Exercises — Altitude compensation methods — nozzle extension, aerospike

What we compare: $p_e$ against $p_a$. Here $p_e=40<101=p_a$. Since exit pressure is below ambient, the outside air pushes back on the exhaust — the gas expanded too much. Answer: over-expanded (the pinched-plume, left picture in the figure above). (Mnemonic: over-expanded means you over-did the spreading, so $p_e$ dropped below $p_a$.)

$p_e=40>0=p_a$. Exit pressure is above ambient ⇒ the gas could still have expanded more. Under-expanded (the flaring-plume, right picture above). This is exactly why one fixed nozzle can never be perfect everywhere: the same $p_e$ flipped from over- to under-expanded just by changing altitude.

$R_e=R_t+L\tan \alpha$: a longer $L$ gives a larger exit radius $R_e$. Since $ϵ=(R_e/R_t{)}^2$ grows with $R_e$, the longer nozzle A has the larger $ϵ$. This is the whole principle behind an extendable nozzle — add length to add $ϵ$.

Step 1 — exit radii (read $R_e=R_t+L\tan \alpha$ straight off the figure). $\tan 15^{\circ }=0.2679$. $R_{e,1}=5+40(0.2679)=5+10.72=15.72\text{cm}$ $R_{e,2}=5+90(0.2679)=5+24.11=29.11\text{cm}$ Step 2 — square the radius ratio. ${ϵ}_1=(15.72/5{)}^2=(3.144{)}^2\approx 9.88$ ${ϵ}_2=(29.11/5{)}^2=(5.821{)}^2\approx 33.9$ Answer: ${ϵ}_1\approx 9.9$, ${ϵ}_2\approx 33.9$. Adding 50 cm of length more than tripled the expansion ratio — because $ϵ$ depends on radius squared.

What tool and why: the isentropic pressure–Mach relation, because it converts "how fast" ($M_e$, set by geometry) into "how much pressure remains." $\frac{p_e}{p_0}={\left(1+\frac{0.2}{2}(3.8{)}^2\right)}^{-1.2/0.2}={\left(1+0.1\cdot 14.44\right)}^{-6}=(2.444{)}^{-6}$ $(2.444{)}^{-6}\approx 0.00470\Rightarrow p_e=3000\text{kPa}\times 0.00470\approx 14.1\text{kPa}$ Answer: $p_e\approx 14.1$ kPa.

Convert first (unit reminder above): $p_e=14.1\text{kPa}=14100\text{Pa}$. $F=\dot{m}{v}_e+(p_e-p_a)A_e=10(3900)+(14100-0)(0.21)$ $F=39000+2961=41961\text{N}$ Answer: $F\approx 42.0$ kN. Note the pressure term (2961 N) is a small $\sim 7.6\%$ of the momentum term (39 000 N) — momentum dominates.

Why this tool: exhaust speed comes from energy conservation, not from $\sqrt{ϵ}$ shortcuts — $ϵ$ only enters by setting $p_e$. Prefactor: $\frac{2\gamma }{\gamma -1}R{T}_0=\frac{2(1.2)}{0.2}(400)(3300)=12\times 400\times 3300=1.584\times 10^7$. Exponent $(\gamma -1)/\gamma =0.2/1.2=0.1667$.

Case A, $p_e/p_0=14.1/3000=4.700\times 10^{-3}$: $(4.700\times 10^{-3}{)}^{0.1667}\approx 0.4004,v_e=\sqrt{1.584\times 10^7(1-0.4004)}=\sqrt{9.498\times 10^6}\approx 3082\text{m/s}$ Case B, $p_e/p_0=0.43/3000=1.433\times 10^{-4}$: $(1.433\times 10^{-4}{)}^{0.1667}\approx 0.2300,v_e=\sqrt{1.584\times 10^7(1-0.2300)}=\sqrt{1.220\times 10^7}\approx 3492\text{m/s}$ Gain: $(3492-3082)/3082=13.3\%$. The lesson: a $\sim 33\times$ drop in $p_e$ gave only $\sim 13\%$ more speed — the bracket $[1-(p_e/p_0{)}^{0.1667}]$ saturates toward 1 as $p_e\to 0$. Diminishing returns, exactly as the parent warned.

Convert to Pa first: $p_e=14100$ Pa, $p_a=101000$ Pa. $(p_e-p_a)A_e=(14100-101000)(0.21)=(-86900)(0.21)=-18249\text{N}$ Answer: $-18.2$ kN. The negative sign means atmospheric pressure is pushing backward on the exhaust column — a ~18 kN penalty. In reality a mismatch this large would also cause the flow to separate from the nozzle wall (edge-case callout), so the true loss is even messier than this clean number. This is why a vacuum-tuned nozzle is terrible at sea level, and why extendable extensions stay stowed until near-vacuum.

With $p_e=p_a$ the pressure term vanishes: $F=\dot{m}{v}_e+\underset{=0}{\underbrace{(p_e-p_a)}}{A}_e=\dot{m}{v}_e$ Why it wins: the fixed nozzle carries a penalty term $(p_e-p_a)A_e$ that is negative low down (over-expanded, even risking separation) and under-uses energy high up; the aerospike keeps that term at zero at every altitude, so no thrust is wasted fighting or ignoring the atmosphere.

