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Fluid dynamics · airfoil pressure workbench

Airfoil Circulation Studio

What sets how much circulation an airfoil carries, and so how much lift it makes? The Kutta condition, given the airfoil's shape and angle of attack. Ideal flow alone would allow any amount of circulation; the Kutta condition picks the one value at which the flow leaves the sharp trailing edge smoothly, at finite speed. Any other value would send the flow whipping around the edge infinitely fast. The circulation chosen this way sets the pressure difference and the lift.

In a three-dimensional wind tunnel you can rotate, the upper and lower surface pressures, a movable pressure probe, a pressure integral around the surface and the lift from circulation all come from the same Joukowski solution. Refine the pressure integration to see how closely it reproduces that lift and zero drag. Turn off the Kutta condition to see the singularity at the trailing edge; the pressures are not clipped to hide it.

Interactive modelAirfoil Circulation Studio
Wall / edge residual (relative to inflow)analytic\text{analytic}
Actual samples / panels128128
Net source-flux residual (normalized)analytic\text{analytic}
Circulation Γ\Gamma0 m2 s−10\,\mathrm{m^2\,s^{-1}}
Theory lift CLC_L00
Pressure-integral lift CL,pC_{L,p}00
Pressure-integral absolute error00
Pressure-integral drag CD,pC_{D,p}00
Edge limit / trailing-panel estimate VTE/U∞V_{\mathrm{TE}}/U_\infty00
Probe pressure difference ΔCp\Delta C_p00
Lift per span L′L'0 N m−10\,\mathrm{N\,m^{-1}}
Model boundary2D potential flow

Physics tutorial

Airfoil Circulation: how a trailing-edge condition selects lift

BackgroundFlow that is incompressible, inviscid and irrotational can be solved around a circle and then carried to an airfoil by a conformal map. Where the circle sits sets the airfoil’s camber and thickness, and the sharp trailing edge comes from a critical point of the map.

Why it mattersThe model cannot predict stall, but it does link geometry, circulation, surface speed, pressure difference and lift in one complete chain of cause and effect.

Start with the essentials

Focus question
Why does real attached flow pick only one of the many circulations the equations allow around an airfoil?
One-sentence intuition
The Kutta condition asks for finite speed and a smooth flow off the trailing edge. That leaves a single circulation Γ\Gamma, and the pressure difference it creates gives the lift.

Core mathematical model

NACA shape and the source/vortex panel solution

yt=5t(0.2969x−0.1260x−0.3516x2+0.2843x3−0.1036x4),V⋅n=0,Vt,1+Vt,N=0y_t=5t\left(0.2969\sqrt{x}-0.1260x-0.3516x^2+0.2843x^3-0.1036x^4\right),\quad \mathbf V\cdot\mathbf n=0,\quad V_{t,1}+V_{t,N}=0

The section uses the four-digit mean line and thickness polynomial from NASA TM 4741, with the last coefficient changed so the trailing edge closes. The Hess–Smith method gives each panel its own source strength and adds one uniform vortex-sheet strength, chosen so that no flow crosses the surface and the Kutta condition holds. Pressure, streamlines and circulation all come from that one solution. The surface residual, the net source-flux error and a pressure-force check recomputed at half resolution are shown separately from the analytic Joukowski mode. A small residual does not prove that the grid has converged.

The trailing-edge speed is finite, and usually not zero

VTEU∞=bR∣cos⁡(α−β)∣,β=arg⁡(b−ζc)\frac{V_{\mathrm{TE}}}{U_\infty}=\frac{b}{R}\left|\cos(\alpha-\beta)\right|,\qquad \beta=\arg(b-\zeta_c)

With the Kutta circulation, the numerator of the complex velocity and the derivative of the map both vanish at the trailing edge. Cancelling that common zero analytically leaves a finite speed, which is usually not zero. If the Kutta condition is off and the common zero disappears, the speed has no finite limit, so the pressure-force check is switched off instead of clipping the singularity.

