2. Prof. Mohammad Saud Afzal

Welcome back students. So, we start with solving some questions that are related to hydraulic jumps, so rapidly varied flow. 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.

In this introduction, the focus is on hydraulic jumps, which are phenomena that occur when a fluid transitions from supercritical to subcritical flow. The lecturer introduces the topic and mentions that the session will tackle questions similar to those found in competitive exams. Understanding this flow behavior is crucial for engineers dealing with open channel hydraulics.

Imagine a river flowing smoothly but hitting a flat area (the apron) where it suddenly slows down and deepens—this is similar to a hydraulic jump. Just as a car must slow down on a steep incline, water must also adjust its speed and depth based on the terrain changes.

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.

Here, we have a specific problem involving water flow over a spillway—a structure used to control water flow. The given dimensions (depth and velocity) set the stage for calculating the depth after the hydraulic jump and the Froude numbers before and after the jump. These parameters are essential for analyzing how the flow changes as it transitions through the jump.

Think of inflating a balloon quickly. Initially, the air moves fast but if you pinch the balloon (like a hydraulic jump), the air may slow down and take up more space within the balloon. Similar adjustments happen in water flow during a hydraulic jump.

The conditions across the jump are determined by the upstream Froude number Fr1, that also we have to find out actually. So, 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.

In fluid mechanics, the Froude number (Fr) is a dimensionless number that helps determine the flow regime. A Fr less than 1 indicates subcritical flow (calm), while Fr greater than 1 indicates supercritical flow (fast and turbulent). Calculating Fr1 shows that the flow is supercritical at this point, confirming that a hydraulic jump will occur downstream.

Imagine riding a skateboard down a steep ramp. As you reach the bottom, if you're going fast enough (like having a high Froude number), you could jump off the ramp—it’s similar to how water behaves during a hydraulic jump.

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. Therefore, y2 is going to be 5.07 into 0.2 and that is 1.01 metre.

The depth after the jump (y2) can be calculated using the depth ratio formula derived from the conservation of momentum and energy principles in hydraulics. Substituting the previously calculated Fr1 value into the formula gives us a new depth that is much greater than the initial depth, meaning that water slows down and spreads out after the jump.

Consider a sports car reaching the end of a racetrack (supercritical flow) where it suddenly hits a wide section (the jump). Just as the car will slow down and spread out over a larger area due to the change in terrain, the water does the same during 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.

By maintaining the continuity of flow, the velocity after the jump (V2) can be calculated using the areas and velocities before and after the jump. Substituting y1 and y2 into the flow rate equation allows us to determine that the water slows down significantly after the jump, reflecting the transition from supercritical to subcritical flow.

Think of a wide river splitting into two smaller streams. As the water distributes itself, it slows down due to the larger area available to flow through, similar to how V2 is calculated after a jump in a spillway.

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 represents energy loss due to turbulence and other factors during the hydraulic jump. By using the energy equation, we can calculate how much potential energy has been converted into other forms of energy (like kinetic and thermal) due to the jump. A head loss of 0.671 meters indicates that some energy has been dissipated.

When you jump off a diving board into water, you experience a splash and turbulence. This 'messy' exit means energy has been lost to the water as turbulence, just as head loss indicates lost energy during a hydraulic jump.

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