Exercises — Area-Mach number relation A - A - = f(M) — isentropic flow

Look at the U-shaped valley in the figure. Its lowest point sits at $A/A^{\ast }=1$, and that only happens at ==$M=1$== (Mach one, the sonic point). So $A/A^{\ast }=1$ and $M=1$. WHY: the throat is the single spot where geometry lets the flow reach the speed of sound.

Draw a horizontal line at height $2$ across the valley in the figure — it crosses the curve in two places. To the left of the throat ($M<1$) is the subsonic root; to the right ($M>1$) is the supersonic root. WHY it isn't unique: the area–Mach curve is U-shaped (goes down then up), so one height corresponds to two Mach numbers. Geometry alone can't tell you which; the back-pressure does.

$a=\sqrt{\gamma RT}=\sqrt{1.4\cdot 287\cdot 300}=\sqrt{120540}\approx 347\text{m/s}$. WHY this formula: sound is a tiny pressure wave; its speed depends on how "stiff" and how "heavy" the gas is, captured by $\gamma RT$. Hotter gas ⇒ faster sound.

WHAT: we use Tool A directly since $M$ is known. Inside the bracket: $1+\frac{\gamma -1}{2}{M}^2=1+0.2\cdot 4=1.8$. Factor $\frac{2}{\gamma +1}=\frac{2}{2.4}=0.8333$, so bracket $=0.8333\cdot 1.8=1.5$. Exponent $\frac{\gamma +1}{2(\gamma -1)}=\frac{2.4}{0.8}=3$. $\frac{A}{A^{\ast }}=\frac{1}{2}(1.5{)}^3=\frac{3.375}{2}=\mathrm{1.6875.}$ So $A/A^{\ast }\approx ==1.688==$. WHY: at supersonic $M$ you climb the right wall of the valley, so the area must be well above the sonic minimum.

WHAT: now use Tool B (Mach → thermodynamics). $\frac{T_0}{T}=1+0.2\cdot 4=1.8\Rightarrow T=\frac{300}{1.8}=166.7\text{K}$. $\frac{p_0}{p}=(1.8{)}^{\gamma /(\gamma -1)}=(1.8{)}^{3.5}=7.824\Rightarrow p=\frac{500}{7.824}=63.9\text{kPa}$. WHY: supersonic flow has expanded and cooled — the gas traded internal energy for speed, so $T$ and $p$ drop far below reservoir values.

Bracket: $1+0.2\cdot 0.25=1.05$; times $0.8333=0.875$. $\frac{A}{A^{\ast }}=\frac{1}{0.5}(0.875{)}^3=2\cdot 0.6699=\mathrm{1.3398.}$ $A/A^{\ast }\approx ==1.340==$. WHY: at $M=0.5$ you're on the left wall of the valley — still above the minimum, because you must widen (relative to the throat) to slow the flow below sonic.

WHAT: $A/A^{\ast }=14/8=1.75$. Tool A cannot be solved algebraically for $M$, so we iterate (or read a table). WHY iterate: $f(M)$ mixes $1/M$ with a power of $M^2$ — no clean inverse.

Subsonic root: guess $M=0.35\Rightarrow A/A^{\ast }=1.778$; $M=0.36\Rightarrow 1.732$. Interpolating, $M\approx ==0.357==$.

Supersonic root: guess $M=2.05\Rightarrow A/A^{\ast }=1.766$; $M=2.03\Rightarrow 1.744$. Interpolating between these, $M\approx ==2.037==$ (which gives $A/A^{\ast }\approx 1.751$). ✔

Both give $A/A^{\ast }\approx 1.75$. WHY two answers: the horizontal line at $1.75$ cuts the valley twice — see the figure.

Bracket $=1+0.2\cdot (2.037{)}^2=1+0.2\cdot 4.149=1.830$. Exponent $1/(\gamma -1)=1/0.4=2.5$. $\frac{{\rho }_0}{\rho }=(1.830{)}^{2.5}=4.531$, so $\rho /{\rho }_0=1/4.531===0.2207==$. WHY: supersonic gas is spread thin — density drops to about a fifth of reservoir density.

At $M=0.6$: bracket $=1+0.2\cdot 0.36=1.072$; times $0.8333=0.8933$; $A/A^{\ast }=\frac{1}{0.6}(0.8933{)}^3=1.6667\cdot 0.7130=1.1883$. So $A^{\ast }=A_{\min }/1.1883=12/1.1883===10.10=={\text{cm}}^2$. Is there a real sonic point? No. The smallest actual area (12 cm²) is still larger than $A^{\ast }$ (10.10 cm²), so the flow never reaches $M=1$. $A^{\ast }$ exists only as a mathematical yardstick here — it is smaller than any real area in the duct. WHY it matters: this steel-mans the "$A^{\ast }$ is always a physical throat" mistake — it isn't.

Step 1 (shape → Mach). $A_e/A^{\ast }=15/6=2.5$. Iterate on the supersonic branch: $M=2.44\Rightarrow A/A^{\ast }=2.494$; $M=2.443\Rightarrow 2.502$. So $M_e\approx ==2.443==$.

Step 2 (Mach → $T$). Bracket $=1+0.2\cdot (2.443{)}^2=1+0.2\cdot 5.968=2.1937$. $T_e=T_0/2.1937=320/2.1937===145.9==\text{K}$.

