Electric pump-fed cycle — modern innovation: Worked examples

This page is the drill hall. The parent note built the formulas; here we hit every kind of number you can throw at them — small and large flows, zero cases, limits, a real-world word problem, and an exam-style twist.

Everything rests on two boxed results from the parent. We restate them so no symbol is used before you can see it:

The scenario matrix

Every problem this topic can pose falls into one of these cells. The examples below are labelled with the cell(s) they cover, so together they fill the whole grid.

Cell	What it stress-tests	Covered by
A — baseline plug-in	one clean pump, all values given	Ex 1
B — unit trap	$Δ p$ given in bar, $e_{b}$ in Wh/kg	Ex 2
C — the $ρ$ effect	denser vs less-dense propellant, same $\overset{m}{˙}$	Ex 3
D — limiting case (long burn)	$t_{b} \to$ large: batteries dominate	Ex 4
E — degenerate / zero inputs	$\overset{m}{˙} = 0$ , $Δ p = 0$ , $η \to 1$	Ex 5
F — real-world word problem	full-engine, drop-battery strategy	Ex 6
G — inverse solve	given battery budget, find max $t_{b}$	Ex 7
H — exam twist (compare cycles)	electric vs pressure-fed break-even	Ex 8

Read the Forecast line and guess before you scroll. Guessing wrong is where learning happens.

Ex 1 — Cell A: baseline plug-in

Worked example Ex 1 — one pump, everything given

$\overset{m}{˙} = 4 kg/s$ , $Δ p = 8 \times 1 0^{6} Pa$ , $ρ = 800 kg/m^{3}$ (RP-1 kerosene), $η_{p} = 0.6$ , $η_{m} = 0.9$ . Find $P_{elec}$ .

Forecast: ideal power is a few tens of kW; losses inflate it by roughly $1/ (0.6 \times 0.9) \approx 1.85$ . Guess a number before computing.

Step 1 — ideal power. Why this step? Start from the loss-free core so we can see where the losses bite. $P_{ideal} = \frac{m ˙ Δ p}{ρ} = \frac{4 \times 8 \times 1 0 ^{6}}{800} = 4.0 \times 1 0^{4} W = 40 kW .$

Step 2 — divide by both efficiencies. Why this step? The pump wastes some as heat/friction, the motor+controller waste some more; the battery must cover all of it. $P_{elec} = \frac{40 kW}{0.6 \times 0.9} = \frac{40}{0.54} \approx 74.1 kW .$

Verify: Units of $P_{ideal}$ : $\frac{( kg/s ) ( Pa )}{kg/m ^{3}} = \frac{kg}{s} \cdot \frac{N/m ^{2}}{kg/m ^{3}} = \frac{N\cdotm}{s} = W .$ ✓ Correct dimensions. And $74.1 > 40$ as expected (real always costs more than ideal).

Ex 2 — Cell B: the unit trap

Worked example Ex 2 — pressure in bar, energy in Wh/kg

$\overset{m}{˙} = 5 kg/s$ , $Δ p = 90 bar$ , $ρ = 1000 kg/m^{3}$ , $η_{p} = 0.65$ , $η_{m} = 0.92$ , burn $t_{b} = 180 s$ , battery rated $e_{b} = 250 Wh/kg$ . Find $m_{batt}$ .

Forecast: the moment you leave bar and Wh unconverted you'll be off by factors of $1 0^{5}$ and $3600$ . Convert first.

Step 1 — convert $Δ p$ to pascals. Why this step? Our formula is SI; $1 bar = 1 0^{5} Pa$ . $Δ p = 90 \times 1 0^{5} = 9.0 \times 1 0^{6} Pa .$

Step 2 — convert $e_{b}$ to J/kg. Why this step? $1 Wh = 3600 J$ , because a watt for an hour is $1 \times 3600$ joules. $e_{b} = 250 \times 3600 = 9.0 \times 1 0^{5} J/kg .$

Step 3 — electrical power. Why this step? Battery mass needs power first. $P_{elec} = \frac{5 \times 9.0 \times 1 0 ^{6}}{1000 \times 0.65 \times 0.92} = \frac{4.5 \times 1 0 ^{4}}{0.598} \approx 7.525 \times 1 0^{4} W .$

Step 4 — battery mass. Why this step? Energy $= P_{elec} t_{b}$ , then divide by joules-per-kg. $m_{batt} = \frac{7.525 \times 1 0 ^{4} \times 180}{9.0 \times 1 0 ^{5}} \approx 15.0 kg .$

Verify: Energy $= 7.525 \times 1 0^{4} \times 180 \approx 1.35 \times 1 0^{7}$ J $= 1.35 \times 1 0^{7} /3600 \approx 3765$ Wh. At 250 Wh/kg that's $3765/250 \approx 15.1$ kg. ✓ Both routes agree.

