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

Everything here needs three ideas from the parent, re-stated so you never use a symbol before it is earned:

The scenario matrix

Every question this topic can throw is one of these cells. Each example below is tagged with the cell it fills.

Cell	Case class	What is special	Example
A	Baseline budget	All losses positive, normal	Ex 1
B	Sign flip — gravity term	Tank below chamber ( $- ρ g h$ vs $+ ρ g h$ )	Ex 2
C	Zero / degenerate input	$Δ p_{inj} \to 0$ : what breaks?	Ex 3
D	Limiting value	$p_{tank} \to p_{c}$ : flow rate $\to 0$	Ex 4
E	Mass-penalty scaling	double $p_{c}$ , what happens to mass?	Ex 5
F	Real-world word problem	full apogee-kick stage sizing	Ex 6
G	Exam twist — units trap	mixed bar / Pa / MPa	Ex 7
H	Blowdown (no regulator)	pressure falls as tank empties	Ex 8

Cell A — Baseline pressure budget

Worked example Example 1 — the everyday case

Chamber pressure $p_{c} = 12$ bar. The injector deliberately eats $30%$ of $p_{c}$ . Pipe friction eats $0.8$ bar, the cooling jacket eats $0.5$ bar. The tank sits at the same height as the chamber (ignore gravity). Find $p_{tank}$ .

Forecast: guess before reading — will the answer be nearer $12$ , $16$ , or $20$ bar?

$Δ p_{inj} = 0.30 \times 12 = 3.6$ bar. Why this step? The injector drop is given as a fraction of chamber pressure, so we scale $p_{c}$ .
Add every downstream loss to $p_{c}$ : $p_{tank} = p_{c} + Δ p_{inj} + Δ p_{feed} + Δ p_{cool} = 12 + 3.6 + 0.8 + 0.5.$ Why this step? Flow only runs downhill in pressure. Every loss downstream must be pre-paid by the tank.
$p_{tank} = 16.9$ bar. Why this step? We finish the sum from step 2 — this is the total pressure the tank must supply, our answer.

Verify: ratio $16.9/12 = 1.41$ , squarely inside the $1.3$ – $1.5$ rule of thumb. Units: all terms in bar, so the sum is in bar. ✔

Cell B — Sign flip in the gravity term

Now bring back the term the baseline ignored. On the launch pad, the liquid column has weight; that weight can help or hurt the feed depending on whether the tank is above or below the chamber.

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

Worked example Example 2 — gravity helps, then hurts

Take Example 1's total loss ( $p_{tank} = 16.9$ bar with no gravity). Now the propellant is dense storable ( $ρ = 1200 kg/m^{3}$ ) with a $3$ m column, and the rocket sits on the pad ( $g = 9.81 m/s^{2}$ ). Find $p_{tank}$ (a) tank above chamber, (b) tank below.

Forecast: one answer is below 16.9 bar, one above. Which is which?

Gravity head $ρ g h = 1200 \times 9.81 \times 3 = 35, 316 Pa = 0.353$ bar. Why this step? We need the column's pressure in the same units (bar) as everything else. $1 bar = 1 0^{5} Pa$ .
Tank above — subtract the head (gravity helps): $p_{tank} = 16.9 - 0.353 = 16.55 bar .$ Why this step? The master equation carries $- ρ g h$ when the column's weight assists the flow, so we take the base 16.9 bar and remove the head.
Tank below — add the head (gravity hurts): $p_{tank} = 16.9 + 0.353 = 17.25 bar .$ Why this step? Now the sign is $+ ρ g h$ : the tank must also lift the column, so the head is added on top of the base 16.9 bar.

Verify: the two differ by exactly $2 \times 0.353 = 0.71$ bar — twice the head, as expected for a pure sign flip. In space ( $g \to 0$ ) both collapse back to $16.9$ bar. ✔

Cell C — Degenerate input: what if the injector drop vanishes?

Worked example Example 3 —

$Δ p_{inj} \to 0$ Suppose a naive designer removes the injector restriction so $Δ p_{inj} = 0$ , keeping $p_{c} = 12$ bar, feed $0.8$ , cooling $0.5$ bar. Compute $p_{tank}$ and explain why the engine is now broken even though the maths gives a valid number.

Forecast: the number looks better (lower tank pressure). Is that good?

$p_{tank} = 12 + 0 + 0.8 + 0.5 = 13.3$ bar. Why this step? Same master equation, with the injector term set to zero.
Interpret physically: with no pressure drop across the injector, the liquid enters as a slow, coherent jet. Why this step? The injector drop is what accelerates and shatters the liquid into fine droplets (atomization). No drop → no atomization → poor, unstable combustion.

