Exercises — Prandtl-Meyer function ν(M)

(a) $\nu =0$ at $M=1$ (sonic). The function is defined to start counting the "turning budget" from the sonic state, so a flow that is exactly sonic has spent zero turning. (b) $\nu$ increases with $M$. Look at the differential relation from the derivation, $d\theta =\frac{\sqrt{M^2-1}}{1+\frac{\gamma -1}{2}{M}^2}\frac{dM}{M}$. For $M>1$ the numerator $\sqrt{M^2-1}$ is real and positive and everything else is positive, so $d\nu >0$ whenever $dM>0$. Accelerate ⇒ $\nu$ grows.

Expansion = flow accelerates = $\nu$ up, static pressure down. Remember the mnemonic E.A.V.: Expansion → $\nu$ Adds → Velocity (Mach) up. Pressure always falls in an expansion because the gas rarefies as it does work speeding up.

Step 1 — get the starting budget. $\nu (M_1)=\nu (2.0)=26.38^{\circ }$. Why: this is the angle already "spent" reaching $M=2$. Step 2 — add the turn. Expansion adds ($\theta >0$): $\nu (M_2)=26.38^{\circ }+10^{\circ }=36.38^{\circ }$. Why: the boxed rule $\theta =\nu (M_2)-\nu (M_1)$. Step 3 — invert. We need the $M$ whose $\nu =36.38^{\circ }$. Why an inversion is needed: $\nu (M)$ has no elementary algebraic inverse, but it is monotonic (Ex. 1.1b), so exactly one $M$ works. Bracket with the table: $\nu (2.0)=26.38$, $\nu (2.5)=39.12$, so $M$ lies in $(2.0,2.5)$. Solving the closed form numerically (Newton from the interpolated guess $2.39$) gives $M_2\approx 2.385$. Answer: $M_2\approx 2.39$.

This is the forward direction — no inversion needed, just evaluate both budgets. $\theta =\nu (3.0)-\nu (1.5)=49.76^{\circ }-11.91^{\circ }=37.85^{\circ }.$ Answer: $\theta \approx 37.85^{\circ }$. Why: the turning angle is literally the difference of the two budgets.

Step 1. The fan is isentropic ⇒ $p_0$ constant ⇒ $\frac{p_2}{p_1}=\frac{p_2/p_0}{p_1/p_0}$. Step 2. $\frac{p_1}{p_0}=(1+0.2\cdot 2^2{)}^{-3.5}=(1.8{)}^{-3.5}=0.1278$. Step 3. $\frac{p_2}{p_0}=(1+0.2\cdot 2.385^2{)}^{-3.5}=(2.1378{)}^{-3.5}=0.07031$. Step 4. $\frac{p_2}{p_1}=0.07031/0.1278\approx \mathrm{0.550.}$ Answer: $p_2/p_1\approx 0.55$ — pressure dropped, as expansion demands.

The fan has a leading Mach wave (set by $M_1$) and a trailing Mach wave (set by $M_2$), both measured from their own local flow direction using $\mu =\arcsin (1/M)$ (valid since both $M\ge 1$). Leading edge: ${\mu }_1=\arcsin (1/2.0)=30.0^{\circ }$ — measured from the incoming flow. Trailing edge: ${\mu }_2=\arcsin (1/2.6)=22.62^{\circ }$ — but this is measured from the outgoing flow, which has itself turned by $\theta =15^{\circ }$ away from the incoming direction. Let ${\varphi }_2$ denote the angle of the trailing wave measured from the incoming flow direction (so both fan edges share a common reference). Since the outgoing flow is tilted by $\theta$ below the incoming flow, we subtract that tilt: ${\varphi }_2={\mu }_2-\theta =22.62^{\circ }-15^{\circ }=7.62^{\circ }.$ Total angular width swept by the fan (leading edge minus trailing edge, both now referenced to the incoming flow): $\Delta ={\mu }_1-{\varphi }_2=30.0^{\circ }-7.62^{\circ }=22.38^{\circ }.$ Answers: leading wave at $30.0^{\circ }$, trailing wave at ${\varphi }_2=7.62^{\circ }$ above the incoming flow line, fan spans $\approx 22.38^{\circ }$.

Rearranging $d\theta =\frac{\sqrt{M^2-1}}{1+\frac{\gamma -1}{2}{M}^2}\frac{dM}{M}$: $\frac{dM}{d\theta }=\frac{M\left(1+\frac{\gamma -1}{2}{M}^2\right)}{\sqrt{M^2-1}}.$ As $M\to 1^{+}$, the numerator $\to 1\cdot (1+\frac{\gamma -1}{2})$ = finite $=1.2$ (for $\gamma =1.4$), while the denominator $\sqrt{M^2-1}\to 0^{+}$. So $dM/d\theta \to +\infty$. Meaning: right at sonic, even the tiniest corner unleashes a burst of acceleration — the fan opens up steeply. That is exactly why $\nu (M)$ has a vertical-tangent (square-root) shape near $M=1$.