Ex 3 — Cell C: the density effect

Worked example Ex 3 — same mass flow, different liquid

Push $\overset{m}{˙} = 6 kg/s$ at $Δ p = 1 0^{7} Pa$ , $η_{p} = 0.6$ , $η_{m} = 0.9$ . Compare liquid oxygen ( $ρ = 1140 kg/m^{3}$ ) vs liquid hydrogen ( $ρ = 71 kg/m^{3}$ ). Which pump needs more power?

Forecast: hydrogen is fluffy (low $ρ$ ). Same mass = far more volume. Pumping raises pressure per unit volume, so guess: hydrogen needs vastly more power.

Step 1 — LOX power. Why this step? Plug the dense case. $P_{LOX} = \frac{6 \times 1 0 ^{7}}{1140 \times 0.54} \approx 9.75 \times 1 0^{4} W \approx 97.5 kW .$

Step 2 — LH₂ power. Why this step? Same formula, tiny $ρ$ . $P_{LH2} = \frac{6 \times 1 0 ^{7}}{71 \times 0.54} \approx 1.565 \times 1 0^{6} W \approx 1565 kW .$

Step 3 — the ratio. Why this step? See the physics cleanly: power $\propto 1/ ρ$ , so $\frac{P _{LH2}}{P _{LOX}} = \frac{1140}{71} \approx 16.1.$

Verify: The ratio equals the density ratio $1140/71 = 16.06$ — independent of the shared factors ( $\overset{m}{˙}, Δ p, η$ cancel). ✓ This is why hydrogen turbopumps are giant power hogs.

Ex 4 — Cell D: limiting case, long burn

Worked example Ex 4 — what happens as burn time grows

Fixed $P_{elec} = 75 kW$ , $e_{b} = 9 \times 1 0^{5} J/kg$ . Tabulate $m_{batt}$ for $t_{b} = 30, 150, 600 s$ .

Forecast: $m_{batt} \propto t_{b}$ , a straight line through the origin. Tripling / quadrupling time does the same to mass.

Step 1 — general form. Why this step? Isolate the time dependence: $m_{batt} = \frac{P _{elec}}{e _{b}} t_{b} = \frac{75000}{9 \times 1 0 ^{5}} t_{b} = 0.0833 t_{b}$ (kg, with $t_{b}$ in s).

Step 2 — evaluate. Why this step? Fill the table. $t_{b} = 30 \Rightarrow 2.5 kg; t_{b} = 150 \Rightarrow 12.5 kg; t_{b} = 600 \Rightarrow 50.0 kg .$

Step 3 — read the limit. Why this step? As $t_{b} \to \infty$ , $m_{batt} \to \infty$ linearly. There is no plateau — batteries never "catch up," they just keep growing. This is the mathematical reason electric pumps are banned from giant, long-burn first stages and thrive on short upper-stage burns. See Tsiolkovsky rocket equation: dead battery mass hurts your mass ratio.

Verify: Slope check: $50/600 = 0.0833 = P_{elec} / e_{b}$ . ✓ Perfectly linear, origin-passing.

Ex 5 — Cell E: degenerate & zero inputs

Worked example Ex 5 — the boundary cases the formula must survive

Evaluate the master equations at each extreme. $Δ p = 1 0^{7}$ , $ρ = 1000$ , $t_{b} = 100$ , $e_{b} = 9 \times 1 0^{5}$ where relevant.

Forecast: zeros should collapse to zero power/mass; efficiency $\to 1$ should collapse $P_{elec}$ down to $P_{ideal}$ ; efficiency $\to 0$ should blow up. Check your intuition.

