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

False. Because $f(M)$ falls then rises (its derivative changes sign at $M=1$), a horizontal line at any $A/A^{\ast }>1$ cuts the U-shaped curve twice — one subsonic, one supersonic root; only $A/A^{\ast }=1$ (the valley floor at $M=1$) is unique.

False. For isentropic flow the mass flow and stagnation state are conserved, so $A^{\ast }$ is a constant — that is exactly why it works as a yardstick.

True. When the nozzle is choked the actual minimum (throat) area is the sonic area, so the geometric minimum coincides with the reference and $M=1$ there.

False. $A^{\ast }$ is a reference (the area the flow would need to hit $M=1$); it exists mathematically as a constant even when no physical point in the duct is sonic.

False. The curve's minimum value is $f(1)=1$; every other Mach gives $A/A^{\ast }>1$, so the ratio never dips below one.

False. As $M\to 0$ the $1/M$ factor in $f(M)$ blows up, so $A/A^{\ast }\to \infty$: to carry a fixed mass flow at nearly-zero speed you'd need an enormous area.

False. They share the same area ratio only. Through $\frac{p}{p_0}=(1+\frac{\gamma -1}{2}{M}^2{)}^{-\gamma /(\gamma -1)}$, higher $M$ means much lower $p/p_0$, so the supersonic root sits at far lower static pressure (and temperature).

False. It comes from continuity ($\rho AV=$ const), the definition of $A^{\ast }$ at $M=1$, and the isentropic stagnation relations — momentum is what gives the separate area–velocity relation.

The area ratio is not the Mach number. Plugging into $f(M)=2$ gives two roots (≈0.31 subsonic, ≈2.2 supersonic); you must solve the equation, not read $M$ off directly.

True only for subsonic flow. From $\frac{dA}{A}=(M^2-1)\frac{dV}{V}$, once $M>1$ the factor $(M^2-1)$ turns positive, so area and speed rise together — you must diverge (widen) the area to accelerate. See Area-Velocity Relation dA-A = (M^2-1) dV-V.

Once choked ($M=1$ at the throat), $\dot{m}$ is fixed by reservoir $p_0,T_0$ and $A^{\ast }$. The throat can't "hear" downstream changes because pressure signals travel only at the speed of sound; see Choked Flow & Maximum Mass Flow.

The smallest actual area equals $A^{\ast }$ only when the flow is sonic there. In a subsonic-only duct the minimum area still has $M<1$, so $f(M)>1$ there and its area exceeds $A^{\ast }$.

In supersonic flow the density and temperature drop fast enough that speed increases as area grows — total (stagnation) enthalpy is conserved; the flow trades thermal energy for kinetic energy, not loses it.

$f(M)$ has no closed-form inverse and is non-monotonic; you must iterate numerically or use tables, and choose the branch (sub- or supersonic) from the physics.

At $M\to 0$ the flow is nearly still ($V\to 0$), so $\rho AV$=const demands huge $A$ (the $1/M$ factor blows up); at $M\to \infty$ the density collapses toward zero (the bracket's power of $M^2$ blows up), so again $A$ must grow enormously to pass the same mass. This is the two rising arms of the U in figure s01.

Because $df/dM\propto (M^2-1)$ (the $\gamma$-terms cancelled to leave exactly $M^2-1$), which is zero only at $M=1$ and changes from negative to positive there — the sole turning point (the red dot in figure s01). Physically it is where the area–velocity sign flips, so the area is stationary (a minimum) exactly at the sonic point.

The sonic point is the unique state fixed for the whole streamtube by two conserved quantities together — mass flow and stagnation state — so its area $A^{\ast }$ is a single constant. The reservoir has $V=0$, giving no throughput per unit area to normalise by; the sonic state is the natural "one flow unit" that makes $\dot{m}={\rho }^{\ast }{A}^{\ast }{a}^{\ast }$ collapse continuity into a clean function of $M$ alone.

Geometry alone gives a symmetric double answer (the two intersections of the green line with the U-curve in figure s01); the back-pressure $p_b$ — the downstream push sketched in figure s02 — sets the pressure gap $p_0-p_b$ that determines whether the flow stayed subsonic or passed through a sonic throat and went supersonic. Only the physics of the ends breaks the tie.

Constant $A^{\ast }$ relies on unchanging stagnation state; friction (viscosity) and shocks raise entropy and drop stagnation pressure, which increases $A^{\ast }$ downstream, so the single-yardstick derivation no longer holds. In a real nozzle, boundary-layer growth and heat transfer make the effective $A^{\ast }$ drift, which is why designers apply discharge-coefficient corrections. See Normal Shock Waves.

Through the Isentropic Stagnation Relations, $M$ alone fixes $T/T_0$, $p/p_0$, and $\rho /{\rho }_0$ relative to the fixed reservoir values; the area ratio is the one hard-to-get piece, and everything thermodynamic follows once $M$ is in hand.

Exactly one — this is the single point where the subsonic and supersonic branches meet at the valley floor (the lone turning point of $f$), so the double-root ambiguity disappears.

The flow is no longer isentropic across the shock: entropy rises, stagnation pressure drops, and the downstream $A^{\ast }$ becomes larger than the upstream one — the yardstick jumps. See Normal Shock Waves.

No. A converging duct can accelerate subsonic flow at most to $M=1$ at its narrowest exit; to go supersonic you need a diverging section afterward, i.e. a Converging-Diverging (de Laval) Nozzle.

Yes for the same $\gamma$ — $f(M)$ depends only on the ratio, not absolute area, so exit Mach is identical (mass flow scales with size, but Mach does not).

Both roots converge toward $M=1$; the subsonic root rises toward 1 and the supersonic root falls toward 1, meeting exactly at the choked throat condition — the valley floor.

No — no point in the duct actually reaches $M=1$, so $A^{\ast }$ is a purely mathematical reference smaller than every real area present.

Never perfectly — viscosity (friction along the walls), boundary-layer growth, and any heat exchange all add entropy, so real flows only approximate the isentropic $f(M)$; the relation is an idealised backbone corrected by empirical factors in practice.