can find the velocity \(V\) by rearranging the flow rate formula to: \ \[ V = \frac{Q}{A} \] \ Substituting the known values, we get: \ \[ V = \frac{280}{30} = 9.33 \text{ m/s}\] \ Thus, the water velocity at the base of the spillway is \(9.33 \text{ m/s}\). Understanding these basics helps in analyzing how the water flows over different surfaces.

The Froude number (Fr) is a dimensionless parameter that helps to describe flow regimes in open channels. It indicates whether the flow is subcritical, critical, or supercritical. \ \[ Fr = \frac{V}{\root 2 \text{g}h} \] \ Where \(V\) is the velocity of flow, \(g\) is the acceleration due to gravity (\(9.81 \text{ m/s}^2\)), and \(h\) is the depth of the flow. In our exercise, the Froude number at the base of the spillway can be calculated using the previously found velocity and given depth: \ \[ Fr = \frac{9.33}{\root 2 \times 9.81 \times 0.6} = \frac{9.33}{2.43} \rightarrow Fr = 3.84 \] \ This value tells us the type of flow: \

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• Fr < 1: Subcritical flow (slow and deep)
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• Fr = 1: Critical flow
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• Fr > 1: Supercritical flow (fast and shallow)
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\ In this problem, the Froude number is \(3.84\), indicating it is supercritical, meaning the flow is fast and shallow. After a hydraulic jump, this flow transitions from supercritical to subcritical, creating a sudden increase in water depth.

Manning's formula is a widely used empirical equation to estimate the velocity of flow in open channels based on channel shape, roughness, and slope. It's represented as: \ \[ V = \frac{1}{n} \times \root S_0 h^{2/3} \] \ Here, \(V\) is the flow velocity, \(n\) is Manning's roughness coefficient, \(S_0\) is the slope of the channel, and \(h\) is the hydraulic radius (typically the depth for wide channels). In our problem, we need to find where the depth transition stabilizes using this formula. The given values are: \

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• n = 0.015
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• Slope \(S_0 = \frac{1}{2500}\)
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\ We can rearrange Manning's formula to find the position \(x\) where the hydraulic jump occurs: \ \[ V = \frac{1}{0.015} \times \root \frac{1}{2500} h^{2/3} \] \ By iterating this formula within computational methods or using real-world telemetry, we can approximate the distance where the flow transitions back to subcritical, helping to determine the exact location of the hydraulic jump. This showcases how fundamental principles like Manning's formula are integral in practical hydraulic engineering.