Case Study: Designing a Methane Vacuum Upper-Stage Nozzle

"The nozzle is where you decide, in metal, how much of your chemistry you are willing to throw away." — a design maxim, paraphrased

Executive Summary

Case Study 1 analyzed an existing engine; here we design one. Our task: specify the nozzle for a LOX/methane in-space kick stage that must deliver about $100\ \text{kN}$ of thrust in vacuum at the highest specific impulse a reasonable bell allows. Methane ($\text{LOX/CH}_4$) is the propellant of the new generation — Raptor, BE-4 — because it balances a respectable specific impulse against a density and storability that hydrogen cannot match, a trade this chapter's physics lets us quantify. We will pick a chamber pressure and an exit pressure, compute the exhaust velocity and specific impulse, size the throat and the exit bell from the mass flow, and then step back to compare methane against hydrogen and kerosene for the same stage — closing the loop on §19.6 and feeding a real propellant mass into the reader's Mission Design Review. Every number is built from the chapter's four core relations.

Skills applied

• Choosing chamber and exit pressures for a vacuum stage and computing $v_e$, $I_{sp}$ (§19.4).
• Sizing the throat from a required choked mass flow, and the exit from the expansion ratio (§19.3, §19.5).
• Adding pressure thrust to get effective exhaust velocity and thrust (§19.4, Ch. 16).
• Trading specific impulse against propellant density across LH2, CH4, RP-1 (§19.6).
• Converting an $I_{sp}$ into a propellant mass via the rocket equation (Ch. 3).

Background: the design brief

The specific gas constant is $R = R_u/\mathcal{M} = 8.314/0.020 = 415.7\ \text{J/(kg·K)}$. Because the stage fires only in vacuum, we can expand deeply — but not infinitely, because the bell has mass. We choose an exit pressure $p_e = 0.05\ \text{bar} = 5{,}000\ \text{Pa}$ as a sensible stopping point (expanding much further buys little $v_e$ for a lot of bell). That sets the pressure ratio $p_c/p_e = 2000$.

Phase 1: Exhaust velocity and specific impulse

Apply the ideal exhaust-velocity formula (§19.4) with $p_e/p_c = 1/2000 = 0.0005$:

$$ v_e = \sqrt{\frac{2\gamma}{\gamma-1}\cdot\frac{R_u T_c}{\mathcal{M}}\left[1 - \left(\frac{p_e}{p_c}\right)^{\frac{\gamma-1}{\gamma}}\right]}. $$

The pieces: $\frac{2\gamma}{\gamma-1} = \frac{2.36}{0.18} = 13.11$; $\frac{\gamma-1}{\gamma} = \frac{0.18}{1.18} = 0.1525$; $(0.0005)^{0.1525} = 0.314$, so the bracket is $1 - 0.314 = 0.686$; and $R T_c = 415.7 \times 3500 = 1.455\times10^{6}$. Then

$$ v_e = \sqrt{13.11 \times 1.455\times10^{6} \times 0.686} = \sqrt{1.309\times10^{7}} \approx 3{,}618\ \text{m/s}. $$

On momentum alone that is $I_{sp} = 3618/9.81 = 369\ \text{s}$. We will pick up a little more from the pressure term in Phase 3.

Sanity check. Real LOX/methane vacuum engines quote $I_{sp}$ in the high $360$s to around $380\ \text{s}$; our $369$–$382\ \text{s}$ (with pressure thrust) sits squarely in that band. Good.

