4. Indian Institute of Technology - Kharagpur

Welcome back students. 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 chunk introduces the concept of hydraulic jumps—a significant phenomenon in fluid mechanics. A hydraulic jump occurs when water flows from a higher velocity (supercritical flow) to a slower velocity (subcritical flow) within an open channel. This phenomenon is critical for civil engineering, especially in designing spillways and other water flow structures. The lecturer emphasizes that this section will focus on practical problems similar to those encountered in competitive exams.

Imagine a kid jumping off a diving board into a swimming pool. As the kid dives down, they are moving rapidly (like supercritical flow). When they hit the water, they slow down and splash, which represents the transition to subcritical flow. Just like the transition from high to low speed causes a splash, a hydraulic jump occurs due to changes in flow velocity.

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 point, so 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 section, a numerical problem is presented involving a spillway, which is a structure that allows water to flow over or around a dam. The problem states the width of the spillway, the depth of the water, and the velocity of the water flowing over it. This information is critical for calculating the effects of the hydraulic jump.

Think of a hose spraying water. If you put your thumb over the end of the hose, the water sprays out more forcefully. In this scenario, the spillway acts like the end of the hose, where the water flow characteristics (depth and velocity) significantly impact the behavior of the water after it flows over.

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, V1 was given 5.5, y1 was given 0.2, you see, 18 feet per second means, 5.5 metres per second and this is 30 meter. So, the Froude number 1 comes out to be 3.92 and this is greater than 1, which means hydraulic jump will occur.

The Froude number (Fr) is a dimensionless value that helps determine flow conditions in open channel hydraulics. It is calculated using the formula Fr = V/sqrt(g*y), where V is the velocity, g is gravity, and y is the depth of flow. A Froude number greater than 1 indicates supercritical flow, while a value less than 1 indicates subcritical flow. In this problem, Fr1 was calculated as 3.92, indicating the flow is supercritical and a hydraulic jump will occur.

Consider a roller coaster that is racing down a steep hill (supercritical) and then suddenly hits a flat section. The speed changes dramatically around that point, just like the water transitioning from supercritical to subcritical in a hydraulic jump.

So, 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. Fr1 whole square which already we found out in the previous slide was 3.92, plug these values here, so what comes out is y2 by y1 is 5.07.

To determine the depth after the hydraulic jump (y2), a formula involving the Froude number is used. This formula calculates the depth ratio (y2/y1), which relates the depth after the jump (y2) to the depth before the jump (y1). By substituting the previously calculated Froude number into the formula, the depth ratio is found to be 5.07, leading to the conclusion that the new depth (y2) will be significantly more than the original depth (0.2 m).

Picture a crowd of people at a concert. Some enter the venue (y1) quickly, but when it hits the main entrance, they scatter and form a larger, more spread out group inside (y2). The density of people decreases as they transition into the venue, similar to how the depth increases in a hydraulic jump.

Now, the way we obtain V2 is by equating the flow rate. So, A1 V1 is equal to A2 V2, V gets cancel out, so V2 will be V1 y1 divided by y2, y1 was known from before, V1 was known from before, y2 we just calculated using that y2 by y1 equation of hydraulic jump, so V2 comes out to be 1.08 meters per second.

To calculate the velocity after the hydraulic jump (V2), the principle of conservation of mass (continuity equation) is applied, stating that the flow rate before the jump must equal the flow rate after the jump. Using the areas and previously calculated values, V2 is computed to be 1.08 m/s. Once V2 is known, the Froude number after the jump (Fr2) can also be calculated. This demonstrates how conditions change as the water flows through the jump.

Think of a water slide: at the beginning, the water moves fast, and when it slides into a pool, the speed changes. The water flow rate must remain consistent, like our calculations, where the inflow and outflow velocities are connected, changing characteristics at the jump.

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, y1 we have, we know from before, V1 we know from before, y2 we have calculated, V2 we have calculated. And after substituting in the values, you see, the head loss, that the energy loss, in terms of head is 0.671 metre.

Head loss refers to the energy lost in the hydraulic jump due to turbulence and dissipation of energy. It is calculated by evaluating the total energy at two sections (before and after the jump). By substituting the known values (depth and velocity) into the energy equation, the head loss of 0.671 meters is determined, providing insight into energy transformation throughout the jump.

Imagine driving a car over a bumpy road. As the car hits bumps, some energy is lost as vibrations and impacts rather than moving smoothly. Similarly, as water experiences turbulence during a hydraulic jump, some energy is lost, which we quantify as head loss.

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