Exercises — Wave drag — transonic and supersonic

Kya poochha gaya hai: drag type aur uska microscopic cause identify karo.

• Shock se paida hone wala drag wave drag hai.
• Woh quantity jo kisi bhi shock ke across badhti hai woh entropy ($s_2>s_1$) hai. Entropy badhne ka matlab hai stagnation pressure girna ($p_{0,2}<p_{0,1}$), jo ek rearward force ke roop mein dikhta hai. Dekho Entropy and stagnation pressure loss. Answer: wave drag, entropy increase ki wajah se.

Yeh formula kyun: Mach cone pure geometry hai — ek weak pulse radius $at$ tak phailta hai jabki body $ut$ move karti hai, aur $\sin \mu =at/ut=1/M$. $\sin \mu =\frac{1}{M}=\frac{1}{2}=0.5\Rightarrow \mu =\arcsin (0.5)=30^{\circ }.$ Answer: $\mu =30^{\circ }$.

Galat. Mach angle $\mu$ infinitesimally weak signals ka envelope hai. Ek real shock flow ko ek finite angle se turn karta hai, jiske liye finite compression chahiye, isliye woh hamesha steeper hota hai: $\beta >\mu$. Equality sirf zero turning ke limit mein hoti hai ($\theta \to 0$). Dekho Normal and oblique shock waves.

$\mu (1.4)=\arcsin \left(\frac{1}{1.4}\right)=\arcsin (0.714)=45.6^{\circ },$ $\mu (4)=\arcsin \left(\frac{1}{4}\right)=\arcsin (0.25)=14.5^{\circ }.$ Yeh kaisa dikhta hai (figure dekho): faster flight ⇒ cone body ke saath tighter wrap hota hai ($\mu$ chhota hota hai). Jab $M\to 1^{+}$, $\mu \to 90^{\circ }$ (cone almost flat wall ban jaata hai). Jab $M\to \infty$, $\mu \to 0$.

Yeh formulas kyun: ek thin section ke liye supersonic flow mein, linearized (Ackeret) theory deta hai $C_p=2\theta /\sqrt{M^2-1}$; flat plate ki surfaces integrate karne se boxed results milte hain. Yahan $\beta \equiv \sqrt{M^2-1}$ compressibility factor hai (shock angle NAHI). $\beta =\sqrt{M^2-1}=\sqrt{4-1}=\sqrt{3}=\mathrm{1.732.}$ $c_l=\frac{4\alpha }{\beta }=\frac{4(0.0524)}{1.732}=\mathrm{0.121.}$ $c_{d,\text{wave}}=\frac{4{\alpha }^2}{\beta }=\frac{4(0.0524{)}^2}{1.732}=\mathrm{0.00634.}$ Answer: $c_l=0.121$, $c_{d,\text{wave}}=0.00634$.

$\frac{L}{D_{\text{wave}}}=\frac{c_l}{c_{d,\text{wave}}}=\frac{4\alpha /\beta }{4{\alpha }^2/\beta }=\frac{1}{\alpha }=\frac{1}{0.0524}=\mathrm{19.1.}$ Itna clean kyun: $\beta$ cancel ho jaata hai aur $\alpha$ ka ek power cancel hota hai, bacha rehta hai $L/D_{\text{wave}}=\cot \alpha \approx 1/\alpha$. Chhota $\alpha$ efficient hai — lekin dhyan raho lift khud bhi chhoti hai, isliye $\alpha$ ko simply chhota nahi kar sakte. Answer: $L/D_{\text{wave}}\approx 19.1$.

Mach angle: $\mu =\arcsin (1/2)=30^{\circ }$. Real shock: $\beta \approx 39.3^{\circ }$ θ–β–M relation se. $\beta >\mu$ kyun: Mach cone zero-strength limit hai. Flow ko finite $10^{\circ }$ turn karne ke liye finite compression chahiye, isliye wave ko flow mein aur steeply lean karna padta hai. Gap $\beta -\mu =9.3^{\circ }$ shock strength ka direct measure hai. Yeh kaisa dikhta hai (figure dekho): wedge tip se do lines — shallow dashed Mach cone ($30^{\circ }$) aur steeper solid shock ($39.3^{\circ }$).

