Expander cycle — hydrogen-cooled nozzle drives turbine: Exercises

Before we start, let us pin down every symbol so nothing is used unexplained.

Definition The symbols, in plain words

$\overset{m}{˙}$ — mass flow rate: how many kilograms of fluid pass a point each second, in $kg/s$ . Picture a hose: $\overset{m}{˙}$ is the "kg per second" gushing out.
$c_{p}$ — specific heat at constant pressure: energy (joules) needed to warm $1 kg$ of the gas by $1 K$ . Big $c_{p}$ = the gas is a greedy sponge for heat.
$T$ — temperature in kelvin ( $K$ ). Kelvin is just Celsius shifted so $0 K$ is absolute cold; we use it because gas laws need an absolute zero.
$p$ — pressure in pascals ( $Pa$ ), or $MPa = 1 0^{6} Pa$ . Push per unit area.
$ρ$ (rho) — density: kilograms packed into one cubic metre, $kg/m^{3}$ .
$γ$ (gamma) — heat-capacity ratio of the gas, a pure number (for H₂ we take $1.4$ ). It controls how temperature falls when pressure falls.
$η$ (eta) — an efficiency, a number between 0 and 1. $η_{t}$ for the turbine, $η_{p}$ for the pump. Real machines waste some energy, so we multiply (turbine) or divide (pump) by $η$ .
$Q$ — heat rate, joules per second = watts ( $W$ ). $1 MW = 1 0^{6} W$ .
$W$ — power (also watts). $W_{t u r b}$ is what the turbine makes; $W_{p u m p}$ is what a pump eats.

Level 1 — Recognition

Can you read the cycle and the symbols?

L1.1

In one sentence, what supplies the energy that spins the turbine in a closed expander cycle?

Recall Solution

The heat conducted through the hot nozzle/chamber walls into the hydrogen coolant. Nothing extra is burned for the turbine. The hydrogen is only burned afterwards, in the main chamber.

L1.2

Match each symbol to its unit: $\overset{m}{˙}$ , $c_{p}$ , $Q$ , $ρ$ .

Recall Solution

$\overset{m}{˙} \to kg/s$
$c_{p} \to J/(kg\cdotK)$ (here $14000 J/(kg\cdotK)$ )
$Q \to W$ (watts)
$ρ \to kg/m^{3}$

L1.3

True or false: in a closed expander cycle, some hydrogen is dumped overboard after the turbine.

Recall Solution

False. That is the open (bleed) expander. In the closed cycle all the heated hydrogen passes through the turbine and then into the chamber — nothing dumped. See Gas generator cycle for the cycle that does dump gas.

Level 2 — Application

Plug into one equation cleanly.

L2.1

Hydrogen flows at $\overset{m}{˙} = 4 kg/s$ , enters the jacket at $T_{in} = 45 K$ , leaves at $T_{o u t} = 420 K$ , with $c_{p} = 14000 J/(kg\cdotK)$ . Find the heat rate $Q$ absorbed.

Recall Solution

$Q = \overset{m}{˙} c_{p} (T_{o u t} - T_{in}) = 4 \times 14000 \times (420 - 45)$ $Q = 4 \times 14000 \times 375 = 2.1 \times 1 0^{7} W = 21 MW$ What we did: multiplied "kg per second" by "joules per kg per kelvin" by "kelvin risen" → joules per second = watts. Why: this is literally the definition of $c_{p}$ run in the forward direction. That $21 MW$ is heat stolen from the flame that both cools the wall and funds the turbine.

L2.2

A pump raises $\overset{m}{˙} = 4 kg/s$ of hydrogen ( $ρ = 71 kg/m^{3}$ ) by $Δ p = 14 MPa$ with $η_{p} = 0.7$ . Find pump power.

Recall Solution

$W_{p u m p} = \frac{m ˙ Δ p}{ρ η _{p}} = \frac{4 \times 14 \times 1 0 ^{6}}{71 \times 0.7}$ $W_{p u m p} = \frac{5.6 \times 1 0 ^{7}}{49.7} \approx 1.127 \times 1 0^{6} W \approx 1.13 MW$ Why the $ρ$ in the bottom: the pump works on volume, not mass. Volume flow is $\overset{m}{˙} / ρ$ ; low density means big volume, so hydrogen's tiny $ρ$ makes it expensive to pump. See Liquid hydrogen properties.

Level 3 — Analysis

Combine two or three steps and reason about the result.

L3.1

Hot hydrogen at $T_{1} = 420 K$ , $\overset{m}{˙} = 4 kg/s$ , expands from $p_{1} = 13 MPa$ to $p_{2} = 9 MPa$ . Take $γ = 1.4$ , $η_{t} = 0.75$ , $c_{p} = 14000$ . Find the turbine power.

Recall Solution

First the exponent: $\frac{γ - 1}{γ} = \frac{0.4}{1.4} = 0.2857$ . Pressure ratio: $\frac{p _{2}}{p _{1}} = \frac{9}{13} = 0.6923$ . Raise to the power: $0.692 3^{0.2857}$ . Using $ln (0.6923) = - 0.3677$ , times $0.2857 = - 0.1051$ , $e^{- 0.1051} = 0.9003$ . $W_{t u r b} = 4 \times 14000 \times 420 \times (1 - 0.9003) \times 0.75$ $= 23520000 \times 0.0997 \times 0.75 \approx 1.759 \times 1 0^{6} W \approx 1.76 MW$ Why the pressure-ratio term: during an isentropic (ideal, no-loss) expansion, temperature and pressure are locked together by $T_{2} / T_{1} = (p_{2} / p_{1})^{(γ - 1) / γ}$ — see Isentropic flow relations. The bracket $1 - (p_{2} / p_{1})^{(γ - 1) / γ}$ is just the fraction of temperature lost, which (since $h = c_{p} T$ ) is the fraction of enthalpy handed to the shaft.

L3.2

Using L3.1's turbine and two pumps — fuel from L2.2 ( $1.13 MW$ ) plus an oxygen pump moving $\overset{m}{˙}_{o x} = 22 kg/s$ ( $ρ_{o x} = 1140 kg/m^{3}$ ) up $Δ p = 14 MPa$ at $η_{p} = 0.7$ — does the cycle close?

Recall Solution

