Specific Energy & Critical Flow — Hydraulic Jump

1. Specific Energy — Definition & Equation

Specific energy (E) at a channel section is the total mechanical energy of the flow per unit weight of fluid, measured above the channel bed (not above an arbitrary datum as in the total energy head).

1.1 Specific Energy vs Total Energy Head

2. Specific Energy Curve

For a fixed discharge Q (or unit discharge q in a rectangular channel), specific energy E is a function of depth y only. Plotting E vs y produces the specific energy curve — one of the most important diagrams in open channel hydraulics.

2.1 Shape and Features of the E–y Curve

From E = y + q²/(2gy²):

• As y → 0 (very shallow): V → ∞ → E → ∞ (velocity head dominates; curve approaches the E-axis asymptotically)
• As y → ∞ (very deep): V → 0 → E → y (depth head dominates; curve approaches the 45° line E = y asymptotically)
• Between these extremes: E has a minimum value E_min at the critical depth y_c

2.2 Two Branches of the E–y Curve

2.3 Effect of Increasing Discharge on the E–y Curve

If Q increases (while E stays fixed), the entire curve shifts to the right — the critical point moves up and to the right, meaning y_c increases and E_min increases. For a given specific energy E, increasing Q means the two possible flow depths (alternate depths) move closer together, meeting at the critical point when Q = Q_max for that E.

3. Critical Flow Conditions

3.1 Critical Flow — General Condition

3.2 Critical Flow — Rectangular Channel Summary

3.3 Critical Slope S_c

The critical slope is the bed slope at which the normal depth equals the critical depth — the channel just sustains critical flow uniformly.

4. Alternate Depths

For a given specific energy E and discharge Q, there are in general two possible flow depths on the specific energy curve — one on the subcritical branch (y > y_c) and one on the supercritical branch (y < y_c). These are called alternate depths.

Key distinction — Alternate depths vs Conjugate (sequent) depths:

5. Channel Transitions — Humps & Width Changes

5.1 Flow Over a Hump (Raised Bed)

When the channel bed is raised by a height Δz (a smooth hump), specific energy at the hump is reduced:

5.2 Flow Through a Channel Contraction (Width Reduction)

When the channel width decreases from b₁ to b₂, for a fixed discharge Q, the unit discharge increases (q₂ = Q/b₂ > Q/b₁ = q₁). The minimum specific energy required increases. If the available specific energy is insufficient, the flow is choked and the upstream depth increases.

6. Hydraulic Jump — Introduction

A hydraulic jump is a rapidly-varied flow phenomenon in which a supercritical flow (high velocity, shallow depth) abruptly transitions to a subcritical flow (low velocity, large depth). It is characterised by intense turbulence, surface rollers, air entrainment, and significant energy dissipation. The hydraulic jump is one of the most practically important concepts in hydraulics engineering.

6.1 Where Hydraulic Jumps Occur in Practice

• Downstream of sluice gates: High-velocity jet below the gate transitions to subcritical channel flow.
• At the toe of spillways and chutes: Supercritical flow down the spillway face must be converted to subcritical before entering the river — stilling basins are designed around the hydraulic jump.
• At slope transitions: Where a steep channel (supercritical flow) meets a mild channel (subcritical flow), a jump forms at the transition.
• Downstream of overflow weirs: The nappe (overfall jet) lands as supercritical flow and may form a jump.

6.2 Why Bernoulli Cannot Be Used for the Hydraulic Jump

The hydraulic jump is a region of intense energy dissipation — kinetic energy is converted to heat and sound by turbulent mixing. Because there are significant unknown head losses, Bernoulli’s energy equation has two unknowns (the downstream depth and the head loss) and cannot be solved alone. Instead, the momentum equation is applied across the jump, where the only external forces are pressure forces (no unknown friction forces, since the jump is assumed to occur over a short distance).

7. Hydraulic Jump — Momentum Equation Derivation

Consider a horizontal rectangular channel of width b. A hydraulic jump occurs between sections 1 (supercritical) and 2 (subcritical). Apply the momentum equation (impulse-momentum theorem) in the flow direction:

7.1 Sequent Depth Ratio — Key Formula

7.2 Momentum Function (Specific Force)

8. Energy Dissipated in Hydraulic Jump

8.1 Height of Jump

9. Classification of Hydraulic Jumps

Practical significance: For stilling basin design below dam spillways (IS 7114), the steady jump (Fr₁ = 4.5–9.0) is preferred because it is stable and dissipates significant energy without the unpredictability of the oscillating jump. Chute blocks, sill blocks, and end sills are added to stabilise the jump and shorten the stilling basin length.

10. Location of Hydraulic Jump

In a real channel, the hydraulic jump does not always form at a fixed point — its location depends on the upstream and downstream control conditions and the channel profile.

10.1 Jump Below a Sluice Gate

A sluice gate on a mild slope produces an M3 GVF profile downstream (supercritical flow building toward normal depth). The jump forms where the M3 profile depth equals the conjugate depth of the subcritical normal depth downstream. If the tailwater depth (y_tw) equals y₂ (sequent depth for y₁ at that location), the jump forms at that exact point.

