What depth do I use in the Froude number for a trapezoidal channel?

The hydraulic depth D = A/T, where A is the cross-sectional area and T is the top width — not the flow depth y. Using y instead of A/T in a non-rectangular section is one of the most common exam traps.

Theory explainer

Critical Depth, Specific Energy & the Froude Number — Explained With Problems

Q: What is critical depth?

Critical depth is the flow depth at which specific energy is minimum for a given discharge. For a rectangular channel it has the closed form yc = (q²/g)^⅓, where q = Q/b is the discharge per unit width. At critical depth the Froude number equals 1.

Q: What is the difference between alternate depths and sequent depths?

Alternate depths are the two depths with the SAME specific energy on one specific-energy curve — one subcritical, one supercritical. Sequent (conjugate) depths are the depths before and after a hydraulic jump; they have different specific energies because the jump dissipates energy. You cannot read sequent depths off the E–y curve.

Q: How do I tell subcritical from supercritical flow?

Compute the Froude number Fr = V/√(gD), where D is the hydraulic depth (A/T). Fr 1 is supercritical (shallow, fast). For a rectangular channel D = y, so Fr = V/√(gy).

Q: What is minimum specific energy at critical depth for a rectangular channel?

Ec = 1.5·yc. This is a handy exam shortcut: once you have yc from (q²/g)^⅓, the minimum specific energy is just three-halves of it.

Q: How is energy loss across a hydraulic jump computed?

Compute the specific energy at each sequent depth, E1 and E2, and take the difference ΔE = E1 − E2. For a rectangular channel there is also the closed form ΔE = (y2 − y1)³/(4·y1·y2). The loss is dissipated as heat and turbulence — it is never recovered.

The three ideas that organize all of open-channel flow: specific energy tells you what energy a depth carries, critical depth is where that energy bottoms out, and the Froude number tells you which side of the minimum you are on. Get these straight and half the open-channel section of the FE and PE exams becomes pattern recognition.

Last reviewed: 2026-10-03. Every worked number below was independently re-solved; steps are shown in full.

Critical depth: where specific energy bottoms out

For a fixed discharge, specific energy E as a function of depth y has a minimum. That minimum is the whole story: differentiate E = y + Q²/(2gA²) with respect to y and set the derivative to zero, and the algebra lands on the critical-flow condition:

Q²T / (gA³) = 1 at critical depth

T	top width of the water surface at depth y
A	cross-sectional area at depth y

In plain steps: E = y + (Q/A)²/2g. The velocity-head term grows as depth shrinks (smaller A → larger V). dE/dy = 0 balances the linear growth of y against the 1/A² decay of the velocity head — and the balance point simplifies to Q²T/(gA³) = 1. For a rectangular channel (T = b, A = b·y, q = Q/b), this collapses to the famous shortcut y_c = (q²/g)^1/3, and substituting back gives E_c = 1.5·y_c.

The specific-energy diagram

Plot E on the horizontal axis and y on the vertical, for a fixed discharge. The curve has two limbs meeting at the critical point — every horizontal slice through the curve cuts it twice, giving the alternate-depth pair. The dashed 45° line is E = y (the limit where the velocity head vanishes, i.e., still water).

Figure: the E–y curve for q = 8 cfs/ft. The horizontal slice at E = 3.0 ft shows the alternate-depth pair from Worked Problem 2 below.

Reading the diagram is an exam skill in itself: above the critical point (upper limb) flow is subcritical — deep, slow, controlled from downstream; below it (lower limb) flow is supercritical — shallow, fast, controlled from upstream. The curve's nose is the minimum-energy point, and no steady flow exists at that discharge with less energy.

The Froude number: which side of the minimum

F_r = V / √(gD) D = A/T

F_r < 1	subcritical — tranquil flow; disturbances travel upstream; y > y_c
F_r = 1	critical — minimum specific energy; the control section
F_r > 1	supercritical — rapid flow; disturbances cannot travel upstream; y < y_c

The Froude number compares inertial forces to gravity forces — the open-channel analogue of the Mach number. Physically, Fr = 1 means a small surface wave travels upstream at exactly the flow speed, so it stands still: that is why control sections (weirs, flumes, channel constrictions) force critical flow.

