Exercises — Electric pump-fed cycle — modern innovation

The turbine (a fan spun by hot gas) is replaced by an electric motor. The motor is powered by a battery pack. No hot gas is tapped or burned to drive the pumps — the plumbing that a turbine needs (gas generator, hot ducts) simply disappears. See Turbopump-fed cycle (gas generator vs staged combustion) for the thing being replaced and Rocket Lab Rutherford engine for the real-world example.

$\dot{m}\Delta p$ has units $\left(\text{kg/s}\right)\times \left(\text{Pa}\right)$. Why start with units: if the units don't reduce to watts, the formula is wrong no matter what number it spits out — units are a free correctness check. A pascal is ${\text{N/m}}^2=\text{kg}{\text{m}}^{-1}{\text{s}}^{-2}$. Why rewrite the pascal: to combine it with kg/s we need every unit in base form (kg, m, s), otherwise we can't cancel anything. Multiplying: $\frac{\text{kg}}{\text{s}}\cdot \frac{\text{kg}}{\text{m}{\text{s}}^2}=\frac{{\text{kg}}^2}{\text{m}{\text{s}}^3}$ A watt is $\text{kg}{\text{m}}^2{\text{s}}^{-3}$. These do not match: our result has two powers of kg and a $1/\text{m}$, but a watt wants one kg and ${\text{m}}^2$. Why this specific mismatch matters: we have one extra kg on top and the metres are inverted — exactly what dividing by a density (${\text{kg/m}}^3$) removes. So divide by $\rho$: $\frac{{\text{kg}}^2}{\text{m}{\text{s}}^3}÷\frac{\text{kg}}{{\text{m}}^3}=\frac{{\text{kg}}^2}{\text{m}{\text{s}}^3}\cdot \frac{{\text{m}}^3}{\text{kg}}=\frac{\text{kg}{\text{m}}^2}{{\text{s}}^3}=\text{W}✓$ Why division by $\rho$ fixes it: dividing by $\rho$ cancels one kg (leaving a single power of mass) and turns $1/\text{m}$ into ${\text{m}}^2$ (because ${\text{m}}^{-1}\times {\text{m}}^3={\text{m}}^2$) — precisely the two edits needed to reach a watt. Physically, dividing $\dot{m}$ by $\rho$ converts a mass flow into a volume flow, and power is pressure × volume-flow. So the correct form is $P_{\text{ideal}}=\dot{m}\Delta p/\rho$.

This is the Pressure-fed cycle. The tank walls must withstand $150\text{bar}$ everywhere, so the tank becomes a thick, heavy pressure vessel. A pump lets the tank stay at $1$–$3\text{bar}$ (thin walls) and raises pressure only in a small pipe just before the chamber.

Convert: $120\text{bar}=120\times 10^5=1.2\times 10^7\text{Pa}$. $P_{\text{ideal}}=\frac{\dot{m}\Delta p}{\rho }=\frac{8\times 1.2\times 10^7}{1000}=\frac{9.6\times 10^7}{1000}=9.6\times 10^4\text{W}=96\text{kW}.$ What we did: plugged straight into $P_{\text{ideal}}=\dot{m}\Delta p/\rho$. Why: ideal power is just the energy per second handed to the fluid, and pressure rise is exactly energy per unit volume.

Here $P_{\text{elec}}$ is the electrical power the battery delivers for this one pump: $P_{\text{elec}}=\frac{P_{\text{ideal}}}{{\eta }_p{\eta }_m}=\frac{96}{0.65\times 0.9}=\frac{96}{0.585}\approx 164.1\text{kW}.$ Why divide: each efficiency below 1 means some supplied energy is lost (friction heats the fluid, resistance heats the wires), so you must feed in more than the ideal to still deliver 96 kW to the fluid.

$P_{\text{ox}}$ is the oxidiser pump's ideal power alone: $P_{\text{ox}}=\frac{18\times 1.2\times 10^7}{1140}=\frac{2.16\times 10^8}{1140}\approx 1.895\times 10^5\text{W}\approx 189.5\text{kW}.$ $P_{\text{total}}$ is the sum of both pumps' ideal powers (fuel from L2.1 = 96 kW): $P_{\text{total}}=P_{\text{ideal,fuel}}+P_{\text{ox}}=96+189.5=285.5\text{kW}.$ Insight: the oxidiser flow is more than double the fuel flow (that's the mixture ratio), so it dominates the pumping power even though it's slightly denser.

