Pressure-fed cycle — simplest, used in upper stages

WHAT is a pressure-fed cycle?

Key players:

Pressurant gas (He/N₂) stored in a small, very-high-pressure bottle.
Regulator: drops the stored gas pressure to a steady feed value.
Propellant tanks: must be strong (they are the "pump").
Combustion chamber at pressure $p_c$ .

WHY does it work? (First-principles pressure budget)

Flow only happens from high pressure to low pressure. Trace the pressure from tank to chamber:

$p_{\text{tank}} \;=\; p_c \;+\; \Delta p_{\text{inj}} \;+\; \Delta p_{\text{feed}} \;+\; \Delta p_{\text{cooling}} \;-\; \rho g h$

Formula Deriving the tank pressure requirement

WHY each term? Energy (Bernoulli-style) is lost or gained at each stage between tank and chamber. Starting from Bernoulli along the feed line for an incompressible liquid: $p_{\text{tank}} + \tfrac12\rho v_{\text{tank}}^2 + \rho g h_{\text{tank}} = p_c + \tfrac12\rho v_{\text{c}}^2 + \rho g h_c + \sum(\text{losses})$

The tank surface is ~stationary, so $v_{\text{tank}}\approx 0$ .
Injector pressure drop $\Delta p_{\text{inj}}$ : the injector is a deliberate restriction so fuel atomizes into fine droplets — this is a real, needed loss (~20–30% of $p_c$ ).
Feed/cooling losses $\Delta p_{\text{feed}},\ \Delta p_{\text{cooling}}$ : friction in pipes and the cooling jacket.
Gravity/acceleration head $\rho g h$ : liquid weight helps or hinders.

Collecting the required minimum: $\boxed{\,p_{\text{tank}} \;=\; p_c + \Delta p_{\text{inj}} + \Delta p_{\text{feed}} + \Delta p_{\text{cool}}\,}$

HOW big is the tank pressure? Since $\Delta p_{\text{inj}}\sim 0.25\,p_c$ plus other losses, typically:

$p_{\text{tank}} \approx (1.3\text{–}1.5)\,p_c$

This is the whole story of why pressure-fed engines run at low chamber pressure: tanks strong enough to hold $1.5\,p_c$ get heavy fast, so designers keep $p_c$ modest (~5–20 bar).

WHY low chamber pressure? (The mass penalty derivation)

Formula Tank wall mass scales with chamber pressure

A thin-walled spherical tank of radius $r$ , wall thickness $t$ , material yield stress $\sigma$ holding internal pressure $p$ satisfies (hoop stress): $\sigma = \frac{p\,r}{2t}\ \Rightarrow\ t = \frac{p\,r}{2\sigma}$ WHY this formula? Cut the sphere in half; internal pressure pushes the two halves apart with force $p\cdot\pi r^2$ ; the wall ring of area $2\pi r t$ resists it with stress $\sigma$ . Balance: $p\pi r^2 = \sigma\,2\pi r t$ .

Wall mass $= \rho_m \cdot (\text{surface area}) \cdot t = \rho_m\,(4\pi r^2)\Big(\dfrac{pr}{2\sigma}\Big)$ : $\boxed{\,m_{\text{tank}} = \frac{3\rho_m}{\sigma}\,p\,V\,}\qquad\Big(V=\tfrac43\pi r^3\Big)$ Tank mass is directly proportional to the pressure it must hold. Since pressure-fed tanks hold $\sim1.5p_c$ , high $p_c$ means brutally heavy tanks. This is the fundamental limit of the cycle.

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

WHERE is it used, and WHY there?

Worked Examples

Worked example Example 2 — Tank mass penalty of raising

$p_c$ Aluminium tank: $\rho_m = 2700\ \text{kg/m}^3$ , $\sigma = 300$ MPa, propellant volume $V=2\ \text{m}^3$ . Compare tank mass at $p_{\text{tank}}=15$ bar vs $150$ bar.

