Worked examples — Staged combustion cycle — full flow, fuel-rich, oxidizer-rich preburners

This page is a drill sheet. The parent note built the two governing equations:

• Pump power $P_{\text{pump}}=\frac{\dot{m}\Delta p}{\rho {\eta }_p}$
• Turbine power $P_{\text{turb}}={\dot{m}}_{pb}{c}_p{T}_{in}{\eta }_t\left[1-{\pi }_t^{-(\gamma -1)/\gamma }\right]$

Here we hit every kind of input these equations can receive — normal, extreme, zero, and degenerate — so no exam or reality can surprise you.

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The scenario matrix

Every question this topic can throw lives in one of these cells. Each example below is tagged with the cell it covers.

Cell	What makes it different	Example
A — Baseline pump	ordinary numbers, one pump	Ex 1
B — Two pumps summed	balance with $\sum$ over fuel + ox	Ex 2
C — Limiting ${\pi }_t\to 1$	pressure ratio near 1 (degenerate turbine)	Ex 3
D — Limiting ${\pi }_t\to \infty$	huge expansion, bracket $\to 1$	Ex 3
E — Zero input	$\Delta p=0$ or $\dot{m}=0$	Ex 4
F — Sign / direction	"pressure drop" vs "pressure rise" trap	Ex 4
G — Temperature control	pick $T_{in}$ so the turbine survives	Ex 5
H — Real-world word problem	Raptor-class FFSC, two turbines	Ex 6
I — Exam twist	solve backwards for $p_c$ from a power budget	Ex 7
J — Cross-cycle contrast	same numbers, gas-generator vs staged waste	Ex 8
K — Degenerate efficiency	${\eta }_p\to 0$ (pump power blows up)	Ex 9

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Example 1 — Baseline single pump (Cell A)

Step 1 — Convert the pressure to SI. $250$ bar $=250\times 10^5$ Pa $=2.5\times 10^7$ Pa. Why this step? The formula wants pascals (J/m³). "bar" is a shortcut; $1\text{ bar}=10^5$ Pa. Mixing units here is the #1 wrong answer.

Step 2 — Plug into $P_{\text{pump}}=\frac{\dot{m}\Delta p}{\rho {\eta }_p}$. $P=\frac{400\times 2.5\times 10^7}{1140\times 0.75}=\frac{1.0\times 10^{10}}{855}\approx 1.17\times 10^7\text{W}$ Why this step? Pressure $\times$ volume-flow $=$ power. Volume flow is $\dot{m}/\rho$; efficiency in the denominator means the real shaft must deliver more than the ideal fluid work.

Step 3 — Read it in human units. $1.17\times 10^7$ W $\approx 11.7$ MW $\approx 15,700$ hp.

Verify: Units: $\frac{(\text{kg/s})(\text{Pa})}{({\text{kg/m}}^3)}=\frac{(\text{kg/s})({\text{J/m}}^3)}{({\text{kg/m}}^3)}=\text{J/s}=\text{W}$. ✓ A dimensionless $\eta$ doesn't change units. The magnitude (tens of MW for one pump) matches the parent's Example 1. ✓

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Example 2 — Two pumps, the real power balance (Cell B)

Step 1 — Fuel pump power. $\Delta p=3\times 10^7$ Pa. $P_f=\frac{100\times 3\times 10^7}{70\times 0.7}=\frac{3\times 10^9}{49}\approx 6.12\times 10^7\text{W}\approx 61.2\text{MW}$ Why this step? LH₂ is feather-light ($\rho =70$). Low density means huge volume flow per kg, and volume flow drives power. The fuel pump is the monster.

Step 2 — Oxidizer pump power. $P_o=\frac{600\times 3\times 10^7}{1140\times 0.7}=\frac{1.8\times 10^{10}}{798}\approx 2.256\times 10^7\text{W}\approx 22.6\text{MW}$ Why this step? Same $\Delta p$, but dense LOX means small volume flow — even 6× the mass costs less power.

Step 3 — Sum (the $\sum$ in the balance). $\sum P_{\text{pump}}=61.2+22.6=83.8\text{MW}$ Why this step? The power balance is $P_{\text{turb}}=\sum P_{\text{pump}}$: the turbines must supply the total.

