Turbopump design — centrifugal pump, axial turbine stages, NPSH: Worked examples

Every symbol, defined before we use it

Nothing below is assumed. Read these once and the whole page is self-contained.

Definition Blade speed, whirl, and head

Blade speed $u = ω r$ = how fast a point on the wheel travels, in metres per second. Here $ω$ (omega) is the spin rate in radians per second (one full turn $= 2 π$ radians), and $r$ is the radius of that point. We write $u_{1} = ω r_{1}$ at the eye and $u_{2} = ω r_{2}$ at the rim.
Whirl $c_{u}$ = the sideways (tangential) part of the liquid's absolute velocity — the part that circles around the axis. Only this part carries angular momentum, so only this part changes the head. $c_{u 1}$ is the whirl going in at the eye; $c_{u 2}$ is the whirl coming out at the rim.
Gravitational acceleration $g = 9.81 m/s^{2}$ — the constant that converts "energy per unit weight" into a height. Every head and NPSH formula on this page divides by this same $g$ .
Head $H$ = energy given to each kilogram-weight of liquid, expressed as a height of that same liquid (metres): $H = \frac{u _{2} c _{u 2} - u _{1} c _{u 1}}{g}$

Definition The NPSH vocabulary (spelled out in full)

NPSH = Net Positive Suction Head — the spare pressure margin at the pump inlet, above the point where the liquid would boil, written as a height of liquid (metres). It comes in two flavours whose subscripts you must not mix up:

$NPSH_{A}$ — the A is for Available: what the plumbing actually delivers to the inlet.
$NPSH_{R}$ — the R is for Required: what the impeller demands to avoid boiling, set by its geometry.
Safe operation needs $NPSH_{A} > NPSH_{R}$ . The available margin is built from three named pieces: $NPSH_{A} = pressure margin \frac{p _{tank} - p _{v}}{ρ g} + elevation z - friction h_{loss}$
$p_{tank}$ = pressure in the propellant tank; $p_{v}$ = vapor pressure, the pressure at which the liquid boils at its temperature. The margin is how far the tank sits above boiling.
$z$ = tank elevation difference, the height of the liquid surface above the pump. Sign convention: $z > 0$ when the tank is above the pump (gravity helps), $z < 0$ when the tank is below (suction lift, gravity fights).
$h_{loss}$ = head loss, the friction the flow loses travelling down the feed line; always subtracted because real pipes only take margin away.

The scenario matrix

Every problem this topic can pose falls in one of these cells. Each example below is tagged with the cell(s) it covers.

Cell	What makes it special	Covered by
A. Standard pump	normal spin, no inlet swirl ( $c_{u 1} = 0$ )	Ex 1
B. Pre-swirl (sign of $c_{u 1}$ )	inlet whirl present, positive vs negative	Ex 2
C. Degenerate: $ω = 0$ or $r_{1} = r_{2}$	wheel stopped / no radius change → zero head	Ex 3
D. Density swap (invariance trap)	same pump, LOX vs LH₂ — head vs pressure	Ex 4
E. Turbine sign flip	Euler run "backwards", work out	Ex 5
F. Power-match coupling	turbine must equal pump demand	Ex 6
G. NPSH normal + limiting	positive margin, and the boiling edge	Ex 7
H. NPSH degenerate	$p_{tank} = p_{v}$ , suction lift ( $z < 0$ )	Ex 8
I. Word problem (real mission)	choose the tool from a story	Ex 9
J. Exam twist	"doubling speed" scaling / hidden unit trap	Ex 10

Ex 1 — Cell A: the plain-vanilla pump

Forecast: Guess — will $H$ be closer to $1, 000$ m or $5, 000$ m? (Recall $H \sim u_{2}^{2} / g$ , and $u_{2}$ will be a few hundred m/s.)

The exit velocity triangle for this example is drawn here — the amber whirl arrow is $c_{u 2}$ , the cyan arrow along the spin direction is the blade speed $u_{2}$ :

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

$ω = 2 π N /60 = 2 π (25000) /60 = 2618 rad/s$ . Why this step? rpm counts whole turns per minute; radians per second is what $u = ω r$ needs (one turn $= 2 π$ radians, one minute $= 60$ s).
$u_{2} = ω r_{2} = 2618 \times 0.08 = 209.4 m/s$ . Why this step? Blade speed = the actual metres-per-second the rim travels; it sets the ceiling for whirl.
$c_{u 2} = 0.75 \times 209.4 = 157.1 m/s$ . Why this step? The blades cannot force more whirl than the rim's own speed; a fraction below $u_{2}$ is realistic.
$H = \frac{u _{2} c _{u 2}}{g} = \frac{209.4 \times 157.1}{9.81} = 3353 m$ . Why this step? Euler head with $c_{u 1} = 0$ — all whirl is added by this stage, and we divide by $g = 9.81$ to turn energy-per-weight into metres.

