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

Intuition How to use this page

Work each problem before opening its Solution fold. The problems climb from recognition (can you name the parts?) to mastery (can you design a full pressure budget and mass tradeoff from scratch?). Every symbol here was built in the parent note the parent topic; if a term feels unfamiliar, re-read that note first.

Symbols we reuse, in plain words:

$p_{c}$ — the pressure inside the combustion chamber (where burning happens). Units: pascals (Pa) or bar; $1 bar = 1 0^{5} Pa$ .
$p_{tank}$ — the pressure the propellant tank must hold.
$Δ p_{inj}$ — the pressure lost squeezing liquid through the injector holes.
$Δ p_{feed}$ — the pressure lost to friction as liquid flows through the feed lines (the pipes from tank to engine).
$Δ p_{cool}$ — the pressure lost as liquid flows through the cooling jacket wrapped around the chamber (friction in those narrow channels).
$ρ$ — density of the liquid (kg/m³). $σ$ — the stress a wall can take before yielding (Pa). $t$ — wall thickness (m). $r$ — tank radius (m). $ρ_{m}$ — density of the tank wall material (kg/m³). $V$ — the propellant volume the tank encloses (m³).
$a$ — the effective acceleration acting on the liquid column (m/s²); on a coasting-then-firing stage this is the stage's own acceleration, replacing ordinary gravity $g$ .
$M$ — molar mass of a gas (kg/mol); $R = 8.314$ J/(mol·K) — the universal gas constant; $T$ — absolute temperature (K); $n$ — number of moles.
$m_{p}$ — propellant mass (kg); $m_{s}$ — structure (mostly tank) mass (kg); $v_{e}$ — exhaust velocity (m/s); $Δ v$ — the velocity change a stage can deliver (m/s).

Key tools we lean on: the Ideal gas law, hoop stress, Bernoulli's equation, and the Tsiolkovsky rocket equation.

The tank-mass formula, built from pictures

Before the mass exercises, let us earn the formula $m_{tank} = \frac{3 ρ _{m} p V}{2 σ}$ instead of quoting it — and pin down where every number comes from.

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

Formula Deriving

$m_{tank} \propto p V / σ$ and its factor $3/2$ Step 1 — wall thickness from hoop stress. For a thin spherical shell holding pressure $p$ , hoop stress gives (see the figure and Hoop stress and thin-walled pressure vessels): $σ = \frac{p r}{2 t} \Rightarrow t = \frac{p r}{2 σ}$ Step 2 — wall mass = surface area × thickness × material density. A sphere's surface area is $A = 4 π r^{2}$ (the blue shell in the figure), so $m_{tank} = ρ_{m} A t = ρ_{m} (4 π r^{2}) (\frac{p r}{2 σ}) = \frac{2 π ρ _{m} p r ^{3}}{σ}$ Step 3 — trade $r^{3}$ for the enclosed volume $V$ . The propellant volume is $V = \frac{4}{3} π r^{3}$ (the amber ball in the figure), so $π r^{3} = \frac{3}{4} V$ . Substitute: $m_{tank} = \frac{2 ρ _{m} p}{σ} \cdot π r^{3} = \frac{2 ρ _{m} p}{σ} \cdot \frac{3}{4} V$ $m_{tank} = \frac{3 ρ _{m} p V}{2 σ}$ Where the "3" and the " $/2$ " come from: the $2$ from surface area $4 π r^{2}$ times thickness $\propto 1/2$ , and the $3/4$ from converting $π r^{3}$ to volume — multiplying $2 \times \frac{3}{4} = \frac{3}{2}$ . So the honest factor is $\frac{3}{2}$ , not $3$ . Mass is still exactly linear in $p$ : that is the physics that matters.

The pressure staircase — read it before you compute

Intuition How to read the staircase figure

The horizontal axis is the journey of the liquid: it starts in the tank (far left) and ends in the chamber (far right). The vertical axis is pressure in bar. Each cyan platform is the pressure at that point; each amber arrow is a drop — a loss you must pre-pay by starting higher in the tank. Sample reading: in the figure the tank sits at $17.0$ bar. Crossing the feed line loses $0.8$ bar (down to $16.2$ ), the cooling jacket loses another $0.6$ (down to $15.6$ ), and then the big amber injector drop of $3.6$ bar drops us onto the chamber platform at $12.0$ bar $= p_{c}$ . Notice the injector step is by far the tallest — that drop is deliberate, because it powers atomization. Every mass and budget problem below is just reading numbers off a staircase like this one.

L1 — Recognition

Recall Solution

What defines it: the absence of turbopumps. The propellants are driven purely by a stored inert gas pressing on the tank tops. Gas: helium (or nitrogen). Why helium: it is inert (won't react with fuel/oxidizer), light (low mass penalty), and won't condense or dissolve into the propellant. See Turbopump-fed cycles for the alternative it replaces.

