Open-channel flow (Manning): practice problems with step-by-step solutions

Free Open-channel flow (Manning) practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: velocity in a channel, flow rate, required slope.

Free practice problemEasy
b = 1 my = 0.8 mn = 0.014

Water flows uniformly 0.8 m deep in a rectangular unfinished concrete channel 1 m wide (Manning n = 0.014). The bed slope is S = 0.0008. Use Manning’s equation in SI units to find the mean velocity V.

b = 1 my = 0.8 mn = 0.014S = 0.0008
m/s
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What you practise

  1. Easyvelocity in a channel
  2. Mediumflow rate
  3. Hardrequired slope

Key formulas

Manning equation
V = (1/n)·R2/3·S1/2
Flow rate
Q = A·V
Hydraulic radius
R = A/P (P = wetted perimeter, free surface excluded)
Rectangle
A = b·y, P = b + 2y
Trapezoid (side slope z:1)
A = (b + z·y)·y, P = b + 2y·√(1 + z²)
Circular pipe, full or half full
R = D/4
Required slope
S = (n·Q/(A·R2/3))²

Learn Open-channel flow (Manning) from scratch

1What open-channel flow is

Water in a river, a canal, a drainage ditch or a sewer that is not full flows with a free surface open to the air. This is open-channel flow.

Nothing pumps the water. Gravity pulls it down the slope of the bed, and friction with the bed and walls holds it back.

When these two balance, the depth and the velocity stay the same along the channel. This is called uniform flow, and it is what Manning’s equation describes.

Engineers use it to answer three questions: how fast does the water flow, how much water can a channel carry, and how steep must the channel be?

Self-check In uniform open-channel flow, which two effects balance each other?
  1. Air pressure and water pressure
  2. Gravity down the slope and friction from the bed and walls
  3. Pump pressure and friction
  4. The weight of the water and its buoyancy

Answer: Gravity down the slope and friction from the bed and walls. The bed slope lets gravity drive the water, and the rough boundary slows it down. When they balance, the depth and velocity stop changing.

2Hydraulic radius and wetted perimeter

Friction acts only where the water touches the channel. That boundary, measured around the cross-section, is the wetted perimeter P. The free surface touches air, so it does not count.

byA = b·yP = b + 2y

The heavy line in the figure is P. The shaded part is the flow area A. Their ratio is the hydraulic radius:

R = A / P
  • Rectangle of width b and depth y: A = b·y and P = b + 2y.
  • Trapezoid with side slopes z:1 (z horizontal for 1 vertical): A = (b + z·y)·y and P = b + 2y·√(1 + z²).
  • Circular pipe flowing full or half full: R = D/4.

A large R means a lot of water for little boundary, so less friction per litre and faster flow.

Self-check A rectangular channel is 4 m wide and the water is 1 m deep. What is the hydraulic radius R?
  1. 0.5 m
  2. 0.667 m
  3. 1 m
  4. 4 m

Answer: 0.667 m. A = 4 · 1 = 4 m² and P = 4 + 2 · 1 = 6 m, so R = 4/6 = 0.667 m. Including the free surface in P (P = 8 m) gives 0.5 m, and 1 m is the depth, not R.

3Manning’s equation

In SI units the mean velocity of uniform flow is:

V = (1/n) · R2/3 · S1/2
  • n is Manning’s roughness coefficient. Smooth finished concrete is about 0.012, brickwork 0.015, clean earth 0.022, grass-lined earth 0.03. Rougher means larger n and slower water.
  • R is the hydraulic radius in m.
  • S is the bed slope in m/m, a pure number. A drop of 1.5 m per kilometre is S = 1.5/1000 = 0.0015.

The flow rate (discharge) is the velocity times the flow area:

Q = A · V

To find the slope a channel needs, combine both and solve for S:

S = (n · Q / (A · R2/3))²
Self-check A channel bed drops 2 m over 1 km. What is the slope S to use in Manning’s equation?
  1. 0.0002
  2. 0.002
  3. 0.02
  4. 2

Answer: 0.002. S = drop/length = 2 m / 1000 m = 0.002. The slope has no unit, so the metres must cancel.

4The method step by step

  1. Sketch the cross-section and write down b, y, z (if trapezoidal), n and S.
  2. Find the flow area A.
  3. Find the wetted perimeter P: bed plus the wetted sides, never the free surface.
  4. Hydraulic radius: R = A/P.
  5. Manning: V = (1/n)·R2/3·S1/2.
  6. If the flow rate is asked: Q = A·V. If the slope is asked, use S = (n·Q/(A·R2/3))².

