Prandtl-Meyer expansion waves — isentropic, supersonic turning: Worked examples

Before any symbol appears, a reminder of the quantities we will juggle, in plain words:

Definition The characters

$M$ = Mach number = (flow speed) ÷ (local speed of sound). $M = 1$ is exactly the speed of sound; $M > 1$ is supersonic. It is a pure number, no units.
$θ$ = deflection angle = how many degrees the wall (and the flow) bends. We measure it in degrees here, but the $ν$ formula produces radians — we convert.
$ν (M)$ = Prandtl–Meyer function = "how many degrees of turning you have already spent accelerating from sonic ( $M = 1$ ) up to the current $M$ ." Think of it as a fuel gauge for turning. $ν (1) = 0$ .
$p$ = static pressure = the pressure a sensor drifting along with the gas would read. Its values $p_{1}, p_{2}$ are the ones that change across the fan and that we compute below.
$p_{0}$ = stagnation (total) pressure = the pressure the gas would reach if brought smoothly to rest. In an isentropic fan $p_{0}$ stays constant; only the static $p$ (and its ratio to $p_{0}$ ) changes. See Isentropic flow relations.

Throughout, air is a perfect gas with $γ = 1.4$ (the ratio of specific heats — the number that tells you how a gas stores energy). With $γ = 1.4$ the useful constants are $\frac{γ + 1}{γ - 1} = 6$ and $\frac{γ - 1}{2} = 0.2$ .

The scenario matrix

Every problem this topic can throw is one of these cells. The examples below are tagged with the cell they cover.

Cell	What makes it special	Covered by
A. Standard turn	Ordinary $M_{1} > 1$ , moderate $θ$ , find $M_{2}$ and $p$ ratio	Ex 1
B. Start at sonic	$M_{1} = 1$ so $ν_{1} = 0$ — degenerate lower limit	Ex 2
C. Vacuum / max turn	$M \to \infty$ , $θ \to ν_{m a x}$ limiting behaviour	Ex 3
D. Wrong sign trap	Someone subtracts $θ$ (compression confusion)	Ex 4
E. Inverse question	Given $M_{1}, M_{2}$ , find the required $θ$	Ex 5
F. Real-world word problem	Airfoil / nozzle lip, physical pressure numbers	Ex 6
G. Over-turn / exam twist	$θ$ so large that $ν_{2} > ν_{m a x}$ → impossible	Ex 7
H. Mach-angle tracking	Follow $μ$ across the fan (first vs last wave)	Ex 8

Two formulas do all the work. Keep them in view:

Cell A — the standard turn

Worked example Example 1 — Air at

$M_{1} = 2.0$ turns through $θ = 1 0^{\circ}$ Forecast: before reading on, guess: does $M$ go up or down? Does $p$ rise or fall? Is the second Mach wave steeper or shallower than the first?

Step 1 — Find $ν (M_{1})$ . Plug $M_{1} = 2$ : $M^{2} - 1 = 3$ , so $ν (2) = 6 arctan 0.5 - arctan 3 = 0.4604 rad = 26.3 8^{\circ}$ . Why this step? $ν (M_{1})$ is the turning already "spent" reaching $M = 2$ . We must know it before we can add the corner's turn.

Step 2 — Add the deflection. Expansion ⇒ add (see the mnemonic "Expand = Away = Add"): $ν (M_{2}) = 26.3 8^{\circ} + 1 0^{\circ} = 36.3 8^{\circ}$ . Why this step? $ν$ measures total turn from sonic; the corner contributes exactly $θ$ more.

Step 3 — Invert to get $M_{2}$ . $ν$ climbs monotonically with $M$ , so exactly one $M$ gives $36.3 8^{\circ}$ . Solving numerically: $M_{2} \approx 2.385$ . Why this step? We need the new Mach number to read off every other property.

