Panel methods for airfoil analysis

Why it matters

Analytical methods handle cylinders, Joukowski shapes and thin airfoils; real sections are thick, arbitrary shapes. Panel methods solve the same potential-flow problem numerically for any shape in a fraction of a second, which is why they sit inside design tools such as XFOIL and why they are still the first step of most airfoil and low-speed aircraft analyses.

Key ideas

Idea. Because Laplace's equation is linear, the flow round a body can be built from a uniform stream plus singularities (sources, doublets, vortices) distributed over the body surface. Their unknown strengths are chosen so that the body surface becomes a streamline.

Discretisation.

1. Divide the surface into N straight panels, ordered from the trailing edge round the lower surface, the leading edge and back along the upper surface.
2. On each panel j put a singularity of uniform (or linearly varying) strength — source strength λ_j or vortex strength γ_j, both per unit length (m/s).
3. Choose one control point on each panel (usually its midpoint).
4. At every control point i, write the flow-tangency condition: the normal component of (free stream + velocity induced by all N panels) is zero.
5. This gives N linear equations Σ_j A_ij·x_j = −V∞·n_i, where A_ij is the geometric influence coefficient of panel j on control point i. Solve for the strengths.
6. Compute the tangential velocity V_s at each control point, then C_p = 1 − (V_s/V∞)², and integrate C_p for forces and moments.

Source panel method. Good for non-lifting bodies. A source panel induces a normal velocity of λ_i/2 on itself (just outside it); the other panels contribute through integrals of ∂(ln r)/∂n. Because sources carry no circulation, a pure source solution gives zero lift. For a closed body, mass conservation gives a check: Σ λ_j S_j = 0.

Vortex panel method. For lifting airfoils. Vortex panels carry circulation, so lift appears naturally: Γ = Σ γ_j S_j and L′ = ρ∞V∞Γ. The system is completed by the Kutta condition at the trailing edge, γ(TE) = 0, approximated as γ₁ + γ_N = 0 for the two trailing-edge panels; one control-point equation is dropped to keep the system square.

Combined (Hess–Smith) method. Sources of varying strength on each panel represent thickness, plus a single vortex strength common to all panels represents circulation; the Kutta condition is imposed as equal tangential velocities on the two trailing-edge panels. This is robust and widely used.

Lumped-vortex (one-panel) model. The simplest vortex panel: put a point vortex at the quarter chord of each panel and the control point at the three-quarter chord. For a single flat panel this reproduces exactly the thin-airfoil result c_l = 2πα — the basis of the vortex-lattice method for wings.

Accuracy and limits. Results converge as N increases (typically 50–200 panels per airfoil), with panels clustered near the leading and trailing edges where curvature and gradients are high. The method remains inviscid and incompressible: no skin friction or separation, so no drag and no stall, and lift slightly over-predicted. Viscous–inviscid coupling (boundary-layer codes) and compressibility corrections extend it.

Formulas

• V∞·cos β_i + λ_i/2 + Σ_{j≠i} (λ_j/2π)·I_ij = 0 — source-panel tangency at control point i; β_i is the angle between V∞ and the outward normal n_i; I_ij = ∫ ∂(ln r_ij)/∂n_i ds_j.
• V_s,i = V∞·sin β_i + Σ_{j} (λ_j/2π)·J_ij — tangential velocity, J_ij = ∫ ∂(ln r_ij)/∂s ds_j.
• Σ λ_j·S_j = 0 — closed non-lifting body; S_j panel length (m), λ_j (m/s).
• C_p,i = 1 − (V_s,i/V∞)² — surface pressure coefficient.
• Γ = Σ γ_j·S_j (m²/s), L′ = ρ∞V∞Γ (N/m), c_l = 2Γ/(V∞c) — vortex panels.
• γ₁ + γ_N = 0 — discrete Kutta condition.
• Lumped vortex, one flat panel: Γ/(2π·c/2) = V∞·α ⇒ Γ = πcV∞α, c_l = 2πα.

Worked examples

Example 1 (standard). A source-panel solution for a closed non-lifting body gives tangential velocity ratios V_s/V∞ = 0.2, 1.5, 1.9 and 1.2 at four control points. Find C_p at each. Three of the body's panels have λ_jS_j = 2.0, −0.5 and −0.8 m²/s; the fourth panel has length 0.35 m. What source strength must it have?

1. C_p = 1 − (V_s/V∞)²: 0.2 → 1 − 0.04 = 0.96; 1.5 → 1 − 2.25 = −1.25; 1.9 → 1 − 3.61 = −2.61; 1.2 → 1 − 1.44 = −0.44.
2. Closure: Σ λ_jS_j = 0 ⇒ λ₄S₄ = −(2.0 − 0.5 − 0.8) = −0.7 m²/s.
3. λ₄ = −0.7/0.35 = −2.0 m/s (a sink panel, near the rear of the body).

Answer: C_p = 0.96, −1.25, −2.61, −0.44; λ₄ = −2.0 m/s. The point with C_p close to 1 is near a stagnation point.

Example 2 (GATE level). Model a flat-plate airfoil of chord 1.2 m at α = 5° in air (ρ = 1.225 kg/m³, V∞ = 50 m/s) with a single lumped vortex Γ at the quarter chord and the control point at the three-quarter chord. Find Γ, c_l and L′.

1. The control point is c/2 = 0.6 m behind the vortex. The vortex (clockwise) induces a downward velocity w = Γ/(2π × 0.6) there.
2. Tangency on the plate: the free-stream normal component V∞ sin α ≈ V∞α must be cancelled: Γ/(2π·c/2) = V∞α ⇒ Γ = πcV∞α.
3. α = 5π/180 = 0.08727 rad; Γ = π × 1.2 × 50 × 0.08727 = 16.45 m²/s.
4. c_l = 2Γ/(V∞c) = 2 × 16.45/(50 × 1.2) = 0.548 (= 2πα ✓).
5. L′ = ρ∞V∞Γ = 1.225 × 50 × 16.45 = 1008 N/m.

Answer: Γ = 16.4 m²/s, c_l = 0.548, L′ ≈ 1.01 kN/m — identical to thin airfoil theory.

Common mistakes

• Expecting lift from a pure source-panel method; sources carry no circulation.
• Forgetting the self-induced term λ_i/2 (the panel's own normal velocity at its control point).
• Adding velocities from different singularities as scalars; induced velocities are vectors and must be resolved into normal and tangential components.
• Omitting the Kutta condition in a vortex-panel method, which leaves the circulation undetermined (the system becomes singular for a closed body).
• Using too few panels near the leading edge, where the suction peak is.
• Treating panel results as including drag — the inviscid pressure drag of a closed body sums to almost zero.

For GATE AE

Expect conceptual MCQs on the boundary conditions (no penetration, Kutta), the role of each singularity type, the closure condition for source panels, and why source panels give no lift; and short numericals: C_p from tangential velocity, lift from Σγ_jS_j, a missing source strength from closure, and the lumped-vortex 1/4–3/4 chord result.

Quick check

1. Why does a source panel method give zero lift?
2. What is Σλ_jS_j for a closed body?
3. Vortex panels give Σγ_jS_j = 12 m²/s on a 1 m chord at V∞ = 40 m/s. Find c_l.
4. Where are the vortex and control point placed in the lumped-vortex model?

Answers: 1. Sources have no circulation, and L′ = ρV∞Γ. 2. Zero. 3. c_l = 2 × 12/(40 × 1) = 0.6. 4. Vortex at the quarter chord, control point at the three-quarter chord.