Pressure-fed cycle — simplest, used in upper stages: Question bank

Before you start, here is the full symbol legend and the two relations everything hangs on.

Recall The pressure budget (open me if rusty)

Flow only runs from high pressure to low pressure, so the tank must "pre-pay" every downstream loss. Including the hydrostatic/acceleration head: $p_{tank} = p_{c} + Δ p_{inj} + Δ p_{feed} + Δ p_{cool} - ρ g h$ Sign convention: $h > 0$ means the liquid surface sits above the injector, so its weight (or thrust-acceleration column) helps push liquid down — hence the minus sign, it lowers the pressure the gas must supply. Coasting ( $g \to 0$ ) or a very short column makes this term negligible.

Recall Why

$p_{tank} \approx 1.5 p_{c}$ (the loss breakdown) The extra half-chamber-pressure is a sum of typical losses, not a magic number:

Injector drop $Δ p_{inj} \approx 0.25 p_{c}$ — the biggest chunk, needed for atomization.
Feed-line friction $Δ p_{feed} \approx 0.1 p_{c}$ .
Cooling-jacket friction $Δ p_{cool} \approx 0.15 p_{c}$ . Summing: $p_{tank} \approx p_{c} (1 + 0.25 + 0.1 + 0.15) = 1.5 p_{c}$ . Different designs land anywhere in the 1.3–1.5 range depending on how these split.

Recall Tank wall mass (corrected derivation)

From hoop stress on a thin sphere, $σ = \frac{p r}{2 t} \Rightarrow t = \frac{p r}{2 σ}$ . Wall mass = (density)(surface area)(thickness): $m_{tank} = ρ_{m} (4 π r^{2}) (\frac{p r}{2 σ}) = \frac{ρ _{m} p}{2 σ} (4 π r^{3}) = \frac{3 ρ _{m} p V}{2 σ}$ using $V = \frac{4}{3} π r^{3} \Rightarrow 4 π r^{3} = 3 V$ . So $m_{tank} = \frac{3 ρ _{m} p V}{2 σ}$ — mass grows straight in step with the pressure it holds.

True or false — justify

A pressure-fed engine has no moving parts at all.

False — it still has valves and a pressure regulator (a moving element). "No turbopumps" is the real defining feature, not "zero moving parts."

If the tank pressure equals the chamber pressure, propellant will still trickle in slowly.

False — at equal pressures the driving pressure difference is zero, so mass flow → 0 and the injector cannot atomize; the engine is effectively dead.

The injector pressure drop is wasted energy the designer should try to eliminate.

False — it is a deliberate, needed loss: the restriction is what breaks the liquid into fine droplets for good atomization. Killing it kills combustion quality.

Doubling the chamber pressure roughly doubles the required tank wall mass.

True — tank mass

\propto p

and

p_{tank} \approx 1.5 p_{c}

, so doubling

p_{c}

roughly doubles

p_{tank}

and hence the wall mass, for a fixed volume.

Helium is chosen mainly because it is cheap.

False — it is chosen because it is inert and light: it won't react with, condense into, or dissolve in the propellants, and low molar mass means less pressurant mass for a given number of moles.

Pressure-fed cycles are always more reliable, so they are the best choice for every rocket.

False — reliability is real, but the tank-mass penalty (

\propto p

) makes them impractical for high-

p_{c}

boosters; they win only where

p_{c}

is low, i.e. upper stages.

A pressure-fed stage can be restarted more easily than a complex pump-fed one.

True — with almost no moving machinery to spin up or prime, reopening valves after a long coast is far simpler, which is exactly why upper stages favour it.

The pressurant gas ends up burning in the chamber and adding to thrust.

False — it is inert by design; its only job is to push liquid. It contributes negligibly to combustion and is often chosen precisely so it won't react.

As the pressurant expands to fill the emptying tank, its temperature stays constant.

False — real expansion is closer to adiabatic: the gas cools as it does work expanding, so its pressure drops faster than an isothermal estimate predicts, meaning you must carry more pressurant than the simple

n = p V / R T

figure.

Spot the error

"To lighten the vehicle, use thin tanks and just raise the pressurant pressure to compensate."

Backwards — from

t = p r /2 σ

, higher pressure demands thicker walls, not thinner. Thin walls at high pressure burst; you cannot trade one for the other freely.

"Since

p_{tank} = p_{c} + losses

, using a bigger injector drop makes the tank lighter."

Wrong sign — a bigger injector drop raises the required

p_{tank}

, which raises wall mass. It improves atomization but costs tank mass; it never lightens the tank.

"Bernoulli says faster flow means higher pressure, so speeding propellant up in the feed line raises its pressure into the chamber."

