• Fuel-rich staged combustion (FRSC): hydrogen-rich turbine gas — SSME / RS-25.
• Oxidizer-rich staged combustion (ORSC): oxygen-rich turbine gas — RD-180 / RD-170.
• Full-flow staged combustion (FFSC): two preburners, one of each — SpaceX Raptor.

The blank: routed into the main combustion chamber and burned again. Why it matters: no propellant is wasted, so both chamber pressure and specific impulse can be pushed higher. This is the "burn it twice" trick, the entire reason staged combustion is efficient. See Specific impulse and chamber pressure.

Stoichiometric LOX/LH₂ burns at ~3300 K, which instantly melts turbine blades. Running rich means one reactant is in excess; that excess mass absorbs heat without adding much energy, so the flame temperature drops to a range (~900–1000 K) the turbine metal survives. The combustion is deliberately incomplete — it is finished in the main chamber. See Combustion stoichiometry and flame temperature.

What we do: plug into $P_{\text{pump}}=\frac{\dot{m}\Delta p}{\rho {\eta }_p}$. Why: pressure is energy per volume; multiplying by the volume flow $\dot{m}/\rho$ gives power, and dividing by ${\eta }_p<1$ accounts for wasted heat. $P=\frac{90\times 3.5\times 10^7}{423\times 0.7}=\frac{3.15\times 10^9}{296.1}\approx 1.064\times 10^7\text{W}\approx 10.6\text{MW}$

What we do: evaluate the exponent first, then the power, then subtract from 1. Why: the bracket is the fraction of the gas enthalpy that an ideal expansion converts to shaft work — a pure number between 0 and 1. Exponent: $\frac{\gamma -1}{\gamma }=\frac{0.3}{1.3}=0.2308$. ${\pi }_t^{-0.2308}=1.8^{-0.2308}$. Since $\ln 1.8=0.5878$, we get $e^{-0.2308\times 0.5878}=e^{-0.1357}=0.8731$. $\left[1-0.8731\right]=0.1269\approx 0.127$

Power per unit flow: $c_p{T}_{in}{\eta }_t\times \text{bracket}$. $=4500\times 800\times 0.75\times 0.127=342,900\text{J/kg}\approx 3.43\times 10^5\text{J/kg}$ What we do next: divide the required power by the power-per-kg. Why: required power ÷ (power each kg/s gives) = how many kg/s we need. ${\dot{m}}_{pb}=\frac{10.6\times 10^6}{3.43\times 10^5}\approx 30.9\text{kg/s}$ In staged combustion these ~31 kg/s are not wasted — they flow into the chamber and burn again.

The balance: the turbine must supply the power of all the pumps, so the right side is a sum over every pump (fuel pump plus oxidizer pump, each contributing its own $\dot{m}\Delta p/(\rho {\eta }_p)$): ${\dot{m}}_{pb}{c}_p{T}_{in}{\eta }_t[\cdot ]=\underset{\text{add up every pump’s demand}}{\underbrace{\sum _{\text{pumps}}\frac{\dot{m}\Delta p}{\rho {\eta }_p}}}$ The $\sum$ (the Greek letter sigma, meaning "add these up") simply says: total the power demand of the fuel pump and the oxidizer pump. Raising $p_c$ raises each pump's $\Delta p$ (every pump must push harder to reach the higher chamber pressure), so the right side grows. To keep the equality, the left side must grow too — meaning more preburner flow ${\dot{m}}_{pb}$ or hotter $T_{in}$.

• Gas-generator: that extra ${\dot{m}}_{pb}$ is dumped overboard producing almost no thrust, so efficiency falls as $p_c$ climbs.
• Staged combustion: the extra ${\dot{m}}_{pb}$ rejoins the chamber and burns again, contributing full thrust — so $p_c$ can climb "for free," reaching 250–300 bar. Contrast with Gas-generator cycle.

Reading the curve: at ${\pi }_t=1$ (no pressure drop, left edge where the arrow points) the bracket is exactly $0$ — no expansion, no work. As ${\pi }_t$ increases, more enthalpy is extractable, so the bracket rises. But the exponent $-(\gamma -1)/\gamma$ is a small negative number, so ${\pi }_t^{-(\gamma -1)/\gamma }$ falls slowly — the bracket saturates (the right-hand arrow marks this flattening), giving diminishing returns. Design meaning: the butter-coloured band shows where real staged engines sit — ${\pi }_t$ deliberately kept small (~1.5–2). A big ${\pi }_t$ would rob pressure from the gas that still has to enter the high-pressure chamber. The flat curve tells you a small pressure ratio already captures most of the available work — so you spend that pressure on the chamber, not the turbine. Relevant to Turbopump design.

