Hydraulic Jump Length
Estimate a hydraulic jump's length, L ≈ 6.9·(y₂ − y₁), by the classic empirical formula, from the upstream y₁ and downstream y₂ sequent depths. Unlike the sequent depths (from momentum), jump length is empirical, from lab tests, since the jump has no mathematically sharp end — its length is the distance from the upstream face to where the surface stabilizes. Several formulas exist (Smetana ≈ 6(y₂−y₁), USBR vs Fr, Elevatorski ≈ 6.9(y₂−y₁)); all give the order of magnitude. Jump length sets the stilling basin size downstream of a spillway: the basin must be long enough to contain the whole jump so dissipation completes within the concrete-lined structure before water returns to the natural bed. Undersizing throws the still-erosive jump tail onto the unprotected bed. Enter the upstream and downstream sequent depths.
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Hydraulic jump length
How long is a hydraulic jump? A classic estimate is L ≈ 6.9·(y₂ − y₁), from the conjugate depths upstream y₁ and downstream y₂. Unlike the conjugate depths themselves (which follow from momentum theory), the length of the jump is empirical, taken from laboratory tests — the jump has no mathematically sharp end, so its length gets defined as the distance from the upstream face to the point where the water surface levels off. Several formulas exist (Smetana ≈ 6(y₂−y₁), USBR as a function of the Froude number, Elevatorski ≈ 6.9(y₂−y₁)), all of the same order of magnitude. This length is what sets the size of the stilling basin downstream of a spillway: the basin has to be long enough to contain the whole jump, making sure that energy dissipation finishes inside the concrete-lined structure built to resist erosion, before the flow returns to the natural channel. Undersizing the basin is a dangerous and costly mistake: it throws the tail of the jump — still turbulent and erosive — onto the unprotected bed downstream, scouring a hole that can grow until it undermines the structure. Impact blocks and end sills help shorten the jump and hold it in place, allowing shorter basins. Enter the conjugate depths upstream and downstream.
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Hydraulic Jump Energy Loss
Calculate the specific energy dissipated in a hydraulic jump, ΔE = (y₂ − y₁)³ ÷ (4·y₁·y₂), from the upstream y₁ (supercritical) and downstream y₂ (subcritical) sequent depths. The hydraulic jump is one of the most efficient energy dissipators in hydraulics: intense turbulence in the transition converts kinetic energy to heat and sound, removing excess flow energy. This head loss ΔE is exactly what is sought downstream of spillways, gates and bottom outlets — water arrives with very high energy (able to scour the riverbed and undermine the structure), and the stilling basin induces the jump to 'burn' that energy in a controlled way. The higher the incoming Froude number, the greater the dissipated fraction — jumps with Fr > 9 dissipate up to 85%. Enter the upstream and downstream sequent depths.
Hydraulic Jump Sequent Depth
Calculate the sequent (conjugate) depth downstream of a hydraulic jump, y₂ = (y₁/2)·(√(1 + 8·Fr₁²) − 1), from the upstream depth y₁ (supercritical) and the incoming Froude number Fr₁. The hydraulic jump is the abrupt transition from fast, shallow (supercritical) to slow, deep (subcritical) flow, with strong turbulence and energy dissipation. This Bélanger equation, from momentum conservation, is the basis for designing stilling basins downstream of spillways and gates: water descending a spillway arrives at very high (supercritical) velocity and must be decelerated before returning to the river, otherwise it erodes the bed catastrophically. The sequent depth y₂ sets the required basin depth for a stable jump. Enter the upstream depth and the Froude number.
Spillway Discharge (Creager/Ogee)
Calculate the discharge over a Creager/ogee dam spillway, Q = C·L·H^1.5, from the discharge coefficient C (typically 2.0-2.2 in SI for well-designed ogee profiles), the crest length L (m) and the head over the crest H (m). The spillway is a dam's most critical safety structure: it releases floods safely, preventing overtopping — the leading cause of dam failure. The ogee profile follows the shape of the underside of a free nappe, maximizing discharge while keeping crest pressure near atmospheric (avoiding cavitation). The coefficient C absorbs gravity and approach effects, exceeding that of a sharp-crested weir. Spillway design starts from the design flood (often the 10,000-year flood or the PMF) and sets the required crest length. Enter the discharge coefficient, crest length and head.
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Calculate the horizontal hydrostatic thrust per metre of length on a dam face, E = ½·γ·H², from the unit weight of water γ (≈ 9.81 kN/m³) and the water depth H (m) at the upstream face. Since hydrostatic pressure grows linearly with depth (p = γ·h), its diagram is triangular and the resultant is its area, ½·γ·H², applied at one third of the height from the base. This thrust is the main action tending to overturn and slide the dam, and the starting point of gravity dam stability analysis: it generates the overturning moment (about the downstream toe) and the horizontal force resisted by base friction. Dam stability depends on its self-weight (providing the stabilizing moment and normal friction force) exceeding these with adequate margin, also accounting for foundation uplift. Enter the unit weight of water and the depth.
Hydraulic Retention Time (HRT)
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Anchor Rode Length (Scope)
Compute the length of anchor rode (chain or line) to pay out, L = scope·(depth + bow roller height), from the scope ratio (typically 5:1 to 7:1), the water depth and the height of the attachment point above the water. A proper ratio makes the rode pull the anchor horizontally, ensuring it sets and the anchored boat stays safe. Enter the depth, the bow roller height and the scope ratio.
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