On a calculator, R2/3 is R^(2/3), with the brackets. Work out R2/3 and √S separately before you multiply.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

Water flows uniformly 1 m deep in a rectangular brickwork channel 3 m wide (Manning n = 0.015). The bed slope is S = 0.001. Find the mean velocity V and the flow rate Q.

1

Look at the section

b = 3 my = 1 mn = 0.015S = 0.001

It is a rectangle, so A = b·y and P = b + 2y.

2

Area and wetted perimeter

A = 3 · 1 = 3 m²
P = 3 + 2 · 1 = 5 m
3

Hydraulic radius

R = A/P = 3/5 = 0.6 m

Then R2/3 = 0.62/3 = 0.7114.

4

Manning’s equation

√S = √0.001 = 0.03162.

V = (1/0.015) · 0.7114 · 0.03162 = 1.50 m/s
5

Flow rate

Q = A · V = 3 · 1.50 = 4.50 m³/s
6

Check the answer

Lined channels usually run at 1 to 3 m/s, so 1.50 m/s is reasonable. A rougher earth lining (n = 0.022) would give a lower velocity for the same slope.

Answer

V = 1.50 m/s and Q = 4.50 m³/s.

6Common mistakes

  • Using the depth y instead of the hydraulic radius R.
  • Counting the free surface in the wetted perimeter.
  • For a trapezoid, taking each side as y long. The wetted side is y·√(1 + z²).
  • Using S instead of √S, or forgetting to square at the end when you solve for S.
  • Using a slope in m/km as if it were m/m.
  • Giving V when Q is asked: multiply by the area.
Self-check Rectangular channel: b = 4 m, y = 1 m, n = 0.02, S = 0.0016. What is the mean velocity V?
  1. 0.061 m/s
  2. 1.26 m/s
  3. 1.53 m/s
  4. 2.00 m/s

Answer: 1.53 m/s. R = 4/6 = 0.667 m, so V = (1/0.02) · 0.667^(2/3) · 0.04 = 1.53 m/s. Using the depth as R gives 2.00 m/s, and including the free surface in P gives 1.26 m/s.

7How exam questions look

  • Easy: velocity in a rectangular channel. Sometimes the slope is given as a drop per kilometre.
  • Medium: the flow rate Q, in a rectangular or trapezoidal channel, or in a pipe flowing full or half full (R = D/4).
  • Hard: the bed slope a channel needs to carry a given Q, or the slope of a sewer for a given velocity.

Trapezoid example: b = 2 m, y = 1 m, side slopes 2:1, n = 0.022, S = 0.001. A = (2 + 2·1)·1 = 4 m² and P = 2 + 2·1·√5 = 6.47 m, so R = 0.618 m.

V = (1/0.022) · 0.6182/3 · √0.001 = 1.04 m/s, Q = 4 · 1.04 = 4.17 m³/s

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
b = 2.5 my = 1 mn = 0.022

Water flows uniformly 1 m deep in a rectangular clean earth channel 2.5 m wide (Manning n = 0.022). The bed drops 4 m per kilometre. Use Manning’s equation in SI units to find the mean velocity V.

b = 2.5 my = 1 mn = 0.022drop = 4 m/km

Find V (m/s)

Show hints
  1. Find the hydraulic radius R = A/P for the wetted cross-section.
  2. The slope is the drop divided by the length: S = 4/1000.
  3. R = 0.5556 m.
Show the answer and the step-by-step solution

Manning’s equation (SI): for uniform flow in an open channel V = (1/n)·R2/3·S1/2 and Q = A·V, with n the roughness coefficient, S the bed slope (m/m) and R = A/P the hydraulic radius: flow area divided by wetted perimeter. Only the boundary in contact with the water counts in P, not the free surface. Rectangle: A = b·y, P = b + 2y. Trapezoid with side slopes z:1 (horizontal:vertical): A = (b + z·y)·y, P = b + 2y·√(1 + z²). A circular pipe flowing full or half full has R = D/4.

Hydraulic radius

A = b·y = 2.5·1 = 2.5 m²
P = b + 2y = 2.5 + 2·1 = 4.5 m
R = A/P = 0.5556 m

Slope

S = 4/1000 = 0.004

Manning’s equation

V = (1/n)·R2/3·S1/2 = (1/0.022)·0.55562/3·0.0041/2 = 1.943 m/s

Answer

V = 1.943 m/s
Example 2Medium
D = 300 mmn = 0.013, S = 0.0005

A concrete storm sewer with an inside diameter of 300 mm flows just full (as an open channel, not under pressure). Manning n = 0.013 and the slope is S = 0.0005. Find the flow rate Q in m³/s.

