6.3. Calculating Froude Number Before 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, 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.

To understand the problem, we recognize that the conditions of flow change and we need to calculate the Froude number (
;Fr1). This number helps us to determine if the flow is supercritical or subcritical. The formula for Fr1 is V1 / sqrt(g*y1), where V1 is the initial velocity of the flow and y1 is the initial depth of the flow. Here, V1 is 5.5 m/s and y1 is 0.2 m. Plugging in these values, we find that Fr1 equals 3.92, which is greater than 1. This indicates that the flow before the hydraulic jump is supercritical, meaning it is faster and has lower depth compared to calmer, slower flows.

Think of a highway with fast-moving cars (supercritical flow) versus a slow-moving parade (subcritical flow). If suddenly the cars encounter a traffic jam (the hydraulic jump), the transition from fast to slow can be abrupt, reflecting the calculations we do with Froude numbers.

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.

Next, we need to find the depth after the hydraulic jump, which is called y2. To do this, we use the equation y2/y1 = (1/2)(-1 + sqrt(1 + 8*(Fr1^2))). We already calculated Fr1 as 3.92. We substitute this into the equation and simplify it to find that y2/y1 equals 5.07. This tells us that y2, the depth after the jump, is 5.07 times the initial depth y1 (0.2 m), leading us to calculate y2 to be 1.01 m.

Imagine a hill where water flows rapidly down a steep slope (representing the supercritical state). As it hits a flat area, the water spreads out (hydraulic jump), resulting in a much deeper layer of water compared to where it started. That's why we calculate how much deeper the water will be after 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, 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.

Next, we find the flow velocity after the jump, V2. To do this, we assume the flow rate (volume per time) before the jump equals the flow rate after the jump: A1V1 = A2V2, where A is the cross-sectional area of the flow and V is the velocity. Because the area changes (depth increases), the velocity will change as well. Thus, we rearrange the equation to find V2 = (V1*y1)/y2. By plugging our known values—V1 = 5.5 m/s, y1 = 0.2 m, y2 = 1.01 m—we find that V2 equals approximately 1.08 m/s.

Think of a hose. When you cover part of the opening, the water speeds up due to the reduced space (like our small depth before the jump). Once the hose opens, the water flows slower but in a broader stream (after the jump). This illustrates how flow velocities change with depth.

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.

Finally, we need to determine the Froude number after the jump, or Fr2. We use the same formula as before: Fr2 = V2/sqrt(g*y2). Using the values we calculated—V2 = 1.08 m/s and y2 = 1.01 m—we find that Fr2 equals approximately 0.343, which is less than 1. This indicates that the flow after the jump has turned subcritical, meaning the water is now flowing slower and is deeper compared to the conditions before the jump.

Returning to our highway analogy, after the fast-moving cars (supercritical flow) slow down significantly at the traffic jam (hydraulic jump) to a normal traffic flow (subcritical flow), we can see that the conditions of the flow have changed markedly.

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.

Finally, we look at head loss, which is the energy lost due to the jump. We determine this using the energy conservation principles, where the total energy before the jump is subtracted from the total energy after the jump. The equation typically used is: head loss = (y1 + V1^2/(2g)) - (y2 + V2^2/(2g)). After substituting all previously found values, we find head loss to be 0.671 m, meaning that energy is lost in the form of turbulence and other effects.

Consider a water slide: when you leap into a pool from the slide, you lose some energy in splashes (head loss). The higher you jump (initial energy), the larger the splash (loss of energy) you'll create when you hit the water. The concept of energy loss helps us understand how flow impacts energy in hydraulic systems.

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