Verify: Units: $(m/s) (m/s) / (m/s^{2}) = m$ . ✓ Magnitude sits between the two guesses, so the $u_{2}^{2} / g$ scaling holds. ✓

Ex 2 — Cell B: inlet pre-swirl, both signs

Forecast: Which case gives more head — swirl with the wheel, or against it?

The figure below shows how the same $c_{u 1}$ term shifts the head up or down from the no-swirl baseline:

Full Euler: $H = \frac{u _{2} c _{u 2} - u _{1} c _{u 1}}{g}$ . Why this step? With $c_{u 1} \neq = 0$ we can no longer drop the second term; it subtracts the angular momentum the fluid already carried in.
Case (i): $H = \frac{209.4 ( 157.1 ) - 78.5 ( + 40 )}{9.81} = \frac{32897 - 3140}{9.81} = 3033 m$ . Why this step? Positive $c_{u 1}$ means the fluid already whirled with the wheel, so the impeller does less new work — head drops.
Case (ii): $H = \frac{32897 - 78.5 ( - 40 )}{9.81} = \frac{32897 + 3140}{9.81} = 3673 m$ . Why this step? Counter-swirl means the impeller must first reverse the fluid's whirl, then drive it forward — more work, more head.

Verify: The no-swirl answer (Ex 1) was $3353$ m; case (i) is below it and case (ii) above it, straddling it symmetrically ( $\pm 320$ m). ✓ That symmetry is exactly the $- u_{1} c_{u 1}$ term flipping sign. ✓ This is why designers choose $c_{u 1} = 0$ — it's the clean middle that maximises head per unit $u_{2}$ without needing counter-swirl hardware.

Ex 3 — Cell C: degenerate inputs (zero head)

Forecast: Do these give a small head or exactly zero?

(a) $u_{2} = ω r_{2} = 0$ , so $H = \frac{0 \cdot c _{u 2} - 0 \cdot c _{u 1}}{g} = 0$ . Why this step? No blade motion = no work done on the fluid. Head is energy added; a still wheel adds none.
(b) $u_{2} = u_{1}$ and $c_{u 2} = c_{u 1}$ , so $H = \frac{u ( c _{u} ) - u ( c _{u} )}{g} = 0$ . Why this step? Euler head is a difference of angular momentum flux. If nothing about the whirl or radius changes between in and out, the difference is zero — a pipe, not a pump.

Verify: Both give exactly $0$ m, and both make physical sense: a pump only works by changing $u c_{u}$ across the impeller. ✓ This is the degenerate boundary of every Euler problem.

Ex 4 — Cell D: the density-swap invariance trap

Here $ρ$ (rho) is the density — mass per cubic metre of the liquid.

Forecast: Same pump, same speed — does the head change between liquids? Does the pressure change?

Head is unchanged: $H = 3353 m$ for both. Why this step? $H = u_{2} c_{u 2} / g$ contains no $ρ$ . Head is energy per unit weight, so it's the same column height regardless of what's flowing.
LOX: $Δ p = ρ g H = 1140 \times 9.81 \times 3353 = 3.75 \times 1 0^{7} Pa = 375 bar$ . Why this step? Multiply head (energy/weight) by $ρ g$ (weight/volume) to get energy/volume = pressure.
LH₂: $Δ p = 71 \times 9.81 \times 3353 = 2.34 \times 1 0^{6} Pa = 23.4 bar$ . Why this step? Same head, but light hydrogen weighs almost nothing per litre, so the same column is a tiny pressure.

Verify: Ratio of pressures $= ρ_{LOX} / ρ_{LH_{2}} = 1140/71 = 16.1$ , and $375/23.4 = 16.0$ . ✓ Head identical, pressure scales with density. Bubbles forming here would be Cavitation — boiling caused by low pressure rather than heat, the very thing NPSH guards against.

Ex 5 — Cell E: Euler run backwards (turbine work)

Forecast: For a pump we added whirl (work in). Here whirl drops across the rotor — will $w$ come out positive (work extracted)?

