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

This page is a drill sheet for the nozzle-geometry topic. We take the three formulas the parent note built — the divergence factor, the cone length, and the bell length — and push them through every kind of input: normal cases, the extreme angles, the "flat" degenerate cone, the limiting bell, a real launch-vehicle word problem, and an exam twist. Nothing new is assumed: every symbol below is re-explained the first time it appears.

Before the first sum, meet the symbols we reuse everywhere:

Look at figure s01: $R_t$ is the narrow throat radius, $R_e$ the wide exit radius, $\alpha$ the tilt of the straight wall, and the amber arrow is the exhaust — tilted by $\alpha$, so only its horizontal shadow (the cyan dashed arrow) actually pushes the rocket.

The next two figures show what changing the two knobs does to the shape and the exhaust vector — study them before the examples so the numbers below have a picture attached.

Figure s02 overlays three cones at the same radii but different $\alpha$ ($15^{\circ }$, $30^{\circ }$, and near-$0^{\circ }$): watch the wall steepen, the length shrink, and the exhaust arrow tilt further off-axis — the amber arrow's forward shadow shrinks, which is $\lambda$ dropping.

Figure s03 fixes $\alpha$ and grows $ϵ$: bigger expansion ratio ⇒ wider exit ⇒ longer nozzle. This is the picture behind Example 6 and the over-expansion trap in Example 8.

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The scenario matrix

Every problem this topic can throw is one of these cells. The worked examples below are tagged with the cell(s) they cover.

Cell	Case class	What makes it tricky	Example
A	Normal cone (12–18°)	plug-and-chug baseline	Ex 1
B	Large half-angle (30°+)	$\lambda$ drops fast — feel the loss	Ex 2
C	Degenerate $\alpha \to 0$	wall parallel to axis: $L\to \infty$, $\lambda \to 1$	Ex 3
D	Degenerate $ϵ=1$	no diverging section at all ($R_e=R_t$)	Ex 3
E	80% bell vs its parent cone	the length-scaling rule, weight saving	Ex 4
F	Limiting bell (${\theta }_e\to 0$)	perfectly axial exit, $\lambda \to 1$	Ex 5
G	Real-world word problem	vacuum upper-stage sizing	Ex 6
H	Exam twist (inverse)	given $L$ and $\lambda$, solve back for $\alpha ,ϵ$	Ex 7
I	Over-expansion trap	more length ≠ more thrust	Ex 8

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Worked examples

Example 1 — Cell A: the standard cone

Forecast: guess before computing — will $L$ be closer to 50 cm or 150 cm? Will $\lambda$ be above or below $0.98$?

1. Exit radius. $R_e=R_t\sqrt{ϵ}=10\sqrt{16}=40$ cm. Why this step? Area scales as radius squared, so the radius only grows by $\sqrt{ϵ}$, not by $ϵ$.
2. Length. The straight wall climbs from $R_t$ to $R_e$ at slope $\tan \alpha$ (rise over run). Rearranging "rise $=\text{run}\times \tan \alpha$": $L=\frac{R_e-R_t}{\tan \alpha }=\frac{R_t(\sqrt{ϵ}-1)}{\tan \alpha }=\frac{10(4-1)}{\tan 15^{\circ }}=\frac{30}{0.2679}\approx 112\text{ cm}.$ Why this step? $\tan \alpha$ (tangent = opposite/adjacent on the wall's right triangle) converts a height we must climb into a horizontal length we must travel. Steeper wall (bigger $\alpha$) ⇒ bigger $\tan \alpha$ ⇒ shorter nozzle.
3. Efficiency. $\lambda =\frac{1+\cos 15^{\circ }}{2}=\frac{1+0.9659}{2}=0.983$.

Verify: $R_e/R_t=40/10=4=\sqrt{16}$ ✓. Units: cm/(dimensionless tan) = cm ✓. $\lambda <1$ as it must be for any real cone ✓. So we lose $1-0.983=1.7\%$ of thrust to sideways leakage.

