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

3.1.30 · D3 · Physics › Compressible Flow & Aerodynamics › Computational aerodynamics — panel method (intro), CFD overv

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The scenario matrix

Har panel-method / CFD question jo tumhe milegi woh in cells mein se kisi ek mein aayegi. Neeche har worked example us cell ke saath tagged hai jo woh fill karta hai.

Cell	Case class	Isme kya khaas hai	Example
A	Angle sign & quadrant of ${\beta }_i$	$\cos {\beta }_i$ ka sign flip hota hai → source andar blow karta hai ya bahar	Ex 1
B	Degenerate: single flat panel	Sirf self-influence, koi neighbour nahi	Ex 2
C	Zero input: ${\lambda }_j=0$ har jagah	"No body" kaisa dikhta hai?	Ex 3
D	Closed body mass balance $\sum {\lambda }_j{\ell }_j=0$	Sign mix conservation se forced hoti hai	Ex 4
E	Full small system $N=2$ haath se solve kiya	Diagonal vs off-diagonal	Ex 5
F	Limiting behaviour $N\to \infty$	Convergence aur cost $N^2$–$N^3$	Ex 6
G	Real-world word problem	Panels vs CFD choose karo, run size karo	Ex 7
H	Exam twist: lift & $d^{\prime }$Alembert	Sources zero lift/drag kyun dete hain; vortex add karo	Ex 8

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Symbols jo hum baar baar use karte rahenge

Woh ek equation jis par poora method tika hai — har panel midpoint par zero air through the wall:

Sab kuch neeche bas yahi equation hai, case by case examine ki gayi.

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Cell A — $\cos {\beta }_i$ ka sign

Yeh figure kaise padhein. Moti cyan horizontal line panel hai (solid wall). Seedha upar point karta white arrow uska outward normal $\hat{n}$ hai. Teen coloured arrows jo panel ki taraf aim kar rahe hain woh wind hai, har ek us normal se alag angle $\beta$ par bana hua hai: amber arrow $\beta =0^{\circ }$ hai (wind dead-on $\hat{n}$ ke saath, toh $\cos \beta =+1$), cyan arrow $\beta =60^{\circ }$ hai ($\cos \beta =+0.5$), aur white arrow $\beta =150^{\circ }$ hai ($\cos \beta =-0.87$). Neeche ka label grazing case $\beta =90^{\circ }$ mark karta hai jahan wind panel ke saath chalta hai aur kabhi use cross nahi karta. Jaise tum Ex 1 padhte ho, har case ko uske arrow se match karo: $\cos \beta$ ka sign (arrow $+\hat{n}$ ke saath ya uske against kitna jhuka hua hai) exactly yahi decide karta hai ki source kis taraf blow kare.

Forecast: guess karo — jab wind panel ke front se takraati hai, source ko wapas bahar push karna hoga ($\lambda >0$). Jab wind peeche se aaye, toh suck in karna hoga ($\lambda <0$). $90^{\circ }$ par wind kabhi wall cross nahi karti, toh koi source ki zaroorat nahi.

1. Har ke liye $V_{\infty }\cos \beta$ compute karo. Yeh step kyun? Yeh term wind ki velocity component outward normal ke along hai — bilkul wahi flow jo humein cancel karni hai. Har ek ko figure mein uske arrow se match karo.

   • (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, neeche ka label)
   • (d) $10\cos 150^{\circ }=-8.66$ m/s (white arrow)

2. Sign padho. Yeh step kyun? Master row kehti hai source term $-V_{\infty }\cos \beta$ ke barabar honi chahiye. Upar ke self-term box se, ek single panel ki sheet apne midpoint par normal velocity $+\lambda /2$ induce karti hai (sprayed air ka aadha har face se exit hota hai), toh $\frac{1}{2}\lambda =-V_{\infty }\cos \beta$, yaani $\lambda =-2V_{\infty }\cos \beta$.

   • (a) $\lambda =-20$ m/s ... ruko — negative? Figure dekho: $\hat{n}$ incoming wind mein point karta hua, $\cos \beta >0$ matlab wind wall ke through bahar flow ho rahi hai, toh source ko cancel karne ke liye suck in karna hoga. Sign honest hai; geometry hi surprise karti hai.
   • (b) $\lambda =-2(10)\cos 60^{\circ }=-2(10)(0.5)=-10$ m/s: same story as (a) par weaker — wind sirf aadha $\hat{n}$ ke along lean karti hai, toh aadha suck-in.
   • (c) $\lambda =-2(10)\cos 90^{\circ }=0$: grazing wind ko koi correction nahi chahiye.
   • (d) $\lambda =-2(10)\cos 150^{\circ }=+17.3$ m/s: wind body mein push karti hai, source wapas bahar push karta hai.