Step 1: Use $m = \dfrac{3\rho_m}{\sigma}pV$ . Why? Derived above; mass ∝ pressure. Step 2: At 15 bar $=1.5\times10^6$ Pa: $m = \frac{3(2700)}{3\times10^8}(1.5\times10^6)(2) = 81\ \text{kg}$ Step 3: At 150 bar: $m = 810$ kg (10× larger). Why? Linear in $p$ . Conclusion: A pump-fed engine at 150 bar would need an insane tank; hence pressure-fed stays at low $p_c$ . ✔

Common Mistakes

Flashcards

What single feature defines a pressure-fed cycle?

No turbopumps — propellants are pushed by stored high-pressure inert gas acting on the tanks.

Why must tank pressure exceed chamber pressure?

To pay for the injector drop plus feed/cooling losses, so liquid flows and atomizes into the chamber.

Typical ratio of tank pressure to chamber pressure?

About 1.3–1.5 × chamber pressure.

Which gas is typically used as pressurant and why?

Helium (or nitrogen) — inert, light, won't react or condense with the propellants.

Why does pressure-fed force low chamber pressure?

Tank wall mass ∝ pressure (

t=pr/2\sigma

), so high

p_c

means impossibly heavy tanks.

Formula for tank wall mass (sphere)?

m_{tank} = \frac{3\rho_m}{\sigma}pV

Why are pressure-fed engines favored on upper stages?

They run at low

p_c

(small penalty) and offer maximum reliability/restartability after long coasts, with almost no moving parts.

Hoop-stress relation for a spherical tank?

\sigma = pr/2t

What is the role of the regulator?

Steps stored high-pressure gas down to a constant feed pressure on the tanks.

Main disadvantage of pressure-fed cycle?

Heavy tanks, limiting it to low chamber pressures / lower performance.

Recall Feynman: explain to a 12-year-old

Imagine a juice box. If you want juice to squirt out the straw fast, you squeeze the box — the harder you squeeze, the faster it comes out. A pressure-fed rocket does exactly that: instead of your hand, it uses a bottle of trapped gas to squeeze the fuel tanks so the fuel shoots into the fire chamber. No spinning pumps, nothing fancy — just a big squeeze. The catch: to squeeze hard, the box (tank) has to be really strong, and strong means heavy. So this trick is only smart when you don't need to squeeze too hard — like little engines used out in space.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, rocket engine ke andar propellant (fuel + oxidizer) ko combustion chamber mein daalna hota hai, lekin chamber ka apna pressure bahut high hota hai. Agar tank ka pressure chamber se kam hua, to jalti hui gas ulta bahar aa jaayegi. To question yeh hai: propellant ka pressure chamber se zyada kaise banayein? Pressure-fed cycle ka jawab sabse simple hai — ek high-pressure inert gas (jaise helium) se tank ko upar se dabao, bilkul juice box ko haath se dabane jaisa. Koi pump nahi, koi turbine nahi — sirf ek squeeze.

Ab tank pressure kitna chahiye? Simple: $p_{tank} = p_c +$ saare downstream losses. Sabse bada loss hai injector drop (jaanbujh ke rakha jaata hai taaki fuel chhote droplets mein toote), plus pipe aur cooling ke friction losses. Isliye typically $p_{tank} \approx 1.4 \times p_c$ . Yaad rakho — agar tank pressure sirf chamber ke barabar hua to flow zero, engine dead.

Iska bada catch hai tank ka weight. Hoop stress se: wall thickness $t = pr/2\sigma$ , aur tank mass pressure ke directly proportional hai. Matlab jitna high $p_c$ , utne bhaari tanks — impossible ho jaata hai. Isiliye pressure-fed sirf low chamber pressure wale engines ke liye theek hai.

Aur yahi wajah hai ki yeh upper stages aur satellite thrusters mein use hota hai — wahan vacuum mein kaam karna hota hai, low $p_c$ chahiye (to penalty chhoti), aur sabse important — reliability. Long space coast ke baad bhi, jab kuch moving parts hi nahi to fail hone ke chances almost zero. Simple, sasta, restartable, robust — perfect for space.