Worked examples — Computational aerodynamics — panel method (intro), CFD overview

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The scenario matrix

Every panel-method / CFD question you will meet falls into one of these cells. Each worked example below is tagged with the cell it fills.

Cell	Case class	What is special about it	Example
A	Angle sign & quadrant of ${\beta }_i$	$\cos {\beta }_i$ flips sign → source blows in or out	Ex 1
B	Degenerate: single flat panel	Self-influence only, no neighbours	Ex 2
C	Zero input: ${\lambda }_j=0$ everywhere	What does "no body" look like?	Ex 3
D	Closed body mass balance $\sum {\lambda }_j{\ell }_j=0$	Sign mix forced by conservation	Ex 4
E	Full small system $N=2$ solved by hand	Diagonal vs off-diagonal	Ex 5
F	Limiting behaviour $N\to \infty$	Convergence and cost $N^2$–$N^3$	Ex 6
G	Real-world word problem	Choose panels vs CFD, size the run	Ex 7
H	Exam twist: lift & $d^{\prime }$Alembert	Why sources give zero lift/drag; add vortex	Ex 8

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Symbols we will keep re-using

The one equation the whole method rests on — zero air through the wall at every panel midpoint:

Everything below is just this equation, examined case by case.

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Cell A — the sign of $\cos {\beta }_i$

How to read this figure. The thick cyan horizontal line is the panel (the solid wall). The white arrow pointing straight up is its outward normal $\hat{n}$. The three coloured arrows aimed at the panel are the wind, each drawn at a different angle $\beta$ to that normal: the amber arrow is $\beta =0^{\circ }$ (wind dead-on along $\hat{n}$, so $\cos \beta =+1$), the cyan arrow is $\beta =60^{\circ }$ ($\cos \beta =+0.5$), and the white arrow is $\beta =150^{\circ }$ ($\cos \beta =-0.87$). The label at the bottom marks the grazing case $\beta =90^{\circ }$ where the wind runs along the panel and never crosses it. As you read Ex 1, match each case to its arrow: the sign of $\cos \beta$ (how much the arrow leans along $+\hat{n}$ vs against it) is exactly what decides which way the source must blow.

Forecast: guess — when the wind hits the front of the panel, the source must push back out ($\lambda >0$). When the wind is behind, it must suck in ($\lambda <0$). At $90^{\circ }$ the wind never crosses the wall, so no source is needed.

1. Compute $V_{\infty }\cos \beta$ for each. Why this step? This term is the wind's velocity component along the outward normal — the very flow we must cancel. Match each to its arrow in the figure.

   • (a) $10\cos 0^{\circ }=+10$ m/s (amber arrow)
   • (b) $10\cos 60^{\circ }=+5$ m/s (cyan arrow)
   • (c) $10\cos 90^{\circ }=0$ m/s (grazing, bottom label)
   • (d) $10\cos 150^{\circ }=-8.66$ m/s (white arrow)

2. Read the sign. Why this step? The master row says the source term must equal $-V_{\infty }\cos \beta$. From the self-term box above, a single panel's sheet induces normal velocity $+\lambda /2$ at its own midpoint (half the sprayed air exits each face), so $\frac{1}{2}\lambda =-V_{\infty }\cos \beta$, i.e. $\lambda =-2V_{\infty }\cos \beta$.

   • (a) $\lambda =-20$ m/s ... wait — negative? Look at the figure: with $\hat{n}$ pointing into the oncoming wind, $\cos \beta >0$ means wind flows out through the wall, so the source must suck in to cancel it. Sign is honest; it is the geometry that surprises.
   • (b) $\lambda =-2(10)\cos 60^{\circ }=-2(10)(0.5)=-10$ m/s: same story as (a) but weaker — the wind leans only halfway along $\hat{n}$, so half as much suck-in.
   • (c) $\lambda =-2(10)\cos 90^{\circ }=0$: grazing wind needs no correction.
   • (d) $\lambda =-2(10)\cos 150^{\circ }=+17.3$ m/s: wind pushes into the body, source pushes back out.

Verify: In every case $V_{\infty }\cos \beta +\frac{1}{2}\lambda =0$. Plug (b): $5+\frac{1}{2}(-10)=0$. ✓ Units: (m/s) + (m/s) since $\lambda$ is a per-length sheet strength carrying velocity units m/s — dimensionally clean.

