2.2. Solution to Problem 1

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. The channel is 10 meter wide and 1.5 meter deep and the velocity of the water that is flowing is 1 meters per second.

In this problem, we need to determine how the water depth in a rectangular channel changes as it flows. We're given specific measurements: the width of the channel is 10 meters, the initial depth is 1.5 meters, and the flow velocity is 1 meter per second. The goal is to calculate the rate of change of depth of the water, denoted as 'dy/dx'. This equation indicates how quickly the water depth changes along the flow direction.

Imagine a water slide at a park. If the angle of the slide (bed slope) is steep, the water flows quickly, whereas a gentle slope allows the water to flow slowly. In this case, we assess how quickly the water level might drop as it travels down the slide (the channel) based on the given measurements.

We have been given S0, the bed slope as 1 in 4000, and the energy line slope, Sf, as 0.00004. The flow area and discharge calculations follow.

To solve this problem, we need to identify key parameters: S0 (the bed slope of the channel) is 1 in 4000, indicating a gentle slope. We also have Sf, which represents the slope of the energy line, defined as 0.00004. These slopes play a crucial role in determining the flow behavior in the channel. Additionally, we can find the area of flow, which is the product of the width and depth: 10 meters x 1.5 meters = 15 square meters. This area will be used in further calculations to determine the discharge, which tells us how much water is flowing through the channel.

Consider a garden hose laid down a gentle hill. The steeper the hill, the faster the water travels down. Here, the bed slope (S0) and the energy line slope (Sf) help us understand how quickly the water flows and how deep it stays in the hose (the channel).

The area of the flow is A = b * y = 10 * 1.5 = 15 m². The discharge Q can be calculated as Q = A * V = 15 * 1 = 15 m³/s.

We calculate the area A by multiplying the width b (10 m) by the depth y (1.5 m), resulting in an area of 15 square meters. The discharge Q represents the volume of water flowing through the channel per second. We find Q by multiplying the area A by the flow velocity V (1 m/s). Thus, the discharge is 15 cubic meters per second. This value is essential as it relates to how much water is moving through the channel at any moment.

Think of a wide river versus a narrow stream. The wider and deeper sections of the river allow more water to flow through per second. In our calculated channel, we determine how much water is moving past a point each second, much like measuring the flow of a river.

The rate of change of depth dy/dx is derived from the equation: dy/dx = (S0 - Sf) / (1 - (Q^2 * T) / (g * A^3)). Calculating this gives us dy/dx = 2.25 x 10^-4, indicating gradual flow.

We use the derived formula for gradually varied flow to find the depth change. Substituting in our values: S0 as 1/4000, Sf as 0.00004, discharge Q as 15 m³/s, top width T as 10 m, gravitational constant g as 9.8 m/s², and area A as 15 m², we calculate dy/dx. After performing the calculations, we find that dy/dx equals 2.25 x 10^-4. A small value indicates that the depth change along the channel is very gradual, which confirms this scenario as gradually varied flow.

Imagine filling a bathtub with a steady stream of water. As the tub fills up slowly, the depth of water changes little by little, which aligns with our result. This slow rate (dy/dx) indicates that while the water is moving, it’s not drastically changing the water level all at once.

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