Verify: Fuel pump power exceeds oxidizer despite 6× less mass — the density ratio $1140/70\approx 16$ overwhelms the mass ratio $6$. Forecast confirmed: fuel wins. ✓

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Example 3 — The two limits of the turbine bracket (Cells C & D)

The turbine's fraction extracted is $B({\pi }_t)=1-{\pi }_t^{-(\gamma -1)/\gamma }$. Let us break it at both extremes. Use $\gamma =1.3$ (methane-rich gas), so the exponent $\frac{\gamma -1}{\gamma }=\frac{0.3}{1.3}\approx 0.2308$.

Step 1 — Cell C: ${\pi }_t=1$ (no expansion). $B=1-1^{-0.2308}=1-1=0$ Why this step? ${\pi }_t=p_{in}/p_{out}=1$ means inlet and outlet pressures are equal — the gas doesn't expand, so it does zero work on the blades. A turbine with no pressure drop is a paperweight. This is the degenerate case — the red dot sitting on the floor of the figure.

Step 2 — Typical staged value: ${\pi }_t=1.5$. $B=1-1.5^{-0.2308}=1-0.9124=0.0876$ Why this step? A modest 1.5× drop extracts only ~9% of the enthalpy — the second dot, barely lifted off the floor. This is the everyday staged-combustion operating point (it sits between Cells C and D), not a matrix corner: staged turbines run low pressure ratio so the gas keeps most of its pressure for the main chamber, which is exactly why so much mass flow ${\dot{m}}_{pb}$ is needed.

Step 3 — Cell D: ${\pi }_t=100$ (huge expansion, the gas-generator regime). $B=1-100^{-0.2308}=1-0.3466=0.6534$ Why this step? Big pressure drop → big extraction (~65%), the top dot in the figure. But $B$ saturates below 1: as ${\pi }_t\to \infty$, ${\pi }_t^{-0.2308}\to 0$, so $B\to 1$ but never reaches it. That is the flattening of the red curve toward the dashed ceiling.

Verify: $B(1)=0$, $B$ is increasing, and $B<1$ for all finite ${\pi }_t$, with the ceiling $B\to 1$ only in the limit. Monotonic and bounded — no unphysical value. ✓

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Example 4 — Zero and sign traps (Cells E & F)

Step 1 — Cell E, $\dot{m}=0$. $P=\frac{0\times \Delta p}{\rho {\eta }_p}=0$ Why this step? No mass crosses the pump → no fluid is being lifted → zero fluid power. (Real pumps still draw idle power from friction, but the ideal formula correctly gives 0 useful output.)

Step 2 — Cell E, $\Delta p=0$. $P=\frac{\dot{m}\times 0}{\rho {\eta }_p}=0$ Why this step? If the pump adds no pressure, it does no useful work no matter how much fluid spins through it — the numerator has a zero factor, so the whole product collapses to 0. Both ways of "doing nothing" (no flow, or no pressure rise) give $P=0$, confirming the formula degrades gracefully.

Step 3 — Cell F, negative $\Delta p$. $\Delta p=-20$ bar $=-2\times 10^6$ Pa gives a negative $P$. Why this step? The sign of $\Delta p$ tells you the direction of energy flow. A negative result means the device is not a pump but a turbine/expander — the fluid is giving energy up, not receiving it. This is exactly why the turbine term in the balance uses a pressure drop (${\pi }_t>1$): energy comes out. Never plug a pressure drop into the pump formula and call it pump power.

Verify: $P=0$ when either factor in the numerator is 0. $P<0\iff \Delta p<0\iff$ energy extracted, i.e. turbine behaviour. Consistent with $P_{\text{turb}}$ carrying the drop ${\pi }_t>1$. ✓

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Example 5 — Choosing $T_{in}$ so the turbine survives (Cell G)

Step 1 — Back out $Q$ from the stoichiometric rise using $Q=m{c}_p\Delta T$. Here $m=1$ kg and the stoichiometric rise is $\Delta T=3300-100=3200$ K, so $Q=1\times 5000\times 3200=1.6\times 10^7\text{J}.$ Why this step? We define $m=1$ kg and use the same $c_p=5000$ J/kg·K here and later, so there is no hidden second heat capacity. The heat law $Q=m{c}_p\Delta T$ lets us recover the fixed energy $Q$ that the chemistry delivers regardless of how much mass we later dilute it into.

