Location Of Hydraulic Jump: Determination, Profile Computations & Solved Examples (2026)

1. Physical Principles Governing Hydraulic Jump Positioning

Determining the precise location of hydraulic jump formation is a fundamental challenge in open channel hydraulic engineering. A hydraulic jump does not form spontaneously at an arbitrary station; rather, it anchors precisely where the upstream supercritical flow depth matches the conjugate sequent depth required by downstream subcritical tailwater controls. If the jump forms too far downstream, high-velocity supercritical flow overshoots protective concrete aprons, causing severe bed scour. Conversely, if it forms too far upstream, it drowns outlet structures and impairs discharge capacity.

                    SPATIAL INTERACTION & JUMP POSITIONING
        Sluice / Chute
              \  Supercritical Profile (M3 Curve)
               \---------------------\     Conjugate Depth Curve y2(x)
                                      \   /============================ Tailwater (y_tw)
                                       \ /   Hydraulic Jump Station
        ================================X===============================================
                                     x_jump

Gradually varied flow (GVF) theory couples upstream boundary conditions (e.g., sluice gate vena contracta, spillway toe) with downstream tailwater rating curves to pinpoint the exact equilibrium station $x_{jump}$.

Core Concepts of Gradually Varied Flow and Hydraulic Controls

2. Interplay Between Upstream Supercritical and Downstream Subcritical Controls

Supercritical flow ($Fr > 1$) is governed entirely by upstream boundary conditions because gravity wave celerity $c = \sqrt{gy}$ is smaller than flow velocity $v$. Information cannot propagate upstream.

Subcritical flow ($Fr < 1$) is governed by downstream boundary conditions (such as weirs, channel transitions, or uniform river normal depth $y_n$) because disturbance waves travel upstream against the flow.

                      HYDRAULIC CONTROL INFLUENCE ZONES
           Upstream Control                     Downstream Control
          (Sluice / Spillway)                     (Tailwater Weir)
                 |                                      |
                 v                                      v
          [Supercritical] =====> [ JUMP ] <===== [Subcritical]
          (Controls y1)          x = x_jump      (Controls y2)

The hydraulic jump acts as the hydrodynamic bridge between these two independent flow controls. The jump occurs at the unique longitudinal coordinate $x$ where the supercritical profile depth $y_1(x)$ satisfies the Bélanger sequent depth equation for the downstream tailwater depth $y_2(x)$:

$$y_2(x) = \frac{y_1(x)}{2} \left( \sqrt{1 + 8 Fr_1(x)^2} – 1 \right)$$

3. Gradually Varied Flow (GVF) Governing Differential Equation

Water surface profiles upstream and downstream of a jump follow the classical 1D GVF differential equation:

$$\frac{dy}{dx} = \frac{S_0 – S_f}{1 – Fr^2}$$

Where:

• $y$ is local flow depth ($\text{ m}$).

• $x$ is longitudinal distance along the channel bed ($\text{ m}$).

• $S_0$ is longitudinal channel bed slope ($-\frac{dz_b}{dx}$).

• $S_f$ is friction slope calculated from Manning’s formula:

$$S_f = \frac{n^2 v^2}{R_h^{4/3}} = \frac{n^2 Q^2}{A^2 R_h^{4/3}}$$

• $Fr$ is local Froude number:

$$Fr = \sqrt{\frac{\alpha Q^2 T}{g A^3}}$$

For rectangular channels of width $B$ with unit discharge $q$:

$$\frac{dy}{dx} = S_0 \frac{1 – \left(\frac{y_n}{y}\right)^3}{1 – \left(\frac{y_c}{y}\right)^3}$$

Where $y_n$ is normal depth and $y_c = \sqrt[3]{q^2/g}$ is critical depth.

