Turbopump design — centrifugal pump, axial turbine stages, NPSH: Visual walkthrough

Step 1 — A blob of liquid on a spinning disk

WHAT. Picture a flat disk spinning about a central axle. Glue a tiny blob of liquid onto the disk at some distance $r$ (its radius) from the axle. As the disk turns, the blob is dragged in a circle.

WHY. A centrifugal pump is exactly this: an impeller (a spinning disk with curved blades) drags liquid around and flings it outward. Before any formula, we need one clean idea: how fast is a point at radius $r$ moving?

PICTURE. Look at the figure. The axle is the black dot. The blob sits at radius $r$ (yellow line). The blue arrow is its velocity — it points sideways, tangent to the circle, because the blob is going around, not in or out.

Figure — Turbopump design — centrifugal pump, axial turbine stages, NPSH

$u = turns/s in rad ω \times how far out r$

Step 2 — Splitting the fluid's velocity into two honest arrows

WHAT. The real liquid does two things at once: it swirls around the axis and it creeps outward toward the rim. We split its velocity into two perpendicular arrows.

WHY. Only one of those two motions carries angular momentum (spin about the axle). To do the bookkeeping cleanly we must isolate that piece — so we decompose.

PICTURE. The pink arrow $c$ is the liquid's actual velocity. We drop it onto two directions: the tangential (whirl) part $c_{u}$ along the circle, and the radial part $c_{r}$ pointing straight out. They form a right angle.

Step 3 — Angular momentum: the quantity the blade actually changes

WHAT. For a spinning object, the "amount of spin" bookkeeping quantity is angular momentum = (mass) × (radius) × (tangential speed). We track how much of it the fluid carries in and out.

WHY. Newton's law for rotation says: to change a thing's angular momentum you must apply a torque (a twisting force). The impeller is the only thing twisting the fluid, so the torque it supplies equals the fluid's change in angular momentum per second.

PICTURE. Two dashed circles: the inner eye at radius $r_{1}$ where fluid enters, and the outer rim at radius $r_{2}$ where it leaves. Each carries its own whirl arrow, $c_{u 1}$ and $c_{u 2}$ .

For a steady flow, mass passes through at a rate $\overset{m}{˙}$ (kg per second). The angular momentum entering per second is $\overset{m}{˙} r_{1} c_{u 1}$ ; leaving is $\overset{m}{˙} r_{2} c_{u 2}$ . The torque supplied is the difference:

$T = \overset{m}{˙} (spin carried OUT at rim r_{2} c_{u 2} - spin carried IN at eye r_{1} c_{u 1})$

Step 4 — Torque becomes power: multiply by $ω$

WHAT. A torque $T$ applied by a shaft turning at $ω$ delivers power $P = T ω$ (energy per second).

WHY. We ultimately want energy given to the fluid, not torque. Power is the bridge: it's torque times how fast you turn, just as straight-line power is force times speed.

PICTURE. The shaft turning at $ω$ , torque arrows on the fluid, and a little energy meter filling up per second. Notice $ω r_{1} = u_{1}$ and $ω r_{2} = u_{2}$ — the radii fold into blade speeds.

$P = T ω = \overset{m}{˙} ω (r_{2} c_{u 2} - r_{1} c_{u 1}) = \overset{m}{˙} (ω r_{2} = u_{2} u_{2} c_{u 2} - ω r_{1} = u_{1} u_{1} c_{u 1})$

Step 5 — From power to head: divide by weight-flow

WHAT. Take the power, split it over the fluid stream, and convert it into a height of liquid column called the head $H$ .

WHY. Energy per unit mass is $P / \overset{m}{˙}$ . Dividing once more by $g$ (gravity's strength, $9.81 m/s^{2}$ ) turns "energy per kilogram" into "metres of liquid" — because lifting 1 kg by 1 m costs exactly $g$ joules. Head is a friendly, density-free way to size a pump.

PICTURE. A tall column of liquid of height $H$ ; the energy the pump adds per kilogram equals the energy that column stores per kilogram. The ruler on the side reads $H$ in metres.

$\frac{P}{m ˙} = u_{2} c_{u 2} - u_{1} c_{u 1} ⟹ H = \frac{1}{g} \cdot \frac{P}{m ˙}$

Step 6 — The clean design choice: no inlet swirl

WHAT. Feed the liquid straight into the eye — pointed along the axis or radially inward, with no pre-spin. Then $c_{u 1} = 0$ .

