SME
Aerodynamics · Problem type
Reynolds number of a wing chord
Use this for “Air at 1.225 kg/m³ with μ = 1.789 × 10⁻⁵ Pa·s flows at 60 m/s past a 1.5 m chord. Find Re.”
This is an example. Type your own values to start a new problem.
Re
=
ρ (kg/m³)
×
V (m/s)
×
c (m)
μ (Pa·s)
Formulas
Re
=
ρ
×
V
×
c
μ
6.163 ×
10
6
=
1.225
×
60
×
1.5
1.789 ×
10
-5
Assumptions
•
Reynolds number Re compares inertia with viscous forces; for a wing the length is the chord c.
•
Sea-level air has μ = 1.789 × 10⁻⁵ Pa·s; μ falls a little in the colder air higher up.
•
Wind-tunnel coefficients carry over to flight only at a matching Reynolds number.
Step-by-step
WE KNOW
ρ = 1.225 kg/m³, V = 60 m/s, c = 1.5 m, μ = 1.789 × 10⁻⁵ Pa·s
FIND
Reynolds number (Re)
Step 1 · Find Reynolds number (Re)
Use Re = ρ × V × c ÷ μ
Density times speed times chord, over the viscosity.
Re
=
1.225
×
60
×
1.5
1.789 ×
10
-5
Re
=
73.5
×
1.5
1.789 ×
10
-5
Re
=
110.25
1.789 ×
10
-5
Re
=
6.163 ×
10
6
Check
Put the numbers back in. Both sides should match.
6.163 ×
10
6
=
1.225
×
60
×
1.5
1.789 ×
10
-5
✓
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