Gas generator cycle — performance penalty vs simplicity: Worked examples

This page is the exhaustive drill sheet for the Gas Generator Cycle topic. Before we solve anything, we lay out a matrix of every kind of situation the topic can throw at you — then we hit every single cell with a fully worked example.

Every symbol you need is defined right here on this page, so you never have to leave to follow a step.

Definition

Every symbol used on this page, in plain words.

$f$ — the turbine flow fraction: the slice of total propellant (a small number like $0.03$ ) that is burned in the gas generator and dumped, $f = \overset{m}{˙}_{g g} / \overset{m}{˙}_{t o t}$ .
$v_{e, c}$ — speed of the main exhaust leaving the big nozzle (fast).
$v_{e, g g}$ — speed of the dumped gas-generator exhaust (slow).
$g_{0} = 9.81 m/s^{2}$ — the reference gravity that turns a speed into seconds of specific impulse.
$Δ p$ — the pressure rise (in pascals) the pump must add to the propellant to push it into the chamber.
$ρ$ — the propellant density (kg/m³); it converts a mass flow $\overset{m}{˙}$ into a volume flow $\overset{m}{˙} / ρ$ .
$η_{p}$ , $η_{t}$ — pump and turbine efficiencies (fractions $\leq 1$ ): what portion of ideal power actually gets delivered.
$c_{p}$ — the gas's specific heat at constant pressure (J/(kg·K)): how many joules one kilogram stores per degree of temperature.
$T_{g g}$ — the gas-generator temperature (K): how hot the turbine-driving gas is.
$γ$ — the ratio of specific heats (a dimensionless gas property, ≈1.2–1.4): it sets how much a gas cools as it expands. It appears only through the bracket explained below.
$p_{in}$ , $p_{o u t}$ — the turbine inlet and outlet pressures: the gas enters the turbine at $p_{in}$ and leaves at the lower $p_{o u t}$ .

The scenario matrix

Every problem in gas-generator land is one of these case classes. Think of it as a checklist: if you can solve one example from each row, nothing on an exam can surprise you.

Cell	Case class	What makes it special	Covered by
A	Ideal-limit $v_{e, g g} = 0$	dumped gas totally wasted → penalty is exactly $f$	Ex 1
B	Non-zero dump $v_{e, g g} > 0$	dumped gas gives a small kick → penalty $< f$	Ex 2
C	Degenerate $f = 0$	closed-cycle limit, no dump	Ex 3
D	Degenerate $f = 1$	everything dumped, no main chamber	Ex 3
E	Power-balance sizing	find $f$ from pump/turbine physics	Ex 4
F	Sign/limit of $Δ p$	high $p_{c}$ → larger $f$ ; the trend, not one number	Ex 5
G	Temperature cap	how hot must GG run to hit a target $f$ ?	Ex 6
H	Real-world word problem	Saturn-V-style thrust bookkeeping	Ex 7
I	Exam twist	given $I_{s p, eff}$ , back out the hidden $f$	Ex 8

The figure below is the mental picture behind the whole matrix: two streams leave the vehicle, one fast (the orange main stream, fraction $1 - f$ , carrying speed $v_{e, c}$ ) and one slow (the plum dump stream, fraction $f$ , carrying $v_{e, g g}$ ). Keep it in view — every example that mentions "the two streams" (especially Ex 2, 3 and 7) is bookkeeping exactly these two arrows.

Figure — Gas generator cycle — performance penalty vs simplicity

Worked examples

Worked example

Example 1 — Cell A: the ideal-limit penalty ( $v_{e, g g} = 0$ ). An engine has ideal $I_{s p} = 340$ s and dumps $f = 5%$ of its flow through the turbine. Treat the dumped gas as contributing zero useful exhaust speed. Find the effective $I_{s p}$ .

Forecast: guess before computing — will the answer be closer to 340, 323, or 170?

Step 1. Use $I_{s p, eff} \approx (1 - f) I_{s p, ideal}$ . Why this step? With $v_{e, g g} = 0$ the whole second term of the weighted average vanishes, so only the surviving fraction $(1 - f)$ of the flow makes momentum.

