Area-velocity relation — dA - A = (M² − 1)(dV - V) — derivation (explains de Laval nozzle): Exercises

These problems walk up a ladder: first just recognising the sign rule, then plugging in, then analysing competing effects, then synthesising a nozzle, and finally mastering the subtle edge cases. Every problem has a full solution hidden inside a collapsible callout — try first, then reveal.

The one tool you need is the area–velocity relation:

Several problems below also use the differential continuity relation, so let us state where it comes from before we lean on it:

Read the master equation like a sign machine:

$(M^{2} - 1) < 0$ (subsonic): $d A$ and $d V$ have opposite signs.
$(M^{2} - 1) > 0$ (supersonic): $d A$ and $d V$ have the same sign.
$(M^{2} - 1) = 0$ (sonic): $d A = 0$ , a throat.

Figure s01 — the sign machine. The curve below plots $(M^{2} - 1)$ against $M$ . Look at where the curve crosses zero (the red dashed line at $M = 1$ ): to its left the shaded region is negative (subsonic, opposite signs), to its right it is positive (supersonic, same signs). This single crossing is the whole story of the page.

Figure — Area-velocity relation — dA - A = (M² − 1)(dV - V) — derivation (explains de Laval nozzle)

Level 1 — Recognition

Exercise 1.1

A gas flows at $M = 0.7$ . You want it to slow down ( $d V < 0$ ). Should the duct converge or diverge?

Recall Solution

WHAT: We need the sign of $d A$ when $d V < 0$ at subsonic speed. WHY: The equation ties the two together through the sign of $(M^{2} - 1)$ . $M = 0.7 \Rightarrow M^{2} - 1 = 0.49 - 1 = - 0.51 < 0$ . So $\frac{d A}{A} = (- 0.51) \frac{d V}{V}$ . With $d V / V < 0$ , the product $(- 0.51) (negative) = positive$ , so $d A > 0$ . Answer: the duct must diverge. (Subsonic gas slows down when the pipe widens — just like water.)

Exercise 1.2

Match each regime to what happens to area when the flow accelerates ( $d V > 0$ ): (a) $M = 0.3$ , (b) $M = 1$ , (c) $M = 3$ .

Recall Solution

Sign of $(M^{2} - 1)$ decides, with $d V / V > 0$ :

(a) $M = 0.3$ : $M^{2} - 1 = - 0.91 < 0 \Rightarrow d A < 0$ → converging.
(b) $M = 1$ : $M^{2} - 1 = 0 \Rightarrow d A = 0$ → throat (stationary area).
(c) $M = 3$ : $M^{2} - 1 = 8 > 0 \Rightarrow d A > 0$ → diverging.

Level 2 — Application

Exercise 2.1

A supersonic flow at $M = 2.5$ gains speed by $2%$ ( $d V / V = 0.02$ ). Find the fractional area change $d A / A$ (as a percent).

Recall Solution

WHAT: Plug numbers into the master equation. WHY: $M$ and $d V / V$ are both given, so $d A / A$ is a direct product. $\frac{d A}{A} = (M^{2} - 1) \frac{d V}{V} = (2. 5^{2} - 1) (0.02) = (6.25 - 1) (0.02) = 5.25 \times 0.02 = 0.105.$ Answer: $d A / A = 0.105 = 10.5%$ — the area must increase by 10.5% to gain 2% speed. Supersonic acceleration is "expensive" in area.

Exercise 2.2

A subsonic flow at $M = 0.4$ passes through a section where the area shrinks by $6%$ ( $d A / A = - 0.06$ ). What is the fractional speed change $d V / V$ ?

Recall Solution

WHAT: Solve the equation for $d V / V$ . WHY: Now $d A / A$ is the input; rearrange. $\frac{d V}{V} = \frac{d A / A}{M ^{2} - 1} = \frac{- 0.06}{0. 4 ^{2} - 1} = \frac{- 0.06}{0.16 - 1} = \frac{- 0.06}{- 0.84} = 0.0714 \dots$ Answer: $d V / V \approx 0.0714 = 7.14%$ — the flow speeds up by about 7.1%. (Squeeze a subsonic gas → it accelerates, as expected.)

Exercise 2.3

At $M = 1.2$ the area increases by $3%$ . Find $d V / V$ .

