Visual walkthrough — Electric pump-fed cycle — modern innovation

We link the parent Rocket Propulsion, and along the way Bernoulli's principle, Battery specific energy, and the Rocket Lab Rutherford engine.

Step 1 — What is "pressure", really?

WHAT. Before any formula, we pin down the one word everything rests on: pressure.

WHY. The whole derivation says "the pump adds energy to the fluid." To use that, we must know what pressure is in energy terms. If we skip this, $\Delta p$ is just a mystery letter.

PICTURE. Look at the figure. A piston of area $A$ pushes a liquid with force $F$. Pressure is force spread over that area:

$p=\underset{\text{push, newtons}}{\underbrace{F}}÷\underset{{\text{area, m}}^2}{\underbrace{A}}$

Now push the piston a tiny distance $d$. The work done (force × distance) is $F\cdot d$. The volume swept is $A\cdot d$. Divide work by that volume:

$\frac{Fd}{Ad}=\frac{F}{A}=p$

Step 2 — The pump raises pressure by $\Delta p$

WHAT. A pump takes liquid at low tank pressure $p_{\text{tank}}$ and hands it out at high pump-exit pressure $p_{\text{exit}}$. The rise is what costs energy:

$\Delta p=\underset{\text{after pump, high}}{\underbrace{p_{\text{exit}}}}-\underset{\text{before pump, low}}{\underbrace{p_{\text{tank}}}}$

WHY. The tank already supplied $p_{\text{tank}}$ for free. The pump only pays for the extra — the difference. Using the full $p_{\text{exit}}$ would double-count the free tank pressure. This "difference is what matters" idea is the same one behind Bernoulli's principle, where only pressure differences push fluid around.

PICTURE. The red column is the exit pressure, the short violet column is the tank pressure, and the orange gap between them is $\Delta p$ — the only part the motor pays for.

Step 3 — From one blob to a steady stream (power)

WHAT. A rocket doesn't pump one blob — it pumps a continuous river of propellant. So we ask for power: energy per second, not energy per blob.

WHY. Thrust is continuous; the motor runs continuously. Power (watts) is what a motor and battery are actually rated in. We convert "energy per blob" into "energy per second" by dividing by the time $dt$ each blob takes.

PICTURE. Watch the pipe: in one second, a slug of liquid of volume $\dot{V}$ (the volume flow rate, m³/s) slides through. Each such volume carried $\Delta p\cdot dV$ of energy, so per second:

$P_{\text{ideal}}=\frac{dW}{dt}=\frac{\Delta pdV}{dt}=\Delta p\underset{{\text{m}}^3\text{/s}}{\underbrace{\dot{V}}}$

Step 4 — Why density $\rho$ walks in (the term everyone forgets)

WHAT. Engineers quote propellant flow in kilograms per second ($\dot{m}$), not cubic metres per second, because thrust depends on mass thrown out. So we trade $\dot{V}$ for $\dot{m}$.

WHY. We need a bridge between "how many m³ per second" and "how many kg per second." That bridge is density $\rho$ — kilograms packed into each cubic metre:

$\rho =\frac{\text{mass}}{\text{volume}}\Rightarrow \dot{m}=\rho \dot{V}\Rightarrow \dot{V}=\frac{\dot{m}}{\rho }$

PICTURE. Two cubes of equal volume: dense liquid (heavy, many kg) vs light liquid (few kg). Same $\dot{V}$, different $\dot{m}$. The figure shows a heavy metre-cube and a light one on a balance.

Substitute $\dot{V}=\dot{m}/\rho$ into Step 3:

Step 5 — Real pumps and motors lose some (efficiencies)

WHAT. Insert two "leak factors": pump efficiency ${\eta }_p$ and motor efficiency ${\eta }_m$, each a number between 0 and 1.

WHY. Real pumps heat the fluid and rub in bearings; real motors and controllers waste heat. To deliver $P_{\text{ideal}}$ to the fluid you must supply more. Dividing by a number less than 1 makes the demand bigger — exactly what a loss should do.

PICTURE. A Sankey-style flow: battery power enters wide, a slice is lost at the motor (${\eta }_m$), another slice at the pump (${\eta }_p$), and the narrow survivor is the useful $P_{\text{ideal}}$ in the fluid.

Step 6 — Power becomes energy becomes kilograms

WHAT. Run the motor for the whole burn time $t_b$. Energy = power × time. Then convert energy to battery mass using specific energy $e_b$ (joules stored per kg).

WHY. A battery is rated by specific energy $e_b$ (J/kg). Mass = total joules ÷ joules-per-kg. This is the step where time enters — and time is what dooms long burns.

PICTURE. A staircase: watts → (× $t_b$) → joules → (÷ $e_b$) → kilograms. Each tread labelled.

$E=P_{\text{elec}}{t}_b,m_{\text{batt}}=\frac{E}{e_b}$

Step 7 — The edge cases (never leave the reader stranded)

WHAT. Check the extreme knobs so no scenario surprises you.

WHY. A formula you trust must behave sanely at its limits, not just at "nice" numbers.

PICTURE. Four mini-panels, each turning one knob to an extreme and showing where $m_{\text{batt}}$ goes.

• $\Delta p\to 0$ (tank already at chamber pressure — a Pressure-fed cycle): $P_{\text{ideal}}\to 0$, $m_{\text{batt}}\to 0$. No pumping work needed. The formula agrees: pumps only cost energy when they raise pressure.
• $t_b\to \infty$ (giant booster, long burn): $m_{\text{batt}}\to \infty$. Batteries swamp the vehicle — precisely why a turbopump wins for boosters.
• $e_b$ tiny (weak 2000s battery): $m_{\text{batt}}$ blows up. This is the historical reason electric pumps waited for ~250 Wh/kg lithium cells before becoming real.
• ${\eta }_p,{\eta }_m\to 1$ (ideal machines): demand collapses to $P_{\text{ideal}}$; $m_{\text{batt}}$ hits its floor. You can never beat this floor, only approach it.

The one-picture summary

One blob of liquid → pressure is energy/volume ($p$) → pump adds $\Delta p$ per blob → flow it steadily ($\dot{V}$) → rewrite via density ($\dot{V}=\dot{m}/\rho$) → account losses (${\eta }_p,{\eta }_m$) → run for $t_b$ → weigh with $e_b$. Two boxes fall out:

$P_{\text{elec}}=\frac{\dot{m}\Delta p}{\rho {\eta }_p{\eta }_m},m_{\text{batt}}=\frac{\dot{m}\Delta p{t}_b}{\rho {\eta }_p{\eta }_m{e}_b}$

Quick reveals: