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

This page pushes the single equation $\frac{d A}{A} = (M^{2} - 1) \frac{d V}{V}$ through every case it can face. Before touching numbers, we build a map of all the scenario types so nothing surprises you. If any symbol here feels unfamiliar, revisit the parent derivation and the prerequisite notes.

Definition When does this relation apply? (the assumptions)

The equation $\frac{d A}{A} = (M^{2} - 1) \frac{d V}{V}$ was derived by combining three physical laws, and each one carries a condition. The relation is valid only when the flow is:

Steady — nothing changes with time at a fixed point (so the mass flow $ρ A V$ is truly constant, from the continuity equation).
Quasi one-dimensional — the gas moves essentially along one axis down the duct, so $A$ , $V$ , $ρ$ take one value per cross-section.
Isentropic and compressible — adiabatic (no heat in/out) and reversible (no friction/shock), so the isentropic link $d p = a^{2} d ρ$ holds and density is allowed to change.
Inviscid — no viscosity, so Euler's momentum equation $d p = - ρ V d V$ applies with no friction term. If any of these break (a shock, heat addition, strong wall friction), the relation no longer holds as written.

The scenario matrix

Every problem this equation can throw at you falls into one of these cells. The two inputs that matter are (a) the Mach regime (which fixes the sign of $M^{2} - 1$ ) and (b) what we want the flow to do (speed up or slow down). Plus the edge/limit cases. The "Example" column points to the matching worked example below (cell C1 ↔ Example 1, and so on).

#	Cell (scenario class)	$M^{2} - 1$ sign	What is asked	Example
C1	Subsonic, want to accelerate	negative	shape?	Example 1
C2	Subsonic, want to decelerate (diffuser)	negative	shape?	Example 2
C3	Supersonic, want to accelerate	positive	shape + magnitude	Example 3
C4	Supersonic, want to decelerate	positive	shape?	Example 4
C5	Exactly sonic $M = 1$ (degenerate: the multiplier $(M^{2} - 1) = 0$ )	zero	what does $d V$ do?	Example 5
C6	Limit $M \to 0$ (incompressible check)	$\to - 1$	recover water-pipe law	Example 6
C7	Real-world word problem (rocket bell sizing)	positive	magnitude	Example 7
C8	Exam twist: given $d A / A$ , find required $M$	solve for $M$	inverse problem	Example 8
C9	Sign/geometry sanity: which way does area move at a throat?	mixed	figure reasoning	Example 9

We now hit every cell.

Reading the sign — one picture to keep in your head

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

How to read Figure 1. The horizontal axis is the Mach number $M$ , running from $0$ (dead calm) to $3$ (three times the speed of sound). The vertical axis is the value of the multiplier $M^{2} - 1$ — the quantity that sits in front of $d V / V$ in our equation. The red curve is this multiplier. Trace it left to right: it starts at $- 1$ when $M = 0$ , climbs, passes through zero exactly at the dashed vertical line $M = 1$ (the red dot), and then rises steeply into positive territory. Below the horizontal axis the multiplier is negative (subsonic region, labelled left); above it the multiplier is positive (supersonic region, labelled right). That one zero-crossing — where the curve pierces the axis — is the entire physics of the de Laval nozzle.

Worked examples — one per matrix cell

(Every " $%$ " below stands in for a small local $d V / V$ ; see the differentials caution above — the answers are local approximations.)

Worked example Example 1 — C1: subsonic, want to speed up

Air at $M = 0.40$ must be accelerated. Should the duct converge or diverge, and by how much (fractionally) to gain $2%$ speed?

Forecast: subsonic → I bet it converges (like a garden hose).

Compute the multiplier: $M^{2} - 1 = 0.4 0^{2} - 1 = 0.16 - 1 = - 0.84$ . Why this step? The sign of this number decides the duct shape and its size sets the magnitude — so it is always the first thing to pin down.
We want $d V / V = + 0.02$ . Plug in: $\frac{d A}{A} = (- 0.84) (0.02) = - 0.0168$ . Why this step? The equation is now fully determined; substituting the target speed change turns it into the required area change.
Negative $d A / A$ means area shrinks → converging. Magnitude $1.68%$ .

Verify: Sign check — subsonic and speeding up should oppose ( $d A < 0$ while $d V > 0$ ). ✅ Units: both sides dimensionless. ✅ Matches the converging inlet.

