Prandtl-Meyer function ν(M): Worked examples — Physics deep dive

This page is the "throw everything at it" companion to the parent ν(M) note. Before we grind numbers, let us name every kind of situation a Prandtl–Meyer problem can be. Then each worked example is tagged with the cell of the matrix it covers, so by the end you have seen all of them.

Everything below uses air, $γ = 1.4$ , unless stated.

We keep one master toolbox of three formulas:

Two words we will keep saying, defined plainly:

Score $= ν (M)$ : the running total of turning a flow has "spent" since it was barely supersonic ( $M = 1$ , score $= 0$ ). Turning away from yourself adds to the score.
Fan edges: an expansion is a wedge of Mach waves. The first (upstream) edge tilts at $μ_{1} = arcsin (1/ M_{1})$ from the incoming flow; the last (downstream) edge tilts at $μ_{2} = arcsin (1/ M_{2})$ from the outgoing flow. Picture a Japanese hand fan opening up.

The scenario matrix

Here is every distinct case class a ν(M) problem can hand you. Each later example fills one or more cells.

#	Cell (case class)	What is special about it	Covered by
A	Forward expansion (find $M_{2}$ from $M_{1}$ , $θ$ )	The bread-and-butter direct problem	Ex 1
B	Pressure/temperature after fan	Chain ν → isentropic ratios	Ex 2
C	Fan geometry (opening angle, edge tilts)	Turns $μ_{1}, μ_{2}$ into a drawn wedge	Ex 3 (figure)
D	Degenerate: $θ = 0$	No turn ⇒ Mach frozen; sanity anchor	Ex 4
E	Limiting: $M_{1} \to 1$	Fan edge $μ \to 9 0^{\circ}$ ; smallest score	Ex 4
F	Sign / direction trap (compression = subtract?)	When ν decreases; when it is illegal	Ex 5
G	Vacuum limit ( $θ$ too big, $M \to \infty$ )	$ν_{m a x}$ ceiling; void forms	Ex 6
H	Real-world word problem (jet exhaust plume)	Under-expanded nozzle turning at lip	Ex 7 (figure)
I	Exam twist: given $p_{2} / p_{1}$ , find the turn	Run the whole chain backwards	Ex 8

Worked examples

Ex 1 — Forward expansion (Cell A)

Worked example Statement

Air at $M_{1} = 2.0$ flows past a convex corner that turns the wall away from the flow by $θ = 1 5^{\circ}$ . Find $M_{2}$ .

Forecast: Will $M_{2}$ be a little above 2, or a lot? Guess before reading. (Hint: 15° is a big chunk of the ~26° score $M = 2$ carries.)

Step 1. Compute the incoming score $ν (M_{1}) = ν (2.0)$ . $ν (2) = 6 arctan \frac{0.4}{2.4} (3) - arctan 3 = 26.3 8^{\circ} .$ Why this step? We need to know how much turning the flow has already spent before this corner, because ν is a running total, not a per-corner quantity.

Step 2. Add the turn: $ν (M_{2}) = ν (M_{1}) + θ = 26.3 8^{\circ} + 1 5^{\circ} = 41.3 8^{\circ}$ . Why this step? Expansion turns the wall away, spreading Mach waves — the boxed rule $θ = ν (M_{2}) - ν (M_{1})$ (stated above) says the score goes up by exactly $θ$ .

Step 3. Invert $ν (M_{2}) = 41.3 8^{\circ} \Rightarrow M_{2} \approx 2.60$ . Why this step? ν is strictly increasing in $M$ , so a score points to exactly one Mach number. We solve the ν equation numerically for $M$ .

Verify: Recompute $ν (2.60)$ directly — it should return $41. 4^{\circ}$ , and $41.4 - 26.38 = 15. 0^{\circ}$ . ✓ Units: all angles in degrees throughout. Physically $M$ rose (expansion speeds flow up), matching Cell A.

Ex 2 — Pressure after the fan (Cell B)

Worked example Statement

Same corner as Ex 1 ( $M_{1} = 2 \to M_{2} = 2.6$ ). Find the static-pressure ratio $p_{2} / p_{1}$ .

Forecast: Expansion accelerates the flow. Does static pressure go up or down? Commit to an answer.

Step 1. Because a fan is isentropic, the stagnation pressure $p_{0}$ (defined above) is unchanged. So write $\frac{p _{2}}{p _{1}} = \frac{p _{2} / p _{0}}{p _{1} / p _{0}}$ . Why this step? We do not know $p_{0}$ , but it cancels. Constancy of $p_{0}$ is exactly what makes ν usable — this is why we could integrate smooth relations at all.

