Worked examples — Expander cycle — hydrogen-cooled nozzle drives turbine

This page is the drill hall for the expander cycle. The parent note built the four equations; here we throw every kind of input at them — normal numbers, zeros, degenerate cases, limiting values, a word problem, and an exam twist — so you never meet a scenario cold.

Before we start, one reminder of the three tools we lean on (all from the parent):

Recall The three equations we will keep reusing

Heat absorbed in the cooling jacket ::: $Q=\dot{m}{c}_p(T_{out}-T_{in})$ Ideal turbine work ::: $W_{turb}=\dot{m}{c}_p{T}_1\left[1-{\left(\frac{p_2}{p_1}\right)}^{\frac{\gamma -1}{\gamma }}\right]{\eta }_t$ Pump power ::: $W_{pump}=\frac{\dot{m}\Delta p}{\rho {\eta }_p}$ The closing condition ::: $W_{turb}\ge W_{pump,fuel}+W_{pump,ox}$

Here $\dot{m}$ is mass flow rate (kilograms passing per second), $c_p$ is specific heat (joules to warm one kilogram by one kelvin), $T$ is temperature in kelvin, $p$ is pressure, $\rho$ (Greek "rho") is density (kilograms per cubic metre), $\gamma$ (Greek "gamma") is the heat-capacity ratio of the gas, and $\eta$ (Greek "eta") is an efficiency between 0 and 1 that discounts a real machine against its ideal. It comes in two flavours here: ${\eta }_t$ is the turbine efficiency (what fraction of the ideal enthalpy drop the turbine actually turns into shaft work), and ${\eta }_p$ is the pump efficiency (the ideal pumping power divided by the real, larger power a pump consumes). Every one of those was defined in the parent — if any feels shaky, re-read it there and come back.

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The scenario matrix

Think of every problem this topic can pose as one cell in this grid. Our examples below tick off every cell.

Cell	What makes it tricky	Example
A. Plain forward calc	Just plug numbers in cleanly	Ex 1 (heat), Ex 2 (turbine)
B. The closing inequality	Two pumps vs one turbine — does it sustain?	Ex 3
C. Zero / degenerate input	$\Delta T=0$, or $p_2=p_1$ (no pressure drop) → what happens?	Ex 4
D. Limiting value	$p_2/p_1\to 0$ (full expansion) — the ceiling of turbine work	Ex 5
E. Scaling / square–cube	Scale the engine by $k$ — supply vs demand race	Ex 6
F. Real-world word problem	Design a wall temperature rise to close the cycle	Ex 7
G. Exam twist	Open (bleed) cycle: only a fraction of flow drives the turbine	Ex 8

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Example 1 — Cell A: Heat picked up (plain forward)

Step 1 — Temperature rise. $\Delta T=T_{out}-T_{in}=400-50=350\text{K}$ Why this step? $c_p$ is defined per kelvin of rise, so we need the rise itself, not the endpoints.

Step 2 — Multiply through. $Q=\dot{m}{c}_p\Delta T=3\times 14000\times 350=1.47\times 10^7\text{W}=14.7\text{MW}$ Why this step? This is literally the definition of specific heat scaled up to a flow: joules-per-kg times kg-per-second gives joules-per-second, i.e. watts.

Verify: Units: $\frac{\text{kg}}{\text{s}}\cdot \frac{\text{J}}{\text{kg⋅K}}\cdot \text{K}=\frac{\text{J}}{\text{s}}=\text{W}$ ✅. And 14.7 MW sits between the 10 and 100 MW guesses — sensible for a real engine that would otherwise melt its nozzle.

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Example 2 — Cell A: Ideal turbine work (plain forward)

Step 1 — The pressure-ratio exponent. $\frac{\gamma -1}{\gamma }=\frac{0.4}{1.4}=0.2857$ Why this step? This exponent comes from the isentropic relation $T_2/T_1=(p_2/p_1{)}^{(\gamma -1)/\gamma }$ — it is what converts a pressure ratio into a temperature ratio. We use isentropic (constant entropy) because an ideal, loss-free turbine neither gains nor loses entropy.

Step 2 — Evaluate the ratio term. ${\left(\frac{8}{12}\right)}^{0.2857}=(0.6667{)}^{0.2857}=0.8900$ So the bracket $\left[1-0.8900\right]=0.1100$. Why this step? The bracket is the fraction of the inlet enthalpy the gas is willing to give up for this pressure drop. Small pressure drop → small fraction.

Step 3 — Assemble the power. $W_{turb}=\dot{m}{c}_p{T}_1[0.1100]{\eta }_t=3\times 14000\times 400\times 0.1100\times 0.75$ $W_{turb}\approx 1.386\times 10^6\text{W}\approx 1.39\text{MW}$ Why this step? $c_p{T}_1$ is the inlet enthalpy per kg; times mass flow gives an enthalpy rate; times the bracket gives the ideal extractable rate; times ${\eta }_t$ discounts for real losses.

Verify: The turbine took only $1.39/14.7\approx 9.4\%$ of the absorbed heat — matching the forecast that a gentle pressure drop yields a small slice. That is deliberate: the gas must stay high-pressure to still enter the combustion chamber afterward.