Mathematical Properties of Sequent Depth Ratios

4. Graphical and Numerical Determination of Jump Location

Pinpointing the location of hydraulic jump involves four systematic engineering steps:

                      STEPWISE COMPUTATIONAL PROCEDURE
      [Step 1: Compute Upstream GVF Profile y1(x) from Upstream Control]
                                    |
                                    v
      [Step 2: Generate Sequent Depth Envelope y2*(x) via Bélanger]
                                    |
                                    v
      [Step 3: Compute Downstream GVF Profile y_tw(x) from Tailwater Control]
                                    |
                                    v
      [Step 4: Intersect Curves: y2*(x_jump) = y_tw(x_jump)]

1. Calculate the Supercritical Profile $y_1(x)$: Integrate downstream from the upstream control (e.g., sluice gate vena contracta $y_0$) along the $M_3$ or $S_2$ curve.
2. Construct the Sequent Depth Profile $y_2^*(x)$: Apply the Bélanger formula at each computed station $x_i$ to calculate required conjugate depth $y_2^*(x_i)$.
3. Calculate the Subcritical Profile $y_{tw}(x)$: Integrate upstream from the downstream control (e.g., river pool or weir) along the $M_1$ or $M_2$ curve.
4. Locate the Intersection: The jump location $x_{jump}$ is the station where the actual tailwater profile intersects the required sequent depth curve: $y_2^*(x_{jump}) = y_{tw}(x_{jump})$.

5. Tailwater Level Classifications: Free, Repelled, and Submerged Jumps

The spatial position of the jump relative to an upstream structure depends on the ratio of actual downstream tailwater depth $y_d$ to required conjugate depth $y_2$:

                       TAILWATER POSITION REGIMES
        Case 1: yd = y2 (Free Jump)
        +-------\~~~~~~~~~~~~~~~+========================================

        Case 2: yd < y2 (Repelled Jump)
        +---------------\~~~~~~~~~~~~~~~+================================

        Case 3: yd > y2 (Submerged Jump)
        +---\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

6. Direct Step Method for Water Surface Profile Integration

The Direct Step Method divides channel reaches into discrete depth increments $\Delta y = y_{i+1} – y_i$ and calculates distance $\Delta x$:

$$\Delta x = \frac{E_{i+1} – E_i}{S_0 – \bar{S}_f} = \frac{\Delta E}{S_0 – \bar{S}_f}$$

Where:

• $E_i = y_i + \frac{v_i^2}{2g}$ is specific energy at station $i$.

• $\bar{S}_f = \frac{S_{f,i} + S_{f,i+1}}{2}$ is average friction slope across the reach.

• Manning’s friction slope: $S_{f} = \frac{n^2 v^2}{R_h^{4/3}}$.

This formulation avoids iterative root-finding for prismatic channels.

7. Solved Numerical Example 1: Location of Jump Downstream of a Sluice Gate on Mild Slope

Problem Statement

A rectangular wide channel ($B = 6.0 \, \text{ m}$, Manning’s $n = 0.015$, bed slope $S_0 = 0.0012$) conveys discharge $Q = 18.0 \, \text{ m}^3/\text{ s}$. A sluice gate produces a vena contracta flow depth $y_0 = 0.35 \, \text{ m}$ at station $x = 0 \, \text{ m}$. Downstream channel conditions maintain uniform normal depth $y_n = 1.95 \, \text{ m}$.

Calculate:
1. Unit discharge $q$ and critical depth $y_c$.
2. Required sequent depth $y_2(0)$ at the gate vena contracta.
3. Distance $x_{jump}$ from the sluice gate to the jump using the Direct Step Method across the $M_3$ profile.

