Exercises — Under-expanded nozzle — Prandtl-Meyer expansion, efficiency loss

What we do: apply test (A). Why: the sign of $p_e-p_{\infty }$ alone fixes the regime. $\frac{p_e}{p_{\infty }}=\frac{1.8}{1.0}=1.8>1\Rightarrow p_e>p_{\infty }$ So the nozzle is under-expanded. The gas is still over-pressured when it reaches the lip, so it keeps expanding outside the nozzle through a Prandtl-Meyer fan, turning outward like a flower opening.

• (a) $0.6<1$: over-expanded — shocks compress and detach the flow (3.3.14-Over-expanded-nozzle-shock-diamonds).
• (b) $1.0$: perfectly expanded — no waves, all momentum axial, maximum thrust.
• (c) $1.4>1$: under-expanded — mild external expansion, small loss.

The perfect case (b) is the peak. Both mismatches lose thrust, but which loses more depends on numbers — here (a) is far from 1 by $0.4$ and involves shocks, which are typically costlier than a mild fan. Ranking by efficiency, worst→best: (a) < (c) < (b).

What/why: plug into formula (C); this is the raw skill of evaluating the Prandtl-Meyer function. First $k=\sqrt{\frac{2.4}{0.4}}=\sqrt{6}=2.4495$, and $\sqrt{M^2-1}=\sqrt{3}=1.7321$. $\nu =2.4495\arctan \left(\frac{1.7321}{2.4495}\right)-\arctan (1.7321)$ $=2.4495\arctan (0.7071)-60^{\circ }=2.4495(35.264^{\circ })-60^{\circ }=86.39^{\circ }-60^{\circ }=26.38^{\circ }$ So $\nu (2.0)\approx 26.4^{\circ }$.

What/why: turning adds to $\nu$ (equation C). We march up the ladder of Mach numbers. $\nu (M_{\text{plume}})=\nu (M_e)+{\theta }_{\text{turn}}=26.38^{\circ }+6^{\circ }=32.38^{\circ }$ Now invert: find the $M$ whose $\nu$ equals $32.38^{\circ }$. Trying $M=2.2$: $\sqrt{M^2-1}=1.9596$, $\nu =2.4495\arctan (1.9596/2.4495)-\arctan (1.9596)=2.4495(38.66^{\circ })-62.96^{\circ }=94.71^{\circ }-62.96^{\circ }=31.75^{\circ }.$ Slightly low; try $M=2.23$: $\sqrt{M^2-1}=1.9917$, $\nu =2.4495(39.11^{\circ })-63.34^{\circ }=95.80^{\circ }-63.34^{\circ }=32.46^{\circ }$. Very close, so $M_{\text{plume}}\approx 2.23$.

What/why: the plume expands until its static pressure equals $p_{\infty }$; equation (B) links that pressure ratio to $M_{\text{plume}}$. With $\gamma =1.3$: exponent $\frac{\gamma }{\gamma -1}=\frac{1.3}{0.3}=4.3333$, and $\frac{\gamma -1}{2}=0.15$. Denominator: $1+0.15(3.5{)}^2=1+0.15(12.25)=2.8375$. $1.5={\left(\frac{1+0.15M_{\text{plume}}^2}{2.8375}\right)}^{4.3333}$ Take the $4.3333$-th root: $1.5^{1/4.3333}=1.0964$, so $1+0.15M_{\text{plume}}^2=1.0964\times 2.8375=3.1110\Rightarrow M_{\text{plume}}^2=\frac{2.1110}{0.15}=14.07\Rightarrow M_{\text{plume}}\approx 3.75.$

At $M=3.5$: $\sqrt{M^2-1}=\sqrt{11.25}=3.3541$. $\nu (3.5)=2.7689\arctan (3.3541/2.7689)-\arctan (3.3541)=2.7689(50.46^{\circ })-73.42^{\circ }=139.72^{\circ }-73.42^{\circ }=66.30^{\circ }.$ At $M=3.75$: $\sqrt{M^2-1}=\sqrt{13.0625}=3.6142$. $\nu (3.75)=2.7689\arctan (3.6142/2.7689)-\arctan (3.6142)=2.7689(52.55^{\circ })-74.53^{\circ }=145.51^{\circ }-74.53^{\circ }=70.98^{\circ }.$ So ${\theta }_{\text{turn}}=70.98^{\circ }-66.30^{\circ }\approx 4.7^{\circ }$.

