Specific Energy

Hydraulics · FE Reference Handbook section

Hydraulics

21 formulas

10 exam-style examples

~60 min

All Hydraulics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Specific Energy within Hydraulics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what specific energy describes physically and when it applies.
State every one of the 21 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: Manning's n is unitless but the constant is 1.486 (US) or 1.0 (SI).

Lecture

Why this section exists. Specific Energy is the part of Hydraulics that lets you connect a channel, culvert or control structure to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as open-channel normal depth, specific energy, or a weir discharge. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. Manning's n is unitless but the constant is 1.486 (US) or 1.0 (SI). Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Row of centrifugal pumps and valved steel piping inside a water pumping station. — Photo 1. Where this shows up in practice: specific energy.
Capstone Studio instructional photograph

Hydraulics — Specific Energy: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a channel, culvert or control structure. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 21 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

V2 αQ	Quantity produced by "V2 αQ" — read its definition and unit from the handbook line directly above the equation.
E	Quantity produced by "E = α 2g + y = +y" — read its definition and unit from the handbook line directly above the equation.
Q	Quantity produced by "Q = discharge" — read its definition and unit from the handbook line directly above the equation.
V	Quantity produced by "V = velocity" — read its definition and unit from the handbook line directly above the equation.
y	Quantity produced by "y = depth of flow" — read its definition and unit from the handbook line directly above the equation.
A	Quantity produced by "A = cross-sectional area of flow" — read its definition and unit from the handbook line directly above the equation.
α	Quantity produced by "α = kinetic energy correction factor, usually 1.0" — read its definition and unit from the handbook line directly above the equation.
Critical Depth	Quantity produced by "Critical Depth = that depth in a channel at minimum specific energy" — read its definition and unit from the handbook line directly above the equation.
g	Quantity produced by "g = T" — read its definition and unit from the handbook line directly above the equation.
T	Quantity produced by "T = width of the water surface" — read its definition and unit from the handbook line directly above the equation.
yc	Quantity produced by "yc = e g o" — read its definition and unit from the handbook line directly above the equation.
q	Quantity produced by "q = unit discharge = B" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

2gA 2
where
Q2 A3
where Q and A are as defined above,
For rectangular channels
2 1 3
where
V Q 2T
gyh gA3
Supercritical flow: Fr > 1
Subcritical flow: Fr < 1

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Manning capacity of a rectangular channel

A 3.0 m wide rectangular channel flows 1.2 m deep on a 0.0016 slope with n = 0.015. Find the discharge.

Given

b = 3.0 m, y = 1.2 m
S = 0.0016
n = 0.015

Find

Q (m³/s)

Start with the thinking

Hydraulic radius uses the wetted perimeter — two walls plus the bed.
Metric Manning uses the 1.00 coefficient, not 1.49.

Step-by-step solution

Area
$A = b y = 3.0 (1.2) = 3.60 m^{2}$
Wetted perimeter
$P = b + 2 y = 3.0 + 2.4 = 5.40 m$
Hydraulic radius
$R = A / P = 3.60/5.40 = 0.667 m$
Manning
$Q = (1/ n) A R^{2/3} S^{1/2}$
Terms
$R^{2/3} = 0.763 an d S^{1/2} = 0.0400$
Substitute
$Q = (1/0.015) (3.60) (0.763) (0.0400)$
Result
$Q = 7.32 m^{3} / s$

Answer: Q ≈ 7.32 m³/s

Why the other options are there

10.9 m³/s (1.49 coefficient used in SI)
5.49 m³/s (R taken as the depth)

Reference: FE Reference Handbook — Hydraulics — Manning's equation

Example 2

Critical depth and flow regime

The same 3.0 m channel carries 7.32 m³/s. Find the critical depth and state whether the 1.2 m normal depth is subcritical or supercritical.

Given

q = Q/b = 7.32/3.0 = 2.44 m²/s
g = 9.81 m/s²
y_n = 1.2 m

Find

y_c and the regime

Start with the thinking

For a rectangle y_c = (q²/g)^{1/3}.
y_n > y_c means subcritical, mild slope.

