SME
Hydrology · Problem type
Channel flow by Manning’s equation
Use this for “A channel 10 m wide flows 2 m deep on a slope of 0.001 with n = 0.035. What are its speed and discharge?”, or the depth that carries 50 m³/s.
Diagram
A = by = 10 × 2 = 20 m².
R
=
A
b + 2y
=
20
14
=
1.43
m.
Q
=
(
1
n
)A
R
2
3
S
1
2
Q
=
(
1
0.035
)
×
20
×
1.43
2
3
×
0.001
1
2
Q = 22.9 m³/s
The side view steepens the slope to show it.
This is an example. Type your own values to start a new problem.
b
Channel width
example · m
y
Flow depth
example · m
A
Flow area
calculated · m²
P
Wetted perimeter
calculated · m
R
Hydraulic radius
calculated · m
n
Manning’s n
example
S
Bed slope
example
v
Mean velocity
calculated · m/s
Q
Discharge
calculated · m³/s
Formulas
A
=
b
×
y
20 = 10 × 2
P
=
b
+
2
×
y
14 = 10 + 2 × 2
R
=
A
P
1.43
=
20
14
v
=
(
1
n
)
×
R
2 ÷ 3
×
S
1 ÷ 2
1.146
=
(
1
0.035
)
×
1.43
2 ÷ 3
×
0.001
1 ÷ 2
Q
=
A
×
v
22.9 = 20 × 1.146
Q
=
(
1
n
)
×
b
×
y
×
(
b
×
y
b
+ 2 ×
y
)
2 ÷ 3
×
S
1 ÷ 2
22.9
=
(
1
0.035
)
×
10
×
2
×
(10 ×
2
10 + 2 × 2
)
2 ÷ 3
×
0.001
1 ÷ 2
Assumptions
•
Steady, uniform flow in a rectangular channel: the water surface runs parallel to the bed.
•
Manning’s equation in SI units (its constant 1 needs meters and seconds).
•
n comes from tables: about 0.013 for concrete, 0.035 for a natural stream, 0.1 for a weedy one.
Step-by-step
WE KNOW
b = 10 m, y = 2 m, n = 0.035, S = 0.001
FIND
flow area (A), wetted perimeter (P), hydraulic radius (R), mean velocity (v), discharge (Q)
Step 1 · Find flow area (A)
Use A = b × y
The flow’s cross-section is a rectangle.
A
=
10
×
2
A
=
20
m²
Step 2 · Find wetted perimeter (P)
Use P = b + 2 × y
The water touches the bed and both walls; the top is open air.
P
=
10
+
2
×
2
P
=
10
+
4
P
=
14
m
Step 3 · Find hydraulic radius (R)
Use R = A ÷ P
Area
per
meter
of
wetted
edge:
a
larger
R
means
less
of
the
water
rubs
on
the
bed.
R
=
20
14
R
=
1.429
m
Step 4 · Find mean velocity (v)
Use v = (1 ÷ n) × R2 ÷ 3 × S1 ÷ 2
Manning’s
equation:
a
steeper
bed
and
a
larger
R
speed
the
water;
a
rougher
bed
(larger
n
)
slows
it.
v
=
(
1
0.035
)
×
1.429
2 ÷ 3
×
0.001
1 ÷ 2
v
=
1
0.035
×
1.2687
×
0.031623
v
=
28.571
×
1.2687
×
0.031623
v
=
36.248
×
0.031623
v
=
1.146
m/s
Step 5 · Find discharge (Q)
Use Q = A × v
The area times the mean speed through it.
Q
=
20
×
1.146
Q
=
22.9
m³/s
Check
Put the numbers back in. Both sides should match.
20 = 10 × 2 ✓
14 = 10 + 2 × 2 ✓
1.429
=
20
14
✓
1.146
=
(
1
0.035
)
×
1.429
2 ÷ 3
×
0.001
1 ÷ 2
✓
22.9 = 20 × 1.146 ✓
22.9
=
(
1
0.035
)
×
10
×
2
×
(10 ×
2
10 + 2 × 2
)
2 ÷ 3
×
0.001
1 ÷ 2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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