$Δ c_{u} = c_{u, in} - c_{u, out} = 520 - 120 = 400 m/s$ . Why this step? The turbine harvests the loss of whirl; that drop is what turns into shaft work.
$w = u Δ c_{u} = 350 \times 400 = 1.40 \times 1 0^{5} J/kg$ . Why this step? Same $u Δ c_{u}$ Euler relation as the pump — only the sign convention flips because the fluid gives momentum to the blade instead of receiving it. (This $w = u Δ c_{u}$ is exactly the Euler turbomachinery relation the parent note derived: work per kg = blade speed times the change in whirl.)

Verify: Units $(m/s)^{2} = m^{2} / s^{2} = J/kg$ . ✓ Positive → work is genuinely extracted, correct for a turbine. ✓

Ex 6 — Cell F: the power-match coupling

Here $\overset{m}{˙}$ (m with a dot) means mass flow rate — kilograms passing per second; $η$ (eta) is an efficiency between 0 and 1.

Forecast: Pump power will be a few MW. Will the gas flow be a large fraction of $150$ kg/s, or a small sip?

$P_{pump} = \frac{m ˙ _{prop} Δ p}{ρ η _{p}} = \frac{150 \times 1.8 \times 1 0 ^{7}}{1140 \times 0.68} = 3.48 \times 1 0^{6} W = 3.48 MW$ . Why this step? $Δ p / ρ$ is the ideal energy per kg; dividing by $η_{p}$ accounts for real pump losses.
Shaft must supply $P_{turb} = P_{pump} / η_{mech} = 3.48/0.98 = 3.55 MW$ . Why this step? Bearing/seal friction on the shared shaft means the turbine must over-produce slightly.
$\overset{m}{˙}_{g} = \frac{P _{turb}}{w} = \frac{3.55 \times 1 0 ^{6}}{1.40 \times 1 0 ^{5}} = 25.4 kg/s$ . Why this step? Every kg of gas yields $w$ joules; divide the power demand by that to get kg/s of gas needed.

Verify: Gas flow $25.4$ kg/s is about $17%$ of propellant flow — plausibly high (this is a work-heavy stage). ✓ Units: $W / (J/kg) = kg/s$ . ✓ In a Gas Generator Cycle this gas is burned in a small side-chamber and dumped overboard once it has spun the turbine; a Staged Combustion Cycle instead feeds it on into the main chamber so none of that mass is wasted.

Ex 7 — Cell G: NPSH normal, then the boiling edge

Forecast: Will the $+ z - h_{loss}$ terms matter much, or does the pressure term dominate?

The margin picture (available bars vs the required amber line) is here:

$\frac{p _{tank} - p _{v}}{ρ g} = \frac{( 4 - 1.8 ) \times 1 0 ^{5}}{1140 \times 9.81} = \frac{2.2 \times 1 0 ^{5}}{11183} = 19.67 m$ . Why this step? This is the driving pressure margin — how far tank pressure sits above boiling, in metres of LOX (using $g = 9.81$ ).
$NPSH_{A} = 19.67 + 1.5 - 0.8 = 20.37 m$ . Why this step? Add gravity head $z$ from the elevated tank, subtract friction loss $h_{loss}$ (energy balance from tank surface to inlet).
Safety: $20.37 > 18$ , margin $= 2.4$ m. Safe, but tight. Why this step? The rule is $NPSH_{A} > NPSH_{R}$ ; below it the eye pressure drops under $p_{v}$ and bubbles form.
(b) Edge: set $NPSH_{A} = 18$ . Then $\frac{p _{tank} - p _{v}}{ρ g} = 18 - 1.5 + 0.8 = 17.3 m$ , so $p_{tank} - p_{v} = 17.3 \times 11183 = 1.935 \times 1 0^{5}$ , giving $p_{tank} = 1.8 \times 1 0^{5} + 1.935 \times 1 0^{5} = 3.73 \times 1 0^{5} Pa = 3.73 bar$ . Why this step? We solve the safety equality backwards to find the lowest tank pressure that still clears $NPSH_{R}$ .

Verify: At $p_{tank} = 3.73$ bar, recompute: $\frac{( 3.73 - 1.8 ) \times 1 0 ^{5}}{11183} = 17.26$ , $+ 1.5 - 0.8 = 17.96 \approx 18$ m. ✓ Any tank pressure below $3.73$ bar → boiling at the inlet. ✓

Ex 8 — Cell H: NPSH degenerate cases

Forecast: In (a) is there any margin left at all? In (b), can a below-pump tank ever be safe?

