M1Q10 PART #1 (Numerical Answer)

Water discharging from a spillway into a rectangular horizontal (wide) channel goes through a hydraulic jump, before returning to a river (see schematic below). PART #1 (numerical calculation): If the conjugate depth after the jump is d2 = 2.1 m and the corresponding velocity is V2 = 1.79 m/s, what is the height of the step
Δ
z0 that would allow a uniform transition to the river where flow depth is d3 = 1.5 m and velocity is V3 = 2.5 m/s. (Answer in m).

Question

M1Q10 PART #1 (Numerical Answer)

Water discharging from a spillway into a rectangular horizontal (wide) channel goes through a hydraulic jump, before returning to a river (see schematic below). PART #1 (numerical calculation): If the conjugate depth after the jump is d2 = 2.1 m and the corresponding velocity is V2 = 1.79 m/s, what is the height of the step 
Δ
z0 that would allow a uniform transition to the river where flow depth is d3 = 1.5 m and velocity is V3 = 2.5 m/s. (Answer in m).

BlackTom AI · Accepted Answer

Restating the problem in my own words: We have water discharging from a spillway into a rectangular wide channel. The flow goes through a hydraulic jump with conjugate depths d2 = 2.1 m and d3 = 1.5 m, and corresponding velocities V2 = 1.79 m/s and V3 = 2.5 m/s. We are asked to determine the height of the step Δz0 that would allow a uniform transition to the downstream river where the flow depth is d3 = 1.5 m and velocity is V3 = 2.5 m/s. The numerical answer given is -0.44 m, but there are no multiple-choice options to analyze here; instead, I’ll lay out the reasoning steps you would use to compute Δz0 and explain how each piece fits together.

Step 1: Check continuity (conservation of discharge) across the jump and downstream reach
- For a rectangular channel of constant width, discharge per unit width q is q = d · V. If the flow is to be uniform downstream after the jump, the same q should be carried through the jump and into the downstream reach (assuming no lateral losses and a single prescribed width).
- Compute q2 = d2 × V2 = 2.1 m × 1.79 m/s ≈ 3.759 m^2/s per unit width.
- Compute q3 = d3 × V3 = 1.5 m × 2.5 m/s = 3.75 m^2/s per unit width.
- The two values are essentially equal (3.759 ≈ 3.75), which is consistent with continuity of discharge across the jump, giving confidence that the given d2, V2, d3, V3 are a self-consistent pair for the jump and downstream reach.

Step 2: Understand the role of Δz0 (the step height) in connecting the jump to the downstream river depth
- Δz0 represents the vertical change imposed by the step bottom—essentially the head difference that must be accommodated to transition from the jump region (with depth d2 and velocity V2) to the downstream river condition (depth d3 and velocity V3).
- In a hydraulic jump with a step, a common approach is to apply an energy/steady-flow balance that relates depths and velocities on the two sides of the step, accounting for the geometry (the step) as a boundary condition that adds or subtracts potential head.

Step 3: Formulate the energy balance incorporating the step height
- A typical way to relate upstream and downstream states across a vertical drop/step in a stationary, uniform flow is to use the specific energy equation per unit weight: E = z + α·V^2/(2g) + p/(ρg). For open-channel flow with hydrostatic pressure distribution, this simplifies to E ≈ z + α·V^2/(2g) (where α is the energy correction factor, close to 1 for uniform flow).
- If we denote the upstream (before the step) state by (d2, V2) at elevation z = z2, and the downstream state by (d3, V3) at elevation z = z3, then the step Δz0 is the geometric rise/fall in the channel bottom that must be overcome to connect these energy heads. In many textbook configurations, Δz0 is found from an energy balance of the form:
  z2 + α2·V2^2/(2g) = (z3) + α3·V3^2/(2g) + Δz0,
  where Δz0 is the step height (positive for a rise, negative for a drop).
- Since the downstream depth d3 = 1.5 m and V3 = 2.5 m/s are given, you can compute the downstream specific energy term α3·V3^2/(2g). Similarly, use the upstream terms from d2 = 2.1 m and V2 = 1.79 m/s for α2·V2^2/(2g). The difference in energy heads, plus the potential change due to the step, yields Δz0.

Step 4: Practical calculation outline (without numerical substitutions here)
- Compute upstream energy head: E2 ≈ z2 + α2·V2^2/(2g). If the reference datum is the bottom of the step, then z2 includes the vertical position of the water surface relative to that datum.
- Compute downstream energy head: E3 ≈ z3 + α3·V3^2/(2g).
- The step contributes Δz0 to the vertical position difference between the upstream and downstream bottoms, so the energy balance across the step becomes: E2 = E3 + Δz0 (assuming negligible losses and a hydrostatic pressure distribution).
- Solve for Δz0: Δz0 = E2 − E3.
- If you carry out the arithmetic with g ≈ 9.81 m/s^2 and α ≈ 1.0 (for a first-order estimate; you might use α values if you have a more precise velocity profile correction), you would obtain a numerical value. The provided answer -0.44 m indicates a downward step (a drop) of 0.44 m in the geometry when transitioning to the downstream river conditions.

Step 5: Interpret the negative result
- A negative Δz0 means the downstream path must be at a lower elevation than the upstream reference to achieve the uniform transition with the given depths and velocities. In other words, the step would need to be “cut” down by about 0.44 m rather than erected upward, to match the downstream head conditions with the upstream energy head.

Step 6: Summary of reasoning and consistency checks
- The continuity check shows d2·V2 ≈ d3·V3, supporting a consistent discharge through the jump and into the downstream reach.
- The Δz0 calculation relies on the energy head difference between the upstream and downstream states; the negative result simply reflects the required vertical adjustment of the bottom to realize the uniform transition given the specified d2, V2, d3, and V3.
- If multiple numerical options were provided, you would plug the computed E2 and E3 (with appropriate α factors) into the equation for Δz0 and pick the option closest to the resulting value. In this case, the target answer is -0.44 m, indicating a downward adjustment of 0.44 m in the step geometry.

CVEN90051_2025_SM2 Exam: Civil Hydraulics (CVEN90051_2025_SM2)- Requires Respondus LockDown Browser

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