11. Worked Examples (GATE CE Level)

Example 1 — Sequent Depth and Energy Loss in Hydraulic Jump (GATE CE 2022 type)

Problem: A hydraulic jump occurs in a rectangular channel 4 m wide. The depth before the jump is 0.4 m and the discharge is 12 m³/s. Find (a) the Froude number before the jump, (b) the sequent depth, (c) the energy dissipated, and (d) the power dissipated.

Answer: Fr₁ = 3.79; y₂ = 1.95 m; ΔE = 1.195 m; P = 140.6 kW

Example 2 — Critical Depth and Minimum Specific Energy (GATE CE 2019 type)

Problem: A rectangular channel 2.5 m wide carries a discharge of 7.5 m³/s with a depth of 0.8 m. Determine (a) whether the flow is subcritical or supercritical, (b) the critical depth, (c) the minimum specific energy, and (d) the specific energy at the current depth.

Answer: Supercritical (Fr = 1.34); y_c = 0.972 m; E_min = 1.458 m; E = 1.517 m

Example 3 — Flow Over a Hump (GATE CE type)

Problem: A rectangular channel 3 m wide carries 9 m³/s at a depth of 1.5 m (subcritical). A smooth hump 0.3 m high is placed in the channel bed. Find the depth of flow over the hump and determine if choking occurs.

Answer: Flow is choked (Δz = 0.30 m > Δz_max = 0.246 m); depth over hump = y_c = 0.972 m; upstream depth rises to ≈ 1.57 m.

Example 4 — Finding Alternate Depths (GATE MCQ type)

Problem: A rectangular channel has a unit discharge q = 2.4 m²/s and a specific energy E = 1.8 m. Find the two alternate depths (subcritical and supercritical).

Answer: Subcritical alternate depth ≈ 1.70 m; Supercritical alternate depth ≈ 0.47 m

Example 5 — Jump from Fr₁ to Classify Type (GATE CE 2021 type)

Problem: Supercritical flow in a 5 m wide rectangular channel has depth 0.5 m and velocity 10 m/s. A hydraulic jump is forced by a downstream obstacle. Find (a) Fr₁, (b) sequent depth y₂, (c) energy dissipated, and (d) classify the jump.

Answer: Fr₁ = 4.52; y₂ = 2.94 m; ΔE = 2.48 m (44.3% energy loss); Steady jump.

Example 6 — Critical Slope Calculation (GATE CE type)

Problem: A wide rectangular channel carries 4 m³/s per metre width with Manning’s n = 0.015. Find the critical slope S_c.

Answer: S_c ≈ 0.00209 (1 in 478)

12. Common Mistakes

Mistake 1 — Confusing Alternate Depths with Conjugate (Sequent) Depths

Error: Using the sequent depth formula y₂/y₁ = ½[√(1+8Fr₁²)–1] to find the alternate depth, or vice versa.
Root Cause: Both pairs involve two depths related to the same discharge; both can be presented as “depth before and after” something. The critical distinction: alternate depths share the same specific energy (related by the E–y curve); conjugate depths share the same momentum function (related by the momentum equation across a hydraulic jump).
Fix: Alternate depth problem → use E₁ = E₂ (energy conservation, no loss). Hydraulic jump problem → use M₁ = M₂ (momentum conservation) → sequent depth ratio formula.

Mistake 2 — Applying the Hydraulic Jump Formula with Fr₂ (Downstream Froude Number) Instead of Fr₁

Error: Writing y₁/y₂ = ½[√(1+8Fr₁²)–1] (getting y₁/y₂ instead of y₂/y₁).
Root Cause: The formula y₂/y₁ = ½[√(1+8Fr₁²)–1] gives the ratio of DOWNSTREAM to UPSTREAM depth in terms of the UPSTREAM Froude number Fr₁. Mixing up subscripts inverts the ratio.
Fix: Always label clearly: subscript 1 = supercritical (upstream of jump), subscript 2 = subcritical (downstream). The result y₂/y₁ must be > 1 always (sequent depth is deeper than initial depth). If you get y₂/y₁ < 1, you’ve used the wrong subscript.

Mistake 3 — Using Bernoulli Instead of the Momentum Equation for Hydraulic Jump

Error: Applying E₁ = E₂ (no head loss) across a hydraulic jump to find the downstream depth.
Root Cause: Forgetting that a hydraulic jump is a highly dissipative phenomenon — energy is not conserved. Bernoulli gives E₁ = E₂, which means the downstream depth is the alternate depth (same energy). But the hydraulic jump gives the conjugate depth (same momentum), which is different.
Fix: For a hydraulic jump, energy is NOT conserved. Use the momentum equation → sequent depth formula. Energy loss ΔE = (y₂–y₁)³/(4y₁y₂) is calculated AFTER finding y₂ from the momentum equation.