Invert the geometry. From ${ϵ}_2=(R_{e,2}/R_t{)}^2$: $R_{e,2}=R_t\sqrt{{ϵ}_2}=5\sqrt{84}=5(9.165)=45.83\text{cm}$ Now solve $R_{e,2}=R_t+L_2\tan \alpha$ for $L_2$: $L_2=\frac{R_{e,2}-R_t}{\tan \alpha }=\frac{45.83-5}{0.2679}=\frac{40.83}{0.2679}\approx 152.4\text{cm}$ Answer: $L_2\approx 1.52$ m. That length under a fairing is exactly why it must be stowed at launch — the synthesis of "high $ϵ$" and "fits in the rocket" is the extendable nozzle.

(a) Propellant $=\dot{m}t=10\times 700=7000$ kg. (b) $I_{sp}=v_e/g_0=4100/9.81=417.9$ s (using standard gravity $g_0=9.81{\text{m/s}}^2$). Answer: 7000 kg burned; $I_{sp}\approx 418$ s. (The larger deployed $ϵ$ nudges $v_e$ up, which is why extendable upper stages chase every extra second of $I_{sp}$.)

Compare fixed $p_e=30$ kPa to each $p_a$:

Optimal altitude: where $p_a=p_e=30$ kPa — roughly ~9 km up. Below it, penalty (and, if the mismatch grows, flow separation); above it, wasted expansion potential. A single fixed nozzle can only be perfect at one line in this table — the exact motivation for aerospikes (perfect at every row) and extendable nozzles (change $ϵ$ mid-flight).

Setup. $F(ϵ)=\dot{m}{v}_e(p_e)+(p_e-p_a)A_e$. As $ϵ$ grows we add a thin ring of exit area $d{A}_e$; this lowers $p_e$ (more expansion) and raises $A_e$. Find the maximum by setting $dF/dϵ=0$.

Step 1 — differentiate term by term. Using the chain rule (since $v_e$ depends on $ϵ$ only through $p_e$): $\frac{dF}{dϵ}=\dot{m}\frac{d{v}_e}{d{p}_e}\frac{d{p}_e}{dϵ}+\frac{d{p}_e}{dϵ}{A}_e+(p_e-p_a)\frac{d{A}_e}{dϵ}.$

Step 2 — use the momentum theorem for the added ring. The one-dimensional momentum balance across the nozzle exit gives the standard differential result $\dot{m}\frac{d{v}_e}{d{p}_e}=-A_e,$ i.e. each drop $d{p}_e$ of exit pressure buys exactly $\dot{m}d{v}_e=-A_{e}d{p}_e$ of momentum (this is just $\dot{m}d{v}_e+A_{e}d{p}_e=0$). Substituting into the first term: $\frac{dF}{dϵ}=\underset{\text{momentum gain}}{\underbrace{(-A_e)\frac{d{p}_e}{dϵ}}}+\underset{\text{pressure-area change}}{\underbrace{\frac{d{p}_e}{dϵ}{A}_e}}+(p_e-p_a)\frac{d{A}_e}{dϵ}.$

Step 3 — watch the cancellation. The first two terms are $-A_e\frac{d{p}_e}{dϵ}+A_e\frac{d{p}_e}{dϵ}=0$ exactly. They cancel because the velocity you gain from dropping $p_e$ is precisely the momentum the pressure term loses. Only the last term survives: $\frac{dF}{dϵ}=(p_e-p_a)\frac{d{A}_e}{dϵ}.$

Step 4 — set the derivative to zero. Since $d{A}_e/dϵ>0$ (more $ϵ$ always means more exit area, so this factor never vanishes), the only way to get $dF/dϵ=0$ is $\boxed{p_e=p_a}$

Step 5 — confirm it is a maximum, and say WHY physically. Look at the sign of $\frac{dF}{dϵ}=(p_e-p_a)\frac{d{A}_e}{dϵ}$:

• While $p_e>p_a$ (nozzle still under-expanded): $dF/dϵ>0$, so adding nozzle raises thrust.
• Once $p_e<p_a$ (nozzle now over-expanded): $dF/dϵ<0$, so adding nozzle lowers thrust.

The derivative flips from $+$ to $-$ exactly at $p_e=p_a$ — that sign change is the signature of a maximum. Physically: each extra sliver of nozzle wall at the exit feels internal pressure $p_e$ pushing outward and ambient $p_a$ pushing inward. While $p_e>p_a$ the net outward push adds thrust; the moment $p_e$ drops below $p_a$ that sliver becomes dead weight the atmosphere is pushing back on. The break-even — the peak — is exactly $p_e=p_a$. This single condition is the north star of all altitude-compensation hardware.

$(p_e-p_a)A_e=(14100-14100)(0.21)=0$ $F=\dot{m}{v}_e=10(3082)=30820\text{N}$ Answer: pressure term $=0$, $F=30.82$ kN — pure momentum thrust, the cleanest a fixed nozzle ever gets, and the state an aerospike mimics everywhere.

Over-expanded means $p_e$ is ::: below $p_a$ (expanded too much) The single condition for maximum thrust at any altitude ::: $p_e=p_a$ $ϵ$ scales with exit radius as ::: $ϵ=(R_e/R_t{)}^2$ (radius squared) Why extendable nozzles stay stowed at launch ::: to save length under the fairing; they fire in near-vacuum, not for sea-level thrust Why doubling $ϵ$ gives diminishing $v_e$ ::: the bracket $1-(p_e/p_0{)}^{(\gamma -1)/\gamma }$ saturates toward 1 as $p_e\to 0$ What goes wrong in a badly over-expanded nozzle at low altitude ::: flow separates from the wall, an internal shock forms, causing thrust loss, side-loads and vibration