Manning's Equation

Hydraulics · FE Reference Handbook section

Hydraulics

10 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 Manning's Equation 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 manning's equation describes physically and when it applies.
State every one of the 10 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. Manning's Equation 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: manning's equation.
Capstone Studio instructional photograph

Hydraulics — Manning's Equation: 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 10 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

Q	Quantity produced by "Q = n AR H2 3 S 1 2" — read its definition and unit from the handbook line directly above the equation.
V	Quantity produced by "V = n R H2 3 S 1 2" — read its definition and unit from the handbook line directly above the equation.
K	Quantity produced by "K = 1.486 for USCS units, 1.0 for SI units" — read its definition and unit from the handbook line directly above the equation.
n	Quantity produced by "n = roughness coefficient" — read its definition and unit from the handbook line directly above the equation.
A	Quantity produced by "A = cross-sectional area of flow (ft2 or m2)" — read its definition and unit from the handbook line directly above the equation.
RH	Quantity produced by "RH = hydraulic radius (ft or m) = P" — read its definition and unit from the handbook line directly above the equation.
P	Quantity produced by "P = wetted perimeter (ft or m)" — read its definition and unit from the handbook line directly above the equation.
S	Quantity produced by "S = slope (ft/ft or m/m)" — 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.

where

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

Manning's equation for a rectangular channel — Manning's Equation

A rectangular channel 15 ft wide flows 4.50 ft deep with n = 0.013 and S = 0.0018. Find the discharge and mean velocity.

Given

b = 15 ft
y = 4.50 ft
n = 0.013
S = 0.0018

Find

Q and V

Start with the thinking

Hydraulic radius is area over wetted perimeter — not depth.
The 1.486 constant belongs to US units only.

Step-by-step solution

Area
$A = b y = 15 (4.50) = 67.50 f t^{2}$
Wetted perimeter
$P = b + 2 y = 15 + 2 (4.50) = 24.00 f t$
Hydraulic radius
$R = A / P = 67.50/24.00 = 2.813 f t$
Manning
$Q = (1.486/ n) A R^{(} 2/3) S^{(} 1/2)$
Substituting
$Q = (1.486/0.013) (67.50) (1.992) (0.0424)$
Evaluate
$Q = 652.2 c f s, V = Q / A = 9.66 f t / s$

Answer: Q ≈ 652.2 cfs; V ≈ 9.66 ft/s

Why the other options are there

438.9 cfs (SI constant used)
1,473 cfs (depth used for R)

Reference: FE Reference Handbook — Hydraulics → Manning's Equation

Example 4

Manning's equation for a rectangular channel — Manning's Equation (2)

A rectangular channel 16 ft wide flows 5.00 ft deep with n = 0.013 and S = 0.0018. Find the discharge and mean velocity.

Given

b = 16 ft
y = 5.00 ft
n = 0.013
S = 0.0018

Find

Q and V

Start with the thinking

Hydraulic radius is area over wetted perimeter — not depth.
The 1.486 constant belongs to US units only.

Step-by-step solution

Area
$A = b y = 16 (5.00) = 80.00 f t^{2}$
Wetted perimeter
$P = b + 2 y = 16 + 2 (5.00) = 26.00 f t$
Hydraulic radius
$R = A / P = 80.00/26.00 = 3.077 f t$
Manning
$Q = (1.486/ n) A R^{(} 2/3) S^{(} 1/2)$
Substituting
$Q = (1.486/0.013) (80.00) (2.115) (0.0424)$
Evaluate
$Q = 820.8 c f s, V = Q / A = 10.26 f t / s$

Answer: Q ≈ 820.8 cfs; V ≈ 10.26 ft/s

Why the other options are there

552.3 cfs (SI constant used)
1,940 cfs (depth used for R)

Reference: FE Reference Handbook — Hydraulics → Manning's Equation

Example 5

Manning's equation for a rectangular channel — Manning's Equation (3)

A rectangular channel 12 ft wide flows 5.00 ft deep with n = 0.015 and S = 0.0052. Find the discharge and mean velocity.

Given

b = 12 ft
y = 5.00 ft
n = 0.015
S = 0.0052

Find

Q and V

Start with the thinking

Hydraulic radius is area over wetted perimeter — not depth.
The 1.486 constant belongs to US units only.

Step-by-step solution

Area
$A = b y = 12 (5.00) = 60.00 f t^{2}$
Wetted perimeter
$P = b + 2 y = 12 + 2 (5.00) = 22.00 f t$
Hydraulic radius
$R = A / P = 60.00/22.00 = 2.727 f t$
Manning
$Q = (1.486/ n) A R^{(} 2/3) S^{(} 1/2)$
Substituting
$Q = (1.486/0.015) (60.00) (1.952) (0.0721)$
Evaluate
$Q = 836.7 c f s, V = Q / A = 13.94 f t / s$

Answer: Q ≈ 836.7 cfs; V ≈ 13.94 ft/s

Why the other options are there

563.1 cfs (SI constant used)
2,143 cfs (depth used for R)

Reference: FE Reference Handbook — Hydraulics → Manning's Equation

Example 6

Manning's equation for a rectangular channel — Manning's Equation (4)

A rectangular channel 12 ft wide flows 4.25 ft deep with n = 0.015 and S = 0.0046. Find the discharge and mean velocity.

Given

b = 12 ft
y = 4.25 ft
n = 0.015
S = 0.0046

Find

Q and V

Start with the thinking

Hydraulic radius is area over wetted perimeter — not depth.
The 1.486 constant belongs to US units only.

