SME
Hydraulics & Hydrology · Problem type
Hydraulic jump: sequent depth and head loss
Use this for “Water 0.3 m deep runs at 6 m/s into a hydraulic jump. Find the depth after it and the head lost.”
Diagram
Fr₁
=
V₁
÷
√
gy₁
=
6
÷
√
9.81 × 0.3
=
3.5:
supercritical,
so
the
flow
jumps.
y₂
=
(
y₁
2
)(
√
1 + 8Fr₁²
−
1)
=
(
0.3
2
)(
√
1 + 8 ×
3.5
2
−
1)
=
1.34
m.
hL
=
(y₂ − y₁)
3
÷
(4y₁y₂)
=
(1.34 − 0.3)
3
÷
(4
×
0.3
×
1.34)
=
0.702
m,
the
energy
line's
drop.
Depths, speeds and the energy line are to scale; the length is shortened.
This is an example. Type your own values to start a new problem.
Fr₁
=
V₁ (m/s)
9.81
×
y₁ (m)
y₂ (m)
=
y₁
2
×
(
1
+
8
×
Fr₁²
−
1)
hL (m)
=
(y₂
−
y₁)³
4
×
y₁
×
y₂
Formulas
Fr₁
=
V₁
÷
√
9.81 ×
y₁
3.497
=
6
÷
√
9.81 × 0.3
y₂
=
(
y₁
2
)
×
(
√
1 + 8 × Fr₁²
−
1)
1.341
=
(
0.3
2
)
×
(
√
1 + 8 ×
3.497
2
−
1)
hL
=
(
y₂
−
y₁
)
3
÷
(4
×
y₁
×
y₂
)
0.7017
=
(1.341 − 0.3)
3
÷
(4
×
0.3
×
1.341)
Assumptions
•
A horizontal, rectangular channel; friction over the jump’s short length is ignored; g = 9.81 m/s².
•
Momentum is kept across the jump; energy is not: the roller turns hL to heat.
•
A jump needs supercritical flow before it: Froude number Fr₁ above 1.
Step-by-step
WE KNOW
y₁ = 0.3 m, V₁ = 6 m/s
FIND
Froude number before (Fr₁), depth after (y₂), head lost (hL)
Step 1 · Find Froude number before (Fr₁)
Use Fr₁ = V₁ ÷ √(9.81 × y₁)
The
Froude
number
before
the
jump:
the
speed
over
a
shallow
wave’s
speed
√
gy₁
.
Fr₁
=
6
÷
√
9.81 × 0.3
Fr₁
=
6
÷
√
2.943
Fr₁
=
6
1.7155
Fr₁ = 3.4975
Step 2 · Find depth after (y₂)
Use y₂ = (y₁ ÷ 2) × (√(1 + 8 × Fr₁²) − 1)
The sequent depth, from the momentum kept across the jump.
y₂
=
(
0.3
2
)
×
(
√
1 + 8 ×
3.4975
2
−
1)
y₂
=
0.15
×
(
√
1 + 97.859
−
1)
y₂
=
0.15
×
(9.9428
−
1)
y₂
=
1.34142
m
Step 3 · Find head lost (hL)
Use hL = (y₂ − y₁)³ ÷ (4 × y₁ × y₂)
The energy the roller turns to heat, as a height of water.
hL
=
(1.34142 − 0.3)
3
÷
(4
×
0.3
×
1.34142)
hL
=
1.0414
3
1.2 × 1.3414
hL
=
1.0414
3
1.6097
hL
=
1.1294
1.6097
hL = 0.7017 m
Check
Put the numbers back in. Both sides should match.
3.4975
=
6
÷
√
9.81 × 0.3
✓
1.34142
=
(
0.3
2
)
×
(
√
1 + 8 ×
3.4975
2
−
1)
✓
0.7017
=
(1.34142 − 0.3)
3
÷
(4
×
0.3
×
1.34142)
✓
Topic check
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