Gas generator cycle — performance penalty vs simplicity: Exercises

Before we start, here is the one toolkit you will reuse. Read it once so no symbol surprises you later.

Formula

The four equations for this whole page $I_{s p} = \frac{v _{e}}{g _{0}} (how fast exhaust leaves \div Earth gravity g_{0} = 9.81 m/s^{2})$ $v_{eff} = (1 - f) v_{e, c} + f v_{e, g g} (flow-weighted average exhaust speed)$ $P_{pump} = \frac{m ˙ _{t o t} Δ p}{ρ η _{p}} (power to pressurize the propellant)$ $\overset{m}{˙}_{g g} = \frac{P _{pump}}{η _{t} c _{p} T _{g g} B}, B = 1 - (\frac{p _{o u t}}{p _{in}})^{\frac{γ - 1}{γ}}$

Here $\overset{m}{˙}_{t o t}$ is the total propellant mass-flow rate (kg/s) — every kilogram per second the engine swallows, main chamber plus gas generator combined. $\overset{m}{˙}_{g g}$ is the slice of that flow sent to the gas generator, so $f = \overset{m}{˙}_{g g} / \overset{m}{˙}_{t o t}$ is the turbine flow fraction. Next, $v_{e, c}$ is main-chamber exhaust speed and $v_{e, g g}$ is the dumped-gas exhaust speed, $ρ$ is propellant density, $η_{p}, η_{t}$ are pump and turbine efficiencies, $c_{p}$ is the gas's heat capacity, and $T_{g g}$ is gas-generator temperature.

Finally $B$ is the pressure-ratio bracket — a pure number between 0 and 1 that says "what fraction of the gas's heat the turbine can actually extract". Inside it: $p_{in}$ is the gas pressure entering the turbine, $p_{o u t}$ is the (lower) pressure leaving it, and $γ$ (the Greek letter "gamma") is the gas's heat-capacity ratio — a property of the exhaust gas, typically about $1.2$ – $1.3$ , that sets how strongly pressure and temperature track each other during expansion. When $p_{o u t}$ is nearly $p_{in}$ the bracket is near 0 (little energy to grab); when $p_{o u t} ≪ p_{in}$ it climbs toward 1.

Level 1 — Recognition

L1.1 In one sentence, why does a gas generator cycle lose specific impulse compared to a closed cycle?

Recall Solution

Because the turbine-drive gas is dumped overboard at low pressure instead of being fed to the main chamber. It expands across almost no pressure ratio, so it leaves with a low exhaust speed and contributes almost no thrust — yet you still carried and burned it, so it drags the flow-weighted average $I_{s p}$ down.

L1.2 An engine burns $\overset{m}{˙}_{t o t} = 300$ kg/s total, of which $\overset{m}{˙}_{g g} = 12$ kg/s goes to the gas generator. What is the turbine flow fraction $f$ ?

Recall Solution

$f = \frac{m ˙ _{g g}}{m ˙ _{t o t}} = \frac{12}{300} = 0.04 = 4%.$ Why this formula? $f$ is defined as "how big a slice of the total flow is spent on the turbine". It is a pure ratio, so units of kg/s cancel.

L1.3 True or false: the gas generator runs at stoichiometric temperature (~3500 K) to maximize turbine power.

Recall Solution

False. At 3500 K the turbine blades would melt. The GG deliberately runs fuel-rich (or ox-rich) to keep $T_{g g} \approx 900$ – $1200$ K. You accept less power per kg in exchange for blades that survive.

Level 2 — Application

L2.1 An engine has ideal $I_{s p} = 340$ s and dumps $f = 3%$ of its flow with negligible useful exhaust speed. Estimate the effective $I_{s p}$ .

Recall Solution

Use the "dump-everything" approximation $I_{s p, eff} \approx (1 - f) I_{s p, ideal}$ . Why? If the dumped gas gives zero thrust, it only adds mass to the denominator, so it scales $I_{s p}$ down by the surviving fraction $(1 - f)$ . $I_{s p, eff} = (1 - 0.03) (340) = 0.97 \times 340 = 329.8 s .$ Penalty $\approx 10.2$ s.

L2.2 Main exhaust $v_{e, c} = 3000$ m/s, dumped gas $v_{e, g g} = 600$ m/s, $f = 0.05$ . Find the effective exhaust speed and the effective $I_{s p}$ ( $g_{0} = 9.81$ ).

Recall Solution

Step 1 — weighted average. Why? Momentum adds up; total thrust is each stream's mass-flow times its speed, so the average speed is flow-weighted. $v_{eff} = (1 - 0.05) (3000) + (0.05) (600) = 2850 + 30 = 2880 m/s .$ Step 2 — convert to $I_{s p}$ . $I_{s p, eff} = \frac{2880}{9.81} = 293.6 s .$

L2.3 For the same engine, what would the ideal $I_{s p}$ be (all flow at $v_{e, c}$ ), and what is the penalty in seconds?

