Exercises — De Laval nozzle geometry — conical, bell (Rao contour), 80% bell

What we do: Expansion ratio is exit area over throat area. Area of a circle $=\pi R^2$, and the $\pi$ cancels: $ϵ=\frac{A_e}{A_t}=\frac{\pi R_e^2}{\pi R_t^2}=\frac{R_e^2}{R_t^2}=\frac{20^2}{5^2}=\frac{400}{25}=16.$ Why: We never need $\pi$ because it divides out — ratios of areas are ratios of radii squared. Answer: $ϵ=16$.

$\lambda =\frac{1+\cos 15^{\circ }}{2}=\frac{1+0.9659}{2}=\mathrm{0.9830.}$ What the number means: $\lambda =0.983$ says 98.3% of the ideal momentum points forward, so the loss is $1-0.983=0.017=1.7\%$. Answer: $\lambda \approx 0.983$, loss $\approx 1.7\%$.

Step 1 — exit radius: $R_e=R_t\sqrt{ϵ}=10\sqrt{16}=10\times 4=40$ cm. Step 2 — length: the wall rises $R_e-R_t=40-10=30$ cm while going outward at slope $\tan 15^{\circ }=0.2679$. Length = rise / slope: $L_{\text{cone}}=\frac{R_t(\sqrt{ϵ}-1)}{\tan \alpha }=\frac{10(4-1)}{0.2679}=\frac{30}{0.2679}\approx 112\text{ cm}.$ Answer: $R_e=40$ cm, $L_{\text{cone}}\approx 112$ cm.

Definition used: an "80% bell" is one whose axial length is $0.8$ times the equivalent conical length at $\alpha =15^{\circ }$. $L_{\text{bell}}=0.8L_{\text{cone}}=0.8\times 112=89.6\text{ cm}.$ Answer: $L_{\text{bell}}\approx 89.6$ cm — 20% shorter than the cone.

Actual thrust $=\lambda \times (\dot{m}{v}_e)$, because $\lambda$ is exactly the fraction of momentum pointing forward.

• Cone: $F=0.983\times 100=98.3$ kN.
• Bell: $F=0.985\times 100=98.5$ kN. Answer: Cone 98.3 kN, bell 98.5 kN — the bell gains 0.2 kN and is 20% shorter.

What we average: each ring of exhaust leaving at angle $\theta$ contributes forward momentum $\propto \cos \theta$, weighted by the ring's area $\propto \sin \theta d\theta$ (a cone's surface ring at angle $\theta$ has circumference $\propto \sin \theta$). So $\lambda =\frac{{\int }_0^{\alpha }\cos \theta \sin \theta d\theta }{{\int }_0^{\alpha }\sin \theta d\theta }.$ Top integral: ${\int }_0^{\alpha }\cos \theta \sin \theta d\theta =\frac{1}{2}{\sin }^2\alpha$. Bottom integral: ${\int }_0^{\alpha }\sin \theta d\theta =1-\cos \alpha$. Combine, then use ${\sin }^2\alpha =(1-\cos \alpha )(1+\cos \alpha )$: $\lambda =\frac{\frac{1}{2}(1-\cos \alpha )(1+\cos \alpha )}{1-\cos \alpha }=\frac{1+\cos \alpha }{2}.$ Limit $\alpha \to 0$: $\cos 0=1$, so $\lambda \to \frac{1+1}{2}=1$. Perfectly axial flow loses nothing — exactly what we expect. Answer: derivation confirmed; $\lambda (0^{\circ })=1$.

$\lambda (15^{\circ })=\frac{1+\cos 15^{\circ }}{2}=0.9830\Rightarrow \text{loss }1.70\%.$ $\lambda (30^{\circ })=\frac{1+\cos 30^{\circ }}{2}=\frac{1+0.8660}{2}=0.9330\Rightarrow \text{loss }6.70\%.$ Is it linear? Doubling $\alpha$ (15→30) multiplied the loss by $6.70/1.70\approx 3.9$, not by 2. Not linear — loss grows roughly like ${\alpha }^2$ for small $\alpha$ (because $1-\cos \alpha \approx {\alpha }^2/2$). This is why designers keep cone angles small and why bells (which turn flow axial) win. Answer: losses 1.70% vs 6.70%; strongly non-linear (near-quadratic).

(a) $L_{\text{cone}}=\frac{8(\sqrt{25}-1)}{\tan 15^{\circ }}=\frac{8(5-1)}{0.2679}=\frac{32}{0.2679}\approx 119.5$ cm. (b) $L_{\text{bell}}=0.8\times 119.5\approx 95.6$ cm. (c) $R_e=8\sqrt{25}=8\times 5=40$ cm. Answer: $L_{\text{cone}}\approx 119.5$ cm, $L_{\text{bell}}\approx 95.6$ cm, $R_e=40$ cm.

Match the exit radius: $R(L)=R_t+aL+b{L}^2=40$. $8+a(95.6)+0.0015(95.6{)}^2=40.$ Compute $0.0015\times 95.6^2=0.0015\times 9139.4=13.71$. So $8+95.6a+13.71=40\Rightarrow 95.6a=18.29\Rightarrow a=0.1913$. Exit wall angle: the wall slope is $R^{\prime }(x)=a+2bx$; at the exit $R^{\prime }(L)=0.1913+2(0.0015)(95.6)=0.1913+0.2868=0.4781$. The angle is ${\theta }_e=\arctan (R^{\prime }(L))$ — arctan answers "which angle has this slope?" ${\theta }_e=\arctan (0.4781)\approx 25.6^{\circ }.$ Reading the result: a $b$ this large leaves the wall still climbing steeply at exit — a real 80% bell would use a larger $b$ (more downward curvature) to bend ${\theta }_e$ toward $\approx 7^{\circ }$. The exercise shows the mechanism: increase $b$ to flatten the exit. Answer: $a\approx 0.191$, exit slope $0.478$, ${\theta }_e\approx 25.6^{\circ }$ (too steep — increase curvature $b$).

Length, thrust, mass for each option:

Why: A longer nozzle expands the gas to even lower exit pressure; when exit pressure falls below ambient (over-expansion), the outside air pushes back and can trigger flow separation / shocks inside the bell that spoil the exit momentum — so thrust drops instead of rising. This is exactly the altitude-matching problem covered in Nozzle Exit Pressure and Altitude Compensation; the design tools live in Method of Characteristics for Nozzle Design and real-wall effects in Manufacturing Tolerances in Nozzles. Answer: over-expansion → separation/shocks → thrust loss; resolved by altitude compensation.