Visual walkthrough — Choked flow — condition M = 1 at throat, maximum mass flow

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Step 1 — What is "mass flow" through a hole?

WHAT. Imagine gas streaming through a slice of pipe of cross-section area $A$. In one second, how many kilograms cross that slice?

WHY. Everything about choking is a statement about this one number — the mass per second, written $\dot{m}$ (the dot means "per second"). Before we can find where it peaks, we must know what it is.

PICTURE. Look at the shaded slug of gas in the figure. In one second the gas moves forward a distance equal to its speed $V$ (metres travelled in one second). So the slug that crosses the slice is a cylinder of length $V$ and face area $A$: its volume is $A\times V$. Multiply by density $\rho$ (kilograms in each cubic metre) and you get kilograms per second.

Three things ($\rho$, $A$, $V$) fight to set $\dot{m}$. As we squeeze the gas faster, $V$ grows — but $\rho$ shrinks (fast gas thins out). That tug-of-war is the whole story. We need a single dial to describe it.

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Step 2 — The single dial: Mach number $M$

WHAT. The Mach number $M$ is the flow speed $V$ divided by the local speed of sound $a$: $M=\frac{V}{a},a=\sqrt{\gamma RT}.$

WHY. Instead of tracking $\rho$ and $V$ separately, we will write both in terms of one variable $M$. Then $\dot{m}$ becomes a function of a single dial, and "find the maximum" becomes a one-variable calculus problem. $M$ is the natural dial because the speed of sound $a$ is exactly the speed at which pressure news travels — and choking is about news that can't get upstream.

PICTURE. The dial below: $M<1$ subsonic (news travels upstream, throat "hears" the exit), $M=1$ sonic (news is frozen at the throat), $M>1$ supersonic (impossible in a purely converging nozzle). Each symbol earned:

• $\gamma$ = ratio of specific heats (for air, $1.4$) — how springy the gas is.
• $R$ = gas constant (for air, $287{\text{J kg}}^{-1}{\text{K}}^{-1}$).
• $T$ = local temperature.

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Step 3 — How temperature falls as the gas speeds up

WHAT. From energy conservation for a steady flowing gas, the moving temperature $T$ and the reservoir ("stagnation") temperature $T_0$ are linked: $\frac{T_0}{T}=1+\frac{\gamma -1}{2}{M}^2.$

WHY. A gas parked in a big tank has all its energy as heat (temperature $T_0$). When it flows, some heat becomes motion. Faster flow (bigger $M$) ⇒ colder gas (smaller $T$). We need this because both $\rho$ and $V$ ride on $T$, and we want them in terms of $M$.

PICTURE. The curve shows $T/T_0$ sliding down from $1$ (still gas, $M=0$) as $M$ climbs. Term by term: $\frac{T_0}{T}=1+\underset{\text{= }0.2\text{ for air}}{\underbrace{\frac{\gamma -1}{2}}}\underset{\begin{array}{c}\text{how fast,}\\ \text{squared}\end{array}}{\underbrace{M^2}}$ The "$+$" says: any motion adds to $T_0/T$, so $T$ can only drop. At $M=1$ for air, $T_0/T=1.2$, i.e. $T=0.833T_0$ — our critical temperature.

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Step 4 — How density falls as the gas speeds up

WHAT. Using isentropic (no-heat-loss, no-friction) flow, density follows temperature by a power law: $\frac{{\rho }_0}{\rho }={\left(1+\frac{\gamma -1}{2}{M}^2\right)}^{\frac{1}{\gamma -1}}.$

WHY. In Step 1 we saw $\dot{m}=\rho AV$: we must know how the density $\rho$ thins out as $M$ rises. Isentropic flow ties $\rho$ to $T$ through $\rho \propto T^{1/(\gamma -1)}$, so the same bracket reappears, just raised to a new power.

PICTURE. Compare the two curves: temperature (Step 3) and density (this step) both slide down, but density drops faster (bigger exponent for air: $1/(\gamma -1)=2.5$). This is the "fast gas thins out" effect that will fight against $V$.

$\rho ={\rho }_0{\left(1+\frac{\gamma -1}{2}{M}^2\right)}^{\stackrel{\text{minus }\Rightarrow \rho \text{shrinks}}{\overbrace{-\frac{1}{\gamma -1}}}}$

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Step 5 — Assemble $\dot{m}$ as a function of $M$ alone

WHAT. Substitute Steps 3–4 into $\dot{m}=\rho AV$, using $V=Ma=M\sqrt{\gamma RT}$ and ${\rho }_0=p_0/(R{T}_0)$. Everything collapses to: $\dot{m}=\underset{\text{fixed constants}}{\underbrace{\frac{A{p}_0}{\sqrt{T_0}}\sqrt{\frac{\gamma }{R}}}}\cdot \underset{f(M)\text{— the only part that moves}}{\underbrace{\frac{M}{{\left(1+\frac{\gamma -1}{2}{M}^2\right)}^{\frac{\gamma +1}{2(\gamma -1)}}}}}$

WHY. For a fixed reservoir ($p_0,T_0$) and a fixed slice area $A$, the leading clump never changes. All the drama lives in $f(M)$. To find peak mass flow we just find the peak of $f(M)$.