Yeh formula kyun: shock ke across flow adiabatic hai lekin irreversible, isliye Gibbs deta hai $p_{0,2}/p_{0,1}=e^{-\Delta s/R}$. $\frac{p_{0,2}}{p_{0,1}}=e^{-0.06}=\mathrm{0.942.}$ Toh stagnation pressure $94.2\%$ tak girta hai — $5.8\%$ loss. Yeh drag kyun hai: body ke around ek control volume pe momentum balance kehta hai ki wake mein stagnation-pressure deficit ek net rearward force ke barabar hai. Woh force wave drag hai — koi viscosity nahi chahiye. Dekho Entropy and stagnation pressure loss. Answer: ratio $\approx 0.942$ (ek $5.8\%$ stagnation-pressure loss).

Kyun: sirf leading edge ke normal velocity component hi section pe compressibility drive karta hai: $M_n=M_{\infty }\cos \Lambda$. $M_{\infty }$ ke liye solve karo. $M_{\infty }=\frac{M_n}{\cos \Lambda }=\frac{0.7}{\cos 30^{\circ }}=\frac{0.7}{0.8660}=\mathrm{0.808.}$ Matlab: sweep se tum faster fly kar sakte ho ($0.808$ vs $0.7$) isse pehle ki section "sonic flow feel kare" — yahi reason hai ki fast wings swept hoti hain. Dekho Critical and drag-divergence Mach number. Answer: $M_{\infty }\approx 0.808$.

Kyun: zero lift aur zero camber pe sirf thickness term bachti hai: $c_{d,\text{wave}}=\frac{4}{\sqrt{M^2-1}}\overline{{\left(\frac{d{y}_t}{dx}\right)}^2}.$ Kyunki slope magnitude $t$ har jagah hai, uska mean square hai $\overline{(d{y}_t/dx{)}^2}=t^2=0.05^2=0.0025$. $c_{d,\text{wave}}=\frac{4}{\sqrt{3}}(0.0025)=\frac{4(0.0025)}{1.732}=\mathrm{0.00577.}$ Insight: wave drag bina lift ke bhi exist karta hai — yeh supersonic flow mein volume hone ki kimat hai. Thickness aadha karne se yeh drag quarter ho jaata hai (yeh square hai). Answer: $c_{d,\text{wave}}\approx 0.00577$.

$c_{d,\text{wave}}=\frac{4}{\sqrt{M^2-1}}({\alpha }^2+\overline{(d{y}_c/dx{)}^2}+\overline{(d{y}_t/dx{)}^2}),\sqrt{M^2-1}=\mathrm{1.732.}$

• Lift term: $\frac{4}{1.732}{\alpha }^2=2.309(0.0349^2)=2.309(0.001218)=\mathrm{0.00281.}$
• Camber term: $2.309(0.0009)=\mathrm{0.00208.}$
• Thickness term: $2.309(0.0025)=\mathrm{0.00577.}$
• Total: $c_{d,\text{wave}}=0.00281+0.00208+0.00577=\mathrm{0.01066.}$ Isko padhna: thickness yahan dominate karti hai — woh volume drag hai jise Area Rule target karta hai. Har term ek square hai, isliye wave drag kabhi negative nahi ho sakta. Answer: total $\approx 0.01066$.

$c_d(M)=\frac{4t^2}{\sqrt{M^2-1}},t^2=\mathrm{0.0025.}$ $c_d(2)=\frac{0.01}{\sqrt{3}}=0.00577,c_d(3)=\frac{0.01}{\sqrt{8}}=\frac{0.01}{2.828}=\mathrm{0.00354.}$ $\frac{c_d(3)}{c_d(2)}=\frac{\sqrt{3}}{\sqrt{8}}=\mathrm{0.612.}$ Matlab: coefficient $1/\sqrt{M^2-1}$ ki tarah girta hai kyunki oblique shocks har unit length pe weaker hote jaate hain jab flow zyada supersonic hoti hai. (Yeh sirf coefficient hai — actual force dynamic pressure par depend karta hai jo speed ke saath badhta hai.) Answer: $M=3$ pe $0.00354$; ratio $\approx 0.612$.