                 SLUICE GATE M3 PROFILE & JUMP LOCATION
              x = 0 m (Gate)                     x = x_jump
              y0 = 0.35 m                        y1 = 0.444 m ----> y2 = 1.95 m
              +-------------------\              ~~~~~~~~~~~~~
              |  M3 Supercritical  \____________/  Subcritical Uniform Flow
              +==========================================================
              |<------------------- x_jump ------------------->|

Step-by-Step Solution

Step 1: Basic Hydraulic Parameters
$$q = \frac{Q}{B} = \frac{18.0}{6.0} = 3.0 \, \text{ m}^2/\text{ s}$$

$$y_c = \sqrt[3]{\frac{q^2}{g}} = \sqrt[3]{\frac{(3.0)^2}{9.81}} = \sqrt[3]{0.9174} = 0.9717 \, \text{ m}$$

Since $y_0 = 0.35 \, \text{ m} < y_c < y_n = 1.95 \, \text{ m}$, the flow follows an $M_3$ water surface curve where depth increases downstream until matching conjugate conditions.

Step 2: Required Sequent Depth at Tailwater ($y_2 = y_n = 1.95 \, \text{ m}$)
To find the required upstream depth $y_1$ that produces sequent depth $y_2 = 1.95 \, \text{ m}$:

$$v_2 = \frac{q}{y_2} = \frac{3.0}{1.95} = 1.5385 \, \text{m/s}$$

$$Fr_2 = \frac{v_2}{\sqrt{g y_2}} = \frac{1.5385}{\sqrt{9.81 \times 1.95}} = \frac{1.5385}{4.3737} = 0.3518$$

Using the inverse Bélanger formula:
$$y_1 = \frac{y_2}{2} \left( \sqrt{1 + 8 Fr_2^2} – 1 \right) = \frac{1.95}{2} \left( \sqrt{1 + 8(0.3518)^2} – 1 \right)$$

$$y_1 = 0.975 \times (\sqrt{1 + 0.9899} – 1) = 0.975 \times (\sqrt{1.9899} – 1) = 0.975 \times (1.4106 – 1) = 0.4003 \, \text{ m}$$

The jump will form when depth along the $M_3$ profile reaches $y_1 = 0.400 \, \text{ m}$.

Step 3: Direct Step Integration from $y = 0.35 \, \text{ m}$ to $y = 0.40 \, \text{ m}$

Section 1 ($y_a = 0.35 \, \text{ m}$):
– Area: $A_a = 6.0 \times 0.35 = 2.10 \, \text{ m}^2$

• Wetted Perimeter: $P_a = 6.0 + 2(0.35) = 6.70 \, \text{ m}$

• Hydraulic Radius: $R_{h,a} = \frac{2.10}{6.70} = 0.3134 \, \text{ m}$

• Velocity: $v_a = \frac{18.0}{2.10} = 8.5714 \, \text{m/s}$

• Specific Energy: $E_a = 0.35 + \frac{(8.5714)^2}{2 \times 9.81} = 0.35 + 3.7447 = 4.0947 \, \text{ m}$

• Friction Slope: $S_{f,a} = \frac{(0.015)^2 \times (8.5714)^2}{(0.3134)^{4/3}} = \frac{0.000225 \times 73.469}{0.2141} = \frac{0.01653}{0.2141} = 0.07721$

Section 2 ($y_b = 0.40 \, \text{ m}$):
– Area: $A_b = 6.0 \times 0.40 = 2.40 \, \text{ m}^2$

• Wetted Perimeter: $P_b = 6.0 + 2(0.40) = 6.80 \, \text{ m}$

• Hydraulic Radius: $R_{h,b} = \frac{2.40}{6.80} = 0.3529 \, \text{ m}$

• Velocity: $v_b = \frac{18.0}{2.40} = 7.500 \, \text{m/s}$

• Specific Energy: $E_b = 0.40 + \frac{(7.500)^2}{2 \times 9.81} = 0.40 + 2.8669 = 3.2669 \, \text{ m}$

• Friction Slope: $S_{f,b} = \frac{(0.015)^2 \times (7.500)^2}{(0.3529)^{4/3}} = \frac{0.000225 \times 56.25}{0.2505} = \frac{0.012656}{0.2505} = 0.05052$