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Cell B — the degenerate single flat panel

Forecast: with only one panel there are no off-diagonal terms — the matrix is a single number. Guess $\lambda \approx -2V_{\infty }$.

1. Write the master row. Why? With $N=1$ the sum has one term. $V_{\infty }\cos {\beta }_1+\frac{{\lambda }_1}{2\pi }{I}_{11}=0$
2. Substitute $I_{11}=\pi$. Why? The self-influence is the diagonal; $\frac{\pi }{2\pi }=\frac{1}{2}$ — exactly the "half out each face" jump. $8(1)+\frac{{\lambda }_1}{2\pi }\pi =8+\frac{{\lambda }_1}{2}=0$
3. Solve. ${\lambda }_1=-16$ (m/s).

Verify: $8+\frac{1}{2}(-16)=0$. ✓ This is exactly why the diagonal of the influence matrix $A$ is $\frac{1}{2}$ — a panel's effect on itself. A single panel is the smallest sanity check of the whole method.

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Cell C — the zero input

Forecast: no sources = no body = just the wind blowing undisturbed.

1. Put ${\lambda }_j=0$ into the total potential. Why? To see what "no body" produces. With all panels off, only the uniform-stream potential survives: $\varphi =V_{\infty }(x\cos \alpha +y\sin \alpha )$
2. Take the gradient to get velocity. Why? By definition of the potential, $\vec{V}=\nabla \varphi$ — differentiating the height-map in each direction gives that velocity component. $\vec{V}=(V_{\infty }\cos \alpha ,V_{\infty }\sin \alpha )$ A perfectly uniform stream — straight parallel streamlines.
3. When is it correct? Only if the wall happens to lie along a streamline already, i.e. the "body" is a flat plate parallel to the wind (${\beta }_i=90^{\circ }$ everywhere, from Ex 1c). Then $\cos {\beta }_i=0$ and indeed $\lambda =0$ solves every row.

Verify: every master row reads $V_{\infty }\cos 90^{\circ }+0=0$. ✓ Zero input is not a bug — it is the answer when the body doesn't disturb the flow.

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Cell D — mass balance forces mixed signs

Forecast: front sprays out to split the incoming air, so rear must suck in to close it up — ${\lambda }_3$ negative.

1. Apply the general closure condition. Why? A solid closed body creates no net air, so the total volume emitted (per unit span) must vanish. Each panel emits ${\lambda }_j{\ell }_j$, so the general requirement is ${\sum }_j{\lambda }_j{\ell }_j=0.$ The per-length strengths ${\lambda }_j$ only sum to zero when every panel length is equal — which is why we must state the length-weighted form as the master condition.
2. Specialise to equal lengths. Why? Here all ${\ell }_j=0.5$, a common factor we can divide out: $0.5({\lambda }_1+{\lambda }_2+{\lambda }_3)=0\Rightarrow 6+2+{\lambda }_3=0.$
3. Solve. ${\lambda }_3=-8$ m/s.
4. Interpret. Why? Negative strength = a sink: the rear panel swallows exactly the volume the front two ejected, so the air smoothly rejoins behind the body.

Verify: length-weighted sum $0.5(6)+0.5(2)+0.5(-8)=0$. ✓ The mix of $+$ and $-$ signs is forced by conservation, not chosen — this is why source-only bodies are front-to-back symmetric in output.

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Cell E — full $N=2$ system by hand

Forecast: panel 1 faces the wind (needs suck-in, ${\lambda }_1<0$ per the Ex-1 convention), panel 2 is sheltered — guess mixed signs.

The figure shows the two-panel wedge with the amber control points at each midpoint. The curved white arrow marks the self-influence ($A_{11}=\frac{1}{2}$, a panel acting on its own midpoint — the "half out each face" jump), and the straight cyan arrow marks the neighbour influence ($A_{12}=\frac{1}{3}$, panel 2 acting on control point 1). Keep those two arrows in mind: the diagonal is a panel talking to itself, the off-diagonal is one panel talking to another.

