SME
Hydraulics & Hydrology · Problem type
Outlet control: headwater from the barrel losses
Use this for “3 m³/s runs full through 40 m of 1.2 m concrete culvert (n = 0.013, kₑ = 0.5) on a 0.01 slope, tailwater 1.5 m. Find the headwater.” Check inlet control too: the higher headwater governs.
Diagram
The
energy
grade
line
falls
by
hL
=
0.7748
m
along
the
pipe;
the
hydraulic
grade
line
runs
V
2
2g
=
0.3586
m
under
it.
Heads are to the scale at the left; the pipe’s length is not.
Roughness, Manning’s n n = 0.013
Entrance loss coefficient kₑ = 0.5
Tailwater depth at the outlet hₒ = 1.5 m
Barrel slope S₀ = 0.01
Headwater above the inlet invert HW = 1.8748 m
This is an example. Type your own values to start a new problem.
V (m/s)
=
4
×
Q (m³/s)
π
×
(
D (m)
)
2
H (m)
=
1
+
+
2g
×
n
2
×
L (m)
(
D (m)
÷
4)
4
3
×
V²
2g
HW (m)
=
H
+
hₒ (m)
−
S₀
×
L
Formulas
V
=
4
×
Q
π ×
D
2
2.6526
=
4
×
3
π ×
1.2
2
H
=
(1
+
kₑ
+
2
×
9.81
×
n
2
×
L
÷
(
D
4
)
4
3
)
×
V
2
2 × 9.81
0.7748
=
(1
+
0.5
+
2
×
9.81
×
0.013
2
×
40
÷
(
1.2
4
)
4
3
)
×
2.6526
2
2 × 9.81
HW
=
H
+
hₒ
−
S₀
×
L
1.8748 = 0.7748 + 1.5 − 0.01 × 40
Assumptions
•
The barrel flows full (outlet control); full, the hydraulic radius is R = D ÷ 4.
•
The tailwater stands above the outlet’s crown, so hₒ is the tailwater depth.
•
The 1 is the exit loss; kₑ is about 0.5 for a square edge, 0.2 bevelled.
•
Manning’s n in SI units; g = 9.81 m/s².
Step-by-step
WE KNOW
Q = 3 m³/s, D = 1.2 m, L = 40 m, n = 0.013, kₑ = 0.5, hₒ = 1.5 m, S₀ = 0.01
FIND
full-barrel velocity (V), head loss through the culvert (H), headwater above the inlet invert (HW)
Step 1 · Find full-barrel velocity (V)
Use V = 4 × Q ÷ (π × D²)
The barrel runs full, so the flow spreads over the whole circle.
V
=
4
×
3
π ×
1.2
2
V
=
4
×
3
1.44 × π
V
=
4
×
3
4.524
V
=
12
4.524
V
=
2.6526
m/s
Step 2 · Find head loss through the culvert (H)
Use H = (1 + kₑ + 2 × 9.81 × n² × L ÷ (D ÷ 4)4/3) × V² ÷ (2 × 9.81)
Exit,
entrance
and
barrel
friction,
each
a
share
of
the
velocity
head
V
2
2g
;
full,
R
=
D
4
.
H
=
(1
+
0.5
+
2
×
9.81
×
0.013
2
×
40
÷
(
1.2
4
)
4
3
)
×
2.6526
2
2 × 9.81
R
=
D
4
=
0.3
m
(full
pipe)
2gn²
L
R
4
3
=
0.6604
V
2
2g
=
0.3586
m
H
=
(1
+
0.5
+
0.6604)
×
0.3586
H
=
0.7748
m
Step 3 · Find headwater above the inlet invert (HW)
Use HW = H + hₒ − S₀ × L
Energy
from
the
pond
to
the
outlet:
the
losses
ride
on
the
tailwater,
and
the
barrel’s
fall
S₀
L
comes
back.
HW = 0.7748 + 1.5 − 0.01 × 40
HW = 0.7748 + 1.5 − 0.4
HW = 1.8748 m
Check
Put the numbers back in. Both sides should match.
2.6526
=
4
×
3
π ×
1.2
2
✓
0.7748
=
(1
+
0.5
+
2
×
9.81
×
0.013
2
×
40
÷
(
1.2
4
)
4
3
)
×
2.6526
2
2 × 9.81
✓
1.8748 = 0.7748 + 1.5 − 0.01 × 40 ✓
Topic check
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