2.1. Problem 1: Rate of change of depth of water

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.

This problem sets up a scenario in a rectangular channel where we need to determine how quickly the depth of water is changing when certain parameters are given. The channel is specified to be 10 meters wide and has a depth of 1.5 meters, with the water flowing at a speed of 1 meter per second. Understanding the rate of change of depth is essential for managing water flow in hydraulic engineering.

Imagine a garden hose flowing water into a rectangular basin. The speed at which the basin fills up (the increase in water depth) depends on the width of the hose and how fast the water flows. Here, we are trying to measure this change in depth in a more scientific manner.

Given is, b is given as 10 meter, depth we have been given 1.5 meter, we also have been given velocity of the flow 1 metres per second, bed slope also has been mentioned; 1 in 4000, so 1 by 4000 is the bed slope, we also have been given the slope of, Sf is also given, that is, 0.00004.

In this chunk, we list out all the parameters and constants provided in the problem statement. We confirm the bed slope (��), velocity of flow (1 m/s), width of the channel (10 m), and depth (1.5 m). This information is necessary for calculations to find the rate of change in water depth.

Think of this like gathering all your measurements before starting a recipe. You need to know the size of your baking dish, the depth of the ingredients, and how fast your mixer works to combine everything effectively.

First, what we need to do? We need to calculate the area of the flow and that is nothing but b into y, so 10 into 1.5 and that is going to be 15 meter square. Discharge is area into velocity, area is 15 meter into velocity is 1 meters per second, so that is 15 meter cube per second.

Here, we calculate the flow area by multiplying the width of the channel (b = 10 m) by the depth of the water (y = 1.5 m), resulting in an area of 15 square meters. Next, we compute the discharge (Q), which represents the volume of water flowing per second, using the formula Q = Area × Velocity. Therefore, the discharge is calculated to be 15 cubic meters per second.

If you think about filling a swimming pool, the area of the pool's surface multiplied by the rate at which the water flows from the hose gives you how quickly the pool fills up. Here, we are performing similar calculations.

dy by dx is equal to S0 - Sf divided by 1 - Q square multiplied by T divided by g into A cube.

In this step, we apply the derived formula related to gradually varied flow, which connects the rate of change of depth (dy/dx) to the slopes (S0 and Sf), discharge (Q), top width (T), gravitational acceleration (g), and the flow area (A). This formula enables us to determine how quickly the depth of water changes along the flow channel.

Imagine a seesaw. The tilt of the seesaw can represent the slope of the bed (S0) and the resistance (Sf). Depending on how steep the seesaw is (the difference between these slopes), the flow of water will respond and change its depth accordingly.

dy by dx will come out to be 2.25 into 10 to the power – 4. Will this have any unit? No unit.

After substituting all the known values into the formula, the rate of change of depth (dy/dx) results in a very small value of 2.25 x 10^-4. This indicates a gradual change in depth, illustrating that in this particular flow scenario, the change is minimal. Also, it's essential to note that dy/dx is a dimensionless quantity, meaning it has no units, further simplifying the interpretation of results.

If we think of a river flowing slowly and steadily, a small change in depth means the water is flowing smoothly without dramatic changes - just a gentle rise and fall with no sudden splashes.

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