Reach Integration:
– $\Delta E = E_b – E_a = 3.2669 – 4.0947 = -0.8278 \, \text{ m}$

• $\bar{S}_f = \frac{0.07721 + 0.05052}{2} = 0.063865$

• $S_0 – \bar{S}_f = 0.0012 – 0.063865 = -0.062665$

• Step Distance:
  $$\Delta x = \frac{\Delta E}{S_0 – \bar{S}_f} = \frac{-0.8278}{-0.062665} = 13.21 \, \text{ m}$$

The hydraulic jump initiates at distance $x_{jump} = 13.21 \, \text{ m}$ downstream from the sluice gate.

8. Solved Numerical Example 2: Spillway Toe Hydraulic Jump Position Computation

Problem Statement

A spillway chute discharges flow with unit discharge $q = 14.0 \, \text{ m}^2/\text{ s}$ at depth $y_{toe} = 0.60 \, \text{ m}$ into a wide horizontal concrete stilling basin ($n = 0.013, S_0 = 0.0$). River tailwater depth downstream is $y_{tw} = 6.20 \, \text{ m}$.

Determine whether the jump is free on the apron or pushed downstream, and compute the toe sequent depth deficit.

Step-by-Step Solution

Step 1: Compute Sequent Depth at Spillway Toe ($y_1 = 0.60 \, \text{ m}$)
$$v_1 = \frac{q}{y_1} = \frac{14.0}{0.60} = 23.333 \, \text{m/s}$$

$$Fr_1 = \frac{23.333}{\sqrt{9.81 \times 0.60}} = \frac{23.333}{2.4261} = 9.617$$

$$y_2 = \frac{0.60}{2} \left( \sqrt{1 + 8(9.617)^2} – 1 \right) = 0.30 \times (\sqrt{1 + 740.0} – 1) = 0.30 \times (27.22 – 1) = 7.866 \, \text{ m}$$

Step 2: Compare Required Conjugate Depth to Tailwater
– Required conjugate subcritical depth: $y_2 = 7.87 \, \text{ m}$

• Actual river tailwater: $y_{tw} = 6.20 \, \text{ m}$

• Tailwater Deficit: $\Delta y = y_2 – y_{tw} = 7.87 – 6.20 = 1.67 \, \text{ m}$

Because $y_{tw} < y_2$, the jump cannot anchor at the spillway toe and is repelled downstream onto the riverbed unless a depressed stilling basin or baffle blocks are constructed.

9. Solved Numerical Example 3: Slope Break Transition (Steep to Mild Slope)

Problem Statement

A rectangular channel changes slope from steep ($S_{01} = 0.035$, normal depth $y_{n1} = 0.45 \, \text{ m}$) to mild ($S_{02} = 0.0015$, normal depth $y_{n2} = 2.20 \, \text{ m}$). Discharge is $q = 4.20 \, \text{ m}^2/\text{ s}$.

Determine whether the jump forms on the steep slope upstream of the break or on the mild slope downstream of the break.

                       SLOPE BREAK TOPOLOGY
             Steep Slope (S01 = 0.035)          Mild Slope (S02 = 0.0015)
             yn1 = 0.45 m (Supercritical)       yn2 = 2.20 m (Subcritical)
             ===========================\
                                         \===============================
                                      Slope Break (Station 0)

Step-by-Step Solution

Step 1: Compute Conjugate Depth for Upstream Steep Uniform Flow ($y_{n1} = 0.45 \, \text{ m}$)
$$v_{n1} = \frac{4.20}{0.45} = 9.333 \, \text{m/s}$$

$$Fr_1 = \frac{9.333}{\sqrt{9.81 \times 0.45}} = \frac{9.333}{2.101} = 4.442$$

$$y_{2,req} = \frac{0.45}{2} \left( \sqrt{1 + 8(4.442)^2} – 1 \right) = 0.225 \times (\sqrt{158.85} – 1) = 0.225 \times (12.603 – 1) = 2.61 \, \text{ m}$$

