SME
Fluid Mechanics
Dimensional analysis
Use this for “A 4 m car drives at 30 m/s. How fast must the air move past a 1 m model (1:4) in the same air?”
Diagram
Same
Reynolds
number:
Vₘ
=
Vₚ
×
(
Lₚ
Lₘ
)
×
(
νₘ
νₚ
)
=
30
×
4
×
1
=
120
m/s.
The model is drawn 0.25 the size; the arrows share one speed scale.
This is an example. Type your own values to start a new problem.
Re
=
Vₚ (m/s)
×
Lₚ (m)
νₚ (m²/s)
=
Vₘ (m/s)
×
Lₘ (m)
νₘ (m²/s)
Formulas
Re
=
Vₚ
×
Lₚ
νₚ
8,000,000
=
30
×
4
1.5 ×
10
-5
Re
=
Vₘ
×
Lₘ
νₘ
8,000,000
=
120
×
1
1.5 ×
10
-5
Assumptions
•
The model has the prototype’s shape, every length scaled by Lₘ ÷ Lₚ.
•
Viscosity and inertia rule the flow, so equal Reynolds numbers make the flows alike.
•
Buckingham Π: n variables in r dimensions give n − r groups; F, V, L, ρ, μ give two.
•
Air much above 100 m/s compresses, which is why small models are often tested in water.
Step-by-step
WE KNOW
Vₚ = 30 m/s, Lₚ = 4 m, νₚ = 1.5 × 10⁻⁵ m²/s, Lₘ = 1 m, νₘ = 1.5 × 10⁻⁵ m²/s
FIND
Reynolds number (Re), model speed (Vₘ)
Step 1 · Find Reynolds number (Re)
Use Re = Vₚ × Lₚ ÷ νₚ
The prototype’s Reynolds number: speed times length, over the kinematic viscosity.
Re
=
30
×
4
1.5 ×
10
-5
Re
=
120
1.5 ×
10
-5
Re = 8,000,000
Step 2 · Find model speed (Vₘ)
Use Re = Vₘ × Lₘ ÷ νₘ
The model has the same Re: multiply it by νₘ, then divide by Lₘ.
Vₘ
=
Re
×
νₘ
Lₘ
Vₘ
=
8,000,000
×
1.5 ×
10
-5
1
Vₘ = 120 m/s
Check
Put the numbers back in. Both sides should match.
8,000,000
=
30
×
4
1.5 ×
10
-5
✓
8,000,000
=
120
×
1
1.5 ×
10
-5
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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