Turbopump design — centrifugal pump, axial turbine stages, NPSH: Question bank

Before we start, here is every symbol this page leans on, defined in plain words so nothing appears unearned. The figures below draw the two geometric objects everything rests on: the impeller velocity triangle and the NPSH head stack.

Definition Symbol glossary (read once, refer back)

$H$ = head, energy given to fluid per unit weight, measured in metres of that fluid's own column.
$u_{2} = ω r_{2}$ = blade speed at the impeller rim (rim radius $r_{2}$ times spin rate $ω$ ).
$c_{u 2}$ = exit whirl, the sideways (tangential) part of the fluid's exit speed; $c_{u 1}$ is the same thing at inlet.
Sign convention: tangential speed $c_{u}$ is counted positive when it points the same way the blade moves (co-rotating swirl) and negative when it points against the blade. So added, forward swirl is a positive $c_{u}$ ; pre-swirl fighting the blade is negative.
$Δ c_{u} = c_{u, in} - c_{u, out}$ = the change in whirl the fluid undergoes across a row. Positive $Δ c_{u}$ means whirl was removed (turbine takes work out).
$Δ p = ρ g H$ = pressure rise, energy per unit volume; $ρ$ = fluid density.
$\overset{m}{˙}$ = mass flow rate, kilograms of fluid passing per second (subscript $g$ for turbine gas, $prop$ for pumped propellant).
$η_{pump}, η_{t}, η_{mech}$ = efficiencies (a fraction between 0 and 1) of the pump, the turbine, and the shaft/bearings — how much of the ideal energy actually gets through.
$c_{p}$ = specific heat of the turbine gas: joules needed to warm 1 kg of it by 1 K at constant pressure. It converts a gas temperature into stored energy per kg.
$T_{0, in}$ = total (stagnation) inlet temperature of the turbine gas — the temperature the gas would have if brought to rest, i.e. its full thermal energy content at entry.
$π_{t}$ = turbine pressure ratio, inlet pressure divided by outlet pressure across the turbine — how much the gas is allowed to expand.
$γ$ = specific-heat ratio of the gas ( $c_{p} / c_{v}$ ), the number that governs how a gas cools as it expands.
$p_{v}$ = vapor pressure, the pressure at which the liquid boils at its current temperature.
$NPSH_{A}$ / $NPSH_{R}$ = suction-head available from the plumbing vs. required by the impeller geometry.

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

The left of the figure is the impeller: the rim moves at $u_{2}$ , the fluid leaves with whirl $c_{u 2}$ , and their product (per $g$ ) is the head. The right is the NPSH stack: tank-pressure margin plus gravity height minus friction gives the head available at the eye.

This second figure shows why $c_{u 2}$ tracks $u_{2}$ : for a fixed blade exit angle $β_{2}$ , the exit velocity triangle simply scales up when the rim spins faster — the whirl side $c_{u 2}$ grows in proportion to $u_{2}$ (allowing for a near-constant "slip" gap). That is the assumption behind writing $H \sim u_{2}^{2} / g$ .

True or false — justify

True or false: A centrifugal pump running at fixed RPM produces the same head (in metres) for water as for liquid oxygen.

True — Euler head

H = u_{2} c_{u 2} / g

contains no density term, so the metres of column are identical; only the pressure rise

Δ p = ρ g H

differs because it scales with

ρ

True or false: Doubling the impeller rim speed roughly doubles the pressure rise.

False — because the exit blade angle is fixed,

c_{u 2}

scales roughly linearly with

u_{2}

, so

H = u_{2} c_{u 2} / g \sim u_{2}^{2} / g

; doubling rim speed therefore quadruples the rise. That quadratic is why turbopumps run at tens of thousands of RPM.

True or false: The turbine and the pump can be sized independently.

False — they share one shaft, so the turbine must produce exactly

P_{turb} = P_{pump} / η_{mech}

; this power-match constraint couples them and sizes the gas generator flow.

True or false: A high

NPSH_{A}

value guarantees no cavitation.