PICTURE. The numerator $M$ pulls the curve up (faster flow). The denominator (density-and-temperature bracket) pulls it down (thinning gas). At small $M$, the top wins and $\dot{m}$ rises. At large $M$, the bottom wins and $\dot{m}$ falls. Somewhere in between is a hilltop.

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Step 6 — Find the hilltop: set the slope to zero

WHAT. The peak is where the curve is momentarily flat — its slope $\frac{d\dot{m}}{dM}=0$. Because $f$ is a product/quotient of powers, we take the logarithm first (turns products into sums, easy to differentiate), then set the derivative to zero: $\frac{1}{M}-\frac{\gamma +1}{2(\gamma -1)}\cdot \frac{(\gamma -1)M}{1+\frac{\gamma -1}{2}{M}^2}=0.$

WHY use the derivative? The derivative is the tool that answers "where is this curve flat?" — exactly the question "where is the maximum?" No other tool gives the hilltop directly. Taking $\ln$ first is a convenience: $\ln f=\ln M-\frac{\gamma +1}{2(\gamma -1)}\ln (1+\frac{\gamma -1}{2}{M}^2)$ differentiates line-by-line.

PICTURE. The tangent line to the hill is horizontal exactly at the top. Clean the equation up: $\frac{1}{M}=\frac{(\gamma +1)M/2}{1+\frac{\gamma -1}{2}{M}^2}\Rightarrow 1+\frac{\gamma -1}{2}{M}^2=\frac{\gamma +1}{2}{M}^2\Rightarrow M^2=1.$

Subtract the $\frac{\gamma -1}{2}{M}^2$ term from both sides: $1=\frac{(\gamma +1)-(\gamma -1)}{2}{M}^2=\frac{2}{2}{M}^2=M^2$. So $M=1$ — the $\gamma$ cancels completely.

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Step 7 — Edge cases: what happens off the hilltop

WHAT. Three regimes, and one degenerate limit.

WHY. The contract: the reader must never meet a scenario we didn't show. So we walk every part of the curve.

PICTURE.

• $M<1$ (left of peak, subsonic): back pressure $p_b$ still controls the flow. Lower $p_b$ ⇒ climb the hill ⇒ more $\dot{m}$. The throat "hears" the exit.
• $M=1$ (the peak): the throat pressure locks at $p^{\ast }=0.528p_0$ for air (from Isentropic Flow Relations at $M=1$). Lowering $p_b$ further does nothing — you're already at the top.
• $M>1$ (right of peak): unreachable in a purely converging duct. Going supersonic needs a diverging section — a Converging–Diverging (de Laval) Nozzle. Past the throat, an over-strong back pressure can force a normal shock.
• Degenerate $M=0$: still gas, $\dot{m}=0$ — the left end of the curve, nothing flows.

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The one-picture summary

This single figure stacks the whole derivation: the two competing effects ($M$ up-pull, density down-pull), their product $f(M)$, and the vertical line at $M=1$ nailing the peak — with the locked throat pressure $p^{\ast }=0.528p_0$ marked.

Recall Feynman retelling — the walkthrough in plain words

We asked: how many kilograms per second squeeze through a slice of pipe? That's density times area times speed. As we push the gas faster, its speed goes up — good, more flow — but the gas also thins out — bad, less flow. To settle the fight we measured speed in "Mach", the flow speed compared to the speed of sound, because sound is exactly how pressure news travels. We wrote both the thinning and the speeding in terms of Mach, multiplied them together, and got one curve that starts at zero, bulges up, and comes back to zero. A curve like that has a top. We found the top by asking "where is the curve flat?" — that's the derivative set to zero — and out popped Mach = 1, and every gas gives the same answer. At that top the throat pressure freezes at about $0.528$ of the tank pressure for air, and the throat goes deaf to whatever happens downstream. That frozen, maximum flow is what we call choking.

Recall Quick self-test

Why write $\dot{m}$ in terms of $M$ instead of $\rho$ and $V$ separately? ::: So the two competing effects (speed up, density down) live in one variable and the maximum becomes a one-variable calculus problem. What tool finds the hilltop, and what question does it answer? ::: The derivative set to zero — it answers "where is the curve flat?", which is where the maximum sits. Does the choke Mach number depend on the gas? ::: No — $\gamma$ cancels; $M=1$ for every perfect gas. Why can't a converging nozzle push past $M=1$? ::: The area–Mach relation needs an increasing area to accelerate beyond sonic; a converging duct only shrinks toward the throat.