Step 1 — pin $p_2/p_0$. Isentropic, so $p_0$ constant. First $p_1/p_0=(1.8{)}^{-3.5}=0.1278$. Target static pressure $p_2=0.30p_1$, so $p_2/p_0=0.30\times 0.1278=0.03834$. Step 2 — invert the pressure relation for $M_2$. Why this inversion is algebraic: raise both sides to $-\frac{\gamma -1}{\gamma }=-\frac{1}{3.5}$ to strip the exponent off the bracket: $1+0.2M_2^2={\left(0.03834\right)}^{-1/3.5}=\mathrm{2.5399.}$ $M_2^2=(2.5399-1)/0.2=7.6995\Rightarrow M_2=\mathrm{2.775.}$ Step 3 — turn angle from $\nu$. $\nu (2.0)=26.38^{\circ }$, $\nu (2.775)=44.87^{\circ }$. $\theta =\nu (M_2)-\nu (M_1)=44.87^{\circ }-26.38^{\circ }=18.49^{\circ }.$ Answers: $M_2\approx 2.78$, jet boundary turns $\theta \approx 18.5^{\circ }$. Why this works: the pressure-match condition fixes $M_2$ through isentropic relations; $\nu$ then reads off how far the boundary had to bend to get there.

$\nu$ is additive, so two turns simply keep adding: $\nu (M_1)=\nu (2.5)=39.12^{\circ }.$ After both corners: $\nu (M_3)=39.12^{\circ }+12^{\circ }+12^{\circ }=63.12^{\circ }.$ Invert (monotonic $\nu$, so unique root; bracket with $\nu (3.5)=58.53$, $\nu (4.0)=65.78$): $M_3\approx \mathrm{3.757.}$ Single-turn check: $\nu (2.5)+24^{\circ }=39.12^{\circ }+24^{\circ }=63.12^{\circ }$ — identical, so $M_3\approx 3.76$ either way. Lesson: because turning angle maps to $\nu$-differences, sequential expansions add exactly like one big expansion. The path doesn't matter, only the total angle.

Step 1 — the ceiling. The maximum $\nu$ (as $M\to \infty$) is ${\nu }_{\max }=130.45^{\circ }$ for $\gamma =1.4$. Step 2 — remaining budget. Starting at $M_1=3.0$, $\nu (3.0)=49.76^{\circ }$, so the flow can still turn at most ${\theta }_{\max }={\nu }_{\max }-\nu (M_1)=130.45^{\circ }-49.76^{\circ }=80.69^{\circ }.$ Step 3 — verdict. The wall asks for $95^{\circ }>80.69^{\circ }$, so the demanded turn exceeds the budget — it is physically impossible for the flow to hug the wall all the way. The gas expands as far as it can, turning the full $80.69^{\circ }$, at which point $M\to \infty$ and the static pressure $p\to 0$. It cannot turn any further because it has run out of pressure to convert into speed. Consequently the streamline separates from the wall: a vacuum (void) opens up in the wedge between the last expanded streamline and the remaining wall surface, and the flow simply coasts straight past that void. Answer: not possible; the maximum turn is ${\theta }_{\max }\approx 80.69^{\circ }$, beyond which the streamline separates and a vacuum forms against the wall.

As $M\to \infty$, both square-root arguments blow up: $\sqrt{\frac{\gamma -1}{\gamma +1}(M^2-1)}\to \infty$ and $\sqrt{M^2-1}\to \infty$. Since $\arctan (x)\to \frac{\pi }{2}$ as $x\to \infty$, each arctan hits $\frac{\pi }{2}$: ${\nu }_{\max }=\sqrt{\frac{\gamma +1}{\gamma -1}}\cdot \frac{\pi }{2}-\frac{\pi }{2}=\frac{\pi }{2}\left(\sqrt{\frac{\gamma +1}{\gamma -1}}-1\right).$ For $\gamma =1.4$: $\sqrt{2.4/0.4}=\sqrt{6}=2.4495$, so ${\nu }_{\max }=\frac{\pi }{2}(1.4495)=2.2770$ rad $=130.45^{\circ }$. ✓

(a) Smooth isentropic compression. Compression subtracts from the budget, so with $\theta =-10^{\circ }$ in the boxed rule $\theta =\nu (M_2)-\nu (M_1)$: $\nu (M_2)=\nu (M_1)+\theta =\nu (2.5)-10^{\circ }=39.12^{\circ }-10^{\circ }=29.12^{\circ }.$ Invert (monotonic $\nu$, unique root; bracket with $\nu (2.0)=26.38$, $\nu (2.5)=39.12$): $M_2\approx 2.125$. The flow slows and its pressure rises, exactly the mirror image of an expansion. This is legitimate because a gradual concave wall keeps the compression Mach waves spread out — they never coalesce, so the flow stays isentropic and $\nu$ still applies. (b) Sharp corner. At a single sharp concave corner the compression Mach waves coalesce into an oblique shock (parent note, "why a fan not a shock"). Across a shock, entropy rises, so the reservoir pressure $p_0$ drops — the flow is no longer isentropic. Since $\nu (M)$ was derived under the isentropic assumption, it silently ignores that entropy jump and would give the wrong downstream state. For a sharp compression you must use the oblique-shock relations, not $\nu$. Answers: (a) $M_2\approx 2.13$ (isentropic, valid); (b) a shock forms, entropy rises, $\nu$ is invalid — use oblique-shock relations.