Pressure-fed cycle — simplest, used in upper stages: Visual walkthrough

Step 1 — Fluid only flows from high pressure to low pressure

WHAT. Put two connected containers side by side, one at high pressure, one at low. Open the valve. The fluid always rushes from the high side to the low side — never the other way.

WHY. Pressure difference is the shove. If both sides push equally, the shoves cancel and nothing moves. If one side pushes harder, the net shove drives fluid toward the weaker side. This one fact is the seed of the whole derivation.

PICTURE. In the figure, the blue arrow shows fluid moving from the high-pressure box (left) toward the low-pressure box (right). Reverse the pressures and the arrow flips.

Figure — Pressure-fed cycle — simplest, used in upper stages

Step 2 — The chamber is a high-pressure enemy at the finish line

WHAT. Inside the combustion chamber the propellants burn, making hot gas at pressure $p_{c}$ ("c" for chamber). This gas pushes outward in all directions — including back up the injector holes toward the tanks.

WHY. From Step 1, fluid flows from high to low. If the chamber pressure $p_{c}$ is higher than the pressure of the liquid trying to enter, the chamber wins and blows the propellant back out. The chamber is not a friendly destination — it is pushing back at you.

PICTURE. The orange arrows show hot gas pressure $p_{c}$ shoving outward. The little upward red arrow at an injector hole is the danger: chamber gas trying to escape backwards into the feed line.

Step 3 — The injector is a deliberate wall you must climb over

WHAT. Right before the chamber, the propellant is forced through the injector and drops in pressure by $Δ p_{inj}$ , typically about a quarter of $p_{c}$ .

WHY. This drop is not a mistake — it is wanted. A big pressure drop across the holes makes the liquid shoot out fast and break into droplets (atomize), which is what lets it burn cleanly. But it means the liquid must arrive at the injector already carrying $p_{c} + Δ p_{inj}$ , so that after the drop it still matches — and beats — the chamber.

PICTURE. The height of the orange step is $Δ p_{inj}$ . The liquid starts on top of the step and must have enough pressure to fall down it and still land above $p_{c}$ .

Step 4 — Pipes and cooling jacket steal a little more pressure

WHAT. Between the tank and the injector the propellant loses two more small chunks of pressure to friction.

WHY. Every rough pipe wall and every bend resists the flow — this is real energy lost to heat, exactly the "losses" term in Bernoulli's equation. You must pre-pay these too, or the liquid won't have enough pressure left when it reaches the injector.

PICTURE. The staircase grows: each new step (green = feed, gray = cooling) is another chunk the tank must supply on top of everything below it.

Step 5 — Bernoulli: bookkeeping the whole pressure staircase

WHAT. Write Bernoulli from the tank surface (top) to the chamber (bottom): $p_{tank} + \frac{1}{2} ρ v_{tank}^{2} + ρ g h_{tank} = p_{c} + \frac{1}{2} ρ v_{c}^{2} + ρ g h_{c} + losses (Δ p_{inj} + Δ p_{feed} + Δ p_{cool})$

WHY each term.

$p_{tank}$ ::: the pressure the pressurant gas supplies — the thing we want to find.
$\frac{1}{2} ρ v_{tank}^{2}$ ::: motion energy at the tank top. The tank surface barely moves, so $v_{tank} \approx 0$ — this term drops out.
$ρ g h$ ::: height (gravity) energy. In coasting upper stages the rocket is nearly weightless, so gravity head $\approx 0$ — these cancel too.
$\frac{1}{2} ρ v_{c}^{2}$ ::: the liquid does gain some speed, but it is tiny next to the injector toll, so we fold it into the losses.
the losses ::: exactly the three steps from Steps 3–4.

PICTURE. The figure lines up the full staircase from tank (top) down to chamber (bottom), with $v_{tank} \approx 0$ and $ρ g h \approx 0$ crossed out in gray.

Step 6 — Collect the terms: the central result

WHAT. Cross out the vanishing terms from Step 5 and gather what's left. The gravity and tank-velocity terms are gone; the chamber-velocity term is small and absorbed. What remains is:

$p_{tank} = p_{c} + Δ p_{inj} + Δ p_{feed} + Δ p_{cool}$

WHY. Read it as a bill: to deliver liquid into a chamber at $p_{c}$ , the tank must pay for the chamber pressure itself plus every downstream toll. Nothing is free; the tank is the sole payer.

PICTURE. The equation shown as stacked bars: the total tank bar equals $p_{c}$ plus each loss bar piled on top. In numbers, with $Δ p_{inj} \approx 0.25 p_{c}$ and small other losses, this stacks up to about $1.3$ – $1.5 p_{c}$ .

Step 7 — The degenerate cases (what happens at the extremes)

Every honest derivation must survive its own edge cases. Here are the three that matter.

Case A — Equal pressures ( $p_{tank} = p_{c}$ ). The pressure difference is zero, flow stops, the injector can't atomize: a dead engine. This is why the boxed equation has strictly more on the right than $p_{c}$ alone.

Case B — Tank runs low ( $p_{tank} < p_{c}$ ). Now the chamber wins Step 1's tug-of-war: hot gas pushes backwards through the injector into the feed line — a dangerous burn-back. The red arrow in the figure reverses.

Case C — Under thrust vs. coasting (the $ρ g h$ term). During firing the rocket accelerates, so the liquid's weight helps push it down (adds to $p_{tank}$ 's effect); in weightless coast this help vanishes. That's why we dropped $ρ g h$ for upper stages — but for a tall, high-acceleration stage you'd keep it.

PICTURE. Three mini-panels: (A) equal bars, no arrow; (B) tank bar shorter, red arrow flips backward; (C) coast vs. thrust showing the gravity term appearing/vanishing.

The one-picture summary

Everything above, compressed into a single staircase: start at the chamber floor ( $p_{c}$ ), climb each loss step, and the height you must stand at is $p_{tank}$ . Beat the chamber by the full stack, or nothing flows.

Recall The whole walkthrough in plain words (Feynman retelling)

Water only flows downhill — from high push to low push. The combustion chamber is a bully at the bottom pushing back at pressure $p_{c}$ . To get liquid in, the tank must push harder than the chamber. But between the tank and the chamber there are tolls: the injector deliberately squeezes the liquid through tiny holes (so it sprays into a fine mist that burns well) and that costs pressure; the pipes and cooling jacket rub off a little more by friction. So the tank has to pay for the chamber pressure and all those tolls stacked on top. Writing Bernoulli's energy-bookkeeping from the still tank surface down to the chamber, the "how fast is the tank moving" term is zero and (for a coasting space stage) the gravity term cancels, leaving exactly $p_{tank} = p_{c} + Δ p_{inj} + Δ p_{feed} + Δ p_{cool}$ . Add it up and the tank sits at roughly $1.3$ – $1.5$ times the chamber pressure. That extra squeeze is the entire reason pressure-fed tanks are heavy — and why the trick is saved for low-pressure upper stages.

Recall Quick self-check

Why can't $p_{tank}$ just equal $p_{c}$ ? ::: Equal pressures mean zero pressure difference, so no flow and no atomization — a dead engine. Which term did we drop because the tank surface barely moves? ::: The tank motion energy $\frac{1}{2} ρ v_{tank}^{2}$ . Which term vanishes in weightless coast? ::: The gravity head $ρ g h$ . What is $Δ p_{inj}$ for? ::: It is a deliberate drop that atomizes the propellant into a fine, well-burning spray.