Step 4 — Static-pressure ratio. With $0.2 M_{1}^{2} = 0.8$ and $0.2 M_{2}^{2} = 0.2 (2.385)^{2} = 1.138$ : $\frac{p _{2}}{p _{1}} = (\frac{1.8}{2.138})^{3.5} \approx 0.55.$ Why this step? Because the fan is isentropic, the stagnation pressure $p_{0}$ stays fixed (as defined above); it is only the static pressures $p_{1}, p_{2}$ — the pressures riding along with the gas — that fall. Their ratio to the common $p_{0}$ is what the isentropic relation gives, and dividing those two ratios leaves the formula above.

Step 5 — Answer the forecast's wave question. The Mach angle is $μ = arcsin (1/ M)$ . First wave: $arcsin (1/2.0) = 30. 0^{\circ}$ ; last wave: $arcsin (1/2.385) = 24. 8^{\circ}$ . Since $M$ rose, $μ$ fell, so the second (last) wave is shallower than the first. Why this step? Closing the loop on the forecast — the fan opens up downstream, exactly as Ex 8 studies in detail.

Verify: $M_{2} > M_{1}$ ✓ (flow sped up), $p_{2} / p_{1} < 1$ ✓ (static pressure fell), $μ_{2} < μ_{1}$ ✓ (last wave shallower) — exactly what "expansion" should do. Sanity on magnitude: a $1 0^{\circ}$ turn near $M = 2$ typically pushes $M$ up by a few tenths; $2.0 \to 2.39$ is reasonable.

Cell B — starting exactly at sonic

Worked example Example 2 — Flow at

$M_{1} = 1$ turns through $θ = 2 0^{\circ}$ Forecast: what is $ν (M_{1})$ when the flow starts at the speed of sound?

Step 1 — Evaluate $ν (1)$ . At $M = 1$ , $M^{2} - 1 = 0$ , so both $arctan$ terms are $arctan 0 = 0$ . Hence $ν (1) = 0$ . Why this step? $M = 1$ is the built-in zero of the $ν$ scale — the degenerate lower edge. Confirming it means our fuel gauge reads empty at the start.

Step 2 — Add the turn. $ν (M_{2}) = 0 + 2 0^{\circ} = 2 0^{\circ}$ . Why this step? With nothing spent, the whole $2 0^{\circ}$ is available to accelerate the flow.

Step 3 — Invert. Solve $ν (M_{2}) = 2 0^{\circ}$ : $M_{2} \approx 1.775$ . Why this step? Reading $M$ straight off $ν = 2 0^{\circ}$ is only legal because $ν_{1} = 0$ — the simplest case.

Verify: $M_{2} \approx 1.78 > 1$ ✓ — a sonic stream turned $2 0^{\circ}$ is now comfortably supersonic. Cross-check with Ex 3's logic: this is exactly the entry point of the whole $ν$ curve.

Cell C — the vacuum limit (maximum turn)

Worked example Example 3 — How far can a sonic stream turn before the gas "runs out"?

Forecast: guess the maximum turn in degrees. More or less than a right angle? More than $9 0^{\circ}$ ?

Step 1 — Take $M \to \infty$ in $ν$ . As $M \to \infty$ , $M^{2} - 1 \to \infty$ , both $arctan$ arguments blow up, and $arctan (\infty) = \frac{π}{2}$ . So $ν_{m a x} = 6 \cdot \frac{π}{2} - \frac{π}{2} = \frac{π}{2} (6 - 1) .$ Why this step? Infinite Mach number is the physical ceiling — the gas has converted all its internal energy to kinetic energy, so pressure $\to 0$ (a vacuum). No further turning is possible.

Step 2 — Numbers. $\frac{π}{2} (6 - 1) = 9 0^{\circ} (6 - 1) \approx 130.4 5^{\circ}$ . Why this step? This single number caps every problem: no expansion starting from $M = 1$ can ever exceed $\approx 130. 5^{\circ}$ of turn.

