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

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What is $A^*$ really?

Note the famous shape fact (steel-manned later):

• Subsonic flow ($M<1$): to speed up, area must decrease (converge).
• Supersonic flow ($M>1$): to speed up, area must increase (diverge).
• At the throat ($A=A^*$, minimum area): $M=1$.

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Derivation from first principles

We use three pillars of 1-D steady isentropic flow.

Pillar 1 — Mass conservation (continuity): $\dot m = \rho A V = \text{const}$

Pillar 2 — Definition of $A^*$: evaluate $\dot m$ at the sonic point where $\rho=\rho^*,\ V=V^*=a^*,\ A=A^*$: $\rho A V = \rho^* A^* a^*$

Rearrange to isolate the area ratio: $\boxed{\dfrac{A}{A^*} = \dfrac{\rho^* a^*}{\rho V}}$

Pillar 3 — Isentropic + stagnation relations. Using stagnation $T_0,\rho_0$:

\frac{\rho_0}{\rho}=\left(1+\frac{\gamma-1}{2}M^2\right)^{\frac{1}{\gamma-1}}$$ Split the ratio through the stagnation state: $$\frac{\rho^*}{\rho}=\frac{\rho^*/\rho_0}{\rho/\rho_0} =\frac{\left(1+\frac{\gamma-1}{2}\right)^{-\frac{1}{\gamma-1}}} {\left(1+\frac{\gamma-1}{2}M^2\right)^{-\frac{1}{\gamma-1}}}$$ For the **speed** term, write $V=Ma$ and $a^*=1\cdot a^*$, with $a=\sqrt{\gamma R T}$: $$\frac{a^*}{V}=\frac{a^*}{M a}=\frac{1}{M}\sqrt{\frac{T^*}{T}} =\frac{1}{M}\sqrt{\frac{T^*/T_0}{T/T_0}} =\frac{1}{M}\sqrt{\frac{\left(1+\frac{\gamma-1}{2}\right)^{-1}}{\left(1+\frac{\gamma-1}{2}M^2\right)^{-1}}}$$ > [!intuition] Why this step? > $V$ depends on both *how fast sound travels here* ($a\propto\sqrt T$) and *how many Mach* > ($M$). So we split $a^*/V$ into a "temperature" piece (handled by stagnation relations) > and an explicit $1/M$ piece. Everything reduces to $M$ and $\gamma$. **Multiply $\dfrac{\rho^*}{\rho}\cdot\dfrac{a^*}{V}$.** The two $(1+\frac{\gamma-1}{2})$ constants combine; collecting exponents $\left(-\frac{1}{\gamma-1}+\frac12\right)=-\frac{\gamma+1}{2(\gamma-1)}$ gives the clean result: > [!formula] Area–Mach relation > $$\left(\frac{A}{A^*}\right)^{2}=\frac{1}{M^{2}} > \left[\frac{2}{\gamma+1}\left(1+\frac{\gamma-1}{2}M^{2}\right)\right]^{\frac{\gamma+1}{\gamma-1}}$$ > or equivalently > $$\frac{A}{A^*}=\frac{1}{M}\left[\frac{2}{\gamma+1}\left(1+\frac{\gamma-1}{2}M^{2}\right)\right]^{\frac{\gamma+1}{2(\gamma-1)}}$$ > Key checks: at $M=1$, $A/A^*=1$. As $M\to0$ or $M\to\infty$, $A/A^*\to\infty$. ![[3.1.06-Area-Mach-number-relation-A-A-=-f(M)-—-isentropic-flow.png]] > [!intuition] Reading the curve (Dual coding) > The plot is a **U-shaped valley** with minimum $A/A^*=1$ at $M=1$. Notice the crucial > twist: each value of $A/A^* > 1$ hits the curve at **two** Mach numbers — one subsonic, > one supersonic. **Geometry alone does not decide which branch.** Boundary/back-pressure > conditions pick the branch (subsonic everywhere, or supersonic after a choked throat). --- ## Worked examples > [!example] Example 1 — Mach from area (forecast first) > A converging–diverging nozzle has throat area $A^*=10\,\text{cm}^2$ and exit area > $A_e=20\,\text{cm}^2$, $\gamma=1.4$. Find exit Mach number(s). > > **Forecast:** $A_e/A^*=2$, comfortably $>1$ ⇒ expect TWO solutions, one $\approx 0.3$, > one $\approx 2.2$. > > Solve $2=\frac{1}{M}\left[\frac{1}{1.2}\left(1+0.2M^2\right)\right]^{3}$ numerically. > **Why this step?