What/why: only the axial component $v\cos {\theta }_{\text{turn}}$ pushes the rocket; the sideways part is wasted. $1-\cos (4.7^{\circ })=1-0.99664=0.00336\approx 0.34\%.$ The extra momentum gained by expanding outside the nozzle loses only about a third of one percent to sideways motion — a small penalty for a mild mismatch. Connect to the thrust coefficient via $C_F$.

Point A: $p_e/p_{\infty }=1.0$ → perfectly expanded, no waves, no loss. Point B: $p_e/p_{\infty }=2.0>1$ → under-expanded.

Plume Mach (equation B, $\gamma =1.4$): exponent $=1.4/0.4=3.5$, $\frac{\gamma -1}{2}=0.2$. Denominator $1+0.2(9)=2.8$. $2.0={\left(\frac{1+0.2M_{\text{plume}}^2}{2.8}\right)}^{3.5}\Rightarrow 2.0^{1/3.5}=1.2190,$ $1+0.2M_{\text{plume}}^2=1.2190\times 2.8=3.4132\Rightarrow M_{\text{plume}}^2=\frac{2.4132}{0.2}=12.07\Rightarrow M_{\text{plume}}\approx 3.47.$

Turning angle ($k=\sqrt{6}=2.4495$): $\nu (3.0)$: $\sqrt{8}=2.8284$, $\nu =2.4495\arctan (2.8284/2.4495)-\arctan (2.8284)=2.4495(49.11^{\circ })-70.53^{\circ }=120.29^{\circ }-70.53^{\circ }=49.76^{\circ }.$ $\nu (3.47)$: $\sqrt{3.47^2-1}=\sqrt{11.04}=3.3227$, $\nu =2.4495\arctan (3.3227/2.4495)-\arctan (3.3227)=2.4495(53.60^{\circ })-73.24^{\circ }=131.30^{\circ }-73.24^{\circ }=58.06^{\circ }.$ ${\theta }_{\text{turn}}=58.06^{\circ }-49.76^{\circ }\approx 8.3^{\circ }.$ So climbing from design altitude to twice-the-ratio point B doubles the pressure mismatch and produces an ~$8.3^{\circ }$ outward turn.

Constants: $\gamma =1.2$, so $\frac{\gamma -1}{2}=0.1$, exponent $=1.2/0.2=6.0$, $k=\sqrt{2.2/0.2}=\sqrt{11}=3.3166$.

(a) Sea level, $p_e/p_{\infty }=1.6$. Denominator $1+0.1(16)=2.6$. $1.6={\left(\frac{1+0.1M_{\text{plume}}^2}{2.6}\right)}^6\Rightarrow 1.6^{1/6}=1.0812,$ $1+0.1M_{\text{plume}}^2=1.0812\times 2.6=2.8111\Rightarrow M_{\text{plume}}^2=18.11\Rightarrow M_{\text{plume}}=\mathrm{4.256.}$ Now the angles: $\nu (4.0)$: $\sqrt{15}=3.8730$, $\nu =3.3166\arctan (3.8730/3.3166)-\arctan (3.8730)=3.3166(49.42^{\circ })-75.50^{\circ }=163.90^{\circ }-75.50^{\circ }=88.40^{\circ }.$ $\nu (4.256)$: $\sqrt{4.256^2-1}=\sqrt{17.11}=4.1364$, $\nu =3.3166\arctan (4.1364/3.3166)-\arctan (4.1364)=3.3166(51.28^{\circ })-76.41^{\circ }=170.07^{\circ }-76.41^{\circ }=93.66^{\circ }.$ ${\theta }_{\text{turn}}=93.66^{\circ }-88.40^{\circ }\approx 5.3^{\circ }.$

(b) Vacuum limit. As $p_{\infty }\to 0$, the ratio $p_e/p_{\infty }\to \infty$, so the plume must expand to infinite Mach. But $\nu (M)$ does not grow without bound — it approaches a ceiling ${\nu }_{\text{max}}$. For $\gamma =1.2$: ${\nu }_{\text{max}}=\frac{\pi }{2}\left(k-1\right)\cdot \frac{180}{\pi }=90^{\circ }(k-1)=90^{\circ }(3.3166-1)=208.5^{\circ }.$ So the largest possible turn from $M_e=4.0$ is ${\nu }_{\text{max}}-\nu (4.0)=208.5^{\circ }-88.40^{\circ }\approx 120.1^{\circ }$. The plume cannot turn more than this — physically it means the gas eventually goes as fast as it possibly can and the fan "runs out" of expansion. The plume boundary becomes a hard maximum-turn cone.