6.4. Depth Ratio Calculation and Depth After the Jump

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. Therefore, the Froude number 1 comes out to be 3.92 and this is greater than 1, which means hydraulic jump will occur. As I said that the upstream flow is supercritical and therefore, it is possible to generate hydraulic jump.

To determine if a hydraulic jump will occur, we calculate the Froude number (Fr1). The formula for Fr1 is the flow velocity (V1) divided by the square root of the product of gravitational acceleration (g) and the upstream depth (y1). In our example, V1 is 5.5 m/s and y1 is 0.2 m. After calculating, we find Fr1 = 3.92, which is greater than 1. This indicates that the flow is supercritical, therefore a hydraulic jump will indeed occur.

Think of a river flowing swiftly (supercritical flow) over rocks. When the water hits a shallow area (like a drop-off), it suddenly slows down and splashes up, just like a hydraulic jump. The Froude number helps us understand when such changes in flow happen.

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.

After we've determined the Froude number, we calculate the ratio of the depths (y2/y1) using the formula: y2/y1 = (1/2) * (−1 + √(1 + 8*(Fr1^2))). By substituting Fr1 (which is 3.92) into this formula, we determine y2/y1 = 5.07. Multiplying this ratio by the original depth y1 (0.2 m), we find the new depth after the jump (y2) is 1.01 m.

Imagine a steep slide in a water park. The water at the top (y1) flows quickly and is shallow, but as it splashes into the pool at the bottom (y2), it becomes deep and turbulent. The depth ratio helps describe how the water's behavior changes through the 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. V1 was known from before, y1 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 find the velocity after the jump (V2), we use the conservation of mass equation, which states that the flow rate before the jump (A1V1) must equal the flow rate after the jump (A2V2). By rearranging, we determine that V2 = (V1*y1)/y2. Substituting our known values, we find that V2 is 1.08 m/s.

Think of water flowing through a garden hose. When you cover the end of the hose partially with your thumb, the water shoots out much faster. Here, as the water goes from a wider area (the hose) to a narrower or deeper area after the jump, its speed changes, just like V2 being less than V1.

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.

To find the Froude number after the jump (Fr2), we use the formula Fr2 = V2 / √(g*y2). Here, we substitute V2 (1.08 m/s) and y2 (1.01 m). Upon completing the calculation, we find that Fr2 equals 0.343, indicating a subcritical flow, meaning the flow is now calmer and more manageable.

If you've ever watched a rapid river slow down as it approaches a lake, you can see this effect. The water slows down and spreads out; the Froude number tells us when water shifts from fast and turbulent to slower and calmer.

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.

To compute the head loss (energy loss) during the hydraulic jump, we apply the energy equation, which calculates the difference in total energy between the two sections. The total energy is composed of the height (y) and kinetic energy (velocity squared). After substituting known values, we find the head loss to be 0.671 m.

Imagine sliding down a slide. At the top, you have maximum potential energy; as you slide down, some of that energy is converted into speed (kinetic energy), but some energy is lost due to friction and turbulence at the bottom, similar to how water loses energy in a hydraulic jump.

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