Step 2. Using the pressure converter, tool (2): $\frac{p _{1}}{p _{0}} = (1 + 0.2 \cdot 2^{2})^{- 3.5} = 0.1278, \frac{p _{2}}{p _{0}} = (1 + 0.2 \cdot 2. 6^{2})^{- 3.5} = 0.0501.$ Why this step? This single formula converts each Mach number into a pressure fraction; higher $M$ means larger bracket, and the negative exponent makes it a smaller fraction.

Step 3. $\frac{p _{2}}{p _{1}} = \frac{0.0501}{0.1278} \approx 0.392$ . Why this step? The ratio of fractions completes the chain ν → M → p.

Verify: $0.392 < 1$ , so pressure fell — correct for expansion (flow does work accelerating, cooling, rarefying). Cell B confirmed.

Ex 3 — Fan geometry: how wide does it open? (Cell C)

Worked example Statement

For Ex 1's fan ( $M_{1} = 2 \to M_{2} = 2.6$ , turn $1 5^{\circ}$ ), find the tilt of the first and last fan edges and the total angular width the fan sweeps in the lab frame.

Forecast: As the flow speeds up, does each successive Mach wave lean more toward the flow direction or away? Guess.

Step 1. First edge tilt from the incoming flow: $μ_{1} = arcsin (1/2) = 30.0 0^{\circ}$ . Why this step? The leading Mach wave sits at the Mach angle for the upstream flow — that is the definition of a Mach wave (Mach Waves and Mach Angle).

Step 2. Last edge tilt from the outgoing flow: $μ_{2} = arcsin (1/2.6) = 22.6 2^{\circ}$ . Why this step? By the last wave the flow already reached $M_{2}$ ; the wave leans at that faster flow's Mach angle. Faster flow ⇒ smaller Mach angle ⇒ wave lies flatter against the flow.

Step 3. Measure both edges from the original flow direction. The first edge is at $μ_{1} = 3 0^{\circ}$ above the incoming flow. The outgoing flow has itself rotated down by $θ = 1 5^{\circ}$ , and the last edge is $μ_{2} = 22.6 2^{\circ}$ above that rotated flow. So the last edge sits at $μ_{2} - θ = 22.62 - 15 = 7.6 2^{\circ}$ above the original flow line. Total fan sweep: $Δ = μ_{1} - (μ_{2} - θ) = 30 - 7.62 = 22.3 8^{\circ} .$ Why this step? We line up both edges against one common reference (the original flow) so their angular gap — the physical width of the fan wedge — is meaningful.

Verify: The fan spans $22.3 8^{\circ}$ , wider than the $1 5^{\circ}$ turn itself, which is right: the fan must contain both Mach cones plus the turning, not just the turning. Cell C ✓.

Ex 4 — Degenerate & limiting inputs (Cells D and E)

Worked example Statement

(a) A "corner" bends the wall by $θ = 0^{\circ}$ at $M_{1} = 2$ . What is $M_{2}$ ? (b) A flow is just barely supersonic, $M_{1} = 1.0$ . What is its score $ν (1)$ , and what is the tilt $μ_{1}$ of the first Mach wave?

Forecast: For (b), guess where a Mach wave points when the flow is exactly sonic — leaning forward, sideways, or straight across?

Step 1 (a). $ν (M_{2}) = ν (M_{1}) + 0 = ν (2)$ , so $M_{2} = M_{1} = 2$ . Why this step? Zero turn adds zero to the score; ν is one-to-one, so the Mach number cannot move. This is the sanity anchor: no wall bend, no expansion.

Step 2 (b). $ν (1) = 6 arctan (0) - arctan (0) = 0^{\circ}$ . Why this step? $M = 1$ is the reference state that defines score $= 0$ ; both arctan arguments contain $M^{2} - 1 = 0$ . This is the very edge of the domain ( $M \geq 1$ ).

Step 3 (b). $μ_{1} = arcsin (1/1) = arcsin (1) = 9 0^{\circ}$ . Why this step? At exactly $M = 1$ the Mach "wave" is a plane standing perpendicular to the flow — it cannot outrun the flow at all, so it lies flat across the stream. As $M$ grows the wave leans forward and $μ$ shrinks.

Verify: (a) $M_{2} = 2$ exactly — degenerate check passes. (b) $ν (1) = 0$ and $μ (1) = 9 0^{\circ}$ ; these are the two boundary values every table starts from. Cells D and E ✓.