Worked problem 1 Critical depth in a rectangular channel (FE pace)

Given: a rectangular channel, b = 10 ft, carries Q = 120 cfs. Find the critical depth y_c, the minimum specific energy E_c, and confirm Fr = 1.

Solution:

Discharge per unit width: q = Q/b = 120/10 = 12.0 cfs/ft
y_c = (q²/g)^1/3 = (144/32.2)^1/3 = (4.4720)^1/3 = 1.6475 ft ≈ 1.65 ft
E_c = 1.5 · y_c = 1.5 × 1.6475 = 2.4713 ft ≈ 2.47 ft
Check: V_c = q/y_c = 12/1.6475 = 7.284 ft/s; Fr = V_c/√(g·y_c) = 7.284/√(32.2 × 1.6475) = 7.284/7.284 = 1.000 ✓

Answer: y_c ≈ 1.65 ft, E_c ≈ 2.47 ft, Fr = 1 confirmed. Trap note: the 1.5·y_c shortcut is rectangular-only — for a trapezoidal section you must use the general condition Q²T/(gA³) = 1.

Worked problem 2 Alternate depths from a given specific energy (PE pace)

Given: a rectangular channel with q = 8 cfs/ft and specific energy E = 3.0 ft. Find both alternate depths, classify each, and verify.

Solution — solve the cubic honestly:

E = y + q²/(2g·y²) → y³ − E·y² + q²/(2g) = 0
q²/(2g) = 64/64.4 = 0.9938 → y³ − 3.0y² + 0.9938 = 0
Roots: y = 2.8802 ft, y = 0.6503 ft, y = −0.5305 ft. Discard the negative root — depth cannot be negative.
y = 2.8802 ft: V = 8/2.8802 = 2.778 ft/s; Fr = 2.778/√(32.2 × 2.8802) = 2.778/9.630 = 0.288 → subcritical ✓
y = 0.6503 ft: V = 8/0.6503 = 12.301 ft/s; Fr = 12.301/√(32.2 × 0.6503) = 12.301/4.577 = 2.688 → supercritical ✓
Energy check on both: 2.8802 + (2.778)²/64.4 = 2.8802 + 0.1198 = 3.0000 ✓; 0.6503 + (12.301)²/64.4 = 0.6503 + 2.3497 = 3.0000 ✓

Answer: alternate depths 2.88 ft (subcritical) and 0.65 ft (supercritical). These are the two points marked on the diagram above. Trap note: these are alternate depths — same energy. If the stem instead describes a jump, you need sequent depths (next problem).

Hydraulic jump: sequent depths and the energy they burn

A hydraulic jump is what happens when supercritical flow is forced to become subcritical — the depth snaps up across a short, turbulent roller. The depths on either side are the sequent (conjugate) depths, related by the Bélanger (momentum) equation for a rectangular channel:

y₂/y₁ = ½(−1 + √(1 + 8F_r1²))

ΔE = E₁ − E₂ (rectangular shortcut: ΔE = (y₂ − y₁)³ / (4·y₁·y₂))

y₁, Fr₁	upstream (supercritical) depth and its Froude number — the equation is directional
ΔE	energy dissipated as heat and turbulence — lost, never recovered downstream

Why sequent depths are not on one E–y curve: the jump destroys energy, so E₂ < E₁. The pre-jump depth sits on the E₁ curve's lower limb and the post-jump depth on the E₂ curve's upper limb. Reading y₂ off the E₁ curve gives the alternate depth — a different number and a wrong answer.

Worked problem 3 Sequent depth and jump energy loss (PE pace)

Given: a rectangular channel, q = 10 cfs/ft, upstream depth y₁ = 0.80 ft (supercritical). Find the sequent depth y₂ and the energy loss across the jump.