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Example 2 — Cell B: the wide-angle warning

Forecast: the nozzle gets shorter — but how much thrust do we pay?

1. Length. $L=\frac{30}{\tan 30^{\circ }}=\frac{30}{0.5774}\approx 51.96$ cm. Why this step? Doubling $\alpha$ more than doubled $\tan \alpha$, so length nearly halved — that is the temptation.
2. Efficiency. $\lambda =\frac{1+\cos 30^{\circ }}{2}=\frac{1+0.8660}{2}=0.9330$. Why this step? $\cos$ falls off faster as angle grows, so the axial fraction drops steeply.
3. Penalty. Loss went from $1.7\%$ to $1-0.933=6.7\%$ — nearly four times worse.

Verify: $\lambda (30^{\circ })=0.933<\lambda (15^{\circ })=0.983$ ✓ (bigger angle, worse aim). $L(30^{\circ })=52\text{cm}<L(15^{\circ })=112\text{cm}$ ✓ (steeper wall, shorter). Trade confirmed: half the length, quadruple the loss. This is exactly why real cones sit at 12–18°. (Look back at figure s02: the $30^{\circ }$ cone is the middle, stubbier funnel with the most tilted exhaust arrow.)

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Example 3 — Cells C & D: the two degenerate limits

Forecast: one of these gives an infinitely long nozzle; one gives no nozzle. Which is which?

1. Cell C, $\alpha \to 0$ (wall parallel to axis). Keep it symbolic first, for a general throat radius $R_t$ and general expansion ratio $ϵ>1$: $\lambda =\frac{1+\cos 0^{\circ }}{2}=\frac{1+1}{2}=1,L=\frac{R_t(\sqrt{ϵ}-1)}{\tan 0^{\circ }}.$ Since $\tan 0^{\circ }=0$ while the numerator $R_t(\sqrt{ϵ}-1)$ is a fixed positive number, we divide a positive constant by something shrinking to zero, so $L\to +\infty$. (For the Example-1 numbers $R_t=10$, $ϵ=16$ the numerator is exactly $30$, i.e. $30/\tan 0^{\circ }\to \infty$.) Why this step? A wall with zero tilt aims the gas perfectly forward ($\lambda =1$), but a zero-slope line rising from $R_t$ to $R_e$ must run forever. Perfect aim costs infinite length — the fundamental tension of nozzle design. This is the rightmost, near-flat cone in figure s02.
2. Cell D, $ϵ=1$ ($A_e=A_t$). Again keep it general: for any $R_t$ and any half-angle $\alpha$, $R_e=R_t\sqrt{1}=R_t,L=\frac{R_t(\sqrt{1}-1)}{\tan \alpha }=\frac{R_t\cdot 0}{\tan \alpha }=0.$ Why this step? If exit area equals throat area, there is nothing to expand — the diverging cone has zero length. The formula returns $L=0$, correctly telling us "there is no diverging section here."

Verify: As $\alpha \to 0^{\circ }$, $\cos \alpha \to 1$ so $\lambda \to 1$ ✓ and $\tan \alpha \to 0$ so $L\to +\infty$ ✓. At $ϵ=1$, $\sqrt{ϵ}-1=0\Rightarrow L=0$ regardless of $\alpha$ ✓. Both limits are physically sensible, not errors.

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Example 4 — Cell E: 80% bell against its parent cone

Forecast: the bell is shorter — does it also give less thrust, or somehow more?

1. Bell length. By definition an "80% bell" has $L_{\text{bell}}=0.8L_{\text{cone}}=0.8\times 112=89.6$ cm. Why this step? The "80%" label is the length rule — it means "as long as 80% of the equivalent 15° cone."
2. Thrust each. Because the nozzle is matched, the pressure term is zero, so the full thrust equation collapses to just the momentum piece: $F=\dot{m}{v}_e\lambda$. $F_{\text{cone}}=100\times 0.983=98.3\text{ kN},F_{\text{bell}}=100\times 0.985=98.5\text{ kN}.$ Why this step? $\lambda$ is literally the fraction of the $100$ kN raw push that survives as axial thrust once sideways leakage is removed.
3. Result. The bell delivers $98.5-98.3=0.2$ kN more thrust while being $112-89.6=22.4$ cm shorter.

