SME
Hydraulics & Hydrology
Culverts
Use this for “A 1.2 m culvert has its headwater 2.5 m above the inlet invert, Cd = 0.6. What flow does the inlet pass?” Check outlet control too: the higher headwater governs.
Diagram
V
=
√
2gh
=
√
2 × 9.81 × 1.9 m
V = 6.106 m/s
Q
=
Cd
×
πD²
4
×
V
=
4.143
m³/s.
As the tank drains, h falls and V falls with √h, so it empties ever more slowly.
This is an example. Type your own values to start a new problem.
h (m)
=
HW (m)
−
D (m)
÷
2
V (m/s)
=
2
×
9.81
×
h
Q (m³/s)
=
Cd
×
A
×
V
HW is the headwater’s height above the culvert’s floor (the invert); the drawn head h is measured from the opening’s centre, D ÷ 2 higher.
A
Opening area
calculated · m²
Formulas
h
=
HW
−
D
2
1.9
=
2.5
−
1.2
2
A
=
π
×
D
2
4
1.131
=
π
×
1.2
2
4
V
=
√
2 × 9.81 ×
h
6.1056
=
√
2 × 9.81 × 1.9
Q
=
Cd
×
A
×
V
4.1431 = 0.6 × 1.131 × 6.1056
Assumptions
•
Inlet control: the barrel could carry more than the entrance lets in.
•
The inlet is submerged (HW at least 1.2D), so it acts as an orifice.
•
Cd is about 0.6 for a square-edged headwall; g = 9.81 m/s².
Step-by-step
WE KNOW
D = 1.2 m, HW = 2.5 m, Cd = 0.6
FIND
head on the opening’s centre (h), opening area (A), orifice speed (V), discharge (Q)
Step 1 · Find head on the opening’s centre (h)
Use h = HW − D ÷ 2
An orifice works on the head over its centre, half a diameter above the invert.
h
=
2.5
−
1.2
2
h
=
2.5
−
0.6
h
=
1.9
m
Step 2 · Find opening area (A)
Use A = π × D² ÷ 4
The barrel runs full at the entrance, so the opening is the whole circle.
A
=
π
×
1.2
2
4
A
=
π
×
1.44
4
A
=
1.44π
4
A
=
1.13097
m²
Step 3 · Find orifice speed (V)
Use V = √(2 × 9.81 × h)
Bernoulli
from
the
still
pond
to
the
jet:
the
head
h
turns
into
1
2
V
2
.
V
=
√
2 × 9.81 × 1.9
V
=
√
19.62 × 1.9
V
=
√
37.278
V
=
6.10557
m/s
Step 4 · Find discharge (Q)
Use Q = Cd × A × V
The opening’s area times the jet speed, cut by Cd for the jet’s narrowing at the edge.
Q
=
0.6
×
1.13097
×
6.10557
Q
=
0.6786
×
6.1056
Q
=
4.1431
m³/s
Check
Put the numbers back in. Both sides should match.
1.9
=
2.5
−
1.2
2
✓
1.13097
=
π
×
1.2
2
4
✓
6.10557
=
√
2 × 9.81 × 1.9
✓
4.1431 = 0.6 × 1.13097 × 6.10557 ✓
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Outlet control: headwater from the barrel losses
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.