Step 3 (Mach → $p$). $p_0/p=(2.1937{)}^{3.5}=15.63$, so $p_e=700/15.63===44.8==\text{kPa}$.

Step 4 (velocity). Local sound speed $a_e=\sqrt{\gamma R{T}_e}=\sqrt{1.4\cdot 287\cdot 145.9}=\sqrt{58622}=242.1\text{m/s}$. Then $V_e=M_e{a}_e=2.443\cdot 242.1===591.5==\text{m/s}$. WHY this chain: geometry gave $M$ (Tool A); $M$ unlocked every thermodynamic property (Tool B); temperature gave the local sound speed, and $V=Ma$ gave the actual speed. This is the "shape → flow" master key promised in the parent note.

WHAT (a): for a choked throat mass flow is fixed by reservoir conditions: $\dot{m}=\frac{p_0{A}^{\ast }}{\sqrt{T_0}}\sqrt{\frac{\gamma }{R}}{\left(\frac{2}{\gamma +1}\right)}^{\frac{\gamma +1}{2(\gamma -1)}}.$ Numbers: $\frac{p_0{A}^{\ast }}{\sqrt{T_0}}=\frac{700000\cdot 6\times 10^{-4}}{\sqrt{320}}=\frac{420}{17.889}=23.48$. $\sqrt{\gamma /R}=\sqrt{1.4/287}=\sqrt{0.004878}=0.06984$. Bracket factor $(0.8333{)}^3=0.5787$. $\dot{m}=23.48\cdot 0.06984\cdot 0.5787===0.949==\text{kg/s}$.

(b) $\dot{m}\propto A^{\ast }$ (everything else unchanged), so doubling $A^{\ast }$ doubles the flow: $\dot{m}=2\cdot 0.949===1.898==\text{kg/s}$. WHY: once the throat is sonic, lowering the exit pressure cannot pull more mass through — the flow is choked (see Choked Flow & Maximum Mass Flow). The only lever left is the throat area itself. See Converging-Diverging (de Laval) Nozzle.

Step 1 (Mach → area ratio). Bracket $=1+0.2\cdot 9=2.8$; times $0.8333=2.3333$. $A_e/A^{\ast }=\frac{1}{3}(2.3333{)}^3=\frac{1}{3}\cdot 12.70=4.2346$. Step 2 (solve for throat). $A^{\ast }=A_e/4.2346=45/4.2346===10.63=={\text{cm}}^2$. Step 3 ($T$, $p$). $T_0/T=2.8\Rightarrow T=500/2.8===178.6==\text{K}$. $p_0/p=(2.8{)}^{3.5}=36.73\Rightarrow p=1000/36.73===27.2==\text{kPa}$. WHY: design runs the chain backwards — the target Mach fixes the area ratio, which sizes the throat for the given exit. Then thermodynamics falls out of $M_e$.

Step 1. Downstream stagnation pressure: $p_{0,2}=0.499\cdot 900===449.1==\text{kPa}$. Step 2 (interpretation). A shock is not isentropic — entropy rises, so $p_0$ drops. Since the choked mass flow $\dot{m}\propto p_0{A}^{\ast }$ is conserved across the shock while $p_0$ falls, the sonic reference area must grow: $A_2^{\ast }/A_1^{\ast }=p_{0,1}/p_{0,2}=1/0.499=2.004$. So $A^{\ast }$ roughly doubles across the shock. WHY it matters: $A^{\ast }$ is only constant along an isentropic run. Cross a shock and the yardstick itself jumps — a key subtlety when matching nozzle sections (see Normal Shock Waves).

As $M\to 0$: the bracket $\to {\left(\frac{2}{\gamma +1}\right)}^{\frac{\gamma +1}{2(\gamma -1)}}$, a finite positive constant, but the prefactor $1/M\to \infty$. So $A/A^{\ast }\to \infty$. Physical: to slow a gas to near-rest at fixed mass flow you need an enormous area (the crowd spreads out infinitely thin and slow).

As $M\to \infty$: the bracket grows like $M^2$ raised to $\frac{\gamma +1}{2(\gamma -1)}$, i.e. like $M^{\frac{\gamma +1}{\gamma -1}}$; dividing by $M$ still leaves $M^{\frac{\gamma +1}{\gamma -1}-1}=M^{2/(\gamma -1)}\to \infty$. Physical: hypersonic flow has vanishing density, so area must blow up to pass the same mass. Both ends → $\infty$; the single minimum sits at $M=1$ — this is exactly the U-valley shape.

At $M=1$, what is $A/A^{\ast }$? ::: Exactly 1 — the minimum of the U-valley. Which exponent sits in the AREA relation for air? ::: $\frac{\gamma +1}{2(\gamma -1)}=3$ (not $\gamma /(\gamma -1)=3.5$, that's the pressure relation). Why do choked-flow problems ignore back-pressure for $\dot{m}$? ::: The sonic throat caps density×velocity, so $\dot{m}=\frac{p_0{A}^{\ast }}{\sqrt{T_0}}(\dots )$ is fixed by reservoir conditions. What happens to $A^{\ast }$ across a normal shock? ::: It increases (by $p_{0,1}/p_{0,2}$), because $p_0$ drops while $\dot{m}\propto p_0{A}^{\ast }$ is conserved. How do you get velocity once $M$ is known? ::: $V=Ma$ with $a=\sqrt{\gamma RT}$, after getting $T$ from the stagnation relation.