Step-by-step solution

Unit discharge
$q = 2.44 m^{2} / s$
Ratio
$q^{2} / g = 5.95/9.81 = 0.607$
Critical depth
$y_{c} = 0.60 7^{1/3} = 0.847 m$
Compare
$y_{n} = 1.2 m > y_{c} = 0.847 m$
Conclusion — subcritical flow on a mild slope; control is downstream

Answer: y_c = 0.847 m; flow is subcritical

Why the other options are there

y_c = 0.607 m (cube root omitted)
Supercritical (comparison reversed)

Reference: FE Reference Handbook — Hydraulics — Critical flow

Example 3

Froude number and critical depth — Specific Energy

A wide channel carries unit discharge q = 22.0 cfs/ft at a depth of 3.00 ft. Find the Froude number and critical depth.

Given

q = 22.0 cfs/ft
y = 3.00 ft

Find

Fr, regime and y_c

Start with the thinking

Fr > 1 is supercritical (fast, shallow).
Critical depth for a wide channel is (q²/g)^(1/3).

Step-by-step solution

Velocity
$V = q / y = 22.0/3.00 = 7.33 f t / s$
Froude
$F r = V / (g y)$
Substituting
$F r = 7.33/ (32.2 \times 3.00) = 0.746$
Regime — subcritical
Critical depth
$y_{c} = (q^{2} / g)^{(} 1/3)$
Substituting
$y_{c} = ((22.0)^{2} /32.2)^{(} 1/3) = 2.468 f t$

Answer: Fr ≈ 0.75 (subcritical); y_c ≈ 2.47 ft

Why the other options are there

Fr = 0.076 (square root omitted)
y_c = 15.03 ft (cube root omitted)

Reference: FE Reference Handbook — Hydraulics → Specific Energy

Example 4

Specific energy and minimum energy in a rectangular channel — Specific Energy

A wide channel carries a unit discharge of 23.5 cfs/ft at a depth of 5.75 ft. Compute the specific energy, the critical depth and the minimum specific energy.

Given

q = 23.5 cfs/ft
y = 5.75 ft

Find

E, y_c and E_min

Start with the thinking

Specific energy is measured from the channel bottom: E = y + V²/2g.
Minimum specific energy occurs at critical depth and equals 1.5 y_c for a rectangular section.

Step-by-step solution

Velocity
$V = q / y = 23.5/5.75 = 4.09 f t / s$
Specific energy
$E = y + V^{2} / (2 g)$
Substituting
$E = 5.75 + (4.09)^{2} /64.4 = 6.009 f t$
Critical depth
$y_{c} = (q^{2} / g)^{(} 1/3)$
Substituting
$y_{c} = ((23.5)^{2} /32.2)^{(} 1/3) = 2.579 f t$
Minimum energy
$E_{m} in = 1.5 y_{c}$
Substituting
$E_{m} in = 1.5 \times 2.579 = 3.868 f t$
Regime — the flow is subcritical because y > y_c

Answer: E ≈ 6.01 ft; y_c ≈ 2.58 ft; E_min ≈ 3.87 ft

Why the other options are there

22.45 ft (2g omitted)
2.58 ft reported as E_min (1.5 factor dropped)

Reference: FE Reference Handbook — Hydraulics → Specific Energy

Example 5

Hydraulic jump: sequent depth and energy dissipated — Specific Energy

A rectangular channel carries a unit discharge of 4.5 m³/s per metre at a supercritical depth of 0.55 m. Compute the upstream Froude number, the sequent depth, and the energy dissipated in the jump.

Given

q = 4.5 m³/s·m
y₁ = 0.55 m

Find

Fr₁, y₂ and the head loss across the jump

Start with the thinking

A jump only forms when the approach flow is supercritical (Fr > 1).
The sequent-depth equation comes from momentum, while the loss comes from the energy equation.