Recall Solution

WHAT: Solve the master equation for the unknown $d V / V$ , given $M$ and $d A / A$ . WHY: The area change is the input here, so we isolate $d V / V$ by dividing by $(M^{2} - 1)$ — the same rearrangement as Ex 2.2, but now the flow is supersonic so the bracket is positive. $M^{2} - 1 = 1.44 - 1 = 0.44$ . $\frac{d V}{V} = \frac{d A / A}{M ^{2} - 1} = \frac{0.03}{0.44} = 0.0682 = 6.82% .$ Answer: $d V / V \approx 6.82%$ . Supersonic, area grows and speed grows (same sign) — consistent with the sign machine.

Level 3 — Analysis

Exercise 3.1

Using the density relation $\frac{d ρ}{ρ} = - M^{2} \frac{d V}{V}$ (from the isentropic combination), at $M = 2$ compare the magnitude of the fractional density drop to the fractional speed rise for $d V / V = + 0.01$ . Which effect dominates, and why does that force the area to grow?

Recall Solution

WHAT: Compute both fractional changes. WHY: The continuity relation stated at the top of the page, $\frac{d ρ}{ρ} + \frac{d A}{A} + \frac{d V}{V} = 0$ , balances three effects; whichever of density/velocity wins forces the sign of $d A$ . $\frac{d ρ}{ρ} = - M^{2} \frac{d V}{V} = - (4) (0.01) = - 0.04 (4% drop) .$ Speed rise is only $+ 0.01$ (1%). So density falls 4× faster than speed rises. Plug into continuity: $\frac{d A}{A} = - \frac{d ρ}{ρ} - \frac{d V}{V} = - (- 0.04) - 0.01 = 0.04 - 0.01 = 0.03.$ Answer: $d A / A = + 0.03$ (3% area increase). Because density is thinning out faster than the gas is speeding up, the product $ρ V$ actually falls; to keep $ρ A V$ constant the area must grow. That is the physical reason supersonic ducts diverge.

Exercise 3.2

Show that at exactly $M = 1$ the density-drop rate equals the speed-rise rate in magnitude, and hence $ρ V$ is momentarily stationary. What does this imply for $d A$ ?

Recall Solution

At $M = 1$ : $\frac{d ρ}{ρ} = - (1) \frac{d V}{V} = - \frac{d V}{V}$ . So the fractional density drop exactly cancels the fractional speed rise: $\frac{d ρ}{ρ} + \frac{d V}{V} = 0 \Rightarrow d (ρ V) / (ρ V) = 0.$ From continuity $\frac{d A}{A} = - (\frac{d ρ}{ρ} + \frac{d V}{V}) = 0$ . Answer: $d A = 0$ — the area is stationary. This is the throat: the exact crossover where density and velocity effects balance. Sonic flow can only occur where $d A = 0$ .

Figure s02 — the hourglass. The nozzle profile below shows the duct half-width (proportional to area) against position. Trace the flow arrow left to right: the walls squeeze in through the subsonic region (green label), pinch to a minimum at the red dashed throat ( $M = 1$ , $d A = 0$ ), then flare open through the supersonic region. Notice the throat is exactly where the two halves are closest — that is the $d A = 0$ point Exercise 3.2 just proved.

Level 4 — Synthesis

Exercise 4.1

Design reasoning: a rocket must accelerate combustion gas from a slow $M = 0.2$ chamber flow all the way to $M = 3$ at exit, always accelerating ( $d V > 0$ ). Describe the required shape of the duct region by region, and justify each transition using the sign of $(M^{2} - 1)$ .

Recall Solution

Trace $(M^{2} - 1)$ as $M$ climbs, always demanding $d V > 0$ :

Region	$M$	$M^{2} - 1$	For $d V > 0$ , need	Shape
Chamber → throat	$0.2 \to 1$	negative	$d A < 0$	Converging
Throat	$1$	zero	$d A = 0$	Minimum area
Throat → exit	$1 \to 3$	positive	$d A > 0$	Diverging
Answer: a converging–diverging (de Laval) nozzle: it narrows to a throat where $M = 1$ , then flares open to reach $M = 3$ . Each transition is forced by the sign of $(M^{2} - 1)$ — you cannot pass through $M = 1$ anywhere except a throat, because only there is $d A = 0$ . See De Laval Nozzle and Choked Flow.

Exercise 4.2

As a sanity estimate of how sharply the supersonic bell must flare, take a single small step of the flow at $M = 3$ where $d V / V = 0.10$ (a $10%$ speed gain). What area change results, and does it confirm the nozzle keeps flaring?