Verify: $6 \approx 2.449$ , so $2.449 - 1 = 1.449$ , times $9 0^{\circ} \approx 130. 4^{\circ}$ ✓. It is indeed larger than $9 0^{\circ}$ — supersonic gas can turn past a right angle when expanding into vacuum.

Cell D — the wrong-sign trap

Worked example Example 4 — Someone writes

$ν (M_{2}) = ν (M_{1}) - θ$ for the Ex 1 turn. What goes wrong? Forecast: if we subtract instead of add, does the predicted $M_{2}$ come out too big or too small?

Step 1 — Apply the wrong rule. $ν (M_{2}) = 26.3 8^{\circ} - 1 0^{\circ} = 16.3 8^{\circ}$ . Why this step? We deliberately follow the mistake to see its consequence, not just be told it.

Step 2 — Invert. Solving $ν = 16.3 8^{\circ}$ gives $M_{2} \approx 1.686$ . Why this step? This exposes the contradiction: the "expanded" flow came out slower than $M_{1} = 2$ .

Step 3 — Detect the absurdity. An expansion (convex corner, flow bending away) must accelerate the flow. $M_{2} < M_{1}$ means we accidentally modelled a compression, which for supersonic flow is a shock — not isentropic, not a fan. Why this step? The sign check is the built-in error alarm: if $M_{2} < M_{1}$ for a stated expansion, the sign is flipped.

Verify: $1.686 < 2.0$ ✓ (the wrong answer really does slow the flow). Correct rule (Ex 1) gave $2.385 > 2.0$ . Rule: Expand = Away = Add.

Cell E — the inverse question

Cell F — real-world word problem

Worked example Example 6 — Wind-tunnel nozzle lip

Air leaves a nozzle at $M_{1} = 2.4$ and static pressure $p_{1} = 40 kPa$ , then flows past a lip that bends the wall $1 5^{\circ}$ away from the flow. Find $M_{2}$ and the pressure $p_{2}$ just downstream. Forecast: will $p_{2}$ be above or below $40 kPa$ ? Above or below atmospheric ( $101 kPa$ )?

Step 1 — $ν (M_{1})$ . $M_{1}^{2} - 1 = 4.76$ , so $ν (2.4) = 0.6265 rad = 35.9 0^{\circ}$ . Why this step? Same first move always: read the starting fuel gauge.

Step 2 — Add the lip turn. $ν (M_{2}) = 35.9 0^{\circ} + 1 5^{\circ} = 50.9 0^{\circ}$ . Why this step? A convex lip is an expansion, so add.

Step 3 — Invert. $ν = 50.9 0^{\circ} \Rightarrow M_{2} \approx 3.061$ . Why this step? Need $M_{2}$ before any pressure calculation.

Step 4 — Static pressure. $0.2 M_{1}^{2} = 1.152$ , $0.2 M_{2}^{2} = 0.2 (3.061)^{2} = 1.874$ : $p_{2} = 40 (\frac{1 + 1.152}{1 + 1.874})^{3.5} = 40 (0.7487)^{3.5} \approx 14.5 kPa .$ Why this step? The isentropic ratio converts the Mach change into a real static pressure, in the same kPa units we were given (the stagnation pressure $p_{0}$ is unchanged across the fan).

Verify: Units: kPa × (dimensionless) $^{3.5}$ = kPa ✓. $14.5 < 40$ ✓ (expansion drops pressure). This gas is now well below atmospheric — a classic overexpanded exit, relevant to Nozzle design and overexpansion.

Cell G — the impossible over-turn (exam twist)

Worked example Example 7 — Air at

$M_{1} = 4.0$ is asked to turn $θ = 12 0^{\circ}$ . Possible? Forecast: we know the ceiling is $ν_{m a x} \approx 130. 5^{\circ}$ . Is $M_{1} = 4$ close to it already?

Step 1 — $ν (M_{1})$ . $M_{1}^{2} - 1 = 15$ , so $ν (4) = 1.1441 rad = 65.5 5^{\circ}$ . Why this step? At $M = 4$ a lot of the $130. 5^{\circ}$ budget is already spent — check how much remains.

