5. Lecture – 37

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 topic of hydraulic jumps and emphasizes the importance of understanding both basic and complex problems related to this phenomenon. Hydraulic jumps are sudden changes in flow conditions, typically from supercritical flow to subcritical flow, which is crucial in hydraulic engineering and often appears in competitive exams.

Think about a water fountain where water shoots high into the air but then suddenly hits a lower basin. This sudden drop can be likened to a hydraulic jump, where water goes from a fast and shallow flow (supercritical) to a slower and deeper flow (subcritical).

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. The question is determining the depth after the jump, the Froude numbers before and after the jump.

This chunk provides the specifics of the problem statement, including the dimensions and the flow conditions before the hydraulic jump occurs. We have a spillway that is 30 meters wide, with a flow depth of 0.20 meters and a velocity of 5.5 m/s. The objective is to determine the depth post-jump and calculate the Froude numbers, which are crucial for analyzing flow conditions.

Imagine a wide slide in a water park. The water flows at a certain speed as it pours down the slide. At the bottom, the water splashes and spreads out, which is akin to what happens during a hydraulic jump as water transitions from fast flow to calm, deeper water.

So, in this question, 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 this section, we calculate the upstream Froude number (Fr1) as a dimensionless quantity indicating the flow regime. Fr1 is calculated using the formula, where V1 is the velocity and y1 is the depth of flow. A Froude number greater than 1 indicates supercritical flow, confirming the occurrence of a hydraulic jump.

It's similar to a car going too fast over a hill. If the speed is too high (akin to Fr1 > 1), the car may go airborne, just like water in supercritical flow. Once it lands (the hydraulic jump), it enters a 'safer' slower zone.

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

This chunk illustrates how to find the depth after the jump (y2) using the depth ratio formula. By applying the previously determined Fr1 value, we derive y2, which indicates how much deeper the water is after the hydraulic jump, resulting in a value of 1.01 meters.

Think of the water level in a swimming pool. When water splashes from a fountain (the jump), it spreads out and creates new, deeper pools at certain spots. Similarly, after the hydraulic jump, the water's depth increases significantly.

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.

In this section, we derive V2, the velocity after the jump, by using the principle of conservation of flow rate. Applying the continuity equation (A1V1 = A2V2), we can solve for V2 using known values of V1 and y1 along with our previously calculated y2.

Imagine a garden hose. When you cover part of the opening, the water speeds up as it comes out. Similarly, after the jump, although the cross-section area changes and deepens (increasing y2), the flow speed adjusts, resulting in a new velocity.

So, therefore, Froude number at location 2 will be V2 by under root gy2. V2 we have calculated here, this was the reason we were calculating V2, for calculating the Froude number and this we calculated in the last slide. So, Froude number 2 comes out to be 0.343, so this means subcritical flow.

Here, we calculate the Froude number after the jump (Fr2) using the newly computed velocity (V2) and the depth (y2). Since Fr2 is less than 1, the flow is confirmed to be subcritical, indicating a significant change in flow characteristics after the jump.

Consider a river. Fast-moving water (upstream) meets a slower stretch downstream. The fast flow loses its energy and slows down, creating calmer water – exactly what happens after a hydraulic 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.

In this chunk, we calculate the energy loss during the hydraulic jump. This is done by comparing the total energies before and after the jump using known values for depths and velocities. The resulting head loss provides insight into the energy dissipated due to the jump.

This can be likened to a car losing speed and energy when going over a speed bump. The car expends energy to rise over the bump, and as it comes down, energy is lost to the surroundings, just like the head loss in the hydraulic jump.

This is the most simplest and the most common type of problems in hydraulic jump, which are the type of questions, you also will be expecting in your assignments and exams and competitive exams especially.

This section emphasizes that understanding and calculating energy loss in hydraulic jumps is a fundamental concept students should master, as it frequently appears in exams and assignments. Mastery of these concepts is essential for students pursuing careers in civil and hydraulic engineering.

Think of training for a race. You focus on mastering specific techniques that are frequently tested in competitions. Similarly, in engineering, understanding essential principles like energy loss in hydraulic jumps prepares students for real-world applications and exams.

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