Step 2 — Re-spread the same $Q$ over the tripled mass. $\Delta T_{\text{rich}}=\frac{Q}{m^{\prime }{c}_p}=\frac{1.6\times 10^7}{3\times 5000}=\frac{1.6\times 10^7}{1.5\times 10^4}\approx 1067\text{K}.$ Why this step? $Q$ is fixed by the chemistry; $c_p$ is unchanged; only the mass grew from 1 kg to $m^{\prime }=3$ kg. Same energy ÷ 3× mass = one-third the rise. The excess H₂ soaks heat without adding much energy of its own — that is why "rich" cools.

Step 3 — Add back the 100 K inlet. $T_{out}\approx 100+1067=1167\text{K}.$ Why this step? $\Delta T$ is a rise above inlet; the gas actually leaves at inlet + rise.

Verify: With $m=1$ kg, $c_p=5000$: $Q=1.6\times 10^7$ J reproduces 3300 K ($100+1.6\times 10^7/5000=3300$ ✓). Tripling mass gives $100+1.6\times 10^7/(3\times 5000)=1167$ K, inside the parent's turbine-survivable band (~900–1200 K), far below the 3300 K that melts blades. Forecast confirmed. ✓

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Example 6 — Real-world FFSC engine (Cell H)

Step 1 — Exponent and bracket. $\frac{\gamma -1}{\gamma }=\frac{0.25}{1.25}=0.2$. $B=1-1.6^{-0.2}=1-0.9099=0.0901$ Why this step? $B$ is the fraction of enthalpy the turbine actually extracts (Ex 3). Low ${\pi }_t$ ⇒ small $B$ ⇒ we'll need lots of flow.

Step 2 — Power per kg of gas. $P/\dot{m}=c_p{T}_{in}{\eta }_{t}B=3200\times 750\times 0.78\times 0.0901\approx 1.687\times 10^5\text{J/kg}$ Why this step? $c_{p}T$ is enthalpy per kg; multiply by turbine efficiency ${\eta }_t$ and the extracted fraction $B$ to get usable work per kg.

Step 3 — Divide the fuel turbine demand by that. ${\dot{m}}_{pb,f}=\frac{60\times 10^6}{1.687\times 10^5}\approx 356\text{kg/s}$ Why this step? Required power ÷ power-per-kg = kg needed per second. In FFSC these 356 kg/s aren't lost — they gasify the fuel and pour into the chamber.

Verify: Units $\frac{\text{W}}{\text{J/kg}}=\frac{\text{J/s}}{\text{J/kg}}=\text{kg/s}$. ✓ A few hundred kg/s of gas is exactly why FFSC pushes "full flow" — nearly all propellant passes the preburners. ✓

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Example 7 — Exam twist: solve backwards for chamber pressure (Cell I)

Step 1 — Rearrange the pump equation for $\Delta p$. From $P=\frac{\dot{m}\Delta p}{\rho {\eta }_p}$: $\Delta p=\frac{P\rho {\eta }_p}{\dot{m}}$ Why this step? The exam gives you power and wants pressure; algebra isolates the unknown. This is the balance run in reverse — power available caps the pressure achievable.

Step 2 — Substitute the numbers. $\Delta p=\frac{30\times 10^6\times 1140\times 0.75}{500}=\frac{2.565\times 10^{10}}{500}=5.13\times 10^7\text{Pa}$ Why this step? We insert the given power budget, LOX density, and efficiency directly into the rearranged formula. Every quantity is already in SI (watts, kg/m³, dimensionless, kg/s), so the result comes out in pascals with no unit juggling — do the multiplication top and bottom, then divide.

Step 3 — Convert to bar. $\Delta p=5.13\times 10^7/10^5=513\text{bar}$ Why this step? $1\text{ bar}=10^5$ Pa, so divide. This is why staged engines reach 250–350 bar — with recovered turbine exhaust, more power budget "buys" more pressure.

Verify: Plug back: $\frac{500\times 5.13\times 10^7}{1140\times 0.75}=\frac{2.565\times 10^{10}}{855}=3.0\times 10^7=30$ MW. ✓ Round-trip closes.

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Example 8 — Cross-cycle contrast: what staged combustion saves (Cell J)

Step 1 — Thrust from each stream ($F=\dot{m}v$, from the Rocket thrust equation). Main flow if all recovered: $F_{\text{ideal}}=520\times 3500=1.82\times 10^6$ N. Why this step? Momentum thrust is mass flow × exhaust speed. Compare the "everything at $v_c$" ideal against reality.

Step 2 — Gas-generator actual thrust. Main $468$ kg/s at $3500$ + dumped $52$ kg/s at $1500$: $F_{gg}=468\times 3500+52\times 1500=1.638\times 10^6+7.8\times 10^4=1.716\times 10^6\text{N}$ Why this step? The dumped stream still makes some thrust, but at a much lower speed — that's the waste.