The pump must lift the fluid from tank pressure up to whatever the chamber plus every downstream loss demands. Build the required outlet pressure: $p_{\text{outlet}}=p_c+\Delta p_{\text{inj}}+\Delta p_{\text{line}}=100+20+5=125\text{bar}.$ Why add the losses: the injector and lines "eat" pressure between pump and chamber, so the pump must start higher to still arrive at $100\text{bar}$ in the chamber. The pump only supplies the rise above the tank: $\Delta p=p_{\text{outlet}}-p_{\text{tank}}=125-3=122\text{bar}=1.22\times 10^7\text{Pa}.$ Corrected ideal power: $P_{\text{ideal}}=\frac{8\times 1.22\times 10^7}{1000}=\frac{9.76\times 10^7}{1000}=9.76\times 10^4\text{W}=97.6\text{kW}.$ Takeaway: the "clean" $120\text{bar}$ of L2.1 hid a chain of losses; sizing real pumps means adding every downstream drop and subtracting the tank head.

First get total electrical power. $P_{\text{elec,total}}$ means the total electrical power for all pumps — take the total ideal power $P_{\text{total}}=285.5\text{kW}$ (from L2.3) and inflate it by the losses: $P_{\text{elec,total}}=\frac{P_{\text{total}}}{{\eta }_p{\eta }_m}=\frac{285.5}{0.65\times 0.9}=\frac{285.5}{0.585}\approx 488.0\text{kW}.$ Energy over the burn ($E$ = total electrical energy): $E=P_{\text{elec,total}}{t}_b=4.880\times 10^5\times 180=8.784\times 10^7\text{J}.$ Battery mass: $m_{\text{batt}}=\frac{E}{e_b}=\frac{8.784\times 10^7}{0.9\times 10^6}\approx 97.6\text{kg}.$ Chain: power ⟶ energy (×time) ⟶ kilograms (÷specific energy) — the "power ⟶ energy ⟶ kilograms" chant from the top of the page.

Battery mass is linear in $t_b$ (all other factors fixed), and passes through the origin — so mass is directly proportional to burn time. The line's slope is $97.6/180=0.542\text{kg/s}$.

• $60\text{s}$: $0.542\times 60\approx 32.5\text{kg}$ (one-third of $180\text{s}$).
• $360\text{s}$: $0.542\times 360\approx 195.2\text{kg}$ (double $180\text{s}$). What the geometry tells you: a straight line through the origin means "double the input, double the output." No need to recompute the whole formula — just scale.

Electric system total at $180\text{s}$: $25+97.6=122.6\text{kg}$. Turbopump: $40\text{kg}$. The turbopump is lighter here. Break-even: set electric mass = turbopump mass. $25+0.542t_b=40⟹0.542t_b=15⟹t_b=\frac{15}{0.542}\approx 27.7\text{s}.$ So electric only wins on mass for burns shorter than $\approx 28\text{s}$. Above that the batteries outweigh the turbopump's fixed machinery. (Electric can still win on cost/simplicity far beyond that — mass isn't the only score.) The linked Battery specific energy is exactly what shifts this break-even later as batteries improve.

Energy needed: $E=P_{\text{elec,total}}{t}_b=4.88\times 10^5\times 200=9.76\times 10^7\text{J}.$ Rearrange $m_{\text{batt}}=E/e_b$ for the minimum $e_b$ (largest allowed mass gives smallest required $e_b$): $e_b\ge \frac{E}{m_{\text{batt}}}=\frac{9.76\times 10^7}{80}=1.22\times 10^6\text{J/kg}.$ Convert to Wh/kg: $\frac{1.22\times 10^6}{3600}\approx 338.9\text{Wh/kg}.$ Synthesis point: today's best cells (~$280\text{Wh/kg}$) fall short, so either the mass cap or burn time must relax — this is precisely why the cycle waited for Battery specific energy to climb.