1. Assemble $A\mathbf{\lambda }=\mathbf{b}$. Why? This is Step 5 of the parent — one row per panel. Diagonal entries are the self-arrow, off-diagonals the neighbour-arrow; the right-hand side is $\mathbf{b}$ from the box above. $\{\begin{array}{l}\frac{1}{2}{\lambda }_1+\frac{1}{3}{\lambda }_2=-10(0.8)=-8\\ \frac{1}{3}{\lambda }_1+\frac{1}{2}{\lambda }_2=-10(-0.6)=+6\end{array}$
2. Clear the fractions by multiplying both rows by 6. Why the number 6? It is the least common multiple of the denominators $2$ and $3$, so multiplying by 6 turns every fraction into a whole number in one stroke — the cleanest way to avoid fraction arithmetic. Row 1 $\times 6$: $3{\lambda }_1+2{\lambda }_2=-48$. Row 2 $\times 6$: $2{\lambda }_1+3{\lambda }_2=36$.
3. Solve. Why? Two equations, two unknowns — standard elimination the computer would do. Multiply row 1 by 3 and row 2 by 2 to match the ${\lambda }_2$ coefficient, then subtract:
   • $9{\lambda }_1+6{\lambda }_2=-144$, $4{\lambda }_1+6{\lambda }_2=72$ → $5{\lambda }_1=-216$ → ${\lambda }_1=-43.2$.
   • Back-substitute: $2(-43.2)+3{\lambda }_2=36\Rightarrow 3{\lambda }_2=122.4\Rightarrow {\lambda }_2=40.8$.

Verify: Row 1: $\frac{1}{2}(-43.2)+\frac{1}{3}(40.8)=-21.6+13.6=-8$ ✓. Row 2: $\frac{1}{3}(-43.2)+\frac{1}{2}(40.8)=-14.4+20.4=6$ ✓. The diagonal $\frac{1}{2}$ (self, white arrow) dominates; the off-diagonal $\frac{1}{3}$ (neighbour, cyan arrow) nudges it — exactly the structure the figure shows.

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Cell F — the limit $N\to \infty$

Forecast: the numbers crowd toward $\approx 0.549$ — accuracy plateaus. Cost, though, explodes.

1. Look at successive changes. Why? Convergence means the change shrinks each refinement. $\Delta =0.040,0.008,0.001$. Each step is roughly $\frac{1}{5}$ of the last: $8/40=0.20$, $1/8=0.125$. The changes are shrinking, so the sequence is converging toward $C_L\approx 0.549$.
2. Define the panel size $h$ and why it matters. Why this step? "$N\to \infty$" is really "$h\to 0$," so we must say what $h$ is. If the body's perimeter is $\mathcal{P}$ and we use $N$ equal panels, then $h=\mathcal{P}/N$ is the length of one panel — the discretisation length. Doubling $N$ halves $h$. For a smooth body, standard panel-method error theory says the error falls like a fixed power of $h\propto 1/N$, so halving $h$ multiplies the error by a roughly constant factor each refinement. That constant factor is exactly what makes the successive changes behave like a geometric series. We use it only as a rough extrapolation, not an exact law.
3. Estimate the plateau (part a). Why? Treating the tail as geometric with ratio $\approx 0.2$, the remaining changes after $0.001$ sum to about $0.001\times \frac{0.2}{1-0.2}=0.00025$, so the limit $\approx 0.549+0.00025\approx 0.5493$. Yes, it is converging.
4. Cost scaling (part b). Why? Dense matrix inversion is $\mathcal{O}(N^3)$. Going $80\to 800$ is $\times 10$ in $N$, so time scales by $10^3=1000$. $t=2.0\times {\left(\frac{800}{80}\right)}^3=2.0\times 1000=2000\text{s}\approx 33\text{min}.$

Verify: $C_L$ change ratios $8/40=0.2$ and $1/8=0.125$ — both $<1$, so bounded/convergent. ✓ Cost: $2.0\times (800/80{)}^3=2.0\times 1000=2000$ s ✓. Lesson: more panels stop helping accuracy long before they stop costing you — the parent's "accuracy plateaus" mistake, quantified.

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Cell G — real-world word problem

Forecast: low speed, attached, no shocks → panel method wins; but drag forces a switch.