** $f(M)=A/A^*$ is not invertible in closed form, so we iterate/use tables. > Roots: $M\approx 0.306$ (subsonic) and $M\approx 2.197$ (supersonic). ✔ matches forecast. > [!example] Example 2 — Properties from $M$ > Take the supersonic root $M=2.197$, $T_0=300\ \text{K}$, $p_0=500\ \text{kPa}$. > $$\frac{T}{T_0}=\frac{1}{1+0.2(2.197)^2}=\frac{1}{1.965}\Rightarrow T=152.7\ \text{K}$$ > $$\frac{p}{p_0}=\left(\frac{T}{T_0}\right)^{\gamma/(\gamma-1)}=(0.509)^{3.5}=0.0939 > \Rightarrow p=46.9\ \text{kPa}$$ > **Why this step?** Once $M$ is known, *every* thermodynamic property follows from the > isentropic stagnation relations — that's the payoff of getting $M$ from $A/A^*$. > [!example] Example 3 — Mass flow through a choked throat > Reservoir $p_0=500$ kPa, $T_0=300$ K, $A^*=10\,\text{cm}^2=10^{-3}\,\text{m}^2$, air > ($R=287$, $\gamma=1.4$). Choked mass flow: > $$\dot m=\frac{p_0 A^*}{\sqrt{T_0}}\sqrt{\frac{\gamma}{R}} > \left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{2(\gamma-1)}}$$ > Here $\frac{2}{\gamma+1}=\frac{2}{2.4}=0.8333$ and the exponent $\frac{\gamma+1}{2(\gamma-1)}=3$, > so the bracket factor is $(0.8333)^{3}\approx 0.579$: > $$=\frac{500000\cdot10^{-3}}{\sqrt{300}}\sqrt{\frac{1.4}{287}}\,(0.579) > \approx 1.16\ \text{kg/s}$$ > **Why this step?** When the throat is sonic ($A=A^*$), $\dot m$ is *fixed* by reservoir > conditions — lowering downstream pressure can't pump more air (the nozzle is **choked**). --- ## Common mistakes (steel-manned) > [!mistake] "Smaller area always means faster flow." > **Why it feels right:** garden-hose / kitchen-tap experience — pinch it, water shoots out. > That's true for **incompressible/subsonic** flow. > **The fix:** the momentum + continuity combo gives $\frac{dA}{A}=(M^2-1)\frac{dV}{V}$. > For $M>1$ the sign flips: to accelerate supersonic flow you must **diverge** the area. > [!mistake] "$A/A^*$ gives a unique Mach number." > **Why it feels right:** $f(M)$ *looks* like a function so inverting should be unique. > **The fix:** $f$ is **U-shaped**, not monotonic. Every $A/A^*>1$ has a subsonic AND a > supersonic root; the operating back-pressure decides which physically occurs. > [!mistake] "$A^*$ is a real physical throat that always exists in the duct." > **Why it feels right:** in a CD nozzle the throat *is* sonic when choked. > **The fix:** $A^*$ is a *reference* area. For a purely subsonic flow the actual minimum > area may never reach $M=1$, yet $A^*$ still exists mathematically as the constant yardstick. --- > [!recall]- Feynman: explain to a 12-year-old > Air flowing through a shaped pipe is like a crowd walking through a hallway that gets > narrow and wide. The narrowest spot is the **throat**. There's a magic speed (the speed > of sound) the crowd can reach exactly at the throat. The formula is a chart: tell it how > wide the hall is compared to that special narrow spot, and it tells you how fast people > are moving. Funny part: the same width can mean a *slow polite crowd* or a *super-fast > sprinting crowd* — you must know if they sped up past the throat or not. > [!mnemonic] Remember it > **"Sub Shrinks to Speed, Super Spreads to Speed; throat = ONE."