Step 1 (normal Mach kyun): criticality section ke normal Mach se govern hoti hai. Swept wing ke liye, free-stream Mach jis pe section $M_n=0.72$ feel karta hai: $M_{\infty }^{crit}=\frac{M_n}{\cos \Lambda }=\frac{0.72}{\cos 35^{\circ }}=\frac{0.72}{0.8192}=\mathrm{0.879.}$ Step 2 (cruise se compare karo):

• Wing A $M_{\infty }=0.72$ pe critical hoti hai — $0.82$ cruise se kaafi neeche ⇒ yeh pehle se $M_{cr}$ se aage hai, drag rise mein deep. Bura.
• Wing B $M_{\infty }=0.879>0.82$ pe critical hoti hai ⇒ cruise pe abhi sonic flow nahi aayi. Acha — margin $0.879-0.82=0.059$. Decision: swept wing B chuno; sweep $M_{cr}$ ko cruise Mach se upar push karta hai, wave drag delay karta hai. Dekho Critical and drag-divergence Mach number aur Prandtl-Glauert compressibility correction. Answer: B (critical at $M_{\infty }\approx 0.879$, $0.82$ cruise se comfortably upar).

${\theta }_{max}$ kyun matter karta hai: $\theta$–$\beta$–$M$ relation ka solution sirf $\theta \le {\theta }_{max}$ ke liye exist karta hai. Nose $25^{\circ }$ ka turn demand karta hai. Kyunki $25^{\circ }>22.97^{\circ }$, koi attached shock flow ko itna sharply turn nahi kar sakta ⇒ shock detach ho jaata hai aur nose ke aage ek curved bow shock ban jaata hai. Drag consequence: centreline ke paas bow shock essentially ek normal shock hai — sabse strong possible ⇒ sabse bada entropy jump ⇒ sabse bada stagnation-pressure loss ⇒ high wave drag. Isliye supersonic noses slender rakhe jaate hain ($\theta <{\theta }_{max}$). Dekho Theta-beta-M relation and detached bow shocks. Answer: shock detach ho jaata hai (bow shock); wave drag high hai kyunki near-centreline shock normal hai.

Step 1 ($\mu$ kyun): boom front Mach cone hai, jiska half-angle $\sin \mu =1/M$ satisfy karta hai. $\mu =\arcsin \left(\frac{1}{1.6}\right)=\arcsin (0.625)=38.68^{\circ }.$ Step 2 (geometry, figure dekho): cone surface flight path ke saath angle $\mu$ banata hai. Ground aircraft ke neeche vertical drop $h$ hai. Aircraft se us jagah tak horizontal lag jahan cone ground se milta hai: $x=\frac{h}{\tan \mu }=\frac{12000}{\tan 38.68^{\circ }}=\frac{12000}{0.8005}=14,991\text{m}\approx 15.0\text{km}.$ Matlab: boom tab sunai deta hai jab jet apne track par tumse roughly $15$ km aage nikal chuka hota hai. Dekho Sonic boom and the Mach cone. Answer: $x\approx 15.0$ km aircraft ki ground position ke peeche.

Is chapter mein "$\beta$" ke teen chehere ::: shock angle (oblique shock tilt), woh letter jo Ackeret mein $\sqrt{M^2-1}$ ke liye overload hai, aur $\Lambda$ sweep hai — inhe alag rakho. Wave drag hamesha $\ge 0$ kyun hota hai ::: har Ackeret term ek square hai (${\alpha }^2$, camber${}^2$, thickness${}^2$). Sweep $M_{cr}$ kyun badhata hai ::: sirf $M_n=M_{\infty }\cos \Lambda$ criticality drive karta hai, isliye free-stream Mach zyada ho sakta hai. Blunt noses mein high wave drag kyun hota hai ::: unka shock near-normal bow shock mein detach ho jaata hai — maximum entropy jump, maximum stagnation-pressure loss.