1. Check the Mach number. Why this step? Compressibility matters when $M=V_{\infty }/a$ approaches $\sim 0.3$; below that, incompressible Laplace is fine. Here $a$ is the speed of sound in the air — how fast a pressure disturbance travels — given as $a=340$ m/s at sea level: $M=\frac{V_{\infty }}{a}=\frac{50}{340}=0.147<\mathrm{0.3.}$ Incompressible assumption holds → Laplace is valid → panel method applicable.
2. Check the Reynolds number (context for whether viscosity is thin). Why? $Re=\frac{V_{\infty }c}{\nu }$ tells us the boundary layer is thin relative to the chord. $Re=\frac{50\times 1.2}{1.5\times 10^{-5}}=4.0\times 10^6.$ High $Re$, attached flow → inviscid pressure prediction is accurate.
3. Choose the tool. Why? Panel method gives ~80% of the lift info for ~1% of the cost (parent's 80/20 rule). Use the panel method for lift/pressure.
4. When drag is needed. Why? Inviscid 2-D ⇒ d'Alembert's Paradox ⇒ zero drag. To get real drag (skin friction + wake) you must add a boundary-layer correction or move to full Navier–Stokes CFD.

Verify: $M=50/340=0.147$ ✓ ($<0.3$). $Re=50\cdot 1.2/1.5\text{e-}5=4\times 10^6$ ✓. Both dimensionless. Decision consistent with the parent's comparison table.

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Cell H — exam twist on lift

Forecast: (a) no circulation → no lift, by Kutta–Joukowski. (b) solve $L^{\prime }=\rho V_{\infty }\Gamma$ for $\Gamma$.

1. Apply Kutta–Joukowski. Why? It is the link between circulation and lift: $L^{\prime }=\rho V_{\infty }\Gamma$. With $\Gamma =0$: $L^{\prime }=\rho V_{\infty }(0)=0$. Sources alone give zero lift — they only shape thickness.
2. Relate $C_L$ to $L^{\prime }$. Why? Lift coefficient nondimensionalises lift by dynamic pressure $\times$ chord. $L^{\prime }=C_L\cdot \frac{1}{2}\rho V_{\infty }^{2}c=0.6\times \frac{1}{2}(1.225)(50^2)(1.2)$ $L^{\prime }=0.6\times \frac{1}{2}(1.225)(2500)(1.2)=1102.5\text{N/m}.$
3. Solve for $\Gamma$. Why? Invert Kutta–Joukowski. $\Gamma =\frac{L^{\prime }}{\rho V_{\infty }}=\frac{1102.5}{1.225\times 50}=18.0{\text{m}}^2/\text{s}.$

Verify: $L^{\prime }=0.6\cdot 0.5\cdot 1.225\cdot 2500\cdot 1.2=1102.5$ N/m ✓. $\Gamma =1102.5/(1.225\cdot 50)=18.0$ m²/s ✓. Reinsert: $\rho V_{\infty }\Gamma =1.225\cdot 50\cdot 18.0=1102.5=L^{\prime }$ ✓. This is why we bolt vortex panels + the Kutta condition onto a source panel code — sources set shape, vortices set lift.

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Recall

Recall Hide answers and test the whole matrix

What sign of $\cos \beta$ means the wind blows through the wall the same way $\hat{n}$ points? ::: Positive $\cos \beta$ (wind component along $+\hat{n}$). Where does the $\frac{1}{2}$ on the matrix diagonal come from? ::: A source sheet sends half its air out each face, so it induces $\lambda /2$ at its own midpoint ($I_{ii}=\pi$). A single flat panel with $I_{11}=\pi$, $V_{\infty }=8$: what is $\lambda$? ::: $-16$. All ${\lambda }_j=0$ describes which physical situation? ::: No body — undisturbed uniform stream (or a plate parallel to the wind). On a closed body why must $\sum {\lambda }_j{\ell }_j=0$ (not just $\sum {\lambda }_j=0$)? ::: Each panel emits ${\lambda }_j{\ell }_j$; only when all lengths are equal do the per-length strengths sum to zero. Cost scales like $N^3$: going from $N=80$ to $N=800$ multiplies runtime by? ::: $1000$. Why does a source-only body have zero lift? ::: Zero circulation → $L^{\prime }=\rho V_{\infty }\Gamma =0$. To get lift you add what, controlled by which condition? ::: Vortex panels, set by the Kutta condition.