False on its own — safety needs

NPSH_{A} > NPSH_{R}

; a huge

NPSH_{A}

with an even larger geometry-driven

NPSH_{R}

still boils at the eye.

True or false: An inducer works by raising

NPSH_{A}

False — the inducer lowers

NPSH_{R}

by gently pre-pressurizing the flow, so the same available head now clears the requirement; the plumbing-side

NPSH_{A}

is untouched.

True or false: The Euler relation

u Δ c_{u}

describes both the pump and the turbine.

True — it is the same angular-momentum bookkeeping (Euler Turbomachinery Equation); the pump adds whirl (work in,

u_{2} c_{u 2} - u_{1} c_{u 1}

) while the turbine removes it (work out,

u Δ c_{u}

), differing only in sign/direction.

True or false: Cavitation's only harmful effect is that it reduces the head.

False — head collapse does hurt performance, but the worse damage is bubbles collapsing on the blades, causing pitting erosion and vibration that physically destroy the impeller. So "only" makes the statement false.

True or false: LH₂ pumps show enormous NPSH values (hundreds of metres) because their inlets are unusually safe.

False — the huge metres come from dividing a modest pressure margin by a tiny density (

ρ_{L H_{2}} \approx 71

); the actual pressure margin is small, which is exactly why LH₂ is cavitation-prone and needs inducers.

Spot the error

"Head

H = u_{2} c_{u 2} / g

, so to raise pressure for a denser propellant we don't need to change anything."

The head stays the same, but the pressure rise

Δ p = ρ g H

grows with density, so the pump must absorb more shaft power (

P = \overset{m}{˙} g H / η

) for the denser fluid at the same head.

"We choose axial stages for the pump because axial rows stack in series and extract more energy."

The reasoning is right but the machine is swapped — turbines are axial (low-density, high-velocity gas, easy to stack). The pump is centrifugal because a single impeller flings dense liquid to high pressure compactly.

"To maximize head we should add lots of inlet swirl, so

c_{u 1}

is large."

Wrong sign of intuition — inlet whirl

c_{u 1}

subtracts in

H = (u_{2} c_{u 2} - u_{1} c_{u 1}) / g

. Designers set

c_{u 1} = 0

(radial/axial entry) so all whirl is added by this stage, maximizing head.

"Bernoulli says fast liquid at the eye has high pressure, so the eye is the safest spot from cavitation."

Backwards — Bernoulli's Principle gives fast flow low static pressure, so the eye is the most likely place to drop below

p_{v}

and boil. That is the whole reason NPSH is defined at the inlet.

"NPSH

_{A}

only involves tank pressure and vapor pressure."

Incomplete —

NPSH_{A} = \frac{p _{tank} - p _{v}}{ρ g} + z - h_{loss}

also adds the static height

z

of tank above pump and subtracts feedline friction

h_{loss}

"The turbine power depends only on the gas mass flow."

No —

P_{turb} = \overset{m}{˙}_{g} c_{p} T_{0, in} η_{t} (1 - π_{t}^{- (γ - 1) / γ})

also scales with inlet temperature and pressure ratio. Why $T_{0, in}$ matters:

c_{p} T_{0}

is the energy stored per kg of hot gas, so hotter gas simply carries more extractable energy. Why $π_{t}$ matters: a larger pressure ratio lets the gas expand further, and it can only surrender work while it is expanding — more expansion, more work per kg.

"Since head has no density, a water test perfectly predicts the LOX pump's shaft power."

The head transfers, but power

P = \overset{m}{˙} g H / η = \overset{m}{˙} Δ p / (ρ η)

depends on mass flow and

Δ p

, both of which change with

ρ

; you must scale the power measurement.

Why questions

Why is turbine work written with

Δ c_{u} = c_{u, in} - c_{u, out}

rather than a single velocity?

Because work comes from the change in the fluid's whirl as it crosses the rotor; the rotor harvests only the angular momentum the gas gives up, which is the difference in tangential speed.

