Kutta condition and Kelvin's circulation theorem

Why it matters

Potential-flow theory gives a valid solution for any circulation round an airfoil, so on its own it cannot predict lift. The Kutta condition picks out the one value nature chooses, and Kelvin's theorem explains how that circulation appears (the starting vortex) and why trailing and shed vortices must exist. Together they make airfoil and wing theory predictive.

Key ideas

The non-uniqueness problem. For flow past a cylinder or an airfoil, any value of Γ produces a flow that satisfies Laplace's equation and the surface boundary condition. For an airfoil with a sharp trailing edge, most of these solutions have the flow turning round the trailing edge from the lower to the upper surface, which requires infinite velocity at the edge — physically impossible in a real (slightly viscous) fluid.

Kutta condition. The circulation round an airfoil with a sharp trailing edge is the value that makes the flow leave the trailing edge smoothly, with finite velocity. Equivalent statements:

• For a trailing edge with a finite included angle, the upper and lower velocities at the edge must both be zero — the trailing edge is the rear stagnation point.
• For a cusped trailing edge (zero angle), the upper and lower velocities at the edge are finite and equal.
• In either case the pressure is continuous across the trailing edge and, in vortex-sheet language, γ(TE) = 0.

It is an empirical condition that reproduces what viscosity does in the real flow; inviscid theory cannot derive it. It fails for blunt or rounded trailing edges (no fixed separation point) and once the flow has separated well upstream of the trailing edge (stall).

Kelvin's circulation theorem. For an inviscid, barotropic fluid (density a function of pressure only — incompressible flow qualifies) with conservative body forces, the circulation round a closed curve that always consists of the same fluid particles does not change with time: DΓ/Dt = 0. Consequences:

• A flow that starts irrotational stays irrotational (outside regions where viscosity acts) — the justification for potential flow round streamlined bodies.
• When an airfoil starts from rest, a large fluid curve surrounding it has Γ = 0 and must keep Γ = 0.

The starting vortex. As the airfoil accelerates, the lower-surface flow briefly tries to turn round the sharp trailing edge; viscosity makes it separate, and a vortex rolls up and is shed downstream. It has strength −Γ, equal and opposite to the bound circulation Γ that remains about the airfoil. Kelvin's theorem is satisfied: the total round a curve enclosing both is zero. Once the Kutta condition is met, shedding stops. The starting vortex moves away downstream and its influence on the airfoil soon becomes negligible.

Unsteady changes. Every change in angle of attack or speed changes the required Γ, so a vortex of strength −ΔΓ is shed into the wake each time. This is the basis of unsteady airfoil theory (Wagner, Theodorsen) and of the delay in lift build-up after a sudden change of α.

Helmholtz's vortex theorems (also inviscid): the strength of a vortex filament is constant along its length, and a filament cannot end in the fluid — it must form a closed loop or reach a boundary. This forces a finite wing's bound vortex to turn downstream at the tips as trailing vortices and close through the starting vortex (horseshoe vortex, later topic).

Formulas

• Γ = ∮ V·ds — circulation (m²/s).
• DΓ/Dt = 0 — Kelvin; inviscid, barotropic, conservative body forces, curve moving with the fluid.
• γ(TE) = 0 — Kutta condition for a vortex-sheet model; γ in m/s.
• Γ_bound + Γ_starting = 0 — airfoil started from rest.
• Γ = ½·c_l·V∞·c — bound circulation from lift coefficient (from L′ = ρ∞V∞Γ).
• ΔΓ_shed = −ΔΓ_bound = −½·Δc_l·V∞·c — vortex shed when the lift changes.
• V_ind = Γ/(2πd) — speed induced by a 2-D vortex at distance d (m).

Worked examples

Example 1 (standard). A symmetric airfoil of chord 1.2 m starts from rest and reaches a steady 40 m/s at α = 5° in air (ρ = 1.225 kg/m³). Take c_l = 2πα (thin airfoil theory). Find the bound circulation, the strength of the starting vortex, the lift per span, and the speed the starting vortex induces at the airfoil when it is 12 m downstream.

1. c_l = 2π × (5π/180) = 2π × 0.08727 = 0.548.
2. Γ = ½·c_l·V∞·c = 0.5 × 0.548 × 40 × 1.2 = 13.16 m²/s.
3. Kelvin: Γ_start = −13.16 m²/s (opposite sense).
4. L′ = ρ∞V∞Γ = 1.225 × 40 × 13.16 = 645 N/m.
5. V_ind = |Γ_start|/(2πd) = 13.16/(2π × 12) = 0.175 m/s, about 0.4 % of V∞ — negligible.

Answer: Γ = 13.2 m²/s, starting vortex −13.2 m²/s, L′ ≈ 645 N/m, induced speed ≈ 0.17 m/s.

Example 2 (GATE level). An airfoil of chord 1.5 m flies at 50 m/s. Its angle of attack is suddenly increased from 2° to 6°. Using c_l = 2πα, find the bound circulation before and after, and the strength and sense of the vortex shed into the wake.

1. Γ = ½(2πα)V∞c = παV∞c.
2. Before: Γ₁ = π × (2π/180) × 50 × 1.5 = 8.22 m²/s.
3. After: Γ₂ = π × (6π/180) × 50 × 1.5 = 24.67 m²/s.
4. ΔΓ_bound = 24.67 − 8.22 = 16.45 m²/s (clockwise for upward lift, flow left to right).
5. Kelvin: the wake must receive −16.45 m²/s, i.e. an anticlockwise vortex of 16.45 m²/s.

Answer: Γ rises from 8.22 to 24.67 m²/s; a vortex of 16.45 m²/s of opposite (anticlockwise) sense is shed.

Common mistakes

• Saying the Kutta condition makes the trailing-edge velocity infinite — it is precisely what keeps it finite.
• Assuming the trailing-edge velocity is always zero. That is true for a finite-angle edge; for a cusp it is finite and equal on both surfaces.
• Applying Kelvin's theorem to a curve fixed in space, or across a region where viscosity is active.
• Thinking the starting vortex "uses up" the lift. Its far-away influence vanishes; the bound circulation remains.
• Forgetting the sign: the shed vortex is opposite to the change in bound circulation.

For GATE AE

Expect conceptual MCQs on the forms of the Kutta condition, the assumptions behind Kelvin's theorem, the starting vortex and Helmholtz's theorems; and numericals giving c_l or lift and asking for bound or shed circulation. Practise stating the assumptions exactly — options often differ by "inviscid" versus "irrotational" or "fixed curve" versus "material curve".

Quick check

1. State the Kutta condition for a finite-angle trailing edge.
2. What are the conditions for Kelvin's theorem?
3. An airfoil started from rest has Γ = 9 m²/s. What is the circulation of the starting vortex?
4. Why can a vortex filament not end in the fluid?

Answers: 1. The trailing edge is a stagnation point; velocities on both surfaces there are zero. 2. Inviscid, barotropic fluid, conservative body forces, closed curve moving with the fluid. 3. −9 m²/s. 4. Helmholtz's theorem: filament strength is constant along its length, so it must close or reach a boundary.