First half ($0$–$90\text{s}$): full $97.6\text{kg}$ aboard. Second half ($90$–$180\text{s}$): $97.6-48.8=48.8\text{kg}$ aboard. Time-averaged mass carried: $\bar{m}=\frac{97.6\times 90+48.8\times 90}{180}=\frac{97.6+48.8}{2}=73.2\text{kg}.$ Why it helps: the rocket equation $\Delta v=I_{sp}{g}_0\ln (m_i/m_f)$ rewards a small final mass $m_f$. Dead battery mass carried all the way to burnout inflates $m_f$ and eats $\Delta v$. Ejecting spent cells shrinks $m_f$, buying back some of the $\Delta v$ the batteries cost.

Battery mass $\propto \frac{1}{{\eta }_p{\eta }_m}$. Compare the fractional reduction from each option.

• Improving ${\eta }_p$: mass scales by $\frac{0.65}{0.72}=0.9028$ → a $9.72\%$ reduction.
• Improving ${\eta }_m$: mass scales by $\frac{0.90}{0.95}=0.9474$ → a $5.26\%$ reduction. The pump-efficiency upgrade wins because it's a larger relative jump ($0.65\to 0.72$ is $+10.8\%$ vs $0.90\to 0.95$ is $+5.6\%$). Synthesis lesson: chase the efficiency term that's currently lowest, since relative gains there are biggest.

(a) Total electrical power $P_{\text{elec,total}}$. Ideal powers: fuel $96\text{kW}$ (L2.1), oxidiser $189.5\text{kW}$ (L2.3), total ideal $P_{\text{total}}=285.5\text{kW}$. Divide by ${\eta }_p{\eta }_m=0.585$: $P_{\text{elec,total}}=\frac{285.5}{0.585}\approx 488.0\text{kW}.$ (b) Battery mass. Energy first, $E=P_{\text{elec,total}}{t}_b=488.0\times 10^3\times 180=8.784\times 10^7\text{J}$, then $m_{\text{batt}}=\frac{E}{e_b}=\frac{8.784\times 10^7}{0.9\times 10^6}\approx 97.6\text{kg}.$ (c) Propellant burned. Total flow $={\dot{m}}_f+{\dot{m}}_{ox}=8+18=26\text{kg/s}$, over $180\text{s}$: $m_{\text{prop}}=26\times 180=4680\text{kg}.$ (d) Ratio. Divide battery mass by propellant mass: $\frac{m_{\text{batt}}}{m_{\text{prop}}}=\frac{97.6}{4680}\approx 0.0209\approx 2.1\%.$ Comment: the battery is only ~2% of propellant mass — this stage is not battery-limited; the batteries are a modest tax, so the electric cycle is a sound choice here. It becomes battery-limited only if burn time or $\Delta p$ grow a lot while propellant stays small.

Power $\propto 1/\rho$. Lower density means each kg occupies more volume, so more volume-per-second must be pressurised — power rises. Ratio: $\frac{P_{\text{new}}}{P_{\text{old}}}=\frac{{\rho }_{\text{old}}}{{\rho }_{\text{new}}}=\frac{1140}{1083}\approx \mathrm{1.0526.}$ So power increases by $\approx 5.26\%$. Mastery point: the density in the denominator isn't a static constant — a real design budgets for the worst-case (lowest) density so the motor never stalls.

Ideal power: $P_{\text{ideal}}=\frac{350\times 1.1\times 10^7}{1050}=\frac{3.85\times 10^9}{1050}\approx 3.667\times 10^6\text{W}=3667\text{kW}.$ Electrical power: $P_{\text{elec,total}}=\frac{3667}{0.585}\approx 6268\text{kW}.$ Energy: $E=6.268\times 10^6\times 160=1.003\times 10^9\text{J}$. Battery mass: $m_{\text{batt}}=\frac{1.003\times 10^9}{0.9\times 10^6}\approx 1114\text{kg}.$ Verdict: over a tonne of dead battery — for comparison the upper-stage case (L5.1) needed under $100\text{kg}$. The huge flow and long-ish burn make energy pile up enormously. This is exactly why electric pumps live on small, short-burn upper stages and turbopumps still rule big boosters. The Rutherford (Rocket Lab Rutherford engine) is a small first-stage engine precisely because its flow is modest and it can jettison spent packs.

Every problem on this page is one idea wearing different clothes: energy per second the pump hands the fluid is $\dot{m}\Delta p/\rho$; losses inflate it; and battery kilograms are just that energy over time divided by how much a kilo of battery holds. Get those three moves and the whole cycle — from a single pump to a whole booster's go/no-go — falls out.