WHY. If the fluid arrives already whirling, some of the exit whirl was not added by this impeller — it was free swirl we brought along. Killing it ( $c_{u 1} = 0$ ) means every bit of whirl at the rim is genuine work this stage did, which both simplifies the formula and maximizes head for a given rim speed.

PICTURE. Left: swirling inflow (messy, part of the head is "borrowed"). Right: clean radial inflow, $c_{u 1} = 0$ , so the whole $u_{2} c_{u 2}$ term is ours.

$H \approx \frac{u _{2} c _{u 2}}{g} (setting c_{u 1} = 0)$

Step 7 — Edge cases: what the formula does at the extremes

WHAT. We test the formula at three corners so the reader never meets an unshown scenario.

WHY. A formula you trust is one you've watched behave at its limits: nothing spinning, no whirl added, and the density question.

PICTURE. Three mini-panels: (a) disk stopped, (b) fluid slips straight out with no whirl, (c) two different liquids on identical pumps.

(a) $ω = 0$ (disk not turning). Then $u_{2} = ω r_{2} = 0$ , so $H = 0$ . A pump that doesn't spin adds no head. ✔ Sensible.
(b) $c_{u 2} = 0$ (fluid leaves with no whirl). Even if the disk spins fast, $u_{2} c_{u 2} = 0 \Rightarrow H = 0$ . Whirl must actually be imparted; a bald spinning disk that lets fluid slide off radially does nothing. This is why blades are curved — they force $c_{u 2} \neq = 0$ .
(c) Two different liquids. The head $H = u_{2} c_{u 2} / g$ contains no density $ρ$ . Run LOX or water at the same speed and you get the same head in metres. Only when you convert to a pressure rise does density enter, via $Δ p = ρ g H$ (heavier liquid $\Rightarrow$ bigger pressure for the same metres). This is the subtle point the parent note steel-mans.

The one-picture summary

Here is the whole derivation on a single board: blade speed $u = ω r$ born at Step 1 → velocity split into whirl $c_{u}$ at Step 2 → angular-momentum torque $T = \overset{m}{˙} Δ (r c_{u})$ at Step 3 → power $P = T ω = \overset{m}{˙} Δ (u c_{u})$ at Step 4 → head $H = Δ (u c_{u}) / g$ at Step 5 → the clean $H \approx u_{2} c_{u 2} / g$ at Step 6.

Recall Feynman retelling — say it back in plain words

A pump is a spinning disk with curved blades. The rim moves fast — that speed is $u = ω r$ , just "how far out" times "how fast it turns." The liquid does two things: it swirls around ( $c_{u}$ ) and it creeps outward ( $c_{r}$ ), but only the swirl carries spin about the axle, so that's all we track. To make the fluid swirl more on the way out than on the way in, the blade must twist it — and twisting means torque, which per second is $\overset{m}{˙}$ times the change in $r c_{u}$ . Multiply torque by turning speed $ω$ and you get power; the $ω r$ 's become blade speeds $u$ , so power per kilogram is the change in $u c_{u}$ . Divide by gravity $g$ and that energy-per-kilogram becomes a height of liquid — the head $H$ . Feed the fluid in with no pre-swirl and only the rim term survives: $H \approx u_{2} c_{u 2} / g$ . Because the exit whirl chases the rim speed, head grows like rim speed squared, so we spin these things insanely fast. And notice: not a single density anywhere — the head in metres is the same for hydrogen or oxygen; density only shows up when you ask for pressure.

Recall Quick self-test

Why does only $c_{u}$ (not $c_{r}$ ) appear in the head? ::: The radial part points through the axle, so it carries zero angular momentum about the axis; only tangential whirl twists the fluid. What replaces $ω r_{2}$ in the power equation and why? ::: The blade speed $u_{2}$ — it is exactly $ω r_{2}$ , the measurable rim-tip speed. With $c_{u 1} = 0$ , what is the head? ::: $H \approx u_{2} c_{u 2} / g$ . If rim speed doubles, head does what? ::: Roughly quadruples, since $H \sim u_{2}^{2} / g$ . Does head depend on the liquid's density? ::: No — head in metres is density-free; only the pressure rise $Δ p = ρ g H$ scales with density.

Prerequisite threads: Bernoulli's Principle (energy bookkeeping in fluids), Cavitation (what limits the inlet), and the broader Euler Turbomachinery Equation this derivation is a special case of. Downstream this head sets the achievable Chamber Pressure and Thrust and thus Specific Impulse, feeding cycle choices like the Gas Generator Cycle and Staged Combustion Cycle.