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Example 3 — Cell B: Does the cycle close?

Step 1 — Fuel pump power. $W_f=\frac{{\dot{m}}_f\Delta p}{{\rho }_f{\eta }_p}=\frac{3\times 13\times 10^6}{71\times 0.7}=7.85\times 10^5\text{W}$ Why this step? Pump power is volume-flow ($\dot{m}/\rho$) times pressure jump, divided by efficiency. Light, fluffy hydrogen has low density, so its volume flow is large — expensive to pump.

Step 2 — Oxidizer pump power. $W_{ox}=\frac{18\times 13\times 10^6}{1140\times 0.7}=2.93\times 10^5\text{W}$ Why this step? Same formula. Even though O₂ has 6× the mass, its density is 16× higher, so its volume flow is smaller → cheaper to pump.

Step 3 — Total demand and the comparison. $W_{pump}=7.85\times 10^5+2.93\times 10^5=1.08\times 10^6\text{W}=1.08\text{MW}$ $W_{turb}=1.39\text{MW}\ge 1.08\text{MW}=W_{pump}✅$ Why this step? This is the parent's self-sustaining condition. Surplus $=1.39-1.08=0.31$ MW is margin for startup and throttling.

Verify: The forecast was wrong — the fuel pump (0.79 MW) dwarfs the ox pump (0.29 MW), because pump work follows volume, not mass. This is a key expander-cycle lesson: low-density hydrogen is the pumping headache. See Turbopump design and Liquid hydrogen properties.

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Example 4 — Cell C: Degenerate inputs (zeros)

Step 1 — Zero temperature rise. $Q=\dot{m}{c}_p(T_{out}-T_{in})=3\times 14000\times (50-50)=0\text{W}$ Why this step? No warming means no heat crossed into the fluid. Physically: the wall isn't hot (engine off, or perfect insulation) — there is no free energy to harvest.

Step 2 — Zero pressure ratio term. ${\left(\frac{p_2}{p_1}\right)}^{\frac{\gamma -1}{\gamma }}={\left(\frac{12}{12}\right)}^{0.2857}=1^{0.2857}=1$ $W_{turb}=\dot{m}{c}_p{T}_1[1-1]{\eta }_t=0\text{W}$ Why this step? A turbine extracts work only from a drop in pressure. Equal pressures = the gas glides through untouched = no shaft work. Physically: a stalled or bypassed turbine.

Verify: Both zero, as forecast. The bracket $[1-(p_2/p_1{)}^{(\gamma -1)/\gamma }]$ elegantly vanishes at $p_2=p_1$ and grows as the ratio shrinks — the formula is self-guarding against the degenerate case. If Example 4b happened in flight, the pumps would receive zero drive and the engine would collapse — the cycle does not close at zero drop.

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Example 5 — Cell D: The limiting value (full expansion)

Step 1 — Take the limit of the ratio term. ${\lim }_{p_2\to 0}{\left(\frac{p_2}{p_1}\right)}^{0.2857}=0\Rightarrow [1-0]=1$ Why this step? Any positive base below 1 raised to a positive power, as the base $\to 0$, tends to 0. So the bracket climbs to its ceiling of exactly 1 — the turbine gives up the entire inlet enthalpy.

Step 2 — Evaluate the ceiling. $W_{max}=\dot{m}{c}_p{T}_1(1){\eta }_t=3\times 14000\times 400\times 0.75=1.26\times 10^7\text{W}=12.6\text{MW}$ Why this step? This is the absolute maximum the 400 K gas stream can deliver: its full enthalpy rate $\dot{m}{c}_p{T}_1=16.8$ MW, discounted by ${\eta }_t=0.75$.

Verify: $12.6\text{MW}<14.7\text{MW}$ of absorbed heat — good, you can never extract more work than the heat you put in (that would break the energy budget). Compare to Example 2's gentle drop (1.39 MW): full expansion could give 9× more, but then the gas would be near-vacuum and useless for feeding the chamber. This is the design tension — expanders sacrifice turbine power to preserve chamber-feed pressure.

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Example 6 — Cell E: Square–cube scaling

Step 1 — Scale each side. Let the baseline supply and demand both equal $P_0$. $\text{Supply}(k)=P_0{k}^2,\text{Demand}(k)=P_0{k}^3$ Why this step? This is the Square-cube law: area grows as the square of size, volume (and the flow it commands) as the cube.

Step 2 — Plug in $k=2$. $\text{Supply}=P_0\cdot 4,\text{Demand}=P_0\cdot 8$ $\frac{\text{Supply}}{\text{Demand}}=\frac{4}{8}=0.5$ Why this step? We compare the two rates directly. At $k=1$ the ratio was 1 (just closing); at $k=2$ it has fallen to 0.5.

Verify: The doubled engine can supply only half the power its pumps demand — it fails to close. The forecast is wrong: bigger is worse for a closed expander. This is exactly why the RL10 stays in the few-hundred-kN class while a Staged combustion cycle engine can be huge.