Recall Solution

Ideal: $I_{s p, ideal} = 3000/9.81 = 305.8$ s. Penalty: $305.8 - 293.6 = 12.2$ s. Why smaller than the " $5%$ of 305.8 = 15 s" you'd guess? Because the dumped gas still leaves at 600 m/s, giving a tiny kick. The real penalty is always a little less than the full $f$ estimate.

Level 3 — Analysis

L3.1 An engine has $v_{e, c} = 3200$ m/s and $v_{e, g g} = 750$ m/s. Mission rules cap the $I_{s p}$ penalty at 8 s. What is the largest turbine fraction $f$ you may use? ( $g_{0} = 9.81$ .)

Recall Solution

Step 1 — write the penalty in seconds. Ideal minus effective: $Δ I_{s p} = \frac{v _{e, c}}{g _{0}} - \frac{( 1 - f ) v _{e, c} + f v _{e, g g}}{g _{0}} = \frac{f ( v _{e, c} - v _{e, g g} )}{g _{0}} .$ Why does $v_{e, c}$ cancel out of the difference? Because the $(1 - f) v_{e, c}$ term differs from full $v_{e, c}$ by exactly $- f v_{e, c}$ , and the $+ f v_{e, g g}$ adds back a bit. The gap is the speed lost per dumped kg, $v_{e, c} - v_{e, g g}$ , times the fraction $f$ . Step 2 — set equal to 8 s and solve for $f$ . $f = \frac{Δ I _{s p} \cdot g _{0}}{v _{e, c} - v _{e, g g}} = \frac{8 \times 9.81}{3200 - 750} = \frac{78.48}{2450} = 0.0320.$ So $f \leq 3.2%$ (about 3.2%).

L3.2 Look at the penalty formula from L3.1. If you could raise the dumped-gas speed $v_{e, g g}$ (e.g. by routing it through a small nozzle), does the allowed $f$ go up or down for the same 8 s budget? Explain using the figure.

Recall Solution

Figure — Gas generator cycle — performance penalty vs simplicity

Raising

v_{e, g g}

shrinks the gap

(v_{e, c} - v_{e, g g})

in the denominator, so for a fixed 8 s budget the allowed

f

goes up — you can afford to dump more flow because each dumped kg now hurts less. Geometrically (red bar in the figure), the "lost speed per kg" bar gets shorter as the yellow

v_{e, g g}

marker climbs toward the blue

v_{e, c}

marker.

Level 4 — Synthesis

L4.1 A GG engine has $\overset{m}{˙}_{t o t} = 250$ kg/s, pressure rise $Δ p = 20$ MPa, propellant density $ρ = 1000$ kg/m³, pump efficiency $η_{p} = 0.75$ , turbine efficiency $η_{t} = 0.65$ , $c_{p} = 2200$ J/(kg·K), $T_{g g} = 1050$ K, and bracket $B = 0.55$ . Find the required turbine flow $\overset{m}{˙}_{g g}$ and the fraction $f$ .

Recall Solution

Step 1 — pump power. Why? The turbine's whole job is to supply the power that pressurizes the propellant. Power = pressure-rise × volume-flow, degraded by pump efficiency. $P_{pump} = \frac{m ˙ _{t o t} Δ p}{ρ η _{p}} = \frac{250 \times 20 \times 1 0 ^{6}}{1000 \times 0.75} = \frac{5 \times 1 0 ^{9}}{750} = 6.667 \times 1 0^{6} W .$ Step 2 — turbine work per kg. Why? Each kg of GG gas can give up this much energy as it drops in temperature. $w = η_{t} c_{p} T_{g g} B = 0.65 \times 2200 \times 1050 \times 0.55 = 8.258 \times 1 0^{5} J/kg .$ Step 3 — divide. Total power needed ÷ work each kg supplies: $\overset{m}{˙}_{g g} = \frac{P _{pump}}{w} = \frac{6.667 \times 1 0 ^{6}}{8.258 \times 1 0 ^{5}} = 8.073 kg/s .$ Step 4 — fraction. $f = \frac{8.073}{250} = 0.0323 = 3.2%.$

L4.2 For that same engine, if the ideal $I_{s p}$ is 335 s and the dumped gas is treated as useless, what effective $I_{s p}$ does this $f$ cost you?