Verify: $0.8\times 112=89.6$ ✓. $\Delta F=0.2$ kN $>0$ ✓ — shorter and stronger, because the curved wall re-aims the exhaust axial while the cone leaks it sideways. Length saving $=22.4/112=20\%$ ✓, matching the "80%" name.

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Example 5 — Cell F: the perfectly-axial limit

Forecast: will the $7^{\circ }$ exit cost us anywhere near the $6.7\%$ we saw for a 30° cone?

1. Ideal exit (${\theta }_e=0^{\circ }$). ${\lambda }_{\text{exit}}=\frac{1+\cos 0^{\circ }}{2}=1$. Why this step? Flow leaving straight down the axis wastes nothing sideways — this is what the bell's curve-back achieves that a cone cannot.
2. Real 80% bell exit (${\theta }_e=7^{\circ }$). ${\lambda }_{\text{exit}}=\frac{1+\cos 7^{\circ }}{2}=\frac{1+0.99255}{2}=0.99627$. Why this step? A gentle $7^{\circ }$ tilt at the very exit is nearly axial, so the exit-plane loss is only $\approx 0.37\%$ — far less than a cone's, because a cone tilts its whole wall by $\alpha$, not just the exit lip.

Verify: $\cos 0^{\circ }=1\Rightarrow \lambda =1$ ✓. $\lambda (7^{\circ })=0.9963>\lambda (15^{\circ })=0.983$ ✓: the bell's near-axial exit beats the cone's 15° everywhere. See Method of Characteristics for Nozzle Design for how the curve is actually solved.

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Example 6 — Cell G: real-world vacuum upper stage

Forecast: exit radius grows as $\sqrt{25}=5\times$ the throat — will the length be under or over a metre?

1. Equivalent cone length at the reference $\alpha =15^{\circ }$: $L_{\text{cone}}=\frac{R_t(\sqrt{ϵ}-1)}{\tan 15^{\circ }}=\frac{8(5-1)}{0.2679}=\frac{32}{0.2679}\approx 119.4\text{ cm}.$ Why this step? The 80% rule is defined relative to the 15° cone, so we must build that reference cone first.
2. Bell length. $L_{\text{bell}}=0.8\times 119.4\approx 95.6$ cm. Why this step? Apply the length-scaling definition of the 80% bell.
3. Exit radius. $R_e=R_t\sqrt{ϵ}=8\times 5=40$ cm. Why this step? Exit area is $25\times$ the throat, so radius is $\sqrt{25}=5\times$ larger. (This large $ϵ$ is the tall funnel on the right of figure s03.)

Verify: $\sqrt{25}=5$ ✓; $8\times 4/0.2679=119.4$ ✓; $0.8\times 119.4=95.6$ ✓; $R_e=40$ cm gives $A_e/A_t=(40/8{)}^2=25=ϵ$ ✓. The big $ϵ$ is correct for vacuum: low ambient pressure ⇒ expand more.

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Example 7 — Cell H: the inverse (exam twist)

Forecast: this cone is shorter than Example 1's for the same radii — so is $\alpha$ bigger or smaller than 15°?

1. Recover the angle. From $L=\frac{R_e-R_t}{\tan \alpha }$, solve for $\tan \alpha$: $\tan \alpha =\frac{R_e-R_t}{L}=\frac{40-10}{80}=0.375\Rightarrow \alpha =\arctan (0.375)=20.56^{\circ }.$ Why this step? $\arctan$ (arctangent) answers "which angle has this tangent?" — it undoes $\tan$, turning our measured slope $0.375$ back into an angle. We need it because the geometry gave us the ratio, not the angle directly.
2. Efficiency. $\lambda =\frac{1+\cos 20.56^{\circ }}{2}=\frac{1+0.9361}{2}=0.9680$. Why this step? Same divergence formula, now with the recovered $\alpha$.