Step-by-step solution

Area
$A = b y = 12 (4.25) = 51.00 f t^{2}$
Wetted perimeter
$P = b + 2 y = 12 + 2 (4.25) = 20.50 f t$
Hydraulic radius
$R = A / P = 51.00/20.50 = 2.488 f t$
Manning
$Q = (1.486/ n) A R^{(} 2/3) S^{(} 1/2)$
Substituting
$Q = (1.486/0.015) (51.00) (1.836) (0.0678)$
Evaluate
$Q = 629.1 c f s, V = Q / A = 12.34 f t / s$

Answer: Q ≈ 629.1 cfs; V ≈ 12.34 ft/s

Why the other options are there

423.4 cfs (SI constant used)
1,456 cfs (depth used for R)

Reference: FE Reference Handbook — Hydraulics → Manning's Equation

Example 7

Manning's equation for a rectangular channel — Manning's Equation (5)

A rectangular channel 11 ft wide flows 3.50 ft deep with n = 0.013 and S = 0.0052. Find the discharge and mean velocity.

Given

b = 11 ft
y = 3.50 ft
n = 0.013
S = 0.0052

Find

Q and V

Start with the thinking

Hydraulic radius is area over wetted perimeter — not depth.
The 1.486 constant belongs to US units only.

Step-by-step solution

Area
$A = b y = 11 (3.50) = 38.50 f t^{2}$
Wetted perimeter
$P = b + 2 y = 11 + 2 (3.50) = 18.00 f t$
Hydraulic radius
$R = A / P = 38.50/18.00 = 2.139 f t$
Manning
$Q = (1.486/ n) A R^{(} 2/3) S^{(} 1/2)$
Substituting
$Q = (1.486/0.013) (38.50) (1.660) (0.0721)$
Evaluate
$Q = 526.8 c f s, V = Q / A = 13.68 f t / s$

Answer: Q ≈ 526.8 cfs; V ≈ 13.68 ft/s

Why the other options are there

354.5 cfs (SI constant used)
1,111 cfs (depth used for R)

Reference: FE Reference Handbook — Hydraulics → Manning's Equation

Example 8

Manning's equation for a rectangular channel — Manning's Equation (6)

A rectangular channel 5 ft wide flows 5.00 ft deep with n = 0.015 and S = 0.0100. Find the discharge and mean velocity.

Given

b = 5 ft
y = 5.00 ft
n = 0.015
S = 0.0100

Find

Q and V

Start with the thinking

Hydraulic radius is area over wetted perimeter — not depth.
The 1.486 constant belongs to US units only.

Step-by-step solution

Area
$A = b y = 5 (5.00) = 25.00 f t^{2}$
Wetted perimeter
$P = b + 2 y = 5 + 2 (5.00) = 15.00 f t$
Hydraulic radius
$R = A / P = 25.00/15.00 = 1.667 f t$
Manning
$Q = (1.486/ n) A R^{(} 2/3) S^{(} 1/2)$
Substituting
$Q = (1.486/0.015) (25.00) (1.406) (0.1000)$
Evaluate
$Q = 348.2 c f s, V = Q / A = 13.93 f t / s$

Answer: Q ≈ 348.2 cfs; V ≈ 13.93 ft/s

Why the other options are there

234.3 cfs (SI constant used)
1,238 cfs (depth used for R)

Reference: FE Reference Handbook — Hydraulics → Manning's Equation

Example 9

Manning's equation for a rectangular channel — Manning's Equation (7)

A rectangular channel 9 ft wide flows 2.75 ft deep with n = 0.022 and S = 0.0100. Find the discharge and mean velocity.

Given

b = 9 ft
y = 2.75 ft
n = 0.022
S = 0.0100

Find

Q and V

Start with the thinking

Hydraulic radius is area over wetted perimeter — not depth.
The 1.486 constant belongs to US units only.

Step-by-step solution

Area
$A = b y = 9 (2.75) = 24.75 f t^{2}$
Wetted perimeter
$P = b + 2 y = 9 + 2 (2.75) = 14.50 f t$
Hydraulic radius
$R = A / P = 24.75/14.50 = 1.707 f t$
Manning
$Q = (1.486/ n) A R^{(} 2/3) S^{(} 1/2)$
Substituting
$Q = (1.486/0.022) (24.75) (1.428) (0.1000)$
Evaluate
$Q = 238.8 c f s, V = Q / A = 9.65 f t / s$

Answer: Q ≈ 238.8 cfs; V ≈ 9.65 ft/s

Why the other options are there

160.7 cfs (SI constant used)
459.7 cfs (depth used for R)

Reference: FE Reference Handbook — Hydraulics → Manning's Equation

Example 10

Manning's equation for a rectangular channel — Manning's Equation (8)

A rectangular channel 5 ft wide flows 3.25 ft deep with n = 0.03 and S = 0.0072. Find the discharge and mean velocity.

Given

b = 5 ft
y = 3.25 ft
n = 0.03
S = 0.0072

Find

Q and V

Start with the thinking

Hydraulic radius is area over wetted perimeter — not depth.
The 1.486 constant belongs to US units only.

Step-by-step solution

Area
$A = b y = 5 (3.25) = 16.25 f t^{2}$
Wetted perimeter
$P = b + 2 y = 5 + 2 (3.25) = 11.50 f t$
Hydraulic radius
$R = A / P = 16.25/11.50 = 1.413 f t$
Manning
$Q = (1.486/ n) A R^{(} 2/3) S^{(} 1/2)$
Substituting
$Q = (1.486/0.03) (16.25) (1.259) (0.0849)$
Evaluate
$Q = 86.00 c f s, V = Q / A = 5.29 f t / s$

Answer: Q ≈ 86.0 cfs; V ≈ 5.29 ft/s

Why the other options are there

57.9 cfs (SI constant used)
222.0 cfs (depth used for R)

Reference: FE Reference Handbook — Hydraulics → Manning's Equation

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

Manning's Equation contains 10 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.