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

Use the boundaries from the parent note: 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 (near 1, mixed pockets)
• $M=3.0$ → supersonic
• $M=7.0$ → hypersonic

$M=\frac{V}{a}=\frac{170}{340}=0.5$ Ratio of your speed to the messenger speed. $M=0.5<0.8$ → subsonic.

Step 1 — find $a$. Why first? Mach needs the local messenger speed. $a=\sqrt{\gamma RT}=\sqrt{1.4\cdot 287\cdot 250}=\sqrt{100,450}\approx 316.9\text{m/s}$ Step 2 — divide. $M=\frac{200}{316.9}\approx 0.631$ Subsonic, but above $M=0.3$ so compressibility already matters (see Compressibility & Bernoulli's Limits).

(a) $a=\sqrt{1.4\cdot 287\cdot 288}=\sqrt{115,718}\approx 340.2\text{m/s}$, so $M=600/340.2\approx 1.76$. (b) $a=\sqrt{1.4\cdot 287\cdot 220}=\sqrt{88,396}\approx 297.3\text{m/s}$, so $M=600/297.3\approx 2.02$. Lesson: same speedometer reading, higher Mach up high because the cold air carries sound more slowly. Mach is about the gas state, not just your speed.

Step 1 — geometry. Why $\sin \mu =1/M$? In time $t$ the jet moves $Vt$ while a sound sphere emitted at the start grows to radius $at$; the wavefront envelope is a cone whose right-triangle gives $\sin \mu =at/Vt=a/V=1/M$ (look at the figure). $\sin \mu =\frac{1}{M}=\frac{1}{2.5}=0.4$ Step 2 — invert. $\mu =\arcsin (0.4)\approx 23.6^{\circ }$ Interpretation: the cone trails at $23.6^{\circ }$ from the flight path. The ground hears nothing until that cone sweeps over — then a sudden sonic boom. See Oblique Shocks & Mach Cone.

Invert $\sin \mu =1/M$: $M=\frac{1}{\sin \mu }=\frac{1}{\sin 40^{\circ }}=\frac{1}{0.6428}\approx 1.556$ Supersonic, as expected — a visible cone only forms when $M>1$.

Step 1 — free-stream speed. $a_{\infty }=\sqrt{1.4\cdot 287\cdot 223}=\sqrt{89,601}\approx 299.3\text{m/s}$, so $V_{\infty }=M_{\infty }{a}_{\infty }=0.85\cdot 299.3\approx 254.4\text{m/s}$. Step 2 — local speed. $V_{\text{local}}=1.15\cdot 254.4\approx 292.6\text{m/s}$. Step 3 — local sound speed (cooler air, slower sound). $a_{\text{local}}=\sqrt{1.4\cdot 287\cdot 205}=\sqrt{82,369}\approx 287.0\text{m/s}$. Step 4 — local Mach. $M_{\text{local}}=\frac{292.6}{287.0}\approx 1.02>1$ Yes — locally supersonic. A subsonic plane grows a supersonic pocket and a local shock. Two effects stack: the flow speeds up AND the air cools (dropping $a$). This is exactly why transonic design is hard. Downstream that pocket usually ends in a normal shock.

Drone: $\frac{1}{2}(0.25{)}^2=0.03125\approx 3.1\%$ — under the $\sim 5\%$ rule of thumb → incompressible Bernoulli is acceptable. Wing pocket: $\frac{1}{2}(1.02{)}^2=0.5202\approx 52\%$ — enormous; density is nowhere near constant. Must use compressible (isentropic) relations. The $M\sim 0.3$ threshold in Compressibility & Bernoulli's Limits is exactly this budget crossing $\sim 5\%$.

Step 1 — solve the isentropic relation for $M$. Why this tool? $T_0/T$ tells us how much of the gas's energy has turned into ordered motion; that fraction is set by $M$. With $\gamma =1.4$, $\frac{\gamma -1}{2}=0.2$. $\frac{300}{150}=2=1+0.2M^2\Rightarrow 0.2M^2=1\Rightarrow M^2=5\Rightarrow M=\sqrt{5}\approx 2.236$ Step 2 — local sound speed at the exit. $a=\sqrt{1.4\cdot 287\cdot 150}=\sqrt{60,270}\approx 245.5\text{m/s}$. Step 3 — flow speed. $V=Ma=2.236\cdot 245.5\approx 549.0\text{m/s}$. Reading it back: cooling the gas from 300 K to 150 K converts thermal energy into a $549\text{m/s}$ supersonic stream at $M\approx 2.24$. Full machinery: Isentropic Flow Relations.

(a) Mach numbers. $a_A=\sqrt{1.4\cdot 287\cdot 250}=\sqrt{100,450}\approx 316.9\text{m/s}$, so $M_A=500/316.9\approx 1.578$. $a_B=\sqrt{1.4\cdot 287\cdot 216}=\sqrt{86,789}\approx 294.6\text{m/s}$, so $M_B=680/294.6\approx 2.308$. (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 has the narrower cone. (c) Why: higher Mach ⇒ smaller $1/M$ ⇒ smaller $\sin \mu$ ⇒ a needle-thinner cone. The faster you outrun your own sound, the tighter the wave sheet wraps behind you. Notice B is faster and in colder (slower-sound) air — both push its Mach up.

• Always compute local $a=\sqrt{\gamma RT}$ (kelvin!) before dividing to get $M$. ::: Mach is a ratio, not a fixed speed.
• Mach cone geometry uses $\sin \mu =1/M$. ::: opposite ($at$) over hypotenuse ($Vt$).
• $d\rho /\rho \approx \frac{1}{2}{M}^2$ decides if Bernoulli survives. ::: $\sim 5\%$ near $M=0.3$.
• Isentropic $T_0/T=1+\frac{\gamma -1}{2}{M}^2$ links a temperature drop to a Mach rise. ::: convert static $T$ → $a$ → $V$.