Recall Solution

Rule: $p_{tank} \approx (1.3 - 1.5) p_{c}$ . Why this range? Tank pressure must cover the injector drop (~25% of $p_{c}$ ) plus feed/cooling losses. $1.3 \times 8 = 10.4 bar 1.5 \times 8 = 12.0 bar$ Answer: roughly $10.4$ – $12.0$ bar.

L2 — Application

Recall Solution

Step 1 — injector drop. $Δ p_{inj} = 0.30 \times 12 = 3.6$ bar. Why fraction of $p_{c}$ ? Injector drop is deliberately sized relative to chamber pressure to guarantee atomization. Step 2 — add all downstream losses. Here $Δ p_{feed}$ is the feed-line friction loss and $Δ p_{cool}$ the cooling-jacket friction loss (both defined in the symbol list). $p_{tank} = p_{c} + Δ p_{inj} + Δ p_{feed} + Δ p_{cool} = 12 + 3.6 + 0.8 + 0.6 = 17.0 bar$ Step 3 — ratio. $17.0/12 = 1.417$ . Answer: $p_{tank} = 17.0$ bar $\approx 1.42 p_{c}$ ✔ (inside the 1.3–1.5 rule). This is exactly the tank pressure drawn in the staircase figure above.

Recall Solution

Tool — why hoop stress? A pressurized sphere tries to burst; the wall carries a tension called hoop stress. Balancing burst force against wall strength (see Hoop stress and thin-walled pressure vessels) gives $σ = \frac{p r}{2 t}$ , valid for a thin wall $t ≪ r$ . Step 1 — solve for $t$ . $t = \frac{p r}{2 σ} = \frac{( 1.7 \times 1 0 ^{6} ) ( 0.5 )}{2 ( 3 \times 1 0 ^{8} )}$ Step 2 — compute. $t = \frac{8.5 \times 1 0 ^{5}}{6 \times 1 0 ^{8}} = 1.417 \times 1 0^{- 3}$ m. Answer: $t \approx 1.42$ mm. Check the assumption: $t / r = 1.42 \times 1 0^{- 3} /0.5 = 0.0028 ≪ 1$ , so thin-wall is valid.

Recall Solution

Tool — why ideal gas law? Helium at these conditions behaves nearly ideally, so $p V = n R T$ where $n$ is moles (see Ideal gas law). Step 1 — moles. $n = \frac{p V}{R T} = \frac{( 1.7 \times 1 0 ^{6} ) ( 1.5 )}{8.314 \times 290} = 1057.7 mol$ Step 2 — mass. $m = n M = 1057.7 \times 0.004 = 4.23$ kg. Answer: $\approx 4.23$ kg of helium in the drained volume (real tanks carry several times this, since gas also remains in the storage bottle).

L3 — Analysis

Recall Solution

Tool — why $m \propto p$ ? Wall thickness $t = p r /2 σ$ (thin-wall, $t ≪ r$ ) grows with pressure, and mass = (area)× $t$ ×density, so tank mass is linear in the pressure it holds. Step 1 — Design A ( $p = 1.8 \times 1 0^{6}$ Pa): $m_{A} = \frac{3 ( 2700 ) ( 1.8 \times 1 0 ^{6} ) ( 2 )}{2 ( 3 \times 1 0 ^{8} )} = 48.6 kg$ Step 2 — Design B ( $p = 1.2 \times 1 0^{7}$ Pa): $m_{B} = \frac{3 ( 2700 ) ( 1.2 \times 1 0 ^{7} ) ( 2 )}{2 ( 3 \times 1 0 ^{8} )} = 324 kg$ Step 3 — ratio. $m_{B} / m_{A} = 120/18 = 6.67$ . Step 4 — check thin-wall for Design B. From $V = \frac{4}{3} π r^{3}$ , $r = (\frac{3 V}{4 π})^{1/3} = (\frac{6}{4 π})^{1/3} = 0.782$ m. Then $t = \frac{p r}{2 σ} = \frac{( 1.2 \times 1 0 ^{7} ) ( 0.782 )}{6 \times 1 0 ^{8}} = 0.0156$ m, so $t / r = 0.0156/0.782 = 0.020 ≪ 1$ ✔ thin-wall still holds even at 120 bar. Answer: the high-pressure tank is $\approx 6.67 \times$ heavier ( $324$ vs $48.6$ kg). This is the core reason pressure-fed engines stay at low $p_{c}$ : their tanks are the pump and pump mass explodes with pressure.

Recall Solution

Tool — why Bernoulli? Bernoulli's equation says pressure energy converts to kinetic energy along a streamline. The injector's pressure drop is spent speeding the liquid up into a fast jet that breaks into droplets. Step 1 — solve for $v$ . $v = \frac{2 Δ p _{inj}}{ρ} = \frac{2 ( 3 \times 1 0 ^{5} )}{1000}$ Step 2 — compute. $v = 600 = 24.49$ m/s. Answer: $v \approx 24.5$ m/s. This fast jet is why we deliberately keep the injector drop large — it powers atomization into fine droplets that burn cleanly.