Recall from the toolbox: $\Delta T=Q/(m{c}_p)$, where $Q$ is the heat released by combustion and $m$ is the mass being heated. Fixed heat spread over more mass gives a smaller temperature rise. What cools: adding more excess oxidizer (moving further from stoichiometric, more oxidizer-rich) adds cold mass — it raises $m$ without much extra $Q$ — so $\Delta T=Q/(m{c}_p)$ falls: cooler flame. Moving toward stoichiometric (b) burns more completely, releasing more $Q$ over less excess mass, making it hotter — the opposite of what you want. The danger: running even more oxygen-rich makes the gas more aggressively oxidising, so the alloy oxidation problem worsens. This is the fundamental ORSC trade — you fight temperature with excess oxygen but pay in corrosion. See Combustion stoichiometry and flame temperature and Regenerative cooling.

(a) Pump power: $P_{\text{pump}}=\frac{120\times 3\times 10^7}{1140\times 0.72}=\frac{3.6\times 10^9}{820.8}\approx 4.386\times 10^6\text{W}\approx 4.39\text{MW}$ (b) Bracket: exponent $=\frac{0.3}{1.3}=0.2308$; $1.6^{-0.2308}=e^{-0.2308\ln 1.6}=e^{-0.2308\times 0.4700}=e^{-0.1085}=0.8972$. Bracket $=1-0.8972=0.1028$. Power per kg/s $=1200\times 750\times 0.72\times 0.1028=66,614\text{J/kg}\approx 6.66\times 10^4$ J/kg. ${\dot{m}}_{pb}=\frac{4.386\times 10^6}{6.66\times 10^4}\approx 65.9\text{kg/s}$ Note how the low $c_p$ of the heavy oxygen-rich gas (1200 vs H₂'s ~4500) forces a large turbine flow — a real reason ORSC turbines are big.

Power per kg/s $=5000\times 750\times 0.72\times 0.1028=277,560\text{J/kg}\approx 2.78\times 10^5$ J/kg. ${\dot{m}}_{pb}=\frac{4.386\times 10^6}{2.78\times 10^5}\approx 15.8\text{kg/s}$ Hydrogen-rich gas delivers the same power with roughly one-quarter the flow because of its huge $c_p$. In FFSC the fuel-rich (H₂/CH₄-rich) turbine is light and efficient, while the ox-rich turbine must move much more mass — which is exactly why FFSC splits the two into separate optimised turbopumps rather than forcing one turbine to do both jobs.

(a) $P_{\text{pump}}=\frac{140\times 3.6\times 10^7}{423\times 0.7}=\frac{5.04\times 10^9}{296.1}\approx 1.702\times 10^7$ W $\approx 17.0$ MW. (b) Exponent $\frac{0.35}{1.35}=0.2593$; $1.7^{-0.2593}=e^{-0.2593\ln 1.7}=e^{-0.2593\times 0.5306}=e^{-0.1376}=0.8714$; bracket $=1-0.8714=0.1286$. (c) Power per kg/s $=3800\times 850\times 0.78\times 0.1286=323,983\text{J/kg}\approx 3.24\times 10^5$ J/kg. ${\dot{m}}_{pb}=\frac{17.02\times 10^6}{3.24\times 10^5}\approx 52.5\text{kg/s}$ (d) Fraction $=52.5/140\approx 0.375$, i.e. about 37.5% of the fuel passes through the preburner. In FFSC that gas is not vented — it exits the turbine still full of unburnt methane and is dumped into the chamber to finish burning ("burned twice"). So the 37.5% still contributes full thrust; the only cost is the plumbing running hot and high-pressure. Ties to Rocket thrust equation and Specific impulse and chamber pressure.

Chain of proportions: pump power scales linearly with $\Delta p$ (from $P_{\text{pump}}=\dot{m}\Delta p/(\rho {\eta }_p)$), and ${\dot{m}}_{pb}$ scales linearly with that power (the bracket, $c_p$, $T_{in}$, ${\eta }_t$ are all unchanged). So the factor is simply the ratio of the two pressures: $\frac{\Delta p_{\text{new}}}{\Delta p_{\text{old}}}=\frac{4.5\times 10^7}{3.6\times 10^7}=1.25$ Required preburner flow therefore rises by 25%. Taking L5.1's value ${\dot{m}}_{pb}\approx 52.5$ kg/s: ${\dot{m}}_{pb,\text{new}}=1.25\times 52.5\approx 65.6\text{kg/s}$ One line: in staged combustion those extra ~13 kg/s rejoin the chamber and produce thrust, so the higher chamber pressure pays off; in a gas-generator cycle they'd be dumped overboard, so the efficiency loss would cancel the gain.

Why does hydrogen-rich turbine gas need less flow than oxygen-rich for the same power? ::: Because ${\dot{m}}_{pb}\propto 1/c_p$, and hydrogen-rich gas has a much larger $c_p$ (~4500–5000 vs ~1200 J/kg·K). Raising chamber pressure raises which term in the power balance first? ::: The pump side, through a larger $\Delta p$. What makes the turbine bracket flatten at high ${\pi }_t$? ::: The small negative exponent $-(\gamma -1)/\gamma$ makes ${\pi }_t^{-(\gamma -1)/\gamma }$ fall slowly, so returns diminish. In FFSC, what fraction of propellant is gasified before the chamber? ::: All of it — that is the definition of "full flow."