** > Subsonic accelerates by shrinking area, Supersonic accelerates by spreading area, and > the sonic point ($M=1$) is the single minimum where $A/A^*=1$. --- ## Active recall #flashcards/physics What does $A^*$ physically represent? ::: The (constant) area at which the flow would be exactly $M=1$ for the same mass flow and stagnation state — a sonic reference yardstick. State the area–Mach relation. ::: $\frac{A}{A^*}=\frac1M\left[\frac{2}{\gamma+1}\left(1+\frac{\gamma-1}{2}M^2\right)\right]^{\frac{\gamma+1}{2(\gamma-1)}}$ What is $A/A^*$ at $M=1$? ::: Exactly 1 (the minimum of the curve). Why does each $A/A^*>1$ give two Mach numbers? ::: The function is U-shaped (non-monotonic): one subsonic root, one supersonic root. Which two conservation laws + relations derive it? ::: Continuity ($\rho A V=$const), the definition of $A^*$ at $M=1$, and isentropic stagnation relations for $\rho/\rho_0$, $T/T_0$. For supersonic flow, how must area change to accelerate? ::: Area must INCREASE (diverge), since $dA/A=(M^2-1)\,dV/V$. What does "choked" mean? ::: Throat reaches $M=1$; mass flow becomes fixed by reservoir $p_0,T_0,A^*$ and cannot increase by lowering back pressure. Limit of $A/A^*$ as $M\to0$ and $M\to\infty$? ::: Both $\to\infty$. --- ## Connections - [[Isentropic Stagnation Relations]] — supplies $\rho/\rho_0$, $T/T_0$, $p/p_0$. - [[Continuity Equation 1-D Compressible Flow]] — Pillar 1 of the derivation. - [[Area-Velocity Relation dA-A = (M^2-1) dV-V]] — explains the shape behavior. - [[Choked Flow & Maximum Mass Flow]] — the $M=1$ throat limit. - [[Converging-Diverging (de Laval) Nozzle]] — primary application. - [[Normal Shock Waves]] — why the supersonic branch may not persist downstream. - [[Speed of Sound a = sqrt(gamma R T)]] — defines $M$ and $a^*$. ## 🖼️ Concept Map ```mermaid flowchart TD MC[Mass conservation rho A V const] AS[Sonic reference area A*] AR[Area ratio A/A* = rho* a* / rho V] ISEN[Isentropic + stagnation relations] DR[rho*/rho from stagnation state] SP[a*/V split into 1/M and temperature] REL[A/A* = f of M and gamma] SUB[Subsonic M<1: converge to speed up] SUP[Supersonic M>1: diverge to speed up] THR[Throat A=A* gives M=1] PROPS[Local p, T, rho, V] MC -->|evaluate at sonic point| AS AS -->|rearrange| AR ISEN -->|gives rho*/rho| DR ISEN -->|gives a*/V| SP AR -->|substitute DR| REL DR --> REL SP --> REL REL -->|predicts| SUB REL -->|predicts| SUP REL -->|minimum area| THR REL -->|solve for M then| PROPS ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, jab hawa ek aise pipe me behti hai jiska cross-section area $A$ jagah-jagah badalta > hai (jaise nozzle), to ek special point hota hai jahan flow exactly **Mach 1** (sound ki > speed) tak pahunchti hai. Us point ka area hum $A^*$ bolte hain — yeh ek constant yardstick > hai. Area–Mach relation humein batata hai ki local area $A$, us sonic area $A^*$ se kitna > bada hai, aur usi se hum local Mach number nikaal lete hain. > > Derivation simple hai: pehle continuity ($\rho A V=$ constant) — mass kahin jam nahi sakta. > Phir sonic point pe yeh equation likho aur ratio bana lo: $A/A^* = (\rho^* a^*)/(\rho V)$. > Density aur temperature ke ratios ko isentropic stagnation relations se $M$ aur $\gamma$ me > likh do, bas final formula aa jata hai. $M=1$ daalo to $A/A^*=1$ aata hai — yeh perfect > sanity check hai. > > Sabse important twist: yeh curve **U-shaped** hai. Matlab koi bhi $A/A^*>1$ ke do answer > aate hain — ek subsonic, ek supersonic. Geometry akele decide nahi karti ki kaunsa branch > chalega; **back-pressure** decide karta hai. Aur yaad rakho: subsonic me speed badhane ke > liye area chhota karo, lekin supersonic me area **bada** karna padta hai (yahi de Laval > nozzle ka secret hai). Jab throat pe $M=1$ ho jaye to flow **choked** ho jati hai — phir > mass flow fix ho jata hai, chahe niche pressure kitna bhi gira do. ![[audio/3.1.06-Area-Mach-number-relation-A-A-=-f(M)-—-isentropic-flow.mp3]]