Step 2. $I_{s p, eff} = (1 - 0.05) (340) = 0.95 \times 340 = 323 s$ . Why? 5% of the weight-flow sits in the denominator (you paid to carry it) but puts ~nothing in the numerator.

Verify: penalty $= 340 - 323 = 17 s$ , which equals $f \times I_{s p, ideal} = 0.05 \times 340 = 17$ s. Consistent, and the answer lands near 323 as forecast — not the wild 170 (that would be $f = 50%$ ).

Worked example

Example 2 — Cell B: dumped gas is NOT useless ( $v_{e, g g} > 0$ ). This is the figure's two arrows made numeric. Main exhaust (orange) $v_{e, c} = 3000$ m/s, dumped-gas (plum) $v_{e, g g} = 700$ m/s, and $f = 0.05$ . Find $I_{s p, eff}$ and compare to the "dump-everything" estimate. Use $g_{0} = 9.81$ .

Forecast: should the real penalty be more or less than Example 1's $5%$ ?

Step 1. Effective exhaust speed $v_{eff} = (1 - f) v_{e, c} + f v_{e, g g} = 0.95 (3000) + 0.05 (700)$ . Why this step? Thrust is a sum of each stream's momentum flux; dividing by total flow gives a flow-weighted exhaust speed.

Step 2. $v_{eff} = 2850 + 35 = 2885 m/s$ . Why this step? We just carry out the weighting arithmetic — the orange stream contributes $0.95 \times 3000 = 2850$ and the plum stream adds only $0.05 \times 700 = 35$ , showing at a glance how little the slow dump arrow matters.

Step 3. $I_{s p, eff} = 2885/9.81 = 294.1 s$ , while ideal $= 3000/9.81 = 305.8 s$ . Why compare? The gap $305.8 - 294.1 = 11.7$ s is the true penalty.

Verify: the naive "dump everything" estimate gives $0.05 \times 305.8 = 15.3$ s of loss. Our real loss $11.7 < 15.3$ s — the dumped gas's small kick recovered part of it, exactly as forecast (less penalty). Units: m/s ÷ (m/s²) = s ✓.

Worked example

Example 3 — Cells C & D: the two degenerate limits ( $f = 0$ and $f = 1$ ). Using the same two figure arrows, $v_{e, c} = 3000$ m/s and $v_{e, g g} = 700$ m/s, evaluate $I_{s p, eff}$ at the two extremes $f = 0$ (no plum arrow) and $f = 1$ (no orange arrow).

Forecast: which limit reproduces an ideal closed-cycle-like number, and which collapses to a pure gas-generator plume?

Step 1 — Case C, $f = 0$ . $v_{eff} = (1 - 0) 3000 + 0 \cdot 700 = 3000$ m/s → $I_{s p} = 3000/9.81 = 305.8$ s. Why this step? $f = 0$ means no propellant is dumped — the plum arrow disappears and the entire flow expands through the main nozzle. This is the "no dump loss" idealization.

Step 2 — Case D, $f = 1$ . $v_{eff} = 0 \cdot 3000 + 1 \cdot 700 = 700$ m/s → $I_{s p} = 700/9.81 = 71.4$ s. Why this step? $f = 1$ means everything goes through the turbine and gets dumped — the orange arrow disappears, there is no main chamber left, so you get the feeble dump-plume $I_{s p}$ .

Verify: the formula is linear in $f$ , so it must interpolate straight between these endpoints. At $f = 0.05$ we got 294.1 s (Ex 2), and indeed $305.8 + 0.05 (71.4 - 305.8) = 305.8 - 11.7 = 294.1$ s ✓. Both degenerate limits are physically sensible.

Worked example

Example 4 — Cell E: size the turbine flow from power balance. $\overset{m}{˙}_{t o t} = 300$ kg/s, $Δ p = 12$ MPa, $ρ = 1000$ kg/m³, pump efficiency $η_{p} = 0.7$ , turbine efficiency $η_{t} = 0.6$ , $c_{p} = 2000$ J/(kg·K), $T_{g g} = 1050$ K, and the pressure-ratio bracket $[1 - (p_{o u t} / p_{in})^{(γ - 1) / γ}] = 0.5$ (recall: "half the GG gas's heat becomes shaft work"). Find $\overset{m}{˙}_{g g}$ and $f$ .