(a) Pressure term $= \frac{p _{v} - p _{v}}{ρ g} = 0$ , so $NPSH_{A} = 0 + 1.5 - 0.8 = 0.7 m$ . Why this step? When $p_{tank} = p_{v}$ the liquid is already at its boiling edge; the only margin left is gravity head minus friction. Almost any $NPSH_{R}$ will fail — this is why tanks are pressurised above $p_{v}$ .
(b) $NPSH_{A} = 3.0 + (- 1.2) - 0.8 = 1.0 m$ . Why this step? A tank below the pump makes $z$ negative (our sign convention) — gravity now fights the flow and eats margin. Only a large pressure term keeps NPSH positive.

Verify: (a) $0.7$ m and (b) $1.0$ m are both small and positive — exactly the marginal region where an inducer (a slender axial screw ahead of the impeller that gently pre-pressurises the flow, lowering $NPSH_{R}$ ) is added. ✓ Sign of $z$ handled correctly: raising the tank helps ( $+$ ), lowering it hurts ( $-$ ). ✓

Ex 9 — Cell I: real-mission word problem

Forecast: Same shaft speed, same target pressure — will the two pumps need the same head, or wildly different?

Head each pump must reach: $H = \frac{Δ p}{ρ g}$ . Why this step? We invert $Δ p = ρ g H$ because the requirement is a pressure, but the pump's job is measured in head.
LH₂: $H = \frac{2.7 \times 1 0 ^{7}}{71 \times 9.81} = 38, 760 m$ . Why this step? Feather-light hydrogen needs a monstrous column to make $270$ bar.
LOX: $H = \frac{2.7 \times 1 0 ^{7}}{1140 \times 9.81} = 2414 m$ . Why this step? Dense oxygen makes the same pressure with a $16 \times$ shorter column.
Ratio $= 38760/2414 = 16.1$ . The LH₂ pump needs $16 \times$ the head → many more stages or a much bigger rim → it is the tall multistage machine. Why this step? Since $H \sim u_{2}^{2} / g$ at fixed $ω$ , needing $16 \times$ head means either stacking $\sim 16$ stages or a larger radius — the physical reason LH₂ turbopumps are the giants.

Verify: Ratio equals the density ratio $1140/71 = 16.1$ . ✓ Head answers scale inversely with $ρ$ , as they must. ✓ This is why a rise in Chamber Pressure and Thrust (needing higher $Δ p$ ) demands ever-taller hydrogen pumps.

Ex 10 — Cell J: the exam twist (scaling + unit trap)

Forecast: Doubling RPM — does head double, or do something more dramatic?

(a) $H \propto u_{2}^{2} \propto N^{2}$ (since $c_{u 2} \propto u_{2} \propto N$ ). Why this step? Euler head with fixed whirl fraction is $H = \frac{k u _{2}^{2}}{g}$ ; both $u_{2}$ and $c_{u 2}$ scale with $N$ , so $H$ scales as $N^{2}$ .
$H_{new} = 2500 \times (40000/20000)^{2} = 2500 \times 4 = 10, 000 m$ . Why this step? Doubling speed quadruples head — the headline non-linearity of centrifugal pumps.
(b) The student forgot to convert rpm → rad/s. True $ω = 2 π \times 40000/60 = 4189 rad/s$ , so $u_{2} = 4189 \times 0.06 = 251.3 m/s$ — not $40000 \times 0.06 = 2400$ m/s (impossibly supersonic). Why this step? $u = ω r$ demands radians per second; using raw rpm inflates blade speed $\approx 9.5 \times$ and head $\approx 90 \times$ — a classic exam-killer.

Verify: (a) factor $4$ from $N^{2}$ . ✓ (b) ratio of wrong to right $ω$ is $40000/4189 = 9.55 = 60/ (2 π)$ . ✓ The unit slip is exactly the missing $2 π /60$ . ✓

Recall Self-test before you move on

Doubling rim speed multiplies head by ::: four (head $\propto u_{2}^{2}$ ) The Euler term that vanishes with no inlet swirl ::: $u_{1} c_{u 1}$ Same pump, LOX vs LH₂ — what stays the same? ::: the head $H$ in metres (pressure differs by $ρ$ ) Turbine vs pump Euler: what flips? ::: the sign — whirl is removed (work out) not added (work in) What does the "A" vs "R" in NPSH stand for? ::: Available (delivered by plumbing) vs Required (demanded by impeller) Tank moved below the pump makes $z$ ::: negative, shrinking $NPSH_{A}$ The fix when $NPSH_{A}$ is marginal ::: add an inducer (lowers $NPSH_{R}$ )