6. Non-Uniform Flow and Hydraulic Jump (Contd.)

So, we start with solving some questions that are related to hydraulic jumps, so rapidly varied flow. So, this entire lecture will be dedicated to solving some of the basics and a little more complex problems, those type of problems which you encountered in exams like GATE or IES.

This section introduces the topic of hydraulic jumps, emphasizing that the lecture will focus on solving questions related to this phenomenon. Hydraulic jumps occur when there is a sudden change in the flow of water, causing it to transition from supercritical to subcritical flow. This can be complex to analyze, which is why practice problems from exams like GATE and IES are used as examples in the lecture.

Imagine a waterfall where water falls rapidly and suddenly meets a pool below. This rapid change in speed creates turbulence and ripples, much like a hydraulic jump in civil engineering, where the flow of water transforms dramatically.

So, we start with one problem as below, it says that water on the horizontal apron of 30 metre wide spillway. So, it is 100 feet or, sorry, that is 30 meter has a depth of 0.06 feet is equivalent to 0.20 metre and a velocity of 18 feet per second that means, 5.5 metres per second.

In this chunk, we are presented with a specific problem involving water flow over a spillway. The parameters given are the width of the spillway (30 meters), the initial depth of water (0.20 meters), and the initial velocity (5.5 meters per second). Understanding these parameters is crucial as they will be used to calculate changes in depth and velocity after a hydraulic jump occurs.

Think of this scenario like a large water slide. The slide represents the spillway, the depth of water is like how much water is filling the slide, and the speed at which people slide down is analogous to our water velocity.

So, we know the conditions here, at number 1, so it is pretty simple to calculate Froude number 1. We will go step by step, the way you should be solving to. Fr1 is given by V1 divided by under root gy1.

Froude Number (Fr) is a dimensionless number used to determine whether the flow is supercritical (Fr > 1) or subcritical (Fr < 1). Here, to find Froude number before the jump (Fr1), we use the formula: Fr1 = V1 / √(g * y1), where V1 is the initial velocity, g is the acceleration due to gravity, and y1 is the depth of flow. This helps us predict the behavior of the flow at different points along the channel.

Consider a car on a highway. The speed at which the car travels relative to the road conditions (e.g., flat or hilly) can be compared to the Froude number, which tells us how the flow is behaving—whether it's speeding through or slowing down due to obstacles.

In the second step, we obtain depth ratio, we had the formula which said y2 by y1 is equal to 1/2 into multiplied by – 1 + 1 + 8 Fr1 square.

We derive the new depth (y2) after the hydraulic jump using the depth ratio formula. By substituting the Froude number calculated earlier (Fr1), we can find out how deep the water becomes after the jump, which is crucial for understanding changes in flow characteristics.

Imagine a highway suddenly leading into a deep valley. The depth of the valley (y2) represents how far the car will drop (the water will deepen after the jump), based on how fast it was going before (indicated by Fr1).

Now, the way we obtain V2 is by equating the flow rate. So, A1 V1 is equal to A2 V2.

To find the velocity after the jump (V2), we compare the area and velocity before and after the jump. This principle of conservation tells us that the product of area (A1) and velocity (V1) before the jump is equal to the product of area (A2) and velocity (V2) after the jump. This helps us ensure that the flow is maintained.

This is similar to squeezing a garden hose. When you cover part of the hose, the water must flow faster through the narrow section to maintain the same amount of flow, just as we equate areas and velocities in our calculations.

Now, we have to also obtain the head loss, that the energy loss. We simply use this equation, total energy at section 1 minus total energy minus section 2.

Head loss refers to the energy lost in the hydraulic jump, and it can be calculated by the difference in total energy at two sections of flow (before and after the jump). Using energy conservation principles, we can effectively determine how much energy is 'lost' due to turbulence and other factors introduced by the jump.

Think of this like a downhill bike ride. You start with a certain amount of energy (potential and kinetic) at the top of the hill, and by the time you reach the bottom, you've lost some kinetic energy due to bumps and friction along the path.

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