Step 1 — no flow, $\overset{m}{˙} = 0$ . Why this step? If nothing moves, the pump does no work. $P_{elec} = \frac{0 \cdot Δ p}{ρ η _{p} η _{m}} = 0 W .$ Physically: motor idles, battery drains nothing. ✓

Step 2 — no pressure rise, $Δ p = 0$ . Why this step? If exit pressure equals inlet, no energy is added per unit volume. $P_{elec} = \frac{m ˙ \cdot 0}{\dots} = 0 W .$ Free flow needs no pump work (ignoring friction). This is the Bernoulli's principle baseline: pressure rise is the energy term.

Step 3 — perfect machine, $η_{p} = η_{m} = 1$ . Why this step? Sanity: the loss factor should vanish. With $\overset{m}{˙} = 4$ : $P_{elec} = \frac{4 \times 1 0 ^{7}}{1000 \times 1 \times 1} = 4 \times 1 0^{4} = 40 kW = P_{ideal}$ . ✓ Electrical demand falls to the ideal floor.

Step 4 — dying motor, $η_{m} \to 0$ . Why this step? Understand the singularity. $P_{elec} = \frac{m ˙ Δ p}{ρ η _{p} η _{m}} \to \infty$ . A useless motor would demand infinite battery power — physically it just can't spin the pump at all. The formula's blow-up is the warning. ✓

Verify: Both zero-cases give $0$ ; $η = 1$ recovers $P_{ideal}$ ; $η \to 0$ diverges. All four match physical expectation.

Ex 6 — Cell F: real-world word problem

Worked example Ex 6 — a Rutherford-class stage with drop-batteries

A stage runs 9 pumps (LOX + RP-1 split), total propellant flow $\overset{m}{˙}_{tot} = 45 kg/s$ , $Δ p = 100 bar$ , effective $ρ = 1030 kg/m^{3}$ (mixture), $η_{p} = 0.6$ , $η_{m} = 0.9$ , burn $t_{b} = 160 s$ , $e_{b} = 0.95 \times 1 0^{6} J/kg$ . Half the batteries are dropped at the burn midpoint. Find (a) total electrical power, (b) battery mass installed, (c) battery mass carried to burnout. Compare to Rocket Lab Rutherford engine.

Forecast: power in the hundreds of kW; installed batteries ~tens of kg; dropping half the drained cells at mid-burn cuts carried mass noticeably.

Step 1 — total power. Why this step? Treat all pumps as one aggregate flow. $P_{elec} = \frac{45 \times 1 0 ^{7}}{1030 \times 0.6 \times 0.9} = \frac{4.5 \times 1 0 ^{9}}{556.2} \approx 8.09 \times 1 0^{5} W \approx 809 kW .$

Step 2 — installed battery mass (full burn). Why this step? You must physically carry enough to power the whole 160 s. $m_{batt} = \frac{P _{elec} t _{b}}{e _{b}} = \frac{8.09 \times 1 0 ^{5} \times 160}{0.95 \times 1 0 ^{6}} \approx 136.2 kg .$

Step 3 — carried-to-burnout mass with mid-burn drop. Why this step? Model: first-half cells (half the pack) are depleted by $t = 80$ s and jettisoned; only the second-half pack ( $\approx 68.1$ kg) is aboard at burnout. $m_{burnout} \approx \frac{1}{2} \times 136.2 \approx 68.1 kg .$

Verify: Energy check: $E = 8.09 \times 1 0^{5} \times 160 \approx 1.294 \times 1 0^{8}$ J. Mass $= 1.294 \times 1 0^{8} /0.95 \times 1 0^{6} \approx 136.2$ kg. ✓ Dropping dead cells is exactly the Rutherford trick that keeps burnout mass ratio favourable — see Tsiolkovsky rocket equation.

Ex 7 — Cell G: inverse solve

Worked example Ex 7 — you have a battery budget, find the max burn

Engineering constraint: only $m_{batt} = 20 kg$ of cells fit. With $\overset{m}{˙} = 5 kg/s$ , $Δ p = 95 bar$ , $ρ = 1000$ , $η_{p} = 0.62$ , $η_{m} = 0.90$ , $e_{b} = 0.9 \times 1 0^{6} J/kg$ , what is the longest burn $t_{b}$ you can power?

Forecast: rearrange the mass formula for $t_{b}$ ; expect a couple of hundred seconds.

