SME
Hydraulics & Hydrology
Open-channel flow
Use this for “A channel 3 m wide runs 1.2 m deep on a slope of 0.001 with n = 0.015. Find Q and Fr.”, or the normal depth that carries 5.79 m³/s.
Diagram
A = by = 3 × 1.2 = 3.6 m².
R
=
A
b + 2y
=
3.6
5.4
=
0.667
m.
Q
=
(
1
n
)A
R
2
3
S
1
2
Q
=
(
1
0.015
)
×
3.6
×
0.667
2
3
×
0.001
1
2
Q = 5.79 m³/s
Fr
=
V
÷
√
gy
=
1.61
÷
√
9.81 × 1.2
=
0.47,
less
than
1:
subcritical,
slow
and
deep.
The side view steepens the slope to show it.
This is an example. Type your own values to start a new problem.
A (m²)
=
b (m)
×
y (m)
R (m)
=
A
÷
(b
+
2
×
y)
Q (m³/s)
=
1
n
×
A
×
R
2
3
×
S
1
2
V
Mean velocity
calculated · m/s
Fr
Froude number
calculated
F
Flow regime
worked out (not typed)
Formulas
A
=
b
×
y
3.6 = 3 × 1.2
R
=
A
b
+ 2 ×
y
0.6667
=
3.6
3 + 2 × 1.2
Q
=
(
1
n
)
×
A
×
R
2 ÷ 3
×
S
1 ÷ 2
5.792
=
(
1
0.015
)
×
3.6
×
0.6667
2 ÷ 3
×
0.001
1 ÷ 2
V
=
Q
A
1.609
=
5.792
3.6
Fr
=
V
÷
√
9.81 ×
y
0.4689
=
1.609
÷
√
9.81 × 1.2
F
=
subcritical
if
Fr
<
1,
supercritical
if
Fr
>
1
0.4689 < 1, so subcritical
Assumptions
•
Uniform, steady flow in a rectangular channel: the depth is the normal depth.
•
Manning’s equation in SI units (constant 1, meters and seconds); g = 9.81 m/s².
•
Froude number Fr below 1 is subcritical (slow, deep), above 1 supercritical (fast, shallow).
Step-by-step
WE KNOW
b = 3 m, n = 0.015, S = 0.001, y = 1.2 m
FIND
flow area (A), hydraulic radius (R), discharge (Q), mean velocity (V), Froude number (Fr), flow regime (F)
Step 1 · Find flow area (A)
Use A = b × y
The flow’s cross-section is a rectangle.
A
=
3
×
1.2
A
=
3.6
m²
Step 2 · Find hydraulic radius (R)
Use R = A ÷ (b + 2 × y)
The
hydraulic
radius
is
the
area
over
the
wetted
perimeter:
the
bed
b
and
both
walls
y
.
R
=
3.6
3 + 2 × 1.2
R
=
3.6
3 + 2.4
R
=
3.6
5.4
R
=
0.666667
m
Step 3 · Find discharge (Q)
Use Q = (1 ÷ n) × A × R2 ÷ 3 × S1 ÷ 2
Manning’s
equation
in
SI
units:
a
rougher
bed
(larger
n
)
carries
less,
a
steeper
one
more.
Q
=
(
1
0.015
)
×
3.6
×
0.666667
2 ÷ 3
×
0.001
1 ÷ 2
Q
=
66.667
×
3.6
×
0.7631
×
0.03162
Q
=
5.7918
m³/s
Step 4 · Find mean velocity (V)
Use V = Q ÷ A
The mean speed is the discharge spread over the area.
V
=
5.7918
3.6
V
=
1.6088
m/s
Step 5 · Find Froude number (Fr)
Use Fr = V ÷ √(9.81 × y)
Compare
the
flow’s
speed
with
a
shallow
wave’s
speed
√
gy
.
Fr
=
1.6088
÷
√
9.81 × 1.2
Fr
=
1.6088
÷
√
11.772
Fr
=
1.6088
3.431
Fr = 0.4689
Step 6 · Find flow regime (F)
Use F = subcritical if Fr < 1, supercritical if Fr > 1
The water moves slower than a surface wave, so a wave can travel upstream.
0.4689 < 1: subcritical
F
=
subcritical
if
0.4689
<
1,
supercritical
if
0.4689
>
1
F
=
subcritical
Check
Put the numbers back in. Both sides should match.
3.6 = 3 × 1.2 ✓
0.666667
=
3.6
3 + 2 × 1.2
✓
5.7918
=
(
1
0.015
)
×
3.6
×
0.666667
2 ÷ 3
×
0.001
1 ÷ 2
✓
1.6088
=
5.7918
3.6
✓
0.4689
=
1.6088
÷
√
9.81 × 1.2
✓
0.4689 < 1, so subcritical ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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