Step 2: Evaluate Slope Break Location Condition
– Required conjugate depth from steep slope: $y_{2,req} = 2.61 \, \text{ m}$

• Actual downstream mild slope normal depth: $y_{n2} = 2.20 \, \text{ m}$

Since $y_{n2} = 2.20 \, \text{ m} < y_{2,req} = 2.61 \, \text{ m}$, the tailwater depth is insufficient to force the jump onto the steep slope. The supercritical flow overshoots the break, creating an $M_3$ profile on the mild slope. Therefore, the jump forms on the mild slope downstream of the slope break.

10. Practical Design Guidelines: USBR Stilling Basin Containment Standards

To guarantee that the jump remains safely positioned within the concrete stilling basin under all operating conditions:

1. Tailwater Safety Margin (USBR Monograph 25): Stilling basins are designed with a conjugate depth safety factor:

$$y_{tw,design} \ge 1.05 \, y_2 \text{ to } 1.10 \, y_2$$

Ensuring the jump remains locked on the apron even during sudden tailwater degradation.

2. Basin Floor Depression: If natural tailwater $y_{tw}$ is deficient, lower the stilling basin floor by depth $d$:

$$d = y_2 – y_{tw} + h_f$$

Where $h_f$ accounts for step expansion losses.

Drag Mechanics and Baffle Block Sizing

11. Location of Hydraulic Jump Engineering Synthesis

Accurately calculating the location of hydraulic jump transitions ensures hydrodynamic equilibrium between upstream supercritical jets and downstream river tailwater. Integrating gradually varied flow profiles with momentum conservation allows civil engineers to anchor energy dissipators reliably, preventing catastrophic foundation undermining across hydraulic infrastructure.

References & Standards Cited

1. US Department of the Interior, Bureau of Reclamation (USBR). (1987). Hydraulic Design of Stilling Basins and Energy Dissipators (Engineering Monograph No. 25). Denver, CO: USBR.
2. US Army Corps of Engineers (USACE). (1993). River Hydraulics (Engineer Manual EM 1110-2-1416). Washington, DC: Department of the Army.
3. Chow, V. T. (1959). Open-Channel Hydraulics. New York: McGraw-Hill.
4. French, R. H. (1985). Open-Channel Hydraulics. New York: McGraw-Hill.
5. Sturm, T. W. (2001). Open Channel Hydraulics. Boston: McGraw-Hill.

Frequently Asked Questions (FAQ)

What primarily determines the spatial location of a hydraulic jump?

The position is determined by the intersection of the upstream supercritical water surface profile (governed by upstream head/gate opening) and the downstream subcritical sequent depth profile (governed by downstream tailwater controls).

What is an $M_3$ water surface curve, and why does a jump frequently terminate it?

An $M_3$ curve occurs in mild-slope channels when actual depth is below critical depth ($y < y_c < y_n$). Flow is supercritical and depth increases downstream until reaching the conjugate depth required to transition into subcritical flow via a hydraulic jump.

How does downstream riverbed degradation affect jump location?

Riverbed erosion or dredging lowers tailwater depth ($y_{tw}$). This causes the jump to migrate downstream (sweep out), exposing unarmored river sections to high-velocity supercritical jets and leading to severe scour.

What engineering measures prevent a jump from sweeping out of a stilling basin?

Engineers construct baffle blocks, chute blocks, solid end sills, or depress the entire stilling basin apron floor below riverbed level to generate extra back-pressure and anchor the jump inside the basin.

Why cannot a hydraulic jump form on a steep slope if tailwater is lower than normal depth?

On steep slopes ($S_0 > S_c$), normal depth is supercritical ($y_n < y_c$). Unless a downstream obstruction (like a dam or high weir) forces subcritical backwater ($S_1$ curve), uniform flow remains supercritical, preventing jump formation.