Step 1 — solve the formula for $t_{b}$ . Why this step? Invert $m_{batt} = \frac{m ˙ Δ p t _{b}}{ρ η _{p} η _{m} e _{b}}$ : $t_{b} = \frac{m _{batt} ρ η _{p} η _{m} e _{b}}{m ˙ Δ p} .$

Step 2 — plug numbers ( $Δ p = 9.5 \times 1 0^{6}$ Pa). Why this step? Direct evaluation. $t_{b} = \frac{20 \times 1000 \times 0.62 \times 0.90 \times 0.9 \times 1 0 ^{6}}{5 \times 9.5 \times 1 0 ^{6}} = \frac{1.0044 \times 1 0 ^{10}}{4.75 \times 1 0 ^{7}} \approx 211.5 s .$

Verify: Forward-check with $t_{b} = 211.5$ : $P_{elec} = \frac{5 \times 9.5 \times 1 0 ^{6}}{1000 \times 0.62 \times 0.9} \approx 8.512 \times 1 0^{4}$ W; $m = \frac{8.512 \times 1 0 ^{4} \times 211.5}{0.9 \times 1 0 ^{6}} \approx 20.0$ kg. ✓ Round-trips to the budget.

Ex 8 — Cell H: exam twist — electric vs pressure-fed break-even

Worked example Ex 8 — when does the pump stop being worth it?

A Pressure-fed cycle tank storing propellant volume $V$ at chamber pressure $p_{c}$ has wall mass roughly $m_{wall} = k p_{c} V$ with $k = 6 \times 1 0^{- 4} kg/(Pa\cdotm^{3})$ (thin-shell scaling). An electric pump lets the tank drop to 2 bar but adds motor+battery mass $m_{add}$ . For $Δ p = 100 bar$ , $V = 1.5 m^{3}$ , and electric add-on $m_{add} = 45 kg$ , does the electric pump win?

Forecast: high-pressure tank mass is $k p_{c} V$ with $p_{c} = 1 0^{7}$ Pa — that's a big number. Guess: pump wins easily.

Step 1 — pressure-fed wall mass. Why this step? Tank must hold the full $1 0^{7}$ Pa. $m_{wall,PF} = 6 \times 1 0^{- 4} \times 1 0^{7} \times 1.5 = 9000 kg .$

Step 2 — pump-fed wall mass. Why this step? Tank now only holds 2 bar $= 2 \times 1 0^{5}$ Pa. $m_{wall,EP} = 6 \times 1 0^{- 4} \times 2 \times 1 0^{5} \times 1.5 = 180 kg .$

Step 3 — total comparison. Why this step? Add the electric penalty to the light tank. $m_{EP,total} = 180 + 45 = 225 kg vs m_{PF} = 9000 kg .$ Electric wins by $8775$ kg — a landslide. The tank-wall term dominates everything.

Step 4 — the twist: find break-even $V$ . Why this step? Exams ask "when would pressure-fed be lighter?" Set $k p_{c} V + 0 = k (2 \times 1 0^{5}) V + 45$ and solve — but note $k p_{c} V$ always exceeds the low-pressure tank, so break-even needs $m_{add}$ to exceed the saving: $k (p_{c} - 2 \times 1 0^{5}) V < m_{add} \Rightarrow V < \frac{45}{6 \times 1 0 ^{- 4} \times 9.8 \times 1 0 ^{6}} \approx 7.65 \times 1 0^{- 3} m^{3} .$ Only for tiny propellant volumes (< 7.65 litres) does the fixed 45 kg add-on outweigh tank savings.

Verify: At $V = 7.65 \times 1 0^{- 3}$ : saving $= 6 \times 1 0^{- 4} \times 9.8 \times 1 0^{6} \times 7.65 \times 1 0^{- 3} \approx 45.0$ kg $= m_{add}$ . ✓ Break-even confirmed; above this the pump always wins.

Recall Self-test: which cell is this?

A problem gives you $Δ p$ in bar, $e_{b}$ in Wh/kg, and asks for battery mass. Which cell, and what's the first move? Answer ::: Cell B (unit trap). First convert $Δ p \to$ Pa ( $\times 1 0^{5}$ ) and $e_{b} \to$ J/kg ( $\times 3600$ ) before touching the formula.

Back to the parent topic · related: Specific impulse (Isp), Turbopump-fed cycle (gas generator vs staged combustion).