Step-by-step solution

Velocity
$v_{1} = q / y_{1} = 4.5/0.55 = 8.182 m / s$
Formula
$F r_{1} = v_{1} / (g y_{1})$
Substituting
$F r_{1} = 8.182/ (9.81 \times 0.55) = 3.522$
Formula
$y_{2} = (y_{1} /2) [(1 + 8 F r_{1}^{2}) - 1]$
Substituting
$y_{2} = (0.55/2) [(1 + 8 \times 3.52 2^{2}) - 1] = 2.479 m$
Energies
$E_{1} = 0.55 + 3.412 = 3.962 m; E_{2} = 2.479 + 0.168 = 2.647 m$
Loss — ΔE = 3.962 − 2.647 = 1.315 m

Answer: Fr₁ = 3.52, y₂ = 2.48 m, ΔE = 1.32 m

Why the other options are there

y₂ = 1.94 m (depth simply scaled by Froude number)
ΔE = 1.93 m (depth change reported as energy loss)

Reference: FE Reference Handbook — Hydraulics → Specific Energy

Example 6

Froude number and critical depth — Specific Energy (2)

A wide channel carries unit discharge q = 37.0 cfs/ft at a depth of 4.80 ft. Find the Froude number and critical depth.

Given

q = 37.0 cfs/ft
y = 4.80 ft

Find

Fr, regime and y_c

Start with the thinking

Fr > 1 is supercritical (fast, shallow).
Critical depth for a wide channel is (q²/g)^(1/3).

Step-by-step solution

Velocity
$V = q / y = 37.0/4.80 = 7.71 f t / s$
Froude
$F r = V / (g y)$
Substituting
$F r = 7.71/ (32.2 \times 4.80) = 0.620$
Regime — subcritical
Critical depth
$y_{c} = (q^{2} / g)^{(} 1/3)$
Substituting
$y_{c} = ((37.0)^{2} /32.2)^{(} 1/3) = 3.490 f t$

Answer: Fr ≈ 0.62 (subcritical); y_c ≈ 3.49 ft

Why the other options are there

Fr = 0.050 (square root omitted)
y_c = 42.52 ft (cube root omitted)

Reference: FE Reference Handbook — Hydraulics → Specific Energy

Example 7

Specific energy and minimum energy in a rectangular channel — Specific Energy (2)

A wide channel carries a unit discharge of 11.5 cfs/ft at a depth of 2.75 ft. Compute the specific energy, the critical depth and the minimum specific energy.

Given

q = 11.5 cfs/ft
y = 2.75 ft

Find

E, y_c and E_min

Start with the thinking

Specific energy is measured from the channel bottom: E = y + V²/2g.
Minimum specific energy occurs at critical depth and equals 1.5 y_c for a rectangular section.

Step-by-step solution

Velocity
$V = q / y = 11.5/2.75 = 4.18 f t / s$
Specific energy
$E = y + V^{2} / (2 g)$
Substituting
$E = 2.75 + (4.18)^{2} /64.4 = 3.022 f t$
Critical depth
$y_{c} = (q^{2} / g)^{(} 1/3)$
Substituting
$y_{c} = ((11.5)^{2} /32.2)^{(} 1/3) = 1.601 f t$
Minimum energy
$E_{m} in = 1.5 y_{c}$
Substituting
$E_{m} in = 1.5 \times 1.601 = 2.402 f t$
Regime — the flow is subcritical because y > y_c

Answer: E ≈ 3.02 ft; y_c ≈ 1.60 ft; E_min ≈ 2.40 ft

Why the other options are there

20.24 ft (2g omitted)
1.60 ft reported as E_min (1.5 factor dropped)

Reference: FE Reference Handbook — Hydraulics → Specific Energy

Example 8

Hydraulic jump: sequent depth and energy dissipated — Specific Energy (2)

A rectangular channel carries a unit discharge of 3.0 m³/s per metre at a supercritical depth of 0.25 m. Compute the upstream Froude number, the sequent depth, and the energy dissipated in the jump.

Given

q = 3.0 m³/s·m
y₁ = 0.25 m

Find

Fr₁, y₂ and the head loss across the jump

Start with the thinking

A jump only forms when the approach flow is supercritical (Fr > 1).
The sequent-depth equation comes from momentum, while the loss comes from the energy equation.