Worked example Example 2 — C2: subsonic diffuser, want to SLOW DOWN

Air at $M = 0.60$ enters a diffuser meant to slow it by $5%$ (recover pressure). What shape?

Forecast: slowing subsonic flow — probably the duct opens out.

Multiplier: $M^{2} - 1 = 0.36 - 1 = - 0.64$ . Why this step? Same first move always — the multiplier's sign is what decides shape, so we get it before anything else.
Now $d V / V = - 0.05$ (a decrease). $\frac{d A}{A} = (- 0.64) (- 0.05) = + 0.032$ . Why this step? Here the target is a slow-down, so $d V / V$ is negative; negative $\times$ negative $=$ positive, which is how a decelerating case flips the outcome versus Example 1.
Positive $d A / A$ → area grows by $3.2%$ → diverging.

Verify: A subsonic diffuser really is a widening pipe; slowing the flow raises its pressure. Sign logic consistent with continuity: slower + fatter keeps $ρ A V$ fixed. ✅

Worked example Example 4 — C4: supersonic, want to SLOW DOWN

Supersonic air at $M = 1.8$ is decelerated by $3%$ . What shape?

Forecast: supersonic is the "backwards" world; slowing down should shrink the area.

Multiplier: $M^{2} - 1 = 3.24 - 1 = 2.24$ . Why this step? As always, the multiplier's sign is what fixes the duct shape, so we compute it first; positive here confirms we are in the supersonic regime.
$d V / V = - 0.03$ : $\frac{d A}{A} = (2.24) (- 0.03) = - 0.0672$ . Why this step? The target is a slow-down, so $d V / V$ is negative, and it carries straight through the positive multiplier to give a negative area change.
$d A / A = - 0.0672$ → area shrinks $6.72%$ → converging.

Verify: In supersonic flow $d A$ and $d V$ share a sign; both went negative here. ✅ This is a supersonic diffuser — it converges to slow supersonic flow, the mirror image of Example 2.

Worked example Example 5 — C5: the degenerate sonic point

$M = 1$ Flow is exactly sonic, $M = 1$ , at a throat where $d A = 0$ . What can $d V$ do?

Forecast: something special must happen — this is the crossover.

Multiplier: $M^{2} - 1 = 1 - 1 = 0$ . Why this step? At $M = 1$ the whole right-hand side is scaled by zero, so we compute it first to expose that collapse.
Equation becomes $\frac{d A}{A} = (0) \frac{d V}{V} = 0$ , i.e. $0 = 0$ . Why this step? With the multiplier zero the relation holds for any value of $d V / V$ — it no longer constrains the speed, which is the crucial degenerate feature.
Interpretation: $d V / V$ is unconstrained by this equation at $M = 1$ . The velocity may keep rising smoothly across the throat even though $d A = 0$ there — this is how flow transitions sub→supersonic.

Verify: $(1 - 1) \cdot x = 0$ is true for every $x$ , so the equation places no limit on $d V$ . ✅ Consistent with the parent note's " $0 \cdot (d V / V)$ indeterminate" remark. Sonic conditions require a throat, but a throat does not force sonic (choking depends on pressure ratio).

Worked example Example 6 — C6: limit

$M \to 0$ recovers the water-pipe law Show that as $M \to 0$ (nearly incompressible), the relation becomes the familiar continuity law for a liquid.

Forecast: at very low speed, density barely changes, so it should reduce to "area up ⇒ speed down."

Take $M \to 0$ : $M^{2} - 1 \to - 1$ . Why this step? We want the leading behaviour of the multiplier when the flow is far below sound speed, so we send $M$ to its extreme.
Equation → $\frac{d A}{A} = (- 1) \frac{d V}{V}$ , i.e. $\frac{d A}{A} + \frac{d V}{V} = 0$ . Why this step? Substituting the limiting multiplier removes the compressibility term and leaves the pure area–speed trade-off.
That is exactly $A V = constant$ (differentiated) — the incompressible continuity result with $ρ$ frozen.

Verify: Numerically at $M = 0.05$ : multiplier $= 0.0025 - 1 = - 0.9975 \approx - 1$ . ✅ So a $1%$ area drop gives $\approx 1%$ speed rise — the garden-hose intuition, recovered as a limit.

Worked example Example 7 — C7: real-world rocket bell

A rocket exhaust must accelerate from $M = 1$ at the throat to $M = 3$ at the exit. Using the equation locally at the exit, roughly how much area change accompanies a final $2%$ speed increase?