Verify: Har case mein $V_{\infty }\cos \beta +\frac{1}{2}\lambda =0$. (b) plug karo: $5+\frac{1}{2}(-10)=0$. ✓ Units: (m/s) + (m/s) kyunki $\lambda$ ek per-length sheet strength hai jo velocity units m/s rakhti hai — dimensionally clean.

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Cell B — degenerate single flat panel

Forecast: sirf ek panel ke saath koi off-diagonal terms nahi hain — matrix ek single number hai. Guess karo $\lambda \approx -2V_{\infty }$.

1. Master row likho. Kyun? $N=1$ ke saath sum mein ek term hai. $V_{\infty }\cos {\beta }_1+\frac{{\lambda }_1}{2\pi }{I}_{11}=0$
2. $I_{11}=\pi$ substitute karo. Kyun? Self-influence diagonal hai; $\frac{\pi }{2\pi }=\frac{1}{2}$ — exactly "half out each face" jump. $8(1)+\frac{{\lambda }_1}{2\pi }\pi =8+\frac{{\lambda }_1}{2}=0$
3. Solve karo. ${\lambda }_1=-16$ (m/s).

Verify: $8+\frac{1}{2}(-16)=0$. ✓ Yahi reason hai ki influence matrix $A$ ki diagonal $\frac{1}{2}$ hai — ek panel ka khud par effect. Ek single panel poore method ka sabse chhota sanity check hai.

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Cell C — zero input

Forecast: no sources = no body = bas wind bina rukawat ke blow karti hai.

1. ${\lambda }_j=0$ total potential mein dalo. Kyun? Dekhne ke liye ki "no body" kya produce karta hai. Jab saare panels off hain, sirf uniform-stream potential bachta hai: $\varphi =V_{\infty }(x\cos \alpha +y\sin \alpha )$
2. Velocity paane ke liye gradient lo. Kyun? Potential ki definition se, $\vec{V}=\nabla \varphi$ — height-map ko har direction mein differentiate karne se woh velocity component milta hai. $\vec{V}=(V_{\infty }\cos \alpha ,V_{\infty }\sin \alpha )$ Bilkul uniform stream — seedhi parallel streamlines.
3. Kab correct hai? Sirf tab agar wall pehle se ek streamline ke along lie karti ho, yaani "body" ek flat plate hai wind ke parallel (${\beta }_i=90^{\circ }$ har jagah, Ex 1c se). Toh $\cos {\beta }_i=0$ aur $\lambda =0$ sach mein har row solve kar leta hai.

Verify: har master row padhti hai $V_{\infty }\cos 90^{\circ }+0=0$. ✓ Zero input ek bug nahi hai — yeh answer hai jab body flow ko disturb nahi karti.

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Cell D — mass balance mixed signs force karta hai

Forecast: front incoming air split karne ke liye bahar spray karta hai, toh rear ko suck in karna hoga close karne ke liye — ${\lambda }_3$ negative.

1. General closure condition apply karo. Kyun? Ek solid closed body net air nahi banata, toh total volume emitted (per unit span) zero hona chahiye. Har panel ${\lambda }_j{\ell }_j$ emit karta hai, toh general requirement hai ${\sum }_j{\lambda }_j{\ell }_j=0.$ Per-length strengths ${\lambda }_j$ sirf tab zero sum karte hain jab har panel length equal ho — yahi reason hai ki hume master condition ke roop mein length-weighted form state karni chahiye.
2. Equal lengths ke liye specialise karo. Kyun? Yahan sab ${\ell }_j=0.5$ hain, ek common factor jo hum divide kar sakte hain: $0.5({\lambda }_1+{\lambda }_2+{\lambda }_3)=0\Rightarrow 6+2+{\lambda }_3=0.$
3. Solve karo. ${\lambda }_3=-8$ m/s.
4. Interpret karo. Kyun? Negative strength = ek sink: rear panel exactly utna volume nig jaata hai jo front do ne eject kiya, toh air smoothly body ke peeche rejoin ho jaati hai.

Verify: length-weighted sum $0.5(6)+0.5(2)+0.5(-8)=0$. ✓ $+$ aur $-$ signs ka mix conservation se forced hai, choose nahi kiya gaya — yahi reason hai ki source-only bodies output mein front-to-back symmetric hoti hain.

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Cell E — full $N=2$ system haath se

Forecast: panel 1 wind face karta hai (suck-in chahiye, ${\lambda }_1<0$ Ex-1 convention se), panel 2 sheltered hai — mixed signs guess karo.

Figure do-panel wedge dikhata hai amber control points ke saath har midpoint par. Curved white arrow self-influence mark karta hai ($A_{11}=\frac{1}{2}$, ek panel apne midpoint par act karta hua — "half out each face" jump), aur seedha cyan arrow neighbour influence mark karta hai ($A_{12}=\frac{1}{3}$, panel 2 control point 1 par act karta hua). Un do arrows ko dhyan mein rakho: diagonal panel ka khud se baat karna hai, off-diagonal ek panel ka doosre se baat karna hai.

