Exercises — Computational aerodynamics — panel method (intro), CFD overview

The equation is Laplace's equation: ${\nabla }^2\varphi =0$ Here $\varphi$ ("phi") is the velocity potential — a single scalar field whose slope (gradient) gives the velocity everywhere, $\vec{V}=\nabla \varphi$. Linearity means: if ${\varphi }_1$ and ${\varphi }_2$ each solve it, so does ${\varphi }_1+{\varphi }_2$. That is superposition. It lets us add up a freestream plus many little sources and still have a valid flow — which is the entire engine of the panel method.

• (a) Uniform stream — air all flowing the same direction at speed $V_{\infty }$; represents the wind far away.
• (b) Source — pushes air out equally in all directions; used to give a body its thickness.
• (c) Vortex — makes air circle around a point; the only brick that produces circulation, and hence lift.
• (d) Doublet — a source and sink squeezed together; the combination that exactly models flow around a cylinder.

WHAT we do: plug the ratio in. WHY: $C_p$ compares local speed to freestream — faster air means lower pressure (that's Bernoulli's Equation). $C_p=1-(1.4{)}^2=1-1.96=-0.96$ The negative value means the pressure here is below freestream — this is a suction region, exactly what happens on the upper surface of a wing where the flow speeds up.

With $V_t=0$: $C_p=1-0^2=1$ Since ${\left(V_t/V_{\infty }\right)}^2\ge 0$ always, the term we subtract is never negative, so $C_p$ can never exceed $1$. Maximum $C_p=1$, achieved only where the flow stops. Every other point has $C_p<1$.

WHAT: project the freestream vector onto the normal. WHY: the boundary condition we ultimately enforce is zero normal velocity; here we measure how much the bare freestream violates it, which is exactly the right-hand side $b_i$ of the panel system. The component of ${\vec{V}}_{\infty }$ along $\hat{n}$ is $V_{\infty }\cos \beta =V_{\infty }\cos 60^{\circ }=0.5V_{\infty }$ So the freestream alone drives $0.5V_{\infty }$ through the wall. The sources must be tuned to cancel exactly this: $b_i=-V_{\infty }\cos {\beta }_i=-0.5V_{\infty }$.

Why the sum is zero: the body is solid and closed — no air is created or destroyed inside it. Every bit of air a source sprays out must be swallowed somewhere by a sink (negative source). So the strengths cancel: $\sum {\lambda }_j=0$. This is just conservation of mass. Consequence for drag: a pure source distribution has no circulation and, being fore-aft symmetric in its pressure pattern, produces zero net drag in 2-D inviscid flow. This is exactly d'Alembert's Paradox: an ideal fluid gives a body no drag at all. Real drag comes from viscosity and the wake, which live in Boundary Layer Theory & Skin Friction Drag, not in this model.

WHAT: build the right-hand side, then solve a $2\times 2$ system. WHY: the diagonal $\frac{1}{2}$ is a panel's effect on itself; the off-diagonal $\frac{1}{4}$ is the neighbour's influence. Solving shows how strengths adjust to enforce flow tangency at both control points. Right-hand sides: $b_1=-\cos 60^{\circ }=-0.5,b_2=-\cos 120^{\circ }=+0.5$ System: $0.5{\lambda }_1+0.25{\lambda }_2=-0.5$ $0.25{\lambda }_1+0.5{\lambda }_2=0.5$ Subtract to exploit symmetry. Multiply the first by 2: ${\lambda }_1+0.5{\lambda }_2=-1$. Multiply the second by 2: $0.5{\lambda }_1+{\lambda }_2=1$. Solve: From the first, ${\lambda }_1=-1-0.5{\lambda }_2$. Substitute: $0.5(-1-0.5{\lambda }_2)+{\lambda }_2=1\Rightarrow -0.5+0.75{\lambda }_2=1\Rightarrow {\lambda }_2=2$ Then ${\lambda }_1=-1-0.5(2)=-2$. $\boxed{{\lambda }_1=-2,{\lambda }_2=2}$ Notice ${\lambda }_1+{\lambda }_2=0$ — mass is conserved automatically, exactly as L3.1 predicted.

WHAT: apply $L^{\prime }=\rho V_{\infty }\Gamma$. WHY: lift in ideal flow is entirely set by circulation; this formula is the bridge from the vortex strength (which the Kutta condition fixed) to a force. $L^{\prime }=\rho V_{\infty }\Gamma =1.225\times 50\times 12=735\text{N/m}$ The source panels only shaped the body's thickness and enforced flow tangency; they carry $\sum {\lambda }_j=0$ and produce no circulation, so they add exactly $0$ to $L^{\prime }$. All the lift came from the vortex part.

Using $C_p=1-(V_t/V_{\infty }{)}^2$:

• Point 1: $V_t=0\Rightarrow C_p=1-0=1.000$ → stagnation point (flow stopped, highest pressure).
• Point 2: $V_t=65\Rightarrow C_p=1-(65/50{)}^2=1-1.69=-0.690$ → suction (fast flow, low pressure — typical upper surface).
• Point 3: $V_t=40\Rightarrow C_p=1-(40/50{)}^2=1-0.64=0.360$ → mild overpressure (slower than freestream — typical lower surface near the front). The pattern (suction on top, overpressure on the bottom) is what produces net lift when integrated around the surface.

Cost factor: cost $\propto N^3$, and $N$ went up by $4\times$, so $\text{factor}=4^3=64.$ The solve becomes 64 times more expensive. Why accuracy plateaus: the panel method solves the inviscid Laplace problem. Once the mesh is fine enough to resolve the geometry, the remaining error is not discretisation error — it's model error: the physics you left out (viscosity, boundary layer, separation, compressibility). No number of panels can recover those. To go further you must switch to Navier–Stokes Equations-based Finite Volume Method CFD.

(a) The code is not broken — it is behaving exactly as an inviscid method must. In 2-D inviscid, incompressible flow, d'Alembert's Paradox guarantees zero drag. The prediction of $0$ is correct for the model. (b) Missing physics: viscosity. Real drag here is skin-friction drag from the thin viscous layer on the surface (see Boundary Layer Theory & Skin Friction Drag), plus any pressure/wake drag. To capture it you solve the Navier–Stokes Equations on a volume mesh with a Finite Volume Method and a turbulence model (Turbulence Modelling — RANS, LES, DNS), or bolt a boundary-layer correction onto the panel solution. The measured $C_d=0.008$ is a purely viscous effect the ideal model cannot see.

Max possible $C_p$ in incompressible flow, and where? ::: $C_p=1$, at a stagnation point. Why must $\sum {\lambda }_j=0$ on a closed body? ::: Conservation of mass — no air created inside. Formula linking lift to circulation? ::: $L^{\prime }=\rho V_{\infty }\Gamma$. Cost scaling of a dense direct panel solve in $N$? ::: About $N^3$. Why does an inviscid airfoil show zero drag? ::: d'Alembert's paradox — no viscosity, no wake.