Verify: Plug back: $L=(40-10)/\tan (20.56^{\circ })=30/0.375=80$ cm ✓. Shorter than Example 1's 112 cm and indeed $\alpha =20.6^{\circ }>15^{\circ }$ ✓ (steeper wall = shorter cone), and $\lambda =0.968<0.983$ ✓ (the price of steepness). This is Manufacturing Tolerances in Nozzles territory: measure, then verify the angle you actually got.

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Example 8 — Cell I: the over-expansion trap

Forecast: the parent note's Mistake 1 is a hint — trust the full thrust equation, not the "longer = better" intuition.

1. Exit area. $A_e=\pi R_e^2=\pi (0.40{)}^2=0.5027{\text{ m}}^2$. Why this step? The pressure term $(p_e-p_a)A_e$ needs the area the pressure acts on, and area $=\pi R^2$.
2. Nozzle X (matched). Pressure term $=(p_e-p_a)A_e=(101,325-101,325)\times 0.5027=0$. So $F_X=\dot{m}{v}_e\lambda +0=100,000\times 0.983=98,300\text{ N}=98.3\text{ kN}.$ Why this step? When $p_e=p_a$ the pressure piece is exactly zero — the whole reason "matched" is the target: no pressure penalty, just the momentum thrust.
3. Nozzle Y (over-expanded). Same momentum piece, but now the pressure piece is negative because $p_e<p_a$: $(p_e-p_a)A_e=(60,000-101,325)\times 0.5027=(-41,325)\times 0.5027=-20,774\text{ N}.$ $F_Y=100,000\times 0.983-20,774=98,300-20,774=77,526\text{ N}\approx 77.5\text{ kN}.$ Why this step? Over-expansion means the outside air presses inward harder than the exhaust presses out, so the pressure term subtracts. The longer nozzle lost about $20.8$ kN.
4. Decision. $F_X=98.3$ kN beats $F_Y=77.5$ kN. The longer nozzle Y is worse at sea level, by roughly $21$ kN. Why this step? This directly quantifies Mistake 1: length only helps until $p_e$ reaches $p_a$; past that, extra length over-expands and shocks eat thrust. Y only wins at altitude, where $p_a$ falls toward $60,000$ Pa and its pressure term climbs back toward zero.

Verify: Geometry alone gives both nozzles $\lambda =0.983$, proving $\lambda$ cannot detect the loss — you must use the pressure term ✓. Sign check: $p_e-p_a=-41,325<0$ ⇒ negative pressure thrust ✓. Units: Pa $\times$ m${}^2$ = N ✓. $F_Y<F_X$ despite Y being longer ✓ — exactly the trap. Full treatment in Nozzle Exit Pressure and Altitude Compensation.

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Recall Self-test (reveal after you try)

Cone with $R_t=10$, $ϵ=16$, $\alpha =15^{\circ }$: length? ::: $112$ cm Same radii but $L=80$ cm — half-angle? ::: $\alpha =\arctan (0.375)=20.56^{\circ }$ $\lambda$ for a $30^{\circ }$ cone? ::: $0.933$ (a $6.7\%$ loss) As $\alpha \to 0^{\circ }$, what happens to $L$ and $\lambda$? ::: $L\to \infty$, $\lambda \to 1$ (perfect aim, infinite length) Length of an 80% bell replacing a 112 cm cone? ::: $89.6$ cm Net thrust of a matched vs over-expanded ($p_e=60$ kPa, $A_e=0.5027$ m²) nozzle at sea level? ::: $98.3$ kN vs $77.5$ kN Why can a longer nozzle give less sea-level thrust? ::: over-expansion makes $p_e<p_a$, so $(p_e-p_a)A_e$ is negative