Bernoulli says the opposite for the pressure term: higher speed trades pressure down (

\frac{1}{2} ρ v^{2}

rises as

p

falls). Feed speed-up consumes pressure head, it doesn't create it.

"The tank mass formula

m = \frac{3 ρ _{m} p V}{2 σ}

shows a smaller tank (small

V

) is always lighter, so shrink the tank."

Shrinking

V

shrinks tank mass but also shrinks propellant carried; by the rocket equation less propellant means less delta-v. You can't cut

V

for free.

"A regulator isn't needed — just fill the bottle to feed pressure and be done."

The stored bottle is at very high pressure and its pressure falls as gas leaves; without a regulator the feed pressure would sag through the burn, starving the engine. The regulator holds it steady.

"Because gas expands to fill the emptied tank volume, you only need enough helium to fill that final volume."

You also need the gas that stays behind in the storage bottle (it can never fully empty) and margin for adiabatic cooling, so real designs carry several times the fill-volume estimate.

"You can drop tank pressure right down to

p_{c}

as long as the average feed pressure is fine."

Ignores cavitation — at low feed pressure the liquid can locally drop below its vapour pressure, boil into bubbles, and choke the flow; the tank pressure must keep the whole feed line comfortably above vapour pressure.

Why questions

Why must the propellant be pushed harder than the chamber pushes back?

Because the burning chamber gases would otherwise flow backward out through the injector; a net inward pressure difference is what keeps propellant moving in and the flame from backing up.

Why does low chamber pressure make the pressure-fed penalty small?

The penalty is heavy tanks, and

m_{tank} \propto p_{tank} \approx 1.5 p_{c}

; at low

p_{c}

that mass is modest, so the simplicity is nearly free.

Why is the same cycle a terrible idea for a sea-level first stage?

Boosters want

p_{c} \sim 100

+ bar for high thrust, so tanks would need to hold

\sim 150

bar — the wall mass (

\propto p

) becomes prohibitive, unlike a pump-fed design that keeps tanks near-empty of pressure.

Why does the hoop-stress balance give the factor of 2 in

σ = p r /2 t

Cut the sphere: pressure pushes the halves apart over area

π r^{2}

, but the wall ring resisting has area

2 π r t

; the "2" comes from that ring circumference, so

p π r^{2} = σ 2 π r t

Why can't self-pressurization replace helium for cryogenic propellants?

Cryogens have low vapour pressure and would need huge, cold volumes to self-pressurize; an inert stored gas gives controllable, non-condensing pressure that the propellant's own vapour cannot.

Why does the ideal gas law enter when sizing the pressurant, not the liquid?

The pressurant is a gas filling a known volume at known pressure and temperature, so

n = p V / R T

predicts its moles; the liquid propellant is treated as incompressible and doesn't obey that relation.

Why does the real pressurant requirement exceed the isothermal

n = p V / R T

estimate?

Expansion is nearly adiabatic, so the gas cools and loses pressure as it works; to keep

p_{tank}

up you must supply extra warm gas beyond the constant-temperature figure.

Edge cases

What happens to required tank pressure if the vehicle is coasting (zero acceleration)?

With

g \to 0

the head term

ρ g h

vanishes, so it drops out; the tank still needs

p_{c}

plus injector and feed losses, just without the ±head contribution.

Under high acceleration with tanks above the chamber (

h > 0

), does the liquid weight help or hurt feed?

It helps — the accelerating column adds head pushing propellant toward the chamber, subtracting

ρ g h

from the pressure the pressurant must supply (the

- ρ g h

term in the budget).

What is

p_{tank}

in the ideal limit of zero feed, cooling, injector losses and zero head?

It collapses to exactly

p_{c}

— but this is unreachable in practice, since zero injector drop means no atomization and thus no usable combustion.

As the tank drains and gas expands, what keeps feed pressure constant despite the gas cooling?

The regulator admits more (relatively warm) gas from the high-pressure bottle to backfill both the emptied volume and the pressure lost to adiabatic cooling, holding

p_{tank}

steady while the bottle has margin.

What limits how small the pressurant bottle can be?

It must supply enough gas for the full drained volume, plus the adiabatic-cooling margin, plus leave itself above regulator pressure at burnout; too small a bottle drops below regulation and the feed pressure sags mid-burn.

What sets the lowest usable tank pressure, independent of the loss budget?

Cavitation — feed-line pressure must stay above the propellant's vapour pressure everywhere, or the liquid boils into bubbles and the flow chokes; this floors

p_{tank}

even if the nominal losses are small.

If a designer wanted higher thrust from a pressure-fed engine, what physically constrains them?

Higher thrust needs higher

p_{c}

(see chamber pressure and thrust), which forces higher

p_{tank}

and heavier tanks — the mass penalty is the hard ceiling, not the plumbing.