L4 — Synthesis

Recall Solution

Step 1 — injector. $Δ p_{inj} = 0.25 \times 15 = 3.75$ bar. Step 2 — gravity/acceleration head. With $a$ the effective acceleration (defined in the symbol list), $ρ ah = 1200 \times 30 \times 1.2 = 43, 200$ Pa $= 0.432$ bar. Why subtract? The accelerating liquid column adds pressure at the bottom (toward the chamber), so the tank needs to supply less. Step 3 — assemble the budget. $p_{tank} = p_{c} + Δ p_{inj} + Δ p_{feed+cool} - ρ ah$ $p_{tank} = 15 + 3.75 + 1.5 - 0.432 = 19.818 bar$ Answer: $p_{tank} \approx 19.82$ bar $\approx 1.32 p_{c}$ . The helpful acceleration head shaves off nearly half a bar.

Recall Solution

Use $p_{tank} = 19.82$ bar $= 1.982 \times 1 0^{6}$ Pa and $m = \frac{3 ρ _{m} p V}{2 σ}$ (derived above, thin-wall $t ≪ r$ ). (a) Aluminium. $m_{A l} = \frac{3 ( 2700 ) ( 1.982 \times 1 0 ^{6} ) ( 3 )}{2 ( 3 \times 1 0 ^{8} )} = 80.27 kg$ (b) Titanium. $m_{T i} = \frac{3 ( 4500 ) ( 1.982 \times 1 0 ^{6} ) ( 3 )}{2 ( 9 \times 1 0 ^{8} )} = 44.60 kg$ Ratio. $m_{T i} / m_{A l} = \frac{4500/9 \times 1 0 ^{8}}{2700/3 \times 1 0 ^{8}} = \frac{5 \times 1 0 ^{- 6}}{9 \times 1 0 ^{- 6}} = 0.556$ . (c) Thin-wall check. $r = (\frac{3 V}{4 π})^{1/3} = (\frac{9}{4 π})^{1/3} = 0.895$ m; $t = \frac{p r}{2 σ} = \frac{( 1.982 \times 1 0 ^{6} ) ( 0.895 )}{6 \times 1 0 ^{8}} = 2.96 \times 1 0^{- 3}$ m, so $t / r = 0.0033 ≪ 1$ ✔. Answer: Al $\approx 80.3$ kg, Ti $\approx 44.6$ kg; titanium is $\sim 0.56 \times$ the mass. Why? What matters is the ratio $ρ_{m} / σ$ (density per unit strength) — titanium's higher strength beats its higher density.

L5 — Mastery

Recall Solution

Tool — why Tsiolkovsky? $Δ v$ is the true performance measure; it weighs exhaust velocity $v_{e}$ against the mass ratio $m_{0} / m_{f}$ (wet over dry). We must see whether the faster exhaust beats the heavier tank. Setup. Wet mass $m_{0} = m_{p} + m_{s} + 500$ . Dry mass $m_{f} = m_{s} + 500$ (propellant burned away).

Case A ( $p_{c} = 15$ bar): $m_{s} = 160$ kg. $m_{0} = 8000 + 160 + 500 = 8660, m_{f} = 160 + 500 = 660$ $Δ v_{A} = 2900 ln \frac{8660}{660} = 2900 \times ln (13.1212) = 2900 \times 2.5743 = 7465 m/s$

Case B ( $p_{c} = 30$ bar): $m_{s} = 320$ kg. $m_{0} = 8000 + 320 + 500 = 8820, m_{f} = 320 + 500 = 820$ $Δ v_{B} = 3100 ln \frac{8820}{820} = 3100 \times ln (10.7561) = 3100 \times 2.3755 = 7364 m/s$

Answer: $Δ v_{A} \approx 7465$ m/s vs $Δ v_{B} \approx 7364$ m/s. Low pressure wins by ~101 m/s: the tank-mass penalty outweighs the modest exhaust-velocity gain. This is exactly why upper stages stay pressure-fed and low- $p_{c}$ — the analysis, not just intuition, confirms it.

Recall Solution

Step 1 — equate. We need $3100 ln \frac{8500 + m _{s}^{B}}{500 + m _{s}^{B}} = 7465$ . Step 2 — isolate the log. $ln \frac{8500 + m _{s}^{B}}{500 + m _{s}^{B}} = 7465/3100 = 2.4081$ , so the ratio $= e^{2.4081} = 11.113$ . Step 3 — clear the fraction. $8500 + m_{s}^{B} = 11.113 (500 + m_{s}^{B}) = 5556.5 + 11.113 m_{s}^{B}$ . Step 4 — collect like terms. Move the $m_{s}^{B}$ terms to one side and the constants to the other: $8500 - 5556.5 = 11.113 m_{s}^{B} - m_{s}^{B} \Rightarrow 2943.5 = 10.113 m_{s}^{B}$ Step 5 — solve. $m_{s}^{B} = 2943.5/10.113 = 291.1$ kg. Answer: Case B breaks even only if its tank could be cut to $\approx 291$ kg (below the $320$ kg it actually needs). Since the real tank is heavier than the break-even, high pressure genuinely loses — confirming Exercise 10.