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

The mnemonic from the parent note is "Expand = Away = Add."

• $M$ increases — the flow accelerates (that is the whole point of an expansion).
• $p$ decreases — static pressure drops as internal energy becomes kinetic energy.
• $T$ decreases — the gas cools for the same reason.
• $\mu =\arcsin (1/M)$: as $M$ rises, $1/M$ falls, so $\mu$ decreases — the waves get shallower downstream.

What it looks like: in figure s01 the bold orange first wave is steep and the bold orange last wave is shallow — that visible shrinking of the wave angle is $\mu$ decreasing, and the teal exit arrow being longer than the entry arrow is $M$ increasing.

Use tool 1: $\mu =\arcsin (1/2.0)=\arcsin (0.5)=30^{\circ }$. Why $\arcsin$? We know the ratio $\sin \mu =1/M$ and ask "which angle has this sine?" — that is exactly what $\arcsin$ (inverse sine) answers.

False. A fan is isentropic — a continuum of infinitesimally weak Mach waves, each contributing entropy $\propto (\text{strength}{)}^3\to 0$. Because entropy is constant, $p_0$ is constant. (Contrast Oblique shock waves, where $p_0$ does fall — see Entropy and the second law.)

With $\gamma =1.4$: $\frac{\gamma +1}{\gamma -1}=\frac{2.4}{0.4}=6$, so $\sqrt{6}\approx 2.4495$, and its reciprocal $\sqrt{1/6}\approx 0.4082$. Also $M^2-1=3$, so $\sqrt{M^2-1}=\sqrt{3}\approx 1.7321$. $\nu =\underset{\sqrt{6}}{\underbrace{2.4495}}\arctan (0.4082\times 1.7321)-\arctan (1.7321)$ $=2.4495\arctan (0.7071)-\arctan (1.7321)=2.4495(0.6155)-1.0472=0.4604\text{rad}.$ Converting: $0.4604\times \frac{180}{\pi }\approx \boxed{26.38^{\circ }}$. Why radians first? Both $\arctan$ terms return radians; I convert only at the end so the two pieces subtract on the same footing.

Tool 3: $\nu (M_2)=\nu (M_1)+\theta =26.38^{\circ }+10^{\circ }=36.38^{\circ }$. Invert (given): $M_2\approx \boxed{2.385}$. Why add? Expansion turns the flow away, so we add $\theta$ — the corner spends $10^{\circ }$ more of "turning budget," pushing the flow to a higher $M$.

Use $1+\frac{\gamma -1}{2}{M}^2=1+0.2M^2$: $1+0.2(2.0{)}^2=1.800,1+0.2(2.385{)}^2=1+0.2(5.688)=\mathrm{2.138.}$ $\frac{p_2}{p_1}={\left(\frac{1.800}{2.138}\right)}^{3.5}={\left(0.8419\right)}^{3.5}\approx \boxed{0.548}.$ Why exponent $3.5$? $\frac{\gamma }{\gamma -1}=\frac{1.4}{0.4}=3.5$. The pressure has dropped to about $55\%$ — expansion lowers static pressure, exactly as predicted.

Because $\nu (1)=0$ by definition (sonic is the zero of the turning score), the turn rule collapses to $\nu (M_2)=0+20^{\circ }=20^{\circ }$, which we read directly. Inverting: $M_2\approx \boxed{1.775}$. What it looks like: the very first Mach wave in the fan stands at $\mu =\arcsin (1/1)=90^{\circ }$ — perpendicular to the flow — the steepest wave a fan can start with.

First wave: ${\mu }_1=\arcsin (1/1.0)=90^{\circ }$. Last wave: ${\mu }_2=\arcsin (1/1.775)=\arcsin (0.5634)=34.28^{\circ }$. But the flow itself has also swung by $\theta =20^{\circ }$. The Mach lines are measured relative to the local flow direction, so the lab-frame angular spread of the fan edges is $\Delta =({\mu }_1)-({\mu }_2-\theta )=90^{\circ }-(34.28^{\circ }-20^{\circ })=90^{\circ }-14.28^{\circ }=\boxed{75.72^{\circ }}.$ Why subtract $\theta$? The last wave sits at ${\mu }_2$ from the already-turned flow, and that flow has rotated $\theta$ toward the wall. Measured from the original flow direction the last edge is at ${\mu }_2-\theta$. Figure s02 draws all three angles explicitly: the teal arrow is the original flow, the plum arrow the turned flow (rotated by $\theta$), the solid orange line the first wave at $90^{\circ }$, and the dashed orange line the last wave — the gap between the orange lines is the boxed $75.72^{\circ }$.

Subtraction is the compression direction: it lowers $\nu$, hence lowers $M$ — a decelerating turn into the flow (a shock family), not an expansion. For a convex corner you must add: ${\nu }_2=26.38^{\circ }+10^{\circ }=36.38^{\circ }\Rightarrow M_2\approx \boxed{2.385}$ (same as L2.2). Rule of thumb: if your "expansion" produced $M_2<M_1$ or $p_2>p_1$, your sign is flipped.