Step 2 — Required $ν (M_{2})$ . $65.5 5^{\circ} + 12 0^{\circ} = 185.5 5^{\circ}$ . Why this step? Straightforward add — but now compare with the ceiling.

Step 3 — Compare with $ν_{m a x}$ . $185.5 5^{\circ} > 130.4 5^{\circ}$ . Impossible. Why this step? $ν$ can never exceed $ν_{m a x}$ ; asking for more means the flow would need $M > \infty$ and negative pressure. Physically the gas separates / voids before completing the turn.

Verify: Remaining budget from $M = 4$ is $130.4 5^{\circ} - 65.5 5^{\circ} = 64.9 0^{\circ}$ . The demanded $12 0^{\circ}$ exceeds this ✓ — the correct exam answer is "no, the flow cannot turn that far; maximum further turn from $M = 4$ is $\approx 64. 9^{\circ}$ ."

Cell H — tracking the Mach angle across the fan

Worked example Example 8 — For the Ex 1 turn (

$M_{1} = 2.0 \to M_{2} \approx 2.385$ ), find the Mach angle of the first and last wave, and confirm the fan opens forward. Forecast: which wave is steeper — the first (at $M_{1}$ ) or the last (at $M_{2}$ )?

Step 1 — Recall the Mach angle. $μ = arcsin (1/ M)$ : the angle each Mach wave makes with the local flow (built in the parent note from the sound-circle picture). Why this step? The fan is bounded by these two waves; their angles define its geometry.

Step 2 — First wave. $μ_{1} = arcsin (1/2.0) = arcsin (0.5) = 30. 0^{\circ}$ . Why this step? This is the leading, steepest edge of the fan.

Step 3 — Last wave. $μ_{2} = arcsin (1/2.385) = arcsin (0.4193) = 24.7 9^{\circ}$ . Why this step? Because $M$ rose, $1/ M$ fell, so $μ$ fell — the trailing wave is shallower.

Verify: $24.7 9^{\circ} < 30. 0^{\circ}$ ✓ — waves get shallower downstream, so the fan spreads open in the flow direction. This matches the mistake-buster " $μ$ decreases through the fan."

The figure below draws exactly this. The upstream wall is horizontal; the corner turns the wall down by $θ = 1 0^{\circ}$ . The red line is the first Mach wave at $μ_{1} = 3 0^{\circ}$ above the incoming flow; the green line is the last wave, drawn at $μ_{2} = 24. 8^{\circ}$ measured from the already-turned downstream flow — so it lies below the red line, and the dashed blue lines in between are intermediate waves of the fan. Notice the red-to-green spread fanning open: that opening is the physical expansion fan, and the fact that green is shallower than red is the $μ$ decrease we just computed. The yellow arc at the corner marks the $1 0^{\circ}$ deflection.

Figure — Prandtl-Meyer expansion waves — isentropic, supersonic turning

Recall Quick self-test on the matrix

Cell B answer — $ν (1)$ equals what? ::: $0$ , because $M^{2} - 1 = 0$ makes both arctangents zero. Cell C answer — $ν_{m a x}$ for $γ = 1.4$ ? ::: $\approx 130.4 5^{\circ} = 9 0^{\circ} (6 - 1)$ . Cell D detector — how do you catch a flipped sign? ::: If the "expanded" flow has $M_{2} < M_{1}$ , you subtracted instead of added. Cell G rule — when is a turn impossible? ::: When $ν (M_{1}) + θ > ν_{m a x}$ ; the flow would need $M = \infty$ , $p = 0$ . Cell H trend — through the fan, $μ$ does what? ::: Decreases, since $M$ rises and $μ = arcsin (1/ M)$ .

See also: Mach waves and Mach cone, Isentropic flow relations, Oblique shock waves, Method of characteristics, Entropy and the second law.