Ex 5 — The sign / direction trap (Cell F)

Worked example Statement

A supersonic flow at $M_{1} = 2.6$ meets a wall that bends toward the flow (concave) by $1 0^{\circ}$ . A student writes $ν (M_{2}) = ν (2.6) - 1 0^{\circ}$ and reads off $M_{2}$ . When is this legal, and what does it give here — and where is the trap?

Forecast: Does bending into the flow speed it up or slow it down? Guess before computing.

Step 1. Compute $ν (2.6) = 41.3 8^{\circ}$ (from Ex 1). Subtracting: $ν (M_{2}) = 41.38 - 10 = 31.3 8^{\circ}$ . Why this step? We are testing the "compression = negative expansion" idea. Turning into the flow would lower the score, so subtraction is the arithmetic that idea predicts.

Step 2. Invert $ν = 31.3 8^{\circ} \Rightarrow M_{2} \approx 2.19$ . So the flow would slow down. Why this step? Lower score ⇒ lower Mach number; ν monotonic. Slowing on compression is directionally correct.

Step 3 — the trap. This is only valid for a smooth, gradual concave wall (an isentropic compression, the mirror of a fan). For a sharp concave corner the Mach waves pile up into an oblique shock — a discontinuity that raises entropy — and ν no longer applies. You must switch to Oblique Shock Waves relations. Why this step? ν was derived assuming isentropic flow. A sharp compression breaks that assumption; using $ν - θ$ there gives a wrong $M_{2}$ and, worse, a wrong (too-low) entropy.

Verify: For a smooth wall the number $M_{2} \approx 2.19$ is self-consistent: $ν (2.19) = 31. 4^{\circ}$ and $41.38 - 31.4 = 1 0^{\circ}$ . ✓ But flag: sharp corner ⇒ shock, not this. Cell F ✓.

Ex 6 — The vacuum ceiling (Cell G)

Worked example Statement

A flow enters at $M_{1} = 3.0$ . (a) What is the largest turn it can physically make by expansion? (b) What happens if a corner demands $9 5^{\circ}$ ?

Forecast: Is there a hard maximum to how much a flow can turn away, or can it turn forever?

Step 1. $ν (3) = 6 arctan \frac{0.4}{2.4} (8) - arctan 8 = 49.7 6^{\circ}$ . Why this step? We need the score already spent; the remaining budget is the ceiling minus this.

Step 2. The absolute ceiling is $ν_{m a x} = \frac{π}{2} (\frac{γ + 1}{γ - 1} - 1) = 130.4 5^{\circ}$ , reached only as $M \to \infty$ . Why this step? $arctan$ tops out at $9 0^{\circ}$ ; feeding $M \to \infty$ into the closed form gives a finite maximum score — the flow cannot spend more turning than this.

Step 3. Largest turn: $θ_{m a x} = ν_{m a x} - ν (3) = 130.45 - 49.76 = 80.6 9^{\circ}$ . Why this step? Any turn up to this drives $M$ toward infinity and $p \to 0$ .

Step 4 (b). A demanded $9 5^{\circ} > 80.6 9^{\circ}$ is impossible: the flow reaches $M = \infty$ , $p = 0$ before finishing the corner. A vacuum void opens between the flow and the wall. Why this step? Physics caps the score; the wall "outruns" the flow's ability to fill the gap.

Verify: $80.6 9^{\circ}$ is less than $ν_{m a x}$ and positive; $9 5^{\circ}$ exceeds it, so the impossibility flag fires. Cell G ✓.

Ex 7 — Real-world: jet exhaust plume at the nozzle lip (Cell H)

Worked example Statement

A supersonic nozzle exits gas at $M_{1} = 2.0$ with static pressure $p_{1}$ that is higher than the ambient (an under-expanded jet). At the sharp lip, the jet boundary must turn until its pressure equals ambient. Ambient pressure is $p_{a} = 0.5 p_{1}$ . By what angle does the plume turn at the lip, and to what Mach number?

Forecast: Under-expanded means the gas is "too fat" for the outside. Does it expand and bend outward, or compress and bend inward at the lip? Guess.

Step 1. Convert the pressure demand into a Mach number. Since $p_{0}$ is fixed and $p_{1} / p_{0} = 0.1278$ (Ex 2), the target is $p_{2} / p_{0} = (p_{a} / p_{1}) (p_{1} / p_{0}) = 0.5 \times 0.1278 = 0.0639$ . Why this step? The plume boundary is a free surface: it keeps turning until its static pressure matches ambient. Pressure is the boundary condition, so we translate it into an $M_{2}$ .

