Question bank — Converging nozzle — subsonic flow, Mach 1 at exit

False — a converging duct accelerates subsonic gas only up to $M=1$; the mass-flow factor $\dot{m}(M)$ peaks exactly at $M=1$, so raising $p_0$ scales the flow up but never pushes $M$ past 1. You need area to increase afterwards (a de Laval nozzle) to go supersonic.

False — choked means $p_e=p^{\ast }=0.528p_0$, set by the nozzle, not the room. The leftover expansion from $p^{\ast }$ down to $p_b$ happens outside as expansion fans, so $p_e$ is usually above $p_b$.

True — a subsonic jet can send pressure signals back upstream, so it continuously adjusts until $p_e=p_b$. This is exactly why the flow "feels" the room in the subsonic regime.

False — below $p_b/p_0=0.528$ the flow is choked and $\dot{m}$ is flat. The sonic exit blocks any pressure signal from travelling upstream, so the reservoir never "hears" that $p_b$ dropped.

$p_0$ on an already-choked nozzle, the mass flow increases. True — $\dot{m}=A{p}_0\sqrt{\gamma /(R{T}_0)}\times (\text{factor at }M=1)$ is directly proportional to $p_0$. Choking caps the Mach number, not the mass flow; more $p_0$ means denser gas at the same sonic speed, hence more $\dot{m}$.

$\sqrt{\gamma R{T}_0}$. False — the exit speed is the local sound speed $a^{\ast }=\sqrt{\gamma R{T}^{\ast }}$, and $T^{\ast }=0.833T_0<T_0$. Because the gas cooled while accelerating, $a^{\ast }$ is smaller than the reservoir sound speed.

False — nothing piles up; the flow is smooth and steady. "Choking" means the information channel is severed: at $M=1$ pressure waves can't outrun the flow to travel upstream, so downstream changes stop influencing $\dot{m}$.

$\gamma =5/3$) the critical ratio is still 0.528. False — 0.528 is specific to $\gamma =1.4$. Plugging $\gamma =5/3$ into ${\left(\frac{2}{\gamma +1}\right)}^{\gamma /(\gamma -1)}$ gives $\approx 0.487$, so the critical ratio depends on the gas.

$p_b/p_0=0.30<0.528$, so the exit pressure is $0.30p_0$." The error: the exit pressure is not $p_b$ here. Since $0.30<0.528$ the nozzle is choked, so $p_e=0.528p_0$, and the gas expands the rest of the way outside the nozzle.

$M=1$, so the gas leaves at the speed of sound measured in the reservoir, about 347 m/s for $T_0=300$ K." The error: $M=1$ uses the local sound speed at the cooled exit ($T^{\ast }=250$ K), giving $a^{\ast }=\sqrt{1.4\cdot 287\cdot 250}\approx 317$ m/s, not the reservoir value 347 m/s.

$p_b$ has zero effect on anything." The error: it has zero effect inside the nozzle and on $\dot{m}$, but it does change the flow outside — a lower $p_b$ means the under-expanded jet expands more strongly downstream. Only the internal state is frozen.

$p_0/p_e=500/101$ and solved the isentropic relation." The error: that only works if unchoked. Here $101/500=0.202<0.528$, so it's choked and $M=1$ automatically; using $p_e=p_b=101$ kPa in the relation gives a fictitious supersonic $M$ that a converging duct can't reach.

$\dot{m}$." The error: $\dot{m}\propto p_0/\sqrt{T_0}$, so raising $T_0$ lowers $\dot{m}$ (hotter gas is less dense at fixed $p_0$). Area is not the only lever.

The error: the internal nozzle flow is (idealized) isentropic, but the external under-expanded jet involves expansion fans and possibly shocks — outside the nozzle you cannot assume isentropy blindly (see Normal Shock Waves for the shock case).

$M=1$ rather than some other Mach number? Because $M=1$ is where pressure signals travel exactly at the flow speed — the boundary between "downstream can talk upstream" and "it can't." It's also where $\dot{m}(M)$ peaks, so it's the natural physical ceiling of a converging duct.

higher than the back pressure when choked, instead of lower? The nozzle locks $p_e$ at $p^{\ast }=0.528p_0$ because the sonic exit can't relay the low $p_b$ back inside. So the gas emerges over-pressured relative to the room and finishes expanding outside — this is called an under-expanded jet.

$V$ into Mach number $M$ in the derivation? Because the special physics — sound-speed signalling, choking — lives at $M=1$, a dimensionless threshold. Writing $V=Ma$ with $a=\sqrt{\gamma RT}$ folds temperature into the story and makes the choking condition a clean "$M=1$" instead of a messy velocity value.

Subsonic flow is slower than its own sound waves, so pressure disturbances can travel upstream and inform the reservoir. At $M=1$ those waves move at exactly the flow speed and are swept downstream — the upstream side is causally cut off.

For steady subsonic flow, conservation of mass ($\rho AV=\text{const}$) with only mild density change forces $V$ up when $A$ goes down. The Isentropic Flow Relations show this trend reverses once supersonic — which is precisely why a converging duct can't push past $M=1$.

$T_0$ reduce the choked mass flow even though the gas is hotter and faster? Hotter reservoir gas is less dense at fixed $p_0$, and the extra sonic speed doesn't compensate: $\dot{m}\propto p_0/\sqrt{T_0}$. Density falls faster than sound speed rises, so $\dot{m}$ drops.

$p_b>p_0$ (back pressure above reservoir pressure)? There is no forward push at all — the pressure gradient points the wrong way. The gas would tend to flow backwards into the reservoir, or simply stay still; the nozzle delivers zero (or reversed) flow until $p_b$ falls below $p_0$.

$p_b=p_0$ exactly? No pressure difference, so no flow: $V=0$, $M=0$, and the "exit" is just the reservoir. $\dot{m}=0$.

at $p_b/p_0=0.528$ (the exact critical ratio)? The exit just reaches $M=1$ with $p_e=p_b=p^{\ast }$ simultaneously. It's the single point where "unchoked, $p_e=p_b$" and "choked, $p_e=p^{\ast }$" agree — the seam between the two regimes.

$p_b\to 0$ (exhausting into a vacuum)? The flow stays choked at $M=1$, $p_e=0.528p_0$ inside, and $\dot{m}$ stays at its ceiling. The jet expands more and more violently outside, but nothing inside the nozzle changes.

$p_b/p_0=1^{-}$ (a whisper below 1)? A tiny $M$ just above 0 — the flow barely trickles and is deeply subsonic, with $p_e=p_b$. You approach the no-flow limit continuously.

$p_e<p^{\ast }$ at its exit plane? No — inside a converging-only duct the exit can't go below $p^{\ast }=0.528p_0$, because that would require $M>1$, which the geometry forbids. Sub-critical exit pressures only occur outside or in a diverging section.

finite tank (not infinite), what changes over time when choked? As gas drains, $p_0$ (and $T_0$) fall, so $\dot{m}\propto p_0/\sqrt{T_0}$ slowly decreases. The nozzle stays choked as long as $p_b/p_0\le 0.528$, but the "constant" mass flow is only constant while $p_0$ is held constant.

Cover the answers: for any $p_b/p_0$ you should instantly say (a) choked or not, (b) what $p_e$ equals, and (c) what $M$ at the exit is. If you can do all three, this bank is done with you.