Step-by-step solution

Velocity
$v_{1} = q / y_{1} = 3.0/0.25 = 12.000 m / s$
Formula
$F r_{1} = v_{1} / (g y_{1})$
Substituting
$F r_{1} = 12.000/ (9.81 \times 0.25) = 7.663$
Formula
$y_{2} = (y_{1} /2) [(1 + 8 F r_{1}^{2}) - 1]$
Substituting
$y_{2} = (0.25/2) [(1 + 8 \times 7.66 3^{2}) - 1] = 2.587 m$
Energies
$E_{1} = 0.25 + 7.339 = 7.589 m; E_{2} = 2.587 + 0.069 = 2.656 m$
Loss — ΔE = 7.589 − 2.656 = 4.934 m

Answer: Fr₁ = 7.66, y₂ = 2.59 m, ΔE = 4.93 m

Why the other options are there

y₂ = 1.92 m (depth simply scaled by Froude number)
ΔE = 2.34 m (depth change reported as energy loss)

Reference: FE Reference Handbook — Hydraulics → Specific Energy

Example 9

Froude number and critical depth — Specific Energy (3)

A wide channel carries unit discharge q = 17.5 cfs/ft at a depth of 5.00 ft. Find the Froude number and critical depth.

Given

q = 17.5 cfs/ft
y = 5.00 ft

Find

Fr, regime and y_c

Start with the thinking

Fr > 1 is supercritical (fast, shallow).
Critical depth for a wide channel is (q²/g)^(1/3).

Step-by-step solution

Velocity
$V = q / y = 17.5/5.00 = 3.50 f t / s$
Froude
$F r = V / (g y)$
Substituting
$F r = 3.50/ (32.2 \times 5.00) = 0.276$
Regime — subcritical
Critical depth
$y_{c} = (q^{2} / g)^{(} 1/3)$
Substituting
$y_{c} = ((17.5)^{2} /32.2)^{(} 1/3) = 2.119 f t$

Answer: Fr ≈ 0.28 (subcritical); y_c ≈ 2.12 ft

Why the other options are there

Fr = 0.022 (square root omitted)
y_c = 9.51 ft (cube root omitted)

Reference: FE Reference Handbook — Hydraulics → Specific Energy

Example 10

Specific energy and minimum energy in a rectangular channel — Specific Energy (3)

A wide channel carries a unit discharge of 18.0 cfs/ft at a depth of 6.75 ft. Compute the specific energy, the critical depth and the minimum specific energy.

Given

q = 18.0 cfs/ft
y = 6.75 ft

Find

E, y_c and E_min

Start with the thinking

Specific energy is measured from the channel bottom: E = y + V²/2g.
Minimum specific energy occurs at critical depth and equals 1.5 y_c for a rectangular section.

Step-by-step solution

Velocity
$V = q / y = 18.0/6.75 = 2.67 f t / s$
Specific energy
$E = y + V^{2} / (2 g)$
Substituting
$E = 6.75 + (2.67)^{2} /64.4 = 6.860 f t$
Critical depth
$y_{c} = (q^{2} / g)^{(} 1/3)$
Substituting
$y_{c} = ((18.0)^{2} /32.2)^{(} 1/3) = 2.159 f t$
Minimum energy
$E_{m} in = 1.5 y_{c}$
Substituting
$E_{m} in = 1.5 \times 2.159 = 3.238 f t$
Regime — the flow is subcritical because y > y_c

Answer: E ≈ 6.86 ft; y_c ≈ 2.16 ft; E_min ≈ 3.24 ft

Why the other options are there

13.86 ft (2g omitted)
2.16 ft reported as E_min (1.5 factor dropped)

Reference: FE Reference Handbook — Hydraulics → Specific Energy

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a channel, culvert or control structure, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Specific Energy contains 21 relations; you must be able to find this page in under 15 seconds.
Exam style: open-channel normal depth, specific energy, or a weir discharge.
Unit rule: Manning's n is unitless but the constant is 1.486 (US) or 1.0 (SI).
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

Manning's n is unitless but the constant is 1.486 (US) or 1.0 (SI)
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.

Specific Energy

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

V2 αQNeeds technical review

E=α2g+y=+y

E=specificenergy

Q=discharge

V=velocity

y=depthofflow

A=cross−sectionalareaofflow

α=kineticenergycorrectionfactor,usually1.0

CriticalDepth=thatdepthinachannelatminimumspecificenergy

g=T

g=accelerationduetogravity

T=widthofthewatersurface

yc=ego

yc=criticaldepth

q=unitdischarge=B

B=channelwidth

g=accelerationduetogravity

FroudeNumber=ratioofinertialforcestogravityforces

Fr = =Needs technical review

whereyh=hydraulicdepth=T

Criticalflow:Fr=1

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section