SME
Fluid Mechanics · Problem type
A ship model: matching the Froude number
Use this for “A 100 m ship sails at 10 m/s. How fast should its 4 m model (1:25) be towed?”
Diagram
Same
Froude
number:
Vₘ
=
Vₚ
×
√(
Lₘ
Lₚ
)
=
10
×
√
0.04
=
2
m/s.
The model is drawn 0.04 the size; the arrows share one speed scale.
This is an example. Type your own values to start a new problem.
Fr
=
Vₚ (m/s)
9.81
×
Lₚ (m)
=
Vₘ (m/s)
9.81
×
Lₘ (m)
Formulas
Fr
=
Vₚ
÷
√
9.81 × Lₚ
0.3193
=
10
÷
√
9.81 × 100
Fr
=
Vₘ
÷
√
9.81 × Lₘ
0.3193
=
2
÷
√
9.81 × 4
Assumptions
•
Waves and gravity rule the drag, so equal Froude numbers V ÷ √(gL) make the flows alike.
•
Model and ship have the same shape; g = 9.81 m/s² for both.
•
The Reynolds numbers then differ; the skin friction is corrected separately.
Step-by-step
WE KNOW
Vₚ = 10 m/s, Lₚ = 100 m, Lₘ = 4 m
FIND
Froude number (Fr), model speed (Vₘ)
Step 1 · Find Froude number (Fr)
Use Fr = Vₚ ÷ √(9.81 × Lₚ)
The
ship’s
Froude
number:
its
speed
over
√
gL
,
the
speed
of
a
long
wave.
Fr
=
10
÷
√
9.81 × 100
Fr
=
10
÷
√
981
Fr
=
10
31.32
Fr = 0.3193
Step 2 · Find model speed (Vₘ)
Use Fr = Vₘ ÷ √(9.81 × Lₘ)
The
model
has
the
same
Fr:
multiply
it
by
√
gLₘ
.
Vₘ
=
Fr
×
√
9.81 × Lₘ
Vₘ
=
0.3193
×
√
9.81 × 4
Vₘ
=
0.3193
×
√
39.24
Vₘ = 0.3193 × 6.264
Vₘ = 2 m/s
Check
Put the numbers back in. Both sides should match.
0.3193
=
10
÷
√
9.81 × 100
✓
0.3193
=
2
÷
√
9.81 × 4
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Dimensional analysis
Main lesson
From the model’s force to the prototype’s
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