13:3 Open Channel Flows - Rapidly Varying Flows, Hydraulic Jumps
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Overview
Derek Elsworth closes his open-channel-flow unit by using specific energy to explain how a bed hump can shift flow between subcritical and supercritical states, then briefly introduces hydraulic jumps as rapidly varying transitions with energy loss. He connects the course’s conservation, drag, and Bernoulli principles to natural and engineered examples, and reviews the three exam topics: pipe networks, external-flow terminal velocities, and uniform open-channel flow using the Chézy formula.
Key takeaways
- For a fixed discharge per unit width, specific energy E = y + q²/(2gy²) has a minimum at critical depth, creating shallow-fast and deep-slow flow branches.
- A bed hump reduces available specific energy; if it forces flow through critical conditions, the downstream state can switch from subcritical to supercritical.
- A hydraulic jump is not an energy-conserving transition: energy drops sharply across it, so momentum balance is needed to relate the upstream and downstream states.
- A 0.1-bar pressure deficit produces only about 1 m of water-level rise, so a roughly 3-m hurricane surge requires additional mechanisms such as wind-driven shear.
- The same drag balance used for engineered flows applies to sports: snowboard speed results from downhill gravity balanced by aerodynamic drag, which scales with speed squared in the turbulent regime.
- The course’s final assessment covers three distinct applications: pipe-network flow, laminar-versus-turbulent external-flow terminal velocity, and uniform open-channel flow using Chézy.
Chapters
- Uniform flow has no depth change along the channel; gradually varying flow changes depth slowly, while rapidly varying flow has a free-surface slope on the order of 1.
- Elsworth reviews Bernoulli’s equation using elevation head, water-depth pressure head, and velocity head, with bed slope adding energy and friction or turbulence removing it.
- For constant discharge per unit width, q, specific energy is E = y + q²/(2gy²), combining depth y and velocity head.
- The worked example uses a unit-width discharge of about 5.75 ft²/s and an upstream depth of 2.3 ft to locate the flow on its specific-energy curve.
- A 0.5-ft bed rise reduces the available specific energy; the flow depth adjusts while discharge remains constant.
- The specific-energy curve has a minimum near critical depth; a sufficiently high hump forces the flow toward this minimum and can shift it onto the shallow, fast branch.
- After the hump, the flow can emerge at a shallower depth and higher velocity, changing from subcritical flow (Froude number below 1) to supercritical flow (above 1).
- A hydraulic jump takes shallow, fast supercritical flow and abruptly raises it into deeper, slower subcritical flow.
- Unlike gradually varying profiles, the free surface changes steeply across the jump, with dy/dx on the order of 1.
- The energy grade line drops across the transition, so upstream and downstream energy cannot be treated as equal.
- Because the jump dissipates energy, analysis must account for head loss and balance momentum between upstream and downstream sections.
- Elsworth recaps the progression from fluid properties and hydrostatic pressure to forces on structures, Bernoulli’s principle, and conservation of mass and momentum.
- Dimensional analysis and Buckingham pi provide a systematic way to check that equations combine quantities with compatible units.
- Pipe flow, external flow, and open-channel flow all use the balance between inertia, viscosity, and drag; open channels add the freedom for water depth to vary.
- Continuity links depth and speed: for a fixed discharge, deeper flow is slower and shallower flow is faster.
- Shallow-water wave speed scales as the square root of gravity times water depth; for a tsunami over roughly 1,000 m of offshore water, Elsworth gives an example speed near 100 miles per hour.
- A 0.1-bar atmospheric-pressure drop, from about 1,000 to 900 millibars, corresponds to only about 1 m of water-level rise by pressure balance.
- A storm surge of around 3 m therefore cannot be explained by that pressure difference alone; wind-driven shear can push water onshore and pile it up.
- Mantle convection helps drive plate motion: heated, buoyant material rises, cooler material sinks, and dense subducting slabs contribute slab pull.
- Glaciers deform over their beds; the base is stationary in the simplified profile while surface ice moves, with velocities on the order of metres per year.
- The 1980 Mount St. Helens eruption illustrates pressure overcoming resistance: gas dissolved in magma expands as the pressurized material moves, sustaining the force.
- A Pitot tube estimates airspeed by comparing stagnation pressure with static pressure; Bernoulli’s equation relates the pressure difference to velocity.
- Pump or turbine power can be related to volumetric flow rate, fluid density, gravity, and head; catchment area multiplied by rainfall rate estimates inflow to a dam.
- A tidal barrage impounds water on the incoming tide and releases it on the outgoing tide, potentially generating power in both flow directions.
- Windmills extract energy from moving air, while hydrofoils reduce boat drag by lifting the hull out of the water.
- Textured swimming suits and golf-ball dimples can trigger a turbulent boundary layer that delays separation and reduces drag over some Reynolds-number ranges.
- Spin changes the flow and pressure distribution around a baseball, producing a net aerodynamic force that curves its path.
- A spinning football spreads the effect of lace-related drag asymmetry around the ball, helping it travel more consistently than an unspun ball.
- Sails act as airfoils as well as catching wind, allowing boats to travel at angles close to the wind.
- For a snowboarder or kayaker, speed follows a force balance between downhill gravity and drag, commonly modeled as one-half times drag coefficient, air density, speed squared, and frontal area.
- The pipe-flow test question concerns a network with parallel or series flows, not branching flow or a reservoir problem.
- The external-flow question requires calculating terminal velocity and determining whether flow is laminar or turbulent before selecting the appropriate drag relation.
- The open-channel question uses the Chézy formula for uniform flow; Elsworth also points students to the review video.
Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Derek Elsworth.