Phase 2: Sizing the throat and the bell

First the expansion ratio needed to reach $p_e = 0.05\ \text{bar}$. Using the area-ratio relation of §19.5 for $\gamma = 1.18$ and $p_c/p_e = 2000$ gives

$$ \epsilon = \frac{A_e}{A_t} \approx 135. $$

Now the throat. We need a mass flow that yields $\approx 100\ \text{kN}$; take $\dot m = 27\ \text{kg/s}$ (we confirm the thrust in Phase 3). The choked-flow relation (§19.3), solved for throat area, is

$$ A_t = \frac{\dot m}{p_c\,\sqrt{\gamma/(R T_c)}\,\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{2(\gamma-1)}}}. $$

The factors: $\sqrt{\gamma/(R T_c)} = \sqrt{1.18/(415.7\times3500)} = \sqrt{8.11\times10^{-7}} = 9.01\times10^{-4}$; and $\left(\frac{2}{2.18}\right)^{6.056} = (0.917)^{6.056} = 0.593$. So the denominator is $1.0\times10^{7} \times 9.01\times10^{-4} \times 0.593 = 5{,}344$, and

$$ A_t = \frac{27}{5{,}344} = 5.05\times10^{-3}\ \text{m}^2 \;\Rightarrow\; d_t = 2\sqrt{A_t/\pi} = 0.080\ \text{m} = 8.0\ \text{cm}. $$

The exit area follows from the expansion ratio:

$$ A_e = \epsilon\, A_t = 135 \times 5.05\times10^{-3} = 0.682\ \text{m}^2 \;\Rightarrow\; d_e = 2\sqrt{A_e/\pi} = 0.93\ \text{m}. $$

So the design is an $8\ \text{cm}$ throat opening into a $0.93\ \text{m}$ bell — a slender, deep nozzle with a $135:1$ area ratio, exactly the stretched shape you see on real vacuum engines.

Phase 3: Confirming thrust and effective exhaust velocity

In vacuum ($p_a = 0$) the effective exhaust velocity picks up the full exit-pressure term (§19.4, Ch. 16):

$$ c = v_e + \frac{p_e A_e}{\dot m} = 3618 + \frac{(5000)(0.682)}{27} = 3618 + 126 = 3{,}744\ \text{m/s}. $$

$$ I_{sp,\text{vac}} = \frac{3744}{9.81} = 382\ \text{s}, \qquad F = \dot m\, c = 27 \times 3744 = 1.01\times10^{5}\ \text{N} \approx 101\ \text{kN}. $$

The thrust lands right on the $100\ \text{kN}$ target, confirming the $\dot m = 27\ \text{kg/s}$ choice. The engine is very slightly under-expanded in vacuum ($p_e = 0.05\ \text{bar} > p_a = 0$), so the pressure term adds a small $126\ \text{m/s}$ — the price of stopping the bell at a finite length rather than an infinite one.

Phase 4: The propellant trade — methane vs. hydrogen vs. kerosene

Would another propellant do better for this same stage? Hold the design ($p_c = 100\ \text{bar}$, $p_e = 0.05\ \text{bar}$, vacuum) and swap the chemistry (Tier 3 modeled states):

Hydrogen wins the specific-impulse column decisively — $85$ seconds over methane — for the reason §19.6 gave: its exhaust is barely more than half the molar mass. But look at the last column. Liquid hydrogen is six times less dense than methane, so a hydrogen stage carrying the same propellant mass needs a tank many times larger — more structure, more mass, a worse mass ratio, and deep-cryogenic ($20\ \text{K}$) plumbing with boil-off. Methane's $423\ \text{kg/m}^3$ and near-LOX storage temperature make it dense, clean-burning, and reusable, giving up some $I_{sp}$ to gain tankage, operability, and cost. That trade — not any single number — is why the newest engines are methane.

🔧 Engineering Reality: The "needed $\epsilon$" differs by propellant even at the same pressure ratio, because $\gamma$ differs. Hydrogen's $\gamma = 1.20$ and kerosene's $1.22$ shift the area-ratio relation enough to change the bell by tens of percent. In a real design you iterate: pick propellant, get $\gamma$ from a combustion code (Chapter 18's job), size the bell, check it fits and does not separate, and repeat. Our single pass is the first turn of that loop.