Verify: the arithmetic is fine (13.3 bar), but the degenerate case is a trap: a smaller tank pressure that produces an engine that won't burn cleanly. The limit tells you the injector drop is a feature, not a loss to minimize. ✔

Cell D — Limiting value: $p_{tank} \to p_{c}$

This is the edge the first [!mistake] on the parent warned about. Let us make it quantitative using Bernoulli's equation for the injector.

Worked example Example 4 — approaching the dead-engine limit

Storable propellant $ρ = 1200 kg/m^{3}$ . Find the jet speed $v$ when the injector drop is (a) $3$ bar, (b) $0.3$ bar, (c) $0$ bar. Watch the trend as $p_{tank} \to p_{c}$ .

Forecast: as the drop shrinks by $10 \times$ , does the speed shrink by $10 \times$ ?

(a) $v = 2 (3 \times 1 0^{5}) /1200 = 500 = 22.4 m/s$ . Why this step? Convert $3$ bar to $3 \times 1 0^{5}$ Pa, plug into Bernoulli.
(b) $v = 2 (0.3 \times 1 0^{5}) /1200 = 50 = 7.07 m/s$ . Why this step? Same formula, $10 \times$ smaller drop — we test how the speed responds to shrinking the drop.
(c) $Δ p_{inj} = 0 \Rightarrow v = 0$ : no flow at all. Why this step? We take the limit $Δ p_{inj} \to 0$ to see the dead-engine endpoint: zero drop forces $v = 0$ .

Verify: a $10 \times$ drop in $Δ p$ gave only $22.4/7.07 = 3.16 \approx 10$ times slower — the square-root law, confirmed. At the limit $p_{tank} = p_{c}$ the drop is zero and $v \to 0$ : the engine is dead, matching the parent's warning. ✔

Cell E — Mass-penalty scaling

Worked example Example 5 — double the chamber pressure, what happens to tank mass?

Aluminium sphere: $ρ_{m} = 2700 kg/m^{3}$ , yield $σ = 300 MPa$ , propellant volume $V = 1.5 m^{3}$ . The tank runs at $p_{tank} = 1.4 p_{c}$ . Compare tank wall mass at $p_{c} = 15$ bar versus $p_{c} = 30$ bar.

Forecast: double $p_{c}$ → mass goes up by a factor of…?

At $p_{c} = 15$ bar: $p_{tank} = 1.4 \times 15 = 21$ bar $= 2.1 \times 1 0^{6}$ Pa. Why this step? Tank holds $1.4 \times$ the chamber pressure (parent rule).
$m = \frac{3 ρ _{m}}{σ} p_{tank} V = \frac{3 ( 2700 )}{3 \times 1 0 ^{8}} (2.1 \times 1 0^{6}) (1.5) = 85.05 kg .$ Why this step? The hoop-stress-derived mass law; mass is linear in the pressure held.
At $p_{c} = 30$ bar: $p_{tank} = 42$ bar $= 4.2 \times 1 0^{6}$ Pa, giving $m = 170.1 kg$ . Why this step? Same law; only the pressure doubled.

Verify: $170.1/85.05 = 2.00$ exactly — mass scales linearly with pressure. This is precisely why pressure-fed engines refuse to run at high $p_{c}$ : the tank mass tracks the pressure one-to-one. ✔

Cell F — Real-world word problem

Worked example Example 6 — sizing an apogee-kick stage

A satellite's apogee engine burns storable propellant. Chamber pressure $p_{c} = 8$ bar, injector drop $25%$ , feed+cooling losses $0.6$ bar, in vacuum (no gravity term). Propellant volume $V = 0.9 m^{3}$ . Pressurant is helium ( $M = 4 g/mol$ ) at $T = 300 K$ . Find (a) $p_{tank}$ , (b) helium mass that must end up filling the drained tank.

Forecast: the helium mass will be a few kilograms — over or under 3 kg?

$Δ p_{inj} = 0.25 \times 8 = 2$ bar; $p_{tank} = 8 + 2 + 0.6 = 10.6$ bar. Why this step? Master equation; vacuum means no $ρ g h$ .
Helium moles filling $V$ at $p_{tank}$ : $n = \frac{p V}{R T} = \frac{( 1.06 \times 1 0 ^{6} ) ( 0.9 )}{8.314 \times 300}$ . Why this step? Ideal gas law $p V = n R T$ rearranged for $n$ ; the gas that ends up inside the tank occupies $V$ at $p_{tank}$ , $T$ , and $R$ is the constant just defined.
$n = \frac{954000}{2494.2} = 382.5 mol$ , so $m = n M = 382.5 \times 0.004 = 1.53 kg$ . Why this step? Convert moles to kilograms with the molar mass $M$ (grams per mole → kg per mole is $\times 0.001$ ).