Mistake 4 — Forgetting (3/2) in the Minimum Specific Energy Formula

Error: Writing E_min = y_c (instead of E_min = 1.5y_c) or E_min = 2y_c.
Root Cause: Rushing through derivation. At critical depth: E_min = y_c + V_c²/(2g) = y_c + y_c/2 = (3/2)y_c. The velocity head at critical depth = y_c/2 — exactly half the depth.
Fix: Derive once: at critical flow, q² = gy_c³ → V_c² = gy_c → V_c²/(2g) = y_c/2 → E_min = y_c + y_c/2 = (3/2)y_c. Also: y_c = (2/3)E_min.

Mistake 5 — Sign Error in Energy Dissipation Formula

Error: Computing ΔE = (y₁ – y₂)³/(4y₁y₂) — using (y₁ – y₂)³ with y₁ < y₂, giving a negative energy loss.
Root Cause: Since y₂ > y₁ in a hydraulic jump, (y₁ – y₂) is negative and (y₁ – y₂)³ is negative, giving ΔE < 0 (energy gained) — physically impossible.
Fix: The correct formula uses (y₂ – y₁)³ which is always positive: ΔE = (y₂ – y₁)³/(4y₁y₂) > 0. Energy is always dissipated in a hydraulic jump (second law of thermodynamics). A positive ΔE = E₁ – E₂ > 0 confirms energy is lost going from 1 (supercritical) to 2 (subcritical).

13. Frequently Asked Questions

Q1. Why is the hydraulic jump analysed using the momentum equation and not the energy equation?

The hydraulic jump involves intense turbulent dissipation — kinetic energy is converted to internal (thermal) energy through viscous dissipation in the rollers and surface agitation. This means total mechanical energy is not conserved across the jump (E₁ ≠ E₂), so the Bernoulli/energy equation cannot be directly applied to find the downstream depth. The momentum equation, however, applies without needing to know the losses — it only requires the external forces (hydrostatic pressure forces at the two sections) and the rate of momentum change, both of which can be computed from known conditions. The key assumption is that the jump occurs over a short enough distance that the friction force on the channel walls is negligible compared to the hydrostatic forces. The energy loss ΔE is not an input to the momentum equation — it is a result that is computed after finding the sequent depth y₂ using momentum. This two-step approach (momentum first, then energy) is a fundamental strategy in hydraulics for flows with unknown dissipation.

Q2. What is a stilling basin, and how does the hydraulic jump concept govern its design?

A stilling basin is a purpose-built structure at the downstream end of a dam spillway, chute, or sluice that forces a hydraulic jump to occur within a controlled location, dissipating the excess kinetic energy of the high-velocity flow before it reaches the unlined riverbed. Without a stilling basin, the supercritical flow would continue downstream, potentially causing severe scour of the river channel, undermining the dam foundations, or damaging downstream infrastructure. The design of a stilling basin (IS 7114 for bucket-type; USBR types I–VI) is based on computing: (a) the incoming Froude number Fr₁ from the spillway velocity and tail depth; (b) the required sequent depth y₂ from the momentum equation; (c) ensuring the tailwater depth in the river equals y₂ (if tailwater is too low, the jump is swept out of the basin — dangerous; if too high, the jump is drowned and no longer dissipates energy efficiently). The basin length is set at 5–6y₂ for steady jumps. Chute blocks, floor blocks (baffle piers), and end sills are added to stabilise the jump, increase energy dissipation per unit length, and reduce basin length.

Q3. Can a hydraulic jump occur in a sloping (non-horizontal) channel?

Yes — hydraulic jumps occur in sloping channels, but the analysis is more complex because the weight component of the fluid in the jump region must be included in the momentum equation. For a channel of bed slope S₀ and jump length L_j, the weight component along the slope is W sinS₀ ≈ ρgAL_jS₀. For mild slopes (S₀ < 0.01 or 1%), this term is small (< 1% of hydrostatic force for typical jumps) and is usually neglected in GATE-level calculations — the horizontal channel formula is used as an approximation. For steep slopes, the weight component is significant. In practice, dam spillways have steep slopes, but the jump is designed to occur in a horizontal stilling basin at the toe where the correction is unnecessary. The GATE CE syllabus treats hydraulic jumps as occurring in horizontal or mildly-sloped channels, so the standard formula y₂/y₁ = ½(√(1+8Fr₁²)–1) applies directly.

Q4. What is the significance of the specific energy concept in irrigation canal design?

The specific energy concept is fundamental to the design of canal falls (drops), transitions, and cross-regulators in irrigation systems. When an irrigation canal needs to descend from a higher elevation to a lower one (due to terrain), a canal fall (IS 6966) is constructed. The specific energy diagram helps determine: (a) whether the flow upstream of the fall is subcritical (M1 profile) or supercritical; (b) the depth-discharge relationship at the fall crest; (c) whether a hydraulic jump will form downstream and, if so, the sequent depth and energy loss. In channel transitions (width changes), the specific energy diagram shows whether the transition will cause choking (critical flow forced at the throat) — which may be intentional (to act as a discharge measuring device) or unintentional (causing backwater and overtopping of canal banks upstream). The concept of minimum specific energy as a physical limit on flow is also used in the design of culverts under roads: if the culvert is too narrow or the approach flow too fast, critical flow at the inlet limits the discharge capacity.