Exercises — Mach number M = V - a — subsonic ( - 1), transonic (~1), supersonic ( - 1), hypersonic ( - 5)

Parent note ke boundaries use karo: subsonic $M<0.8$, transonic $\approx 0.8$–$1.2$, supersonic $\approx 1.2$–$5$, hypersonic $M>5$.

• $M=0.6$ → subsonic
• $M=0.95$ → transonic (1 ke paas, mixed pockets)
• $M=3.0$ → supersonic
• $M=7.0$ → hypersonic

$M=\frac{V}{a}=\frac{170}{340}=0.5$ Apni speed ka messenger speed se ratio. $M=0.5<0.8$ → subsonic.

Step 1 — $a$ nikalo. Pehle kyun? Mach ko local messenger speed chahiye. $a=\sqrt{\gamma RT}=\sqrt{1.4\cdot 287\cdot 250}=\sqrt{100,450}\approx 316.9\text{m/s}$ Step 2 — divide karo. $M=\frac{200}{316.9}\approx 0.631$ Subsonic hai, lekin $M=0.3$ se upar hai isliye compressibility already matter karti hai (dekho Compressibility & Bernoulli's Limits).

(a) $a=\sqrt{1.4\cdot 287\cdot 288}=\sqrt{115,718}\approx 340.2\text{m/s}$, isliye $M=600/340.2\approx 1.76$. (b) $a=\sqrt{1.4\cdot 287\cdot 220}=\sqrt{88,396}\approx 297.3\text{m/s}$, isliye $M=600/297.3\approx 2.02$. Lesson: same speedometer reading, zyada Mach upar kyunki thandi air mein sound zyada dheere travel karti hai. Mach gas ki state ke baare mein hai, sirf aapki speed ke baare mein nahi.

Step 1 — geometry. $\sin \mu =1/M$ kyun? Time $t$ mein jet $Vt$ move karta hai jabki shuru mein emit hua ek sound sphere radius $at$ tak grow karta hai; wavefront envelope ek cone hai jiska right-triangle deta hai $\sin \mu =at/Vt=a/V=1/M$ (figure dekho). $\sin \mu =\frac{1}{M}=\frac{1}{2.5}=0.4$ Step 2 — invert karo. $\mu =\arcsin (0.4)\approx 23.6^{\circ }$ Interpretation: cone flight path se $23.6^{\circ }$ par trail karta hai. Zameen tab tak kuch nahi sunti jab tak woh cone sweep nahi kar jaata — phir achanak ek sonic boom aata hai. Dekho Oblique Shocks & Mach Cone.

$\sin \mu =1/M$ ko invert karo: $M=\frac{1}{\sin \mu }=\frac{1}{\sin 40^{\circ }}=\frac{1}{0.6428}\approx 1.556$ Supersonic hai, jaisi umeed thi — ek visible cone tabhi banta hai jab $M>1$.

Step 1 — free-stream speed. $a_{\infty }=\sqrt{1.4\cdot 287\cdot 223}=\sqrt{89,601}\approx 299.3\text{m/s}$, isliye $V_{\infty }=M_{\infty }{a}_{\infty }=0.85\cdot 299.3\approx 254.4\text{m/s}$. Step 2 — local speed. $V_{\text{local}}=1.15\cdot 254.4\approx 292.6\text{m/s}$. Step 3 — local sound speed (thandi air, dheemi sound). $a_{\text{local}}=\sqrt{1.4\cdot 287\cdot 205}=\sqrt{82,369}\approx 287.0\text{m/s}$. Step 4 — local Mach. $M_{\text{local}}=\frac{292.6}{287.0}\approx 1.02>1$ Haan — locally supersonic hai. Ek subsonic plane ek supersonic pocket aur ek local shock develop karta hai. Do effects stack hote hain: flow speed up bhi hoti hai AUR air cool bhi hoti hai ($a$ drop karta hai). Isliye transonic design itna mushkil hota hai. Downstream woh pocket usually ek normal shock par khatam hoti hai.

Drone: $\frac{1}{2}(0.25{)}^2=0.03125\approx 3.1\%$ — $\sim 5\%$ rule of thumb se neeche → incompressible Bernoulli acceptable hai. Wing pocket: $\frac{1}{2}(1.02{)}^2=0.5202\approx 52\%$ — bahut zyada; density kahin bhi constant ke paas nahi. Compressible (isentropic) relations use karni padegi. Compressibility & Bernoulli's Limits mein $M\sim 0.3$ threshold exactly yahi budget $\sim 5\%$ cross karna hai.

Step 1 — $M$ ke liye isentropic relation solve karo. Yeh tool kyun? $T_0/T$ batata hai ki gas ki kitni energy ordered motion mein convert ho gayi hai; woh fraction $M$ se set hoti hai. $\gamma =1.4$ ke saath, $\frac{\gamma -1}{2}=0.2$. $\frac{300}{150}=2=1+0.2M^2\Rightarrow 0.2M^2=1\Rightarrow M^2=5\Rightarrow M=\sqrt{5}\approx 2.236$ Step 2 — exit par local sound speed. $a=\sqrt{1.4\cdot 287\cdot 150}=\sqrt{60,270}\approx 245.5\text{m/s}$. Step 3 — flow speed. $V=Ma=2.236\cdot 245.5\approx 549.0\text{m/s}$. Isko samjhein: gas ko 300 K se 150 K tak cool karna thermal energy ko ek $549\text{m/s}$ supersonic stream mein $M\approx 2.24$ par convert karta hai. Poori machinery: Isentropic Flow Relations.

(a) Mach numbers. $a_A=\sqrt{1.4\cdot 287\cdot 250}=\sqrt{100,450}\approx 316.9\text{m/s}$, isliye $M_A=500/316.9\approx 1.578$. $a_B=\sqrt{1.4\cdot 287\cdot 216}=\sqrt{86,789}\approx 294.6\text{m/s}$, isliye $M_B=680/294.6\approx 2.308$. (b) Cone angles. ${\mu }_A=\arcsin (1/1.578)=\arcsin (0.6338)\approx 39.3^{\circ }$; ${\mu }_B=\arcsin (1/2.308)=\arcsin (0.4333)\approx 25.7^{\circ }$. Plane B ki cone narrow hai. (c) Kyun: zyada Mach ⇒ chhota $1/M$ ⇒ chhota $\sin \mu$ ⇒ ek needle jaisi patchli cone. Jitna zyada aap apni khud ki sound se aage bhaago, wave sheet utni hi tightly aapke peeche wrap hoti hai. Dhyaan do ki B tez bhi hai aur thandi (slower-sound) air mein bhi hai — dono iska Mach upar push karte hain.

• $M$ nikalne ke liye divide karne se pehle hamesha local $a=\sqrt{\gamma RT}$ (kelvin mein!) compute karo. ::: Mach ek ratio hai, fixed speed nahi.
• Mach cone geometry $\sin \mu =1/M$ use karti hai. ::: opposite ($at$) over hypotenuse ($Vt$).
• $d\rho /\rho \approx \frac{1}{2}{M}^2$ decide karta hai ki Bernoulli survive karta hai ya nahi. ::: $M=0.3$ ke paas $\sim 5\%$.
• Isentropic $T_0/T=1+\frac{\gamma -1}{2}{M}^2$ ek temperature drop ko Mach rise se link karta hai. ::: static $T$ → $a$ → $V$ convert karo.