Verify: $10.6/8 = 1.33$ , inside the $1.3$ – $1.5$ band. Helium mass $1.53$ kg is a small fraction of the propellant — the pressurant is "cheap", as the parent claimed. (Real designs carry several × this because gas also stays in the storage bottle.) ✔

Cell G — Exam twist: the units trap

Worked example Example 7 — mixed units, one wrong conversion fails you

Given directly: chamber pressure $p_{c} = 1.5 MPa$ , injector drop $Δ p_{inj} = 400 kPa$ , feed loss $Δ p_{feed} = 0.3 bar$ . Find $p_{tank}$ in bar.

Forecast: the sneaky term is the one in a different unit — spot it before computing.

Convert all to Pa: $p_{c} = 1.5 \times 1 0^{6}$ ; $Δ p_{inj} = 4 \times 1 0^{5}$ ; $Δ p_{feed} = 0.3 \times 1 0^{5} = 3 \times 1 0^{4}$ . Why this step? You cannot add MPa, kPa and bar directly — bring everything to one base unit first.
Sum: $p_{tank} = 1.5 \times 1 0^{6} + 4 \times 1 0^{5} + 3 \times 1 0^{4} = 1.93 \times 1 0^{6} Pa .$ Why this step? The master equation is a straight sum once units agree.
Convert to bar: $1.93 \times 1 0^{6} /1 0^{5} = 19.3 bar .$ Why this step? Final answer requested in bar; $1 bar = 1 0^{5} Pa$ .

Verify: ratio $19.3/15 = 1.29$ — just at the low edge of the $1.3$ – $1.5$ rule (feed loss here is tiny). If you had forgotten the kPa→Pa step and added $400$ raw, you'd get nonsense; the sanity ratio catches it. ✔

Cell H — Blowdown mode (no regulator)

Every earlier example assumed a regulator holding $p_{tank}$ steady. Cheapest thrusters skip the regulator: the gas simply expands as liquid leaves, so pressure falls over the burn. This is blowdown.

Worked example Example 8 — pressure at end of a blowdown burn

A blowdown tank starts with gas volume $V_{1} = 0.5 m^{3}$ at $p_{1} = 20 bar$ (regulator-free). As propellant drains, the gas expands to $V_{2} = 1.5 m^{3}$ . Assume isothermal. Find the final tank pressure $p_{2}$ , and check it still beats a chamber at $p_{c} = 12$ bar with a $2.5$ bar total loss.

Forecast: the blowdown ratio is $1.5/0.5 = 3$ . Does the pressure drop by $3 \times$ ?

Isothermal ideal gas: $p_{1} V_{1} = p_{2} V_{2} \Rightarrow p_{2} = p_{1} \frac{V _{1}}{V _{2}}$ . Why this step? Fixed gas, fixed temperature → the product $p V$ is conserved.
$p_{2} = 20 \times \frac{0.5}{1.5} = 20/3 = 6.67 bar$ . Why this step? Plug the blowdown ratio in; pressure and volume are inversely proportional.
Required minimum to feed: $p_{need} = p_{c} + losses = 12 + 2.5 = 14.5 bar$ . Why this step? The master equation still applies at every instant of the burn.

Verify: $p_{2} = 6.67 bar < 14.5 bar$ — the engine starves long before the tank empties! A blowdown design must start high enough that even the final pressure clears the budget. So the blowdown ratio (here $3$ ) is limited: pick $V_{2}$ so $p_{2} \geq 14.5$ bar. This is the extra constraint that regulated systems (Ex 1–7) never worry about. ✔

Recall Scenario checklist — can you place each case in the matrix?

Which cell involves a sign flip, and what flips? ::: Cell B — the $\pm ρ g h$ gravity term flips sign (tank above: minus; tank below: plus). Which cell is a degenerate input, and why is the valid number a trap? ::: Cell C — $Δ p_{inj} \to 0$ gives lower tank pressure but no atomization, so the engine won't burn cleanly. Which cell shows flow → 0 at a limit? ::: Cell D — as $p_{tank} \to p_{c}$ the injector drop → 0 and jet speed $v \to 0$ . Which cell is the real-world word problem, and what does it size? ::: Cell F — the apogee-kick stage, sizing tank pressure and helium pressurant mass. Which cell is the units trap? ::: Cell G — mixed MPa / kPa / bar must all be converted before summing. In blowdown mode (Cell H), why must the tank start over-pressured? ::: Because $p V$ is constant, pressure falls as liquid drains; the final pressure must still exceed $p_{c}$ + losses.