Recall Solution

WHAT: Find $d A / A$ for a supersonic step at $M = 3$ given $d V / V = 0.10$ . WHY: The master equation directly gives $d A / A$ from $M$ and $d V / V$ ; evaluating it at high Mach shows how large the bracket $(M^{2} - 1)$ becomes, which tells us how fast the bell must open. At $M = 3$ : $M^{2} - 1 = 9 - 1 = 8$ . $\frac{d A}{A} = (M^{2} - 1) \frac{d V}{V} = 8 \times 0.10 = 0.80 = 80% .$ Answer: an $80%$ area increase for a $10%$ speed gain. The bracket $(M^{2} - 1)$ is huge at high Mach, so the bell must flare very rapidly — exactly why real supersonic nozzles have wide exit bells. This confirms continual divergence.

Level 5 — Mastery

Exercise 5.1

A duct is converging ( $d A < 0$ ) but the flow is measured to be decelerating ( $d V < 0$ ). What can you conclude about the Mach regime? Explain fully using the sign logic, and identify the physical setting where this occurs.

Recall Solution

WHAT: Both $d A < 0$ and $d V < 0$ → they have the same sign. WHY: Same-sign $d A$ and $d V$ requires $(M^{2} - 1) > 0$ . $\frac{d A}{A} = (M^{2} - 1) \frac{d V}{V}, ( neg ) = (M^{2} - 1) (neg) \Rightarrow (M^{2} - 1) > 0.$ Answer: the flow must be supersonic ( $M > 1$ ). A converging supersonic duct decelerates the flow — this is a supersonic diffuser (used in supersonic engine intakes and wind-tunnel diffusers). Subsonically the same shrinking duct would accelerate; the sign flip is the whole point.

Exercise 5.2

At the throat $(d A = 0)$ of a choked de Laval nozzle, the flow is sonic ( $M = 1$ ). The equation gives $0 = (M^{2} - 1) d V / V = 0 \cdot d V / V$ . Explain why $d V / V$ is not forced to zero here, and why this indeterminate form is exactly what permits a smooth subsonic→supersonic transition.

Recall Solution

WHAT: At $M = 1$ the bracket $(M^{2} - 1) = 0$ , so the equation reads $0 = 0 \cdot (d V / V)$ . WHY: Any value of $d V / V$ satisfies $0 = 0$ . The equation places no constraint on the speed change at the throat. The physics: because $d V / V$ is unconstrained, the flow may keep accelerating ( $d V > 0$ ) right through the throat even though the area is momentarily stationary. If $(M^{2} - 1)$ were nonzero, $d A = 0$ would force $d V = 0$ (flow neither speeds up nor slows) — a dead point. Only at $M = 1$ does the indeterminate $0 \cdot (d V / V)$ open a "gate" for the velocity to slide continuously from below sound to above sound. Answer: $d V / V$ is indeterminate (not zero) at the sonic throat; this indeterminacy is precisely the mathematical door through which flow passes from subsonic to supersonic.

Exercise 5.3

Degenerate check: what does the relation predict for a duct of constant area ( $d A = 0$ ) carrying subsonic flow ( $M < 1$ , not sonic)? Is acceleration possible?

Recall Solution

WHAT: Impose $d A = 0$ with $M < 1$ (so $M^{2} - 1 \neq = 0$ ) and solve the master equation for $d V / V$ . WHY: Setting the area change to zero isolates what the equation permits for the speed; with a nonzero bracket the only way the product can vanish is for $d V / V$ itself to vanish — the opposite situation to the sonic throat of Ex 5.2, where the bracket was zero. $0 = (M^{2} - 1) \frac{d V}{V}, (M^{2} - 1) \neq = 0 \Rightarrow \frac{d V}{V} = 0.$ Answer: $d V = 0$ — a constant-area subsonic duct produces no speed change (ignoring friction/heat). Acceleration is impossible without an area change. This is the opposite of the sonic throat: with the bracket nonzero, $d A = 0$ rigidly forces $d V = 0$ . It highlights that the throat's special behaviour is entirely due to $(M^{2} - 1) = 0$ , not merely $d A = 0$ .

Recall Master recall — the sign machine

The area–velocity relation is one sign rule ::: $d A / A = (M^{2} - 1) d V / V$ ; sign of $(M^{2} - 1)$ links area and velocity changes. Subsonic accelerate needs ::: converging duct ( $d A < 0$ ). Supersonic accelerate needs ::: diverging duct ( $d A > 0$ ). Constant-area subsonic duct gives ::: $d V = 0$ (no acceleration). Converging supersonic duct does ::: decelerates the flow (supersonic diffuser). Sonic throat allows subsonic→supersonic because ::: $0 \cdot (d V / V)$ is indeterminate, leaving $d V / V$ unconstrained.