Solution:

V₁ = q/y₁ = 10/0.80 = 12.50 ft/s
Fr₁ = V₁/√(g·y₁) = 12.50/√(32.2 × 0.80) = 12.50/5.0755 = 2.463 (supercritical, as stated)
y₂ = (y₁/2)(−1 + √(1 + 8·Fr₁²)) = 0.40 × (−1 + √(1 + 8 × 6.0654)) = 0.40 × (−1 + 7.0375) = 2.4150 ft
E₁ = y₁ + V₁²/2g = 0.80 + 156.25/64.4 = 0.80 + 2.4262 = 3.2262 ft
V₂ = q/y₂ = 10/2.4150 = 4.141 ft/s; E₂ = 2.4150 + (4.141)²/64.4 = 2.4150 + 0.2662 = 2.6812 ft
ΔE = E₁ − E₂ = 3.2262 − 2.6812 = 0.5450 ft
Shortcut check: (y₂ − y₁)³/(4·y₁·y₂) = (1.6150)³/(4 × 0.80 × 2.4150) = 4.2134/7.728 = 0.5452 ✓

Answer: y₂ ≈ 2.42 ft, energy loss ≈ 0.545 ft. The shortcut agreeing to the third decimal is your arithmetic check — but on the exam, show the ΔE = E₁ − E₂ route; it is the one that generalizes to non-rectangular sections.

Common exam traps

Alternate vs sequent swap. Same-energy pair = alternate depths (one curve). Jump pair = sequent depths (two curves, energy lost). The stem saying "hydraulic jump" while you read depths off the E–y curve is the trap.
Depth instead of hydraulic depth in Fr. Fr = V/√(gD) with D = A/T. For a trapezoid or triangle, D ≠ y — using y misclassifies the regime.
Forgetting q = Q/b. The rectangular shortcuts (y_c, Bélanger) need discharge per unit width. Plugging total Q into (q²/g)^1/3 is the single most common arithmetic slip on this topic.
The 1.486-style unit slip, open-channel edition: g = 32.2 ft/s² in US units, 9.81 m/s² in SI. A critical depth computed with the wrong g is off by (32.2/9.81)^1/3 ≈ 1.49 — close enough to look plausible, wrong enough to miss.
Sign of the jump loss. ΔE = E₁ − E₂ is positive — energy is dissipated. A negative loss means the depths were swapped or the arithmetic slipped.

Frequently asked questions

What is critical depth?

Critical depth is the flow depth at which specific energy is minimum for a given discharge. For a rectangular channel: y_c = (q²/g)^⅓ with q = Q/b. At critical depth the Froude number equals 1.

What is the difference between alternate depths and sequent depths?

Alternate depths share the same specific energy on one E–y curve (one subcritical, one supercritical). Sequent depths are the pre- and post-jump depths of a hydraulic jump — they have different specific energies because the jump dissipates energy. You cannot read sequent depth off the E–y curve.

How do I tell subcritical from supercritical flow?

Compute Fr = V/√(gD). Fr < 1 is subcritical (deep, slow), Fr = 1 is critical, Fr > 1 is supercritical (shallow, fast). For rectangular channels D = y, so Fr = V/√(gy).

What is the minimum specific energy at critical depth?

For a rectangular channel, E_c = 1.5·y_c. Once you have y_c from (q²/g)^⅓, the minimum specific energy is just three-halves of it — a useful exam shortcut.

What depth goes in the Froude number for a trapezoidal channel?

The hydraulic depth D = A/T (area over top width), not the flow depth y. Using y in a non-rectangular section is one of the most common exam traps on this topic.

How is energy loss across a hydraulic jump computed?

Compute specific energy at each sequent depth and take the difference: ΔE = E₁ − E₂. For a rectangular channel, the shortcut ΔE = (y₂ − y₁)³/(4·y₁·y₂) gives the same answer — useful as an arithmetic check.

E	specific energy — energy per unit weight measured from the channel bed (ft or m)
F_r	Froude number: < 1 subcritical · = 1 critical · > 1 supercritical
D = A/T	hydraulic depth = area / top width (equals y only for rectangular channels)
q = Q/b	discharge per unit width, for the rectangular-channel shortcuts