D = 300 mmn = 0.013S = 0.0005

Find Q (m³/s)

Show hints
  1. Flowing full: A = π·D²/4 and P = π·D.
  2. In both cases R = A/P = D/4 = 0.075 m.
  3. Find V with Manning, then Q = A·V.
Show the answer and the step-by-step solution

Manning’s equation (SI): for uniform flow in an open channel V = (1/n)·R2/3·S1/2 and Q = A·V, with n the roughness coefficient, S the bed slope (m/m) and R = A/P the hydraulic radius: flow area divided by wetted perimeter. Only the boundary in contact with the water counts in P, not the free surface. Rectangle: A = b·y, P = b + 2y. Trapezoid with side slopes z:1 (horizontal:vertical): A = (b + z·y)·y, P = b + 2y·√(1 + z²). A circular pipe flowing full or half full has R = D/4.

Geometry

A = 0.07069 m², R = D/4 = 0.075 m

Velocity

V = (1/0.013)·0.0752/3·0.00051/2 = 0.3059 m/s

Flow rate

Q = A·V = 0.07069·0.3059 = 0.02162 m³/s

Answer

Q = 0.02162 m³/s
Example 3Hard
b = 2.5 my = 1.8 mn = 0.03, Q = 2.7 m³/s

A rectangular grass-lined earth channel 2.5 m wide must carry 2.7 m³/s in uniform flow at a depth of 1.8 m. Manning n = 0.03. Find the required bed slope S (in m/m).

Q = 2.7 m³/sb = 2.5 my = 1.8 mn = 0.03

Find S (m/m)

Show hints
  1. Combine Q = A·V with Manning: Q = (1/n)·A·R2/3·S1/2.
  2. A = 4.5 m², R = 0.7377 m.
  3. Solve for S: S = (n·Q/(A·R2/3))².
Show the answer and the step-by-step solution

Manning’s equation (SI): for uniform flow in an open channel V = (1/n)·R2/3·S1/2 and Q = A·V, with n the roughness coefficient, S the bed slope (m/m) and R = A/P the hydraulic radius: flow area divided by wetted perimeter. Only the boundary in contact with the water counts in P, not the free surface. Rectangle: A = b·y, P = b + 2y. Trapezoid with side slopes z:1 (horizontal:vertical): A = (b + z·y)·y, P = b + 2y·√(1 + z²). A circular pipe flowing full or half full has R = D/4.

Geometry

A = b·y = 2.5·1.8 = 4.5 m²
P = b + 2y = 2.5 + 2·1.8 = 6.1 m
R = A/P = 0.7377 m

Solve Manning for S

S1/2 = n·Q/(A·R2/3) = 0.03·2.7/(4.5·0.8164) = 0.02205
S = 0.02205² = 0.0004861

Answer

S = 0.0004861

That is a fall of 0.4861 m per kilometre.

Questions and answers

What is Open-channel flow (Manning)?

Water in a river, a canal, a drainage ditch or a sewer that is not full flows with a free surface open to the air. This is open-channel flow.

Nothing pumps the water. Gravity pulls it down the slope of the bed, and friction with the bed and walls holds it back.

How do I solve Open-channel flow (Manning) problems step by step?

Manning’s equation (SI): for uniform flow in an open channel V = (1/n)·R2/3·S1/2 and Q = A·V, with n the roughness coefficient, S the bed slope (m/m) and R = A/P the hydraulic radius: flow area divided by wetted perimeter. Only the boundary in contact with the water counts in P, not the free surface. Rectangle: A = b·y, P = b + 2y. Trapezoid with side slopes z:1 (horizontal:vertical): A = (b + z·y)·y, P = b + 2y·√(1 + z²). A circular pipe flowing full or half full has R = D/4.

What are common mistakes in Open-channel flow (Manning)?
  • Use the hydraulic radius R = A/P, not the depth. The depth is only a good guess for very wide channels.
  • The free surface is not part of the wetted perimeter: P = b + 2y.
  • The slope is dimensionless: 4 m per 1000 m gives S = 0.004.
What do the Open-channel flow (Manning) exercises cover?
  • Easy: velocity in a channel
  • Medium: flow rate
  • Hard: required slope

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Calculators

Manning equation calculator

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Hydraulic EngineeringFE Civil exam practice

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Printable worksheet

Open-channel flow (Manning) worksheet with answers
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