2. Problems and Solutions

The question is; find the rate of change of depth of water in a rectangular channel, which is 10 meter wide and 1.5 meter deep, when the water is flowing with a velocity of 1 meters per second.

In this problem, we are given a rectangular channel with specific dimensions: a width of 10 meters and a depth of 1.5 meters. The water flows through the channel at a velocity of 1 meter/second. We are tasked with calculating the rate of change of water depth, denoted as dy/dx, which indicates how much the water depth changes as we move along the channel. To begin, we need to gather all the relevant information and apply it systematically using known hydraulic equations.

Imagine a water slide at a park where the angle of the slide changes. When you slide down, the depth of water underneath you varies based on how steep the slide is. The calculation we are doing helps us understand how quickly the depth changes as you move along the water slide.

Given: b = 10 meters, depth = 1.5 meters, velocity = 1 m/s, bed slope = 1 in 4000, slope of energy line = 0.00004. First, calculate area (A = b × depth = 10 × 1.5 = 15 m²) and discharge (Q = A × velocity = 15 m² × 1 m/s = 15 m³/s). The equation relating dy/dx is dy/dx = (S0 - Sf) / (1 - (Q² × T) / (g × A³)). Substituting known values gives dy/dx = 2.25 × 10⁻⁴.

First, we identify our known quantities. We calculate the area of flow (A) by multiplying the width (b) by the depth of the channel. This gives us a flow area of 15 m². Next, we calculate the discharge (Q) by multiplying the flow area (A) by the flow velocity (1 m/s), resulting in a discharge of 15 m³/s. We then substitute these values into the derived formula for dy/dx, which relates the rate of change of depth with the slope of the channel and energy line. Upon simplifying, we find dy/dx to be approximately 2.25 × 10⁻⁴, indicating a gradual change in depth along the channel.

Using the analogy of a garden hose with a nozzle, if you squeeze the hose, the water comes out faster and at a different height to water the plants. Similarly, our calculation shows how the change in slope and flow velocity affects how 'deep' the water feels at different points in the channel.

A rectangular channel with a bottom width of 4 meters and a bottom slope of 0.0008 has a discharge of 1.5 m³/s. In a gradually varied flow in this channel, the depth at a certain location is found to be 0.30 meters. Assuming the Manning's n = 0.016, determine the type of gradually varied profile.

In this problem, we need to analyze another rectangular channel with specific characteristics. We begin by noting the bottom width (b), slope (S0), discharge (Q), and the depth at a location (y). We then calculate the critical depth (yc) using the formula, yc = (q²/g)^(1/3), where q is the discharge per width. Using the values, we find yc to be 0.243 meters. We then apply Manning's equation to find the normal depth (y0). For normal depth, we use the Manning’s equation, and further analysis reveals that y0 is greater than yc, indicating a mild slope channel.

Think of a winding river where the slope of the river bank changes. When there’s a decrease in slope, the river might spread out and deepen, while a steeper slope allows the river to flow quickly and remain shallow. This problem analyzes those variations in depth along the channel.

Given: b = 4 meters, y = 0.3 meters, Q = 1.5 m³/s, S0 = 0.0008, Manning's n = 0.016. Calculating small q gives 0.0375 m³/s/m. Critical depth yc is calculated as 0.243 m. Solving with Manning's equation yields y0 = 0.426 m. Since y0 > yc, we identify the profile as an M2 curve.

We organize the information provided and compute the discharge intensity (small q), which simplifies our calculations. We then calculate the critical depth (yc) and analyze the geometry of the flow using Manning's equation, which helps us determine the normal depth (y0). By comparing the values of y and y0, we see that y0 is greater than yc, classifying it as an M2 curve, indicating a mild slope channel with certain characteristics.

Consider a ski slope. At the top, it might be steep and quick, representing a shallow 'y' effect. But as you go down, if the slope flattens and widens, you experience a longer glide, similar to how the water behaves in our channel. Here, our analysis helps us predict that gentle transition.

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