Step 1 — find $M_2$ from the pressure match. Since $p_0$ is fixed (isentropic), use tool 4 with $p_2=p_a=25$: $\frac{p_2}{p_1}=\frac{25}{40}=0.625={\left(\frac{1+0.2M_1^2}{1+0.2M_2^2}\right)}^{3.5}.$ With $1+0.2(2.4{)}^2=2.152$: ${\left(\frac{2.152}{1+0.2M_2^2}\right)}^{3.5}=0.625\Rightarrow \frac{2.152}{1+0.2M_2^2}=0.625^{1/3.5}=\mathrm{0.8783.}$ $1+0.2M_2^2=\frac{2.152}{0.8783}=2.450\Rightarrow M_2^2=\frac{1.450}{0.2}=7.252\Rightarrow M_2\approx \boxed{2.693}.$ Step 2 — find $\theta$ from the turning scores. $\nu (2.693)$: with $M^2-1=6.252$, $\sqrt{M^2-1}=2.500$, $\nu =2.4495\arctan (0.4082\times 2.500)-\arctan (2.500)=2.4495\arctan (1.0206)-\mathrm{1.1903.}$ Now $\arctan (1.0206)=0.7955\text{rad}$, so the product is $2.4495\times 0.7955=1.9486$ (not $2.95$ — watch that multiplication): $\nu =1.9486-1.1903=0.7583\text{rad}=43.45^{\circ }.$ Then $\theta =\nu (M_2)-\nu (M_1)=43.45^{\circ }-36.75^{\circ }=\boxed{6.70^{\circ }}$. Why this order? The pressure condition fixes the downstream state $M_2$; the turning score difference then tells us how far the jet boundary swings. See Nozzle design and overexpansion.

Turning scores simply accumulate because $\nu$ is defined from a common origin (sonic): $\nu (M_3)=\nu (M_1)+{\theta }_a+{\theta }_b=14.86^{\circ }+8^{\circ }+12^{\circ }=34.86^{\circ }.$ Invert: $M_3\approx \boxed{2.334}$. Why can we just add both angles? Each fan is isentropic and shares the same $p_0,T_0$, so the intermediate state never needs to be computed separately — only the total turn matters. This is the discrete cousin of Method of characteristics.

$\frac{T_2}{T_1}=\frac{1+0.2(2.4{)}^2}{1+0.2(2.693{)}^2}=\frac{2.152}{2.450}=\boxed{0.878}.$ The gas cools to about $88\%$ of its temperature (in kelvin) — modest, because the turn is only $\sim 7^{\circ }$. Since the fan is isentropic, this drop is fully recoverable: bring the flow back to rest and $T_0$ returns unchanged.

As $M\to \infty$: $\sqrt{M^2-1}\to \infty$, so both $\arctan$ terms approach $\pi /2$:

Write $\nu =k\arctan (u)-\arctan (w)$ with $k=\sqrt{\frac{\gamma +1}{\gamma -1}}$, $u=\sqrt{\frac{\gamma -1}{\gamma +1}(M^2-1)}$, $w=\sqrt{M^2-1}$. WHY differentiate at all? $\nu$ was built by integrating $d\nu /dM$; the only honest check that the closed form is correct is to run the derivative backwards and land on the same integrand.

Using $\frac{d}{dM}\arctan (f)=\frac{f^{\prime }}{1+f^2}$:

Second term. $w=\sqrt{M^2-1}\Rightarrow w^{\prime }=\frac{M}{\sqrt{M^2-1}}$, and $1+w^2=M^2$, so $\frac{d}{dM}\arctan (w)=\frac{M/\sqrt{M^2-1}}{M^2}=\frac{1}{M\sqrt{M^2-1}}.$

First term. Let $c=\frac{\gamma -1}{\gamma +1}$ so $u=\sqrt{c(M^2-1)}$, $u^{\prime }=\frac{cM}{\sqrt{c(M^2-1)}}$, and $1+u^2=1+c(M^2-1)$. Then

$\mu =\arcsin (1/1)=\arcsin (1)=\boxed{90^{\circ }}$ — the first Mach wave is normal to the flow. This is the geometric edge case, not a failure: $\nu (1)=0$ is finite and well-defined, and $d\nu /dM=\frac{\sqrt{M^2-1}}{\cdots }\to 0$ as $M\to 1^{+}$, so $\nu$ leaves $1$ smoothly (with zero slope). The flow simply spends its first bit of turning building up from a right-angle wave. This is the leftmost, vertical solid-orange wave drawn in figure s02.

$\frac{\gamma +1}{\gamma -1}=\frac{8/3}{2/3}=4$, so $\sqrt{4}=2$. ${\nu }_{\max }=\frac{\pi }{2}(2-1)=\frac{\pi }{2}=\boxed{90.00^{\circ }}.$ Helium can be turned at most $90^{\circ }$ from sonic — far less than air's $130.5^{\circ }$. Why? A larger $\gamma$ means a stiffer gas (less internal energy per unit temperature), so less speed is available for a given expansion, limiting the total turn.

Expansion ⇒ add $\theta$, $M\uparrow$, $p\downarrow$, $T\downarrow$, $\mu \downarrow$, $p_0$ fixed. If any of those came out the other way, re-read your sign.