SME
Hydraulics & Hydrology · Problem type
Critical depth and specific energy
Use this for “A channel 3 m wide carries 5.79 m³/s. Find the critical depth, Eₘᵢₙ, and E at 1.2 m deep.”, or the alternate depth with the same E.
Diagram
q
=
Q
b
=
5.79
3
=
1.93
m²/s.
yc
=
(
q
2
g
)
1
3
=
(
1.93
2
9.81
)
1
3
=
0.724
m,
where
E
is
least:
Eₘᵢₙ
=
1.5yc
=
1.09
m.
At
y
=
1.2
m,
E
=
y
+
q
2
2gy²
=
1.33
m,
subcritical;
the
open
dot
is
the
alternate
depth
0.469
m
with
the
same
E.
This is an example. Type your own values to start a new problem.
q (m²/s)
=
Q (m³/s)
b (m)
yc (m)
=
q²
9.81
1
3
E (m)
=
y (m)
+
q²
2
×
9.81
×
y²
Eₘᵢₙ
Least specific energy
calculated · m
B
Branch
calculated
Formulas
q
=
Q
b
1.93
=
5.79
3
yc
=
(
q
2
9.81
)
1 ÷ 3
0.7241
=
(
1.93
2
9.81
)
1 ÷ 3
Eₘᵢₙ = 1.5 × yc
1.086 = 1.5 × 0.7241
E
=
y
+
q
2
2 × 9.81 ×
y
2
1.332
=
1.2
+
1.93
2
2 × 9.81 ×
1.2
2
B
=
subcritical
if
y
≥
yc,
else
supercritical
1.2 ≥ 0.7241, so subcritical
Assumptions
•
A rectangular channel; specific energy E is measured from the bed, and g = 9.81 m/s².
•
Each E above Eₘᵢₙ has two alternate depths: subcritical above yc, supercritical below it.
•
Typing E, pick the branch; the depth is then found by trial.
Step-by-step
WE KNOW
Q = 5.79 m³/s, b = 3 m, y = 1.2 m
FIND
unit discharge (q), critical depth (yc), least specific energy (Eₘᵢₙ), branch (B), specific energy (E)
Step 1 · Find unit discharge (q)
Use q = Q ÷ b
The discharge per meter of width.
q
=
5.79
3
q
=
1.93
m²/s
Step 2 · Find critical depth (yc)
Use yc = (q² ÷ 9.81)1 ÷ 3
The
critical
depth,
where
the
specific
energy
is
least:
the
cube
root
of
q
2
g
.
yc
=
(
1.93
2
9.81
)
1 ÷ 3
yc
=
∛(
3.7249
9.81
)
yc = ∛0.3797
yc = 0.7241 m
Step 3 · Find least specific energy (Eₘᵢₙ)
Use Eₘᵢₙ = 1.5 × yc
At
critical
depth
the
velocity
head
is
half
the
depth,
so
E
is
1.5
times
yc.
Eₘᵢₙ = 1.5 × 0.7241
Eₘᵢₙ = 1.086 m
Step 4 · Find branch (B)
Use B = subcritical if y ≥ yc, else supercritical
Deeper
than
yc,
the
flow
is
slow
and
subcritical:
the
upper
limb
of
the
E
–
y
curve.
1.2 ≥ 0.7241: subcritical
B
=
subcritical
if
1.2
≥
0.7241,
else
supercritical
B
=
subcritical
Step 5 · Find specific energy (E)
Use E = y + q² ÷ (2 × 9.81 × y²)
The
depth
plus
the
velocity
head
V
2
2g
,
with
V
=
q
y
.
1.2 ≥ 0.7241, on the subcritical branch: subcritical
E
=
1.2
+
1.93
2
2 × 9.81 ×
1.2
2
E
=
1.2
+
3.7249
28.253
E
=
1.2
+
0.1318
E
=
1.332
m
Check
Put the numbers back in. Both sides should match.
1.93
=
5.79
3
✓
0.7241
=
(
1.93
2
9.81
)
1 ÷ 3
✓
1.086 = 1.5 × 0.7241 ✓
1.332
=
1.2
+
1.93
2
2 × 9.81 ×
1.2
2
✓
1.2 ≥ 0.7241, so subcritical ✓
Topic check
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