SME
Aerodynamics · Problem type
Lift per span from the circulation (Kutta–Joukowski)
Use this for “Air at 1.225 kg/m³ and 40 m/s; the circulation round a 1.5 m chord is 20 m²/s. Find L′ and cₗ.”
Diagram
Γ = 20 m²/s turns clockwise round the section: L′ = ρ∞V∞Γ, lift ⟂ the wind.
This is an example. Type your own values to start a new problem.
L′ (N/m)
=
ρ∞ (kg/m³)
×
V∞ (m/s)
×
Γ (m²/s)
cₗ
=
L′ (N/m)
q∞ (Pa)
×
c (m)
Formulas
L′
=
ρ∞
×
V∞
×
Γ
980 = 1.225 × 40 × 20
q∞ = 0.5 × ρ∞ × V∞²
980
=
0.5
×
1.225
×
40
2
cₗ
=
L′
q∞ ×
c
0.6667
=
980
980 × 1.5
Assumptions
•
Steady, inviscid flow past a two-dimensional section: lift per meter of span.
•
The circulation Γ is the flow’s net swirl round the section; the Kutta condition fixes it.
•
Γ is positive clockwise with the flow from the left: then the lift is up.
Step-by-step
WE KNOW
ρ∞ = 1.225 kg/m³, V∞ = 40 m/s, Γ = 20 m²/s, c = 1.5 m
FIND
lift per span (L′), dynamic pressure (q∞), lift coefficient (cₗ)
Step 1 · Find lift per span (L′)
Use L′ = ρ∞ × V∞ × Γ
Kutta–Joukowski: density times speed times circulation.
L′ = 1.225 × 40 × 20
L′ = 49 × 20
L′ = 980 N/m
Step 2 · Find dynamic pressure (q∞)
Use q∞ = 0.5 × ρ∞ × V∞²
Dynamic pressure: half the density times the speed squared.
q∞
=
0.5
×
1.225
×
40
2
q∞ = 0.5 × 1.225 × 1,600
q∞ = 0.6125 × 1,600
q∞ = 980 Pa
Step 3 · Find lift coefficient (cₗ)
Use cₗ = L′ ÷ (q∞ × c)
The lift per span in units of q∞ times the chord.
cₗ
=
980
980 × 1.5
cₗ
=
980
1,470
cₗ = 0.6667
Check
Put the numbers back in. Both sides should match.
980 = 1.225 × 40 × 20 ✓
980
=
0.5
×
1.225
×
40
2
✓
0.6667
=
980
980 × 1.5
✓
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