Step 3 — Fractional loss vs the fully-recovered (staged) ideal. $\frac{F_{\text{ideal}}-F_{gg}}{F_{\text{ideal}}}=\frac{1.82\times 10^6-1.716\times 10^6}{1.82\times 10^6}=\frac{1.04\times 10^5}{1.82\times 10^6}\approx 0.0571=5.7\%$ Why this step? We take the shortfall $F_{\text{ideal}}-F_{gg}$ (the thrust the dumped stream failed to deliver by leaving at 1500 instead of 3500 m/s) and divide by the ideal to express it as a fraction. Staged combustion recovers that 52 kg/s into the chamber, so it exits at the full $3500$ m/s — reclaiming this ~5.7%, and enabling higher $p_c$ on top. Contrast with Gas-generator cycle and Specific impulse and chamber pressure.

Verify: Loss fraction $=\frac{52(3500-1500)}{520\times 3500}=\frac{104000}{1820000}=0.0571=5.71\%$. ✓ Units: velocities cancel, mass flows cancel — a pure dimensionless fraction, as a fraction must be. A few-percent $I_{sp}$ gain that decides whole missions — the reason FFSC exists. ✓

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Example 9 — Degenerate efficiency: what happens as ${\eta }_p\to 0$ (Cell K)

Step 1 — Good pump, ${\eta }_p=0.75$. $P=\frac{400\times 2.5\times 10^7}{1140\times 0.75}=\frac{1.0\times 10^{10}}{855}\approx 1.17\times 10^7\text{W}\approx 11.7\text{MW}$ Why this step? This is the Ex 1 baseline — our reference point. ${\eta }_p$ sits in the denominator, so shrinking it will grow the answer.

Step 2 — Bad pump, ${\eta }_p=0.25$ (one-third as efficient). $P=\frac{1.0\times 10^{10}}{1140\times 0.25}=\frac{1.0\times 10^{10}}{285}\approx 3.51\times 10^7\text{W}\approx 35.1\text{MW}$ Why this step? Cutting ${\eta }_p$ to one-third triples the required power ($P\propto 1/{\eta }_p$). A wasteful pump forces the turbine — and therefore the preburner flow ${\dot{m}}_{pb}$ — to work three times harder. This is why turbopump efficiency is fought for so fiercely.

Step 3 — The degenerate limit ${\eta }_p\to 0$. ${\lim }_{{\eta }_p\to 0}P={\lim }_{{\eta }_p\to 0}\frac{\dot{m}\Delta p}{\rho {\eta }_p}=+\infty$ Why this step? A fixed nonzero numerator divided by a denominator heading to zero blows up without bound. Physically: a pump that wastes all its input as heat (${\eta }_p=0$) delivers no pressure rise for any finite power — so demanding a real $\Delta p$ from it needs infinite shaft power. This mirrors Ex 4's zero cases but from the opposite side: there the numerator went to zero giving $P=0$; here the denominator goes to zero giving $P\to \infty$. Between them they bracket the whole behaviour of the formula.

Verify: $P(0.75)=1.17\times 10^7$ W and $P(0.25)=3.51\times 10^7$ W: the ratio is exactly $0.75/0.25=3$, confirming $P\propto 1/{\eta }_p$. ✓ As ${\eta }_p\to 0^{+}$, $P\to +\infty$ (denominator $\to 0^{+}$, numerator fixed positive). ✓ Units unchanged from Ex 1 (W). ✓

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Recall Self-test

Bracket $B$ at ${\pi }_t=1$ ::: $B=0$ — no expansion, no work. As ${\pi }_t\to \infty$, $B\to$ ::: 1 (but never reaches it for finite ${\pi }_t$). Sign of pump power when $\Delta p<0$ ::: negative — it's acting as a turbine, extracting energy. Why is the LH₂ pump power larger than the LOX pump despite less mass? ::: Its density is ~16× lower, so volume flow (which sets power) is far higher. In FFSC, are the preburner kg/s wasted? ::: No — they gasify propellant and burn again in the chamber. As pump efficiency ${\eta }_p\to 0$, the required power $\to$ ::: $+\infty$ ($P\propto 1/{\eta }_p$).

See also: Turbopump design · Combustion stoichiometry and flame temperature · Regenerative cooling · Expander cycle.