Force from integrating surface pressure

Cx=−∮Cp d(y/c),Cy=∮Cp d(x/c),CL,p=−Cxsin⁡α+Cycos⁡αC_x=-\oint C_p\,\mathrm{d}(y/c),\quad C_y=\oint C_p\,\mathrm{d}(x/c),\quad C_{L,p}=-C_x\sin\alpha+C_y\cos\alpha

Integrate the unclipped surface pressure counterclockwise around the airfoil, then take the force component perpendicular to the free stream. Compare the result with the lift from circulation and with zero drag. The half-resolution error comes from repeating the calculation, not from an assumed accuracy. The contour integral follows MIT 18.354J, Lecture 20.

Joukowski conformal map

z=ζ+b2ζz=\zeta+\frac{b^2}{\zeta}

An offset circle in the preimage plane maps to an airfoil with a rounded leading edge and a sharp trailing edge.

Complex potential outside the circle

W(ζ)=U∞(e−iαη+eiαR2η)+iΓ2πlog⁡η,η=ζ−ζcW(\zeta)=U_\infty\left(e^{-i\alpha}\eta+e^{i\alpha}\frac{R^2}{\eta}\right)+\frac{i\Gamma}{2\pi}\log\eta,\qquad \eta=\zeta-\zeta_c

The exact flow around the circle is the sum of a uniform stream, the cylinder’s image term and a point vortex that carries the circulation.

Pressure and lift

Cp=1−∣VU∞∣2,L′=ρU∞Γ,CD=0C_p=1-\left\lvert\frac{V}{U_\infty}\right\rvert^2,\qquad L'=\rho U_\infty\Gamma,\qquad C_D=0

Bernoulli’s equation turns surface speed into pressure. When the speed stays finite at the trailing edge, the Kutta–Joukowski theorem gives the lift per unit span and the closed pressure integral gives zero drag. That zero drag shows what the model leaves out: viscosity. The pressure-integral check is not applied to singular solutions.

Common difficulties

The Kutta condition adds no force

Typical misconceptionTurning on the Kutta condition secretly adds a source of lift.

Better mental modelIt applies no external force. Among all the circulations the mathematics allows, it picks the one where the flow leaves the trailing edge smoothly at finite speed.

Real airfoils do have drag

Typical misconceptionIf the pressure field gives lift but no drag, a real airfoil should also fly without losses.

Better mental modelInviscid potential flow leaves out boundary layers, wakes and skin friction, so it must miss drag and stall. This mismatch is D’Alembert’s paradox.

Run the experiment

  1. 01

    Start with zero lift

    Choose Symmetric · zero lift and leave the Kutta condition on.

    What to observe: The streamlines and pressures above and below the airfoil are nearly symmetric, and circulation and lift are close to zero.
  2. 02

    Change only the angle of attack

    Choose Cruise section and watch the upper-surface pressure taps and the lift vector.

    What to observe: Flow over the upper surface speeds up and its pressure coefficient drops. The circulation picked by the Kutta condition rises, and the lift per span rises with it.
  3. 03

    Turn off the trailing-edge condition

    Choose Kutta off and compare the flow around the same airfoil.

    What to observe: With zero circulation the theory gives zero lift, but the speed at the trailing edge becomes singular. The pressure curve can run off the plot, and the raw data is kept. The pressure-force check becomes unavailable, so a clipped solution is never reported as valid.
  4. 04

    Compare the two lift calculations

    Return to Cruise section, raise the quadrature samples from N=32N=32 to N=128N=128, then move the pressure probe.

    What to observe: The lift from the pressure integral approaches the lift from circulation, and the numerical drag approaches zero. In the analytic Joukowski mode the probe reads the exact surface. In panel mode it interpolates between computed panel-midpoint pressures, so its reading can change as the panels are refined.

Conceptual inspiration: FluidMotion. Original simulations and graphics; the model limits are described above.

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