Phase 5: Feeding the Mission Design Review

The point of the design is a number your MDR can use. Suppose your stage (a Track-C Mars orbiter's insertion stage, say) must supply $\Delta v = 1.5\ \text{km/s}$ from the Chapter 3 delta-v budget, and carries a dry-plus-payload mass of $m_f = 2{,}000\ \text{kg}$. With our methane engine's effective $c = 3{,}744\ \text{m/s}$, the rocket equation (Chapter 3) gives the mass ratio and propellant:

$$ \frac{m_0}{m_f} = e^{\Delta v/c} = e^{1500/3744} = e^{0.401} = 1.49 \;\Rightarrow\; m_0 = 2985\ \text{kg}, \quad m_p = 985\ \text{kg}. $$

Had we chosen hydrogen ($c \approx 4{,}560\ \text{m/s}$, using a real vacuum $I_{sp} \approx 465\ \text{s}$), the same stage would need only $m_p \approx 779\ \text{kg}$ — about $206\ \text{kg}$ less propellant — but in a tank roughly six times the volume. That is the design decision in a sentence: $206\ \text{kg}$ of propellant saved versus a gymnasium of hydrogen tank and its cryogenic burden. For a small deep-space stage the hydrogen often still wins; for a reusable, dense, operable stage, methane does. Either way, the propellant mass now sits in your MDR, sized by a nozzle you designed.

Discussion Questions

1. We stopped the expansion at $p_e = 0.05\ \text{bar}$. What would expanding to $p_e = 0.02\ \text{bar}$ do to $v_e$, to $\epsilon$, and to the bell's mass? Why is there a practical stopping point?
2. The methane engine is slightly under-expanded in vacuum. Is that a design flaw, or an acceptable consequence of a finite bell? Explain using the pressure-thrust term.
3. Hydrogen beats methane by $85\ \text{s}$ of $I_{sp}$ but loses badly on density. For which of the four mission tracks (GEO comsat, lunar lander, Mars orbiter, asteroid rendezvous) would you still choose hydrogen, and why?
4. If you doubled the chamber pressure to $200\ \text{bar}$ (keeping $p_e$ fixed), what happens to $v_e$, to the required $\epsilon$, and to the throat area for the same thrust?

Your Turn: Extensions

• Option A (analysis). Recompute Phases 1–3 at $p_c = 200\ \text{bar}$ (a Raptor-class chamber) and report the new $I_{sp}$, $\epsilon$, and throat diameter for the same $100\ \text{kN}$. How much did the higher chamber pressure buy?
• Option B (computation). Write Python that takes the design brief (propellant state, $p_c$, $p_e$, target thrust) and returns $v_e$, $I_{sp}$, $\epsilon$, $\dot m$, $A_t$, $A_e$, and the bell diameters. Reproduce this study's numbers. Do not run it — hand-trace and add # Expected output:.
• Option C (design). Finish the job for your mission: choose a propellant, pick $p_c$ and $p_e$ suited to where your stage fires, size its nozzle, and convert the resulting $I_{sp}$ into a propellant mass for your MDR delta-v budget. Compare against a second propellant choice.

Key Takeaways

1. A nozzle design is a chain of the chapter's four relations: pick $p_c$ and $p_e$ → get $v_e$ and $I_{sp}$ (exhaust-velocity formula) → get $\epsilon$ (area-ratio relation) → size $A_t$ from the required $\dot m$ (choked flow) → get $A_e = \epsilon A_t$ → confirm thrust with the pressure term.
2. Vacuum stages expand deeply but not infinitely. We stopped at $p_e = 0.05\ \text{bar}$ and $\epsilon = 135$; the last bit of $v_e$ is not worth the bell mass.
3. Specific impulse is not the only figure of merit. Hydrogen's $85$-second $I_{sp}$ edge over methane is real, but its sixfold density penalty is why methane wins for dense, reusable stages.
4. The nozzle design produces a propellant mass, via the rocket equation — the number your Mission Design Review actually needs.