Recall Solution

Step 1 — pick the right model. Why? The problem says the dumped gas is useless, so its exhaust speed contributes nothing to the numerator of the weighted average — only its mass survives in the denominator. That collapses the full weighted average into the simple scaling $I_{s p, eff} = (1 - f) I_{s p, ideal}$ . Step 2 — substitute. Why these numbers? $f = 0.0323$ came straight from L4.1's power balance, and $I_{s p, ideal} = 335$ s is the value all the flow would give if none were dumped. $I_{s p, eff} = (1 - f) I_{s p, ideal} = (1 - 0.0323) (335) = 0.9677 \times 335 = 324.2 s .$ Penalty $\approx 10.8$ s — a typical GG-cycle tax.

Level 5 — Mastery

L5.1 You must choose between two designs for a fixed mission with $v_{e, c} = 3300$ m/s, $g_{0} = 9.81$ :

Design A (GG cycle): $f = 0.045$ , dumped-gas speed $v_{e, g g} = 700$ m/s.

Design B (closed cycle): no dump, but the extra plumbing adds dry mass that effectively lowers usable $I_{s p}$ by a flat 6 s from the ideal.

Which design delivers the higher effective $I_{s p}$ , and by how much?

Recall Solution

Step 1 — Design A effective speed (weighted average). Why start here? Design A dumps a real fraction of flow at a real speed, so we need the flow-weighted average of the two streams before converting to $I_{s p}$ — a simple $(1 - f) v_{e, c}$ scaling would ignore the small kick the 700 m/s gas still gives. $v_{eff, A} = (1 - 0.045) (3300) + (0.045) (700) = 3151.5 + 31.5 = 3183.0 m/s .$ Step 2 — turn Design A's speed into $I_{s p}$ . Why divide by $g_{0}$ ? That is the definition $I_{s p} = v_{e} / g_{0}$ ; it converts a speed in m/s into the "seconds" that specific impulse is measured in. $I_{s p, A} = \frac{3183.0}{9.81} = 324.5 s .$ Step 3 — Design B. Why just subtract 6 s? Design B dumps nothing, so its exhaust reaches the full ideal $I_{s p}$ ; the only loss is the flat 6 s the problem assigns to its heavier plumbing, so we subtract it directly. $I_{s p, B} = \frac{3300}{9.81} - 6 = 336.4 - 6 = 330.4 s .$ Step 4 — compare. Why subtract the two? The question asks "by how much", i.e. the gap between the two effective values. $330.4 - 324.5 = 5.9 s .$ Design B wins by about 5.9 s. The GG dump penalty here (~11.9 s) is bigger than the closed-cycle's complexity tax (6 s), so the closed cycle is the better performer — at the cost of the higher complexity the Staged Combustion Cycle note describes.

L5.2 For Design A above, at what dumped-gas speed $v_{e, g g}$ would the two designs tie?

Recall Solution

Step 1 — state the tie condition. Why? "Tie" means the two effective specific impulses are equal, so we set $I_{s p, A} = I_{s p, B} = 330.4$ s and read off the exhaust speed Design A would need. Step 2 — convert that target $I_{s p}$ back to a speed. Why multiply by $g_{0}$ ? $I_{s p} = v / g_{0}$ rearranges to $v = I_{s p} g_{0}$ ; we invert the definition to get the required $v_{eff}$ . $v_{eff, A}^{tie} = 330.4 \times 9.81 = 3241.2 m/s .$ Step 3 — solve the weighted average for $v_{e, g g}$ . Why isolate $v_{e, g g}$ ? Everything else in the weighted average is fixed ( $f = 0.045$ , $v_{e, c} = 3300$ ), so the one unknown is the dumped-gas speed; we peel the known term off and divide by its coefficient $f$ . $3241.2 = (0.955) (3300) + 0.045 v_{e, g g} = 3151.5 + 0.045 v_{e, g g},$ $0.045 v_{e, g g} = 3241.2 - 3151.5 = 89.7 \Rightarrow v_{e, g g} = \frac{89.7}{0.045} = 1993 m/s .$ So the GG cycle would need to recover its dumped gas to ~1993 m/s (nearly a real nozzle's worth) to match the closed cycle — which is why, in practice, closed cycles win on pure performance. See Specific Impulse and Exhaust Velocity and Nozzle Expansion and Pressure Ratio for why recovering that much dumped-gas speed is so hard.

Recall

One-line self-check: penalty in seconds ::: $Δ I_{s p} = f (v_{e, c} - v_{e, g g}) / g_{0}$ One-line self-check: turbine flow ::: $\overset{m}{˙}_{g g} = P_{pump} / (η_{t} c_{p} T_{g g} B)$ , with $P_{pump} = \overset{m}{˙}_{t o t} Δ p / (ρ η_{p})$

See also: Turbopump Fundamentals · Rocket Thrust Equation · Specific Impulse and Exhaust Velocity.