1. $A\mathbf{\lambda }=\mathbf{b}$ assemble karo. Kyun? Yeh parent ka Step 5 hai — ek row per panel. Diagonal entries self-arrow hain, off-diagonals neighbour-arrow; right-hand side upar ke box se $\mathbf{b}$ hai. $\{\begin{array}{l}\frac{1}{2}{\lambda }_1+\frac{1}{3}{\lambda }_2=-10(0.8)=-8\\ \frac{1}{3}{\lambda }_1+\frac{1}{2}{\lambda }_2=-10(-0.6)=+6\end{array}$
2. Dono rows ko 6 se multiply karke fractions clear karo. Number 6 kyun? Yeh denominators $2$ aur $3$ ka least common multiple hai, toh 6 se multiply karne se ek hi stroke mein har fraction whole number ban jaata hai — fraction arithmetic avoid karne ka sabse clean tarika. Row 1 $\times 6$: $3{\lambda }_1+2{\lambda }_2=-48$. Row 2 $\times 6$: $2{\lambda }_1+3{\lambda }_2=36$.
3. Solve karo. Kyun? Do equations, do unknowns — standard elimination jo computer karta. ${\lambda }_2$ coefficient match karne ke liye row 1 ko 3 se aur row 2 ko 2 se multiply karo, phir subtract karo:
   • $9{\lambda }_1+6{\lambda }_2=-144$, $4{\lambda }_1+6{\lambda }_2=72$ → $5{\lambda }_1=-216$ → ${\lambda }_1=-43.2$.
   • Back-substitute karo: $2(-43.2)+3{\lambda }_2=36\Rightarrow 3{\lambda }_2=122.4\Rightarrow {\lambda }_2=40.8$.

Verify: Row 1: $\frac{1}{2}(-43.2)+\frac{1}{3}(40.8)=-21.6+13.6=-8$ ✓. Row 2: $\frac{1}{3}(-43.2)+\frac{1}{2}(40.8)=-14.4+20.4=6$ ✓. Diagonal $\frac{1}{2}$ (self, white arrow) dominant hai; off-diagonal $\frac{1}{3}$ (neighbour, cyan arrow) use nudge karta hai — exactly wahi structure jo figure dikhata hai.

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Cell F — limit $N\to \infty$

Forecast: numbers $\approx 0.549$ ki taraf crowd kar rahe hain — accuracy plateau hoti hai. Cost, though, explode karta hai.

1. Successive changes dekho. Kyun? Convergence matlab change har refinement par sihrta hai. $\Delta =0.040,0.008,0.001$. Har step roughly pichle ka $\frac{1}{5}$ hai: $8/40=0.20$, $1/8=0.125$. Changes shrink ho rahe hain, toh sequence $C_L\approx 0.549$ ki taraf converge kar rahi hai.
2. Panel size $h$ define karo aur kyun matter karta hai. Yeh step kyun? "$N\to \infty$" actually "$h\to 0$" hai, toh humein batana hoga $h$ kya hai. Agar body ka perimeter $\mathcal{P}$ hai aur hum $N$ equal panels use karte hain, toh $h=\mathcal{P}/N$ ek panel ki length hai — discretisation length. $N$ double karne se $h$ half ho jaata hai. Ek smooth body ke liye, standard panel-method error theory kehti hai ki error ek fixed power of $h\propto 1/N$ ki tarah girta hai, toh $h$ halving se error roughly constant factor se multiply hoti hai har refinement mein. Yahi constant factor exactly wahi hai jo successive changes ko geometric series ki tarah behave karaata hai. Hum ise sirf rough extrapolation ke roop mein use karte hain, exact law nahi.
3. Plateau estimate karo (part a). Kyun? Tail ko geometric with ratio $\approx 0.2$ treat karte hue, $0.001$ ke baad remaining changes sum hote hain roughly $0.001\times \frac{0.2}{1-0.2}=0.00025$ mein, toh limit $\approx 0.549+0.00025\approx 0.5493$. Haan, yeh converge ho raha hai.
4. Cost scaling (part b). Kyun? Dense matrix inversion $\mathcal{O}(N^3)$ hai. $80\to 800$ jaana $N$ mein $\times 10$ hai, toh time $10^3=1000$ se scale karta hai. $t=2.0\times {\left(\frac{800}{80}\right)}^3=2.0\times 1000=2000\text{s}\approx 33\text{min}.$

Verify: $C_L$ change ratios $8/40=0.2$ aur $1/8=0.125$ — dono $<1$, toh bounded/convergent. ✓ Cost: $2.0\times (800/80{)}^3=2.0\times 1000=2000$ s ✓. Lesson: zyada panels accuracy mein madad karna band kar dete hain bahut pehle se, aur cost zyada dino tak explode karti rehti hai — parent ki "accuracy plateaus" mistake, quantified.