Forecast: will $f$ land in the "few percent" band the parent note predicted?

Step 1. Pump power $P_{p u m p} = \frac{m ˙ _{t o t} Δ p}{ρ η _{p}} = \frac{300 \times 12 \times 1 0 ^{6}}{1000 \times 0.7}$ . Why this step? Power = (pressure rise)×(volume flow), and volume flow is $\overset{m}{˙} / ρ$ ; divide by efficiency for the real shaft demand.

Step 2. $P_{p u m p} = \frac{3.6 \times 1 0 ^{9}}{700} = 5.143 \times 1 0^{6}$ W. Why this step? Just evaluating the numbers — this is the hydraulic power the turbine must supply.

Step 3. Turbine specific work $w = η_{t} c_{p} T_{g g} \times 0.5 = 0.6 \times 2000 \times 1050 \times 0.5 = 6.3 \times 1 0^{5}$ J/kg. Why this step? Each kilogram of GG gas holds $c_{p} T_{g g}$ of heat; the bracket ( $0.5$ ) is the share that expansion turns into work, and $η_{t}$ trims for real-turbine losses.

Step 4. $\overset{m}{˙}_{g g} = P_{p u m p} / w = 5.143 \times 1 0^{6} /6.3 \times 1 0^{5} = 8.16 kg/s$ . Why this step? Power needed ÷ work-per-kilogram = kilograms-per-second of GG gas required.

Step 5. $f = \overset{m}{˙}_{g g} / \overset{m}{˙}_{t o t} = 8.16/300 = 0.0272 = 2.72%$ .

Verify: $f \approx 2.7%$ sits squarely in the 2–5% band — a few-percent penalty, as forecast. See Turbopump Fundamentals for why the pump term $Δ p / ρ$ dominates this bookkeeping.

Worked example

Example 5 — Cell F: the $Δ p$ trend (why high chamber pressure hurts). Take Example 4's engine but double the chamber pressure demand to $Δ p = 24$ MPa, everything else fixed. What happens to $f$ ?

Forecast: $f$ grows, shrinks, or stays the same?

Step 1. $\overset{m}{˙}_{g g} \propto Δ p$ (it sits alone on top of the fraction, all else constant). Why this step? In $\overset{m}{˙}_{g g} = \frac{m ˙ _{t o t} Δ p}{ρ η _{p} η _{t} c _{p} T _{g g} [ \dots ]}$ , doubling $Δ p$ doubles the numerator only.

Step 2. New $f = 2 \times 0.0272 = 0.0544 = 5.44%$ . Why this step? Linear scaling: double the pressure you must pump against → double the turbine flow you must burn and dump.

Verify: recompute from scratch — $P_{p u m p} = \frac{300 \times 24 \times 1 0 ^{6}}{700} = 1.0286 \times 1 0^{7}$ W, $w$ unchanged at $6.3 \times 1 0^{5}$ , so $\overset{m}{˙}_{g g} = 16.32$ kg/s and $f = 16.32/300 = 0.0544$ ✓. This is exactly why gas generator cycles lose their edge at very high $p_{c}$ — the penalty climbs with pressure. High- $p_{c}$ engines lean toward Staged Combustion Cycle instead.

Worked example

Example 6 — Cell G: temperature cap (how hot to hit a target $f$ ?). You want to hold $f$ down to $0.02$ on Example 4's engine (original $Δ p = 12$ MPa). All else fixed, what $T_{g g}$ is required — and is it safe?

Forecast: hotter or cooler than the 1050 K we used?

Step 1. From $\overset{m}{˙}_{g g} = P_{p u m p} / (η_{t} c_{p} T_{g g} \times 0.5)$ and $\overset{m}{˙}_{g g} = f \overset{m}{˙}_{t o t}$ , solve for $T_{g g}$ : $T_{g g} = \frac{P _{p u m p}}{f m ˙ _{t o t} η _{t} c _{p} \times 0.5} .$ Why this step? $\overset{m}{˙}_{g g}$ is inversely proportional to $T_{g g}$ , so demanding a smaller $f$ (smaller $\overset{m}{˙}_{g g}$ ) forces a larger $T_{g g}$ .