The figure below plots both curves against the scale factor $k$. Follow the chalk-blue supply line ($k^2$) and the chalk-pink demand line ($k^3$): they cross exactly at $k=1$ (the pale-yellow dot where the engine just closes), and past that point the pink demand curve pulls away above the blue — the shaded pink region is the power deficit. The second pair of yellow dots marks $k=2$, where supply has reached 4 but demand has raced to 8.

Figure s01 — Square–cube law: heat supply ($\propto k^2$, blue) versus pump demand ($\propto k^3$, pink). They meet at $k=1$; beyond it demand outruns supply (pink shaded deficit), so a scaled-up closed expander can no longer power its own pumps.

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Example 7 — Cell F: Real-world design problem

Step 1 — Reuse Example 2's bracket (same pressures and $\gamma$): the ratio term $=0.8900$, so the bracket $=0.1100$. Why this step? Nothing changed about the pressure ratio, so this dimensionless factor is identical — no need to recompute.

Step 2 — Solve the turbine equation for $T_1$. $W_{turb}=\dot{m}{c}_p{T}_1[0.1100]{\eta }_t\ge 2.0\times 10^6$ $T_1\ge \frac{2.0\times 10^6}{4\times 14000\times 0.1100\times 0.75}=\frac{2.0\times 10^6}{4620}=432.9\text{K}$ Why this step? We invert the forward formula — this is reverse-engineering a requirement. Every factor except $T_1$ is fixed, so $T_1$ is forced.

Step 3 — Translate to a wall heat requirement. If H₂ enters the jacket at 50 K, the wall must add $\Delta T=432.9-50=382.9\text{K}$, absorbing $Q=\dot{m}{c}_p\Delta T=4\times 14000\times 382.9=2.144\times 10^7\text{W}\approx 21.4\text{MW}.$ Why this step? Closes the loop back to the cooling jacket: the required turbine temperature dictates how much heat the walls must dump into the coolant.

Verify: $432.9\text{K}>400\text{K}$ — higher, as forecast, because we asked for more power than the modest flow increase alone provides. Sanity: $21.4$ MW of wall heat $>$ the $2.0$ MW extracted (turbine takes only ~9%), consistent with Example 2's ~9.4% extraction fraction. Feasible.

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Example 8 — Cell G: The exam twist (open / bleed cycle)

Step 1 — Turbine sees only the bled flow. ${\dot{m}}_{turb}=f\dot{m}=0.6\times 3=1.8\text{kg/s}$ Why this step? In an open cycle the turbine's working mass is only the bled fraction — the rest bypasses the turbine and goes straight to the chamber.

Step 2 — Apply the turbine formula to ${\dot{m}}_{turb}$. The bracket is still $0.1100$ (same pressures and $\gamma$ as Example 2). $W_{turb}={\dot{m}}_{turb}{c}_p{T}_1[0.1100]{\eta }_t=1.8\times 14000\times 400\times 0.1100\times 0.75$ $W_{turb}\approx 8.32\times 10^5\text{W}\approx 0.83\text{MW}$ Why this step? Turbine power is linear in mass flow, so scaling the working flow by $f=0.6$ scales the power by 0.6. This is the whole arithmetic of the open cycle.

Step 3 — Why accept the loss (part b)? The bled gas is dumped, not fed to the chamber, so it need not stay high-pressure afterward. That frees the designer to use a much bigger pressure drop across the turbine — a smaller $p_2$ pushes the bracket $[1-(p_2/p_1{)}^{(\gamma -1)/\gamma }]$ up toward its ceiling of 1 (recall Example 5), so each kilogram of bled hydrogen yields far more work. A small engine can therefore raise enough turbine power to drive its pumps even when a closed expander of the same size would fail to close. Why this step? It resolves the trade: you sacrifice a slice of specific impulse (the dumped hydrogen never burns, so its energy leaves unused) in exchange for more pump-driving muscle. Compare Gas generator cycle, which dumps combustion products for exactly the same reason.

Verify: $0.83=0.6\times 1.39$ ✅ — exactly the forecast fraction, confirming turbine power's linearity in $\dot{m}$. Physical sanity check: the dumped 40% of hydrogen (1.2 kg/s) exits unburned, so the chamber gets less fuel per second than a closed cycle would deliver — that is precisely the $I_{sp}$ penalty being traded away.

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Recall Self-test: which cell?

Which matrix cell asks "what if $p_2=p_1$?" ::: Cell C — degenerate/zero input (Ex 4). Which cell shows bigger engines fail to close? ::: Cell E — square–cube scaling (Ex 6). In Ex 3, which pump dominates the demand and why? ::: The fuel (H₂) pump — hydrogen's low density means huge volume flow. Full expansion ($p_2\to 0$) caps turbine work at what fraction of inlet enthalpy? ::: ${\eta }_t$ times the whole enthalpy $\dot{m}{c}_p{T}_1$ (bracket $\to 1$). In the open (bleed) cycle of Ex 8, why can it use a bigger turbine pressure drop? ::: The bled gas is dumped overboard, so it needn't stay high-pressure to feed the chamber.