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Cell G — real-world word problem

Forecast: low speed, attached, no shocks → panel method jeetega; par drag ek switch force karega.

1. Mach number check karo. Yeh step kyun? Compressibility matter karti hai jab $M=V_{\infty }/a$ $\sim 0.3$ ke paas aaye; neeche, incompressible Laplace fine hai. Yahan $a$ air mein speed of sound hai — kitni tezi se pressure disturbance travel karti hai — sea level par $a=340$ m/s diya gaya hai: $M=\frac{V_{\infty }}{a}=\frac{50}{340}=0.147<\mathrm{0.3.}$ Incompressible assumption hold karti hai → Laplace valid hai → panel method applicable.
2. Reynolds number check karo (viscosity thin hai ya nahi iska context). Kyun? $Re=\frac{V_{\infty }c}{\nu }$ batata hai ki boundary layer chord ke relative thin hai. $Re=\frac{50\times 1.2}{1.5\times 10^{-5}}=4.0\times 10^6.$ High $Re$, attached flow → inviscid pressure prediction accurate hai.
3. Tool choose karo. Kyun? Panel method ~1% cost par ~80% lift info deta hai (parent ka 80/20 rule). Lift/pressure ke liye panel method use karo.
4. Jab drag chahiye. Kyun? Inviscid 2-D ⇒ d'Alembert's Paradox ⇒ zero drag. Real drag paane ke liye (skin friction + wake) tumhe boundary-layer correction add karni hogi ya full Navier–Stokes CFD par jaana hoga.

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

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Cell H — exam twist on lift

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

1. Kutta–Joukowski apply karo. Kyun? Yeh circulation aur lift ke beech the link hai: $L^{\prime }=\rho V_{\infty }\Gamma$. $\Gamma =0$ ke saath: $L^{\prime }=\rho V_{\infty }(0)=0$. Sources akele zero lift dete hain — woh sirf thickness shape karte hain.
2. $C_L$ ko $L^{\prime }$ se relate karo. Kyun? Lift coefficient lift ko dynamic pressure $\times$ chord se nondimensionalise karta hai. $L^{\prime }=C_L\cdot \frac{1}{2}\rho V_{\infty }^{2}c=0.6\times \frac{1}{2}(1.225)(50^2)(1.2)$ $L^{\prime }=0.6\times \frac{1}{2}(1.225)(2500)(1.2)=1102.5\text{N/m}.$
3. $\Gamma$ ke liye solve karo. Kyun? Kutta–Joukowski invert karo. $\Gamma =\frac{L^{\prime }}{\rho V_{\infty }}=\frac{1102.5}{1.225\times 50}=18.0{\text{m}}^2/\text{s}.$

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 ✓. Reinset karo: $\rho V_{\infty }\Gamma =1.225\cdot 50\cdot 18.0=1102.5=L^{\prime }$ ✓. Yahi reason hai ki hum source panel code par vortex panels + Kutta condition bolt karte hain — sources shape set karte hain, vortices lift set karte hain.

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Recall

Recall Answers chhupao aur poora matrix test karo

$\cos \beta$ ka kaunsa sign matlab hai wind wall ke through usi taraf blow kar rahi hai jis taraf $\hat{n}$ point karta hai? ::: Positive $\cos \beta$ (wind component along $+\hat{n}$). Matrix diagonal par $\frac{1}{2}$ kahaan se aata hai? ::: Ek source sheet apni air aadhi aadhi har face se baahir bhejta hai, toh apne midpoint par $\lambda /2$ induce karta hai ($I_{ii}=\pi$). $I_{11}=\pi$, $V_{\infty }=8$ ke saath single flat panel: $\lambda$ kya hai? ::: $-16$. Saare ${\lambda }_j=0$ kaunsi physical situation describe karta hai? ::: No body — undisturbed uniform stream (ya ek plate jo wind ke parallel hai). Closed body par $\sum {\lambda }_j{\ell }_j=0$ (sirf $\sum {\lambda }_j=0$ nahi) kyun hona chahiye? ::: Har panel ${\lambda }_j{\ell }_j$ emit karta hai; sirf tab jab sab lengths equal hon per-length strengths zero sum karti hain. Cost $N^3$ ki tarah scale karti hai: $N=80$ se $N=800$ jaane par runtime kitni baar multiply hoti hai? ::: $1000$. Source-only body ka lift zero kyun hota hai? ::: Zero circulation → $L^{\prime }=\rho V_{\infty }\Gamma =0$. Lift paane ke liye kya add karte hain, kaunsi condition se control kiya? ::: Vortex panels, Kutta condition se set kiya.