Step 2. $T_{g g} = \frac{5.143 \times 1 0 ^{6}}{0.02 \times 300 \times 0.6 \times 2000 \times 0.5} = \frac{5.143 \times 1 0 ^{6}}{3600} = 1428.6 K$ . Why this step? Straight substitution of the known $P_{p u m p}$ and target $f$ into the rearranged formula.

Step 3. Safety check: $1429$ K exceeds the practical $900$ – $1200$ K blade limit. Why this step? The parent note's second mistake — run too hot and you melt the turbine.

Verify: plug $T_{g g} = 1428.6$ K back: $w = 0.6 \times 2000 \times 1428.6 \times 0.5 = 8.571 \times 1 0^{5}$ J/kg, $\overset{m}{˙}_{g g} = 5.143 \times 1 0^{6} /8.571 \times 1 0^{5} = 6.0$ kg/s, $f = 6.0/300 = 0.02$ ✓. Conclusion: you cannot safely reach $f = 0.02$ this way — the blades would melt, so you accept the larger $f = 0.027$ from Example 4.

Worked example

Example 7 — Cell H: real-world thrust bookkeeping (Saturn-V style). A single gas-generator engine runs $\overset{m}{˙}_{t o t} = 2600$ kg/s with $f = 0.03$ , main exhaust (figure's orange arrow) $v_{e, c} = 2650$ m/s, dumped exhaust (plum arrow) $v_{e, g g} = 600$ m/s. Ignoring pressure terms, compute the total thrust two ways: (a) stream-by-stream momentum, (b) via $v_{eff}$ . They must match.

Forecast: roughly how many meganewtons — near 3, 7, or 20 MN?

Step 1 — main stream flow. $\overset{m}{˙}_{c} = (1 - f) \overset{m}{˙}_{t o t} = 0.97 \times 2600 = 2522$ kg/s; $\overset{m}{˙}_{g g} = 0.03 \times 2600 = 78$ kg/s. Why this step? Split the total into the two physical streams (the two figure arrows) before summing momentum.

Step 2 — momentum thrust (method a). $F = \overset{m}{˙}_{c} v_{e, c} + \overset{m}{˙}_{g g} v_{e, g g} = 2522 (2650) + 78 (600)$ . Why? Thrust from expelling mass is $\sum \overset{m}{˙} v_{e}$ (Rocket Thrust Equation, pressure-matched case).

Step 3. $F = 6, 683, 300 + 46, 800 = 6, 730, 100 N \approx 6.73 MN$ . Why this step? Just adding the two momentum fluxes — the orange arrow supplies almost all of it.

Step 4 — method b, via $v_{eff}$ . $v_{eff} = 0.97 (2650) + 0.03 (600) = 2570.5 + 18 = 2588.5$ m/s; $F = \overset{m}{˙}_{t o t} v_{eff} = 2600 \times 2588.5 = 6, 730, 100$ N. Why? $v_{eff}$ was defined as the flow-weighted average precisely so that $\overset{m}{˙}_{t o t} v_{eff}$ reproduces the stream sum.

Verify: both methods give $6.73$ MN — near the "7 MN" forecast, in the right family for a single big first-stage engine. Units: (kg/s)(m/s)=N ✓.

Recall

When $v_{e, g g} = 0$ , the penalty on $I_{s p}$ equals what quantity? ::: The turbine flow fraction $f$ itself: $I_{s p, eff} = (1 - f) I_{s p, ideal}$ . As $f \to 1$ , what does $I_{s p, eff}$ approach? ::: The feeble dump-plume value $v_{e, g g} / g_{0}$ — no main chamber left. Doubling $Δ p$ (chamber pressure demand) does what to $f$ ? ::: Doubles it — $\overset{m}{˙}_{g g} \propto Δ p$ , so high- $p_{c}$ engines pay a bigger dump penalty. Why can't you just raise $T_{g g}$ arbitrarily to shrink $f$ ? ::: Blade material melts above ~1200 K; running fuel-rich keeps $T_{g g} \approx 900$ – $1200$ K. What does the bracket $[1 - (p_{o u t} / p_{in})^{(γ - 1) / γ}]$ represent? ::: The fraction of the GG gas's heat that expansion (from $p_{in}$ to $p_{o u t}$ ) converts into turbine shaft work.