Why does a turbopump exist at all instead of just pressurizing the tanks?

Pressure-fed tanks would need thick, heavy walls to hold chamber-level pressure; the turbopump lets tanks stay light and low-pressure while a small turbine spends a little propellant to raise pressure 100× at the pump.

Why is only the tangential fluid velocity used in the Euler head, not the radial part?

Angular momentum about the shaft axis is carried only by motion around the axis; the radial component points through the axis and contributes zero moment, so it cannot exchange torque with the impeller.

Why does the power-match constraint size the gas generator?

The turbine must supply exactly the pump's demanded power; fixing pump power fixes the required

\overset{m}{˙}_{g} c_{p} T_{0, in} η_{t} (1 - π_{t}^{- (γ - 1) / γ})

, which sets how much propellant the gas generator must burn.

Why do staged-combustion engines tolerate higher chamber pressures than gas-generator engines?

Staged combustion routes the turbine exhaust into the chamber rather than dumping it, so no propellant is wasted and more pump power can be justified — enabling higher pump

Δ p

and higher chamber pressure for better Specific Impulse.

Why does dividing power by

g

(not just by mass) give "head" in metres?

Power/

\overset{m}{˙}

is energy per unit mass; dividing further by

g

converts to energy per unit weight, which has units of length — the height of a fluid column with that same potential energy.

Why is

Δ p = ρ g H

obtained by multiplying head by

ρ g

Head is energy per unit weight; multiplying by weight-per-unit-volume (

ρ g

) converts to energy per unit volume, and energy per volume has the units of pressure.

Edge cases

What is the head if the impeller is not spinning (

ω = 0

Then

u_{2} = ω r_{2} = 0

, so

H = u_{2} c_{u 2} / g = 0

— a stationary impeller adds no energy; the "pump" is just a stationary passage.

What happens to

NPSH_{A}

if the tank pressure exactly equals the vapor pressure?

The pressure margin term vanishes, leaving

NPSH_{A} = z - h_{loss}

; only gravity head minus friction remains, so unless the tank is well above the pump the inlet is on the edge of boiling.

What if friction loss

h_{loss}

exceeds the tank pressure margin plus

z

NPSH_{A}

goes negative, meaning the inlet pressure is already below vapor pressure — the liquid boils before it even reaches the impeller and the pump fails.

If exit whirl equaled blade speed exactly (

c_{u 2} = u_{2}

), what would the head become?

H = u_{2}^{2} / g

, the theoretical maximum for that rim speed; real blades give

c_{u 2} < u_{2}

(slip and finite blade angle), so achievable head is somewhat lower.

What limits how fast

u_{2}

(and thus head) can grow in practice?

Material strength — centrifugal stress in the impeller rises with

u_{2}^{2}

, so the disc would burst before head grows without bound; blade-tip speed is capped by the metal, not by the fluid.

What does the turbine work approach as the pressure ratio

π_{t} \to 1

(no expansion)?

The factor

(1 - π_{t}^{- (γ - 1) / γ}) \to 0

, so extracted work goes to zero — with no pressure drop across it, the gas gives up no energy and the shaft receives no power.

What is the sign of

Δ c_{u}

if a rotor row adds whirl instead of removing it?

With the convention that

c_{u}

is positive along blade motion, adding whirl makes

c_{u, out} > c_{u, in}

, so

Δ c_{u} = c_{u, in} - c_{u, out}

turns negative — the machine is now doing work on the fluid (a pump/compressor), not extracting it.

Recall One-line summary of the whole trap set

Head is density-free but pressure and power are not; Euler ( $u Δ c_{u}$ ) runs both machines with opposite sign; and NPSH safety is $NPSH_{A} > NPSH_{R}$ — inducers fix it by lowering the required side. Confirm the three traps ::: (1) $H$ has no $ρ$ ; (2) pump adds whirl, turbine removes it; (3) inducer lowers NPSH $_{R}$ , not raises NPSH $_{A}$ .