Step 2. Solve $(1 + 0.2 M_{2}^{2})^{- 3.5} = 0.0639 \Rightarrow M_{2} \approx 2.43$ . Why this step? Invert the isentropic pressure relation (tool 2) to find the Mach number that produces ambient pressure.

Step 3. Turning angle: $θ = ν (M_{2}) - ν (M_{1}) = ν (2.43) - ν (2.0)$ . Numerically $ν (2.43) = 37.5 5^{\circ}$ , so $θ = 37.55 - 26.38 = 11.1 7^{\circ}$ . Why this step? The plume expands (pressure drops toward ambient), so the boundary bends outward by exactly the ν gap (the boxed rule).

Verify: $p_{2} / p_{1} = 0.0639/0.1278 = 0.50 = p_{a} / p_{1}$ ✓ (matches ambient). Flow accelerated $2.0 \to 2.43$ and bent outward $11.1 7^{\circ}$ — a textbook under-expanded plume. Cell H ✓.

Ex 8 — Exam twist: run the chain backwards (Cell I)

Worked example Statement

A fan drops the static pressure to $p_{2} / p_{1} = 0.30$ starting from $M_{1} = 1.8$ . Find the turning angle $θ$ required.

Forecast: You are given the pressure result and asked for the cause (the turn). Which quantity do you find first — $M_{2}$ or $ν_{2}$ ?

Step 1. Get $M_{1}$ 's pressure fraction with tool (2): $p_{1} / p_{0} = (1 + 0.2 \cdot 1. 8^{2})^{- 3.5} = 0.1740$ . Why this step? We must anchor to $p_{0}$ (constant across the isentropic fan) before we can locate $M_{2}$ .

Step 2. Target fraction: $p_{2} / p_{0} = 0.30 \times 0.1740 = 0.05219$ . Invert to get $M_{2}$ : $(1 + 0.2 M_{2}^{2})^{- 3.5} = 0.05219 \Rightarrow M_{2} \approx 2.575$ . Why this step? The pressure ratio, times $p_{1} / p_{0}$ , pins $p_{2} / p_{0}$ , and the isentropic relation converts that to $M_{2}$ . This is the "backwards" leg the exam is testing.

Step 3. Now use the boxed rule: $θ = ν (M_{2}) - ν (M_{1}) = ν (2.575) - ν (1.8)$ . Numerically $ν (1.8) = 20.7 3^{\circ}$ and $ν (2.575) = 40.9 5^{\circ}$ , so $θ = 40.9 5^{\circ} - 20.7 3^{\circ} = 20.2 2^{\circ} .$ Why this step? With both Mach numbers known, the turn is just the score difference — the boxed rule, used last rather than first.

Step 4. State the answer: the fan needed a wall turn of $θ \approx 20. 2^{\circ}$ to produce that pressure drop. Why this step? This is the quantity the exam asked for; everything before it was scaffolding to get from a pressure ratio to an angle.

Verify: Forward-check: from $M_{1} = 1.8$ turning $20.2 2^{\circ}$ gives $ν_{2} = 40.9 5^{\circ} \Rightarrow M_{2} = 2.575 \Rightarrow p_{2} / p_{0} = 0.0522$ , and $0.0522/0.1740 = 0.30$ ✓. The whole chain closes on itself. Cell I ✓.

Recall

Recall Did every cell get covered?

Forecast test: name which example handled the vacuum ceiling. Which example? ::: Ex 6 (Cell G) — $θ_{m a x} = ν_{m a x} - ν (M_{1})$ . Which cell frozen-Mach lives in ::: Cell D, Ex 4a — $θ = 0 \Rightarrow M_{2} = M_{1}$ . Backwards problem (pressure → turn) is which example ::: Ex 8 (Cell I). When is $ν_{2} = ν_{1} - θ$ legal ::: only a smooth isentropic concave wall; a sharp corner makes a shock (Ex 5, Cell F). Domain of ν(M) ::: only $M \geq 1$ (supersonic); below sonic it is undefined.

Connections

Prandtl-Meyer function ν(M) — the parent this page drills.
Isentropic Flow Relations — the $p / p_{0}$ conversions in Ex 2, 7, 8.
Mach Waves and Mach Angle — the edge tilts $μ$ in Ex 3, 4.
Oblique Shock Waves — the correct tool when Ex 5's compression is sharp.
Expansion Fan / Centered Rarefaction — the physical fan Ex 3 draws.
Nozzle Design (Supersonic) — the plume of Ex 7.
Method of Characteristics — uses $ν \pm θ$ as invariants downstream of these fans.