Forecast: far supersonic → area change should be huge.

At $M = 3$ : $M^{2} - 1 = 9 - 1 = 8$ . Why this step? We evaluate the multiplier at the local exit Mach number because the equation is a point relation — its coefficient changes as $M$ grows along the nozzle.
$\frac{d A}{A} = (8) (0.02) = 0.16$ . Why this step? With the local multiplier fixed, substituting the desired $d V / V = + 0.02$ gives the area change needed right there at the exit.
A mere $2%$ speed gain needs a $16%$ area increase — hence the wide flaring bell.

Verify: $8 \times 0.02 = 0.16$ . ✅ Larger $M$ ⇒ larger multiplier ⇒ steeper flare, matching real nozzle geometry (bells widen fastest near the exit).

Worked example Example 8 — C8: exam twist, solve for

$M$ A duct segment measures $d A / A = + 0.09$ while the flow accelerates by $d V / V = + 0.03$ . What is the local Mach number, and is the flow sub- or supersonic?

Forecast: area and speed both rose together → must be supersonic; let's find $M$ .

Rearrange: $M^{2} - 1 = \frac{d A / A}{d V / V} = \frac{0.09}{0.03} = 3$ . Why this step? The equation is linear in the multiplier, so dividing the two measured fractional changes isolates $(M^{2} - 1)$ — the inverse of the usual plug-in.
$M^{2} = 4 \Rightarrow M = 2$ . Why this step? Add $1$ and take the positive square root, because Mach number is a positive physical speed ratio.
$M = 2 > 1$ → supersonic, consistent with $d A$ , $d V$ sharing a sign.

Verify: Back-substitute: $(2^{2} - 1) (0.03) = 3 \times 0.03 = 0.09 = d A / A$ . ✅ Both fractional changes positive at $M > 1$ — correct regime.

Worked example Example 9 — C9: geometry sanity at a throat

Look at Figure 2. On the converging side ( $M < 1$ ), on the diverging side ( $M > 1$ ), and at the throat, state whether $d A$ is negative, positive, or zero as we move downstream in an accelerating de Laval nozzle.

Forecast: shrink, then flat, then grow.

Converging side: subsonic accelerating → from Example 1 logic $d A < 0$ (area shrinking). Why this step? We reuse the subsonic sign result to label the upstream half without recomputing — the physics is identical to C1.
Throat: $M = 1$ → $d A = 0$ (area minimum, momentarily flat). Why this step? At $M = 1$ the multiplier vanishes (Example 5), forcing $d A = 0$ ; geometrically that is a minimum-area point.
Diverging side: supersonic accelerating → from Example 3 logic $d A > 0$ (area growing). Why this step? Past the throat the flow is supersonic, so the C3 sign result applies and area must grow to keep accelerating.

Verify: The nozzle outline in Figure 2 pinches to a single narrowest red point (the throat) then flares — exactly the shape our three signs predict. ✅

How to read Figure 2. The black curves are the top and bottom walls of a de Laval nozzle seen in cross-section; the horizontal axis is distance travelled along the nozzle and the vertical axis is the duct half-width (how far each wall sits from the centre-line). The gas flows left to right (black arrow, labelled $d V > 0$ ). The walls pinch inward on the left — this is the converging, subsonic region where $d A < 0$ . They reach their narrowest gap at the vertical red line, the throat, where $M = 1$ and $d A = 0$ (the red dot marks it on the centre-line). Past the throat the walls flare outward — the diverging, supersonic region where $d A > 0$ . Reading the three regions left-to-right reproduces exactly the shrink → flat → grow pattern of Example 9.

Recall Quick self-test

Supersonic flow, area shrinks by $6%$ — is it speeding up or slowing down? ::: Slowing down. $M > 1$ means $d A$ and $d V$ share a sign, so $d A < 0 \Rightarrow d V < 0$ (Example 4 logic). You measure $d A / A = - 0.005$ and $d V / V = + 0.005$ . What does this tell you about $M$ ? ::: The ratio gives $M^{2} - 1 = - 1$ , i.e. $M^{2} \to 0$ , so $M \to 0$ : the flow is essentially incompressible (very low Mach). This is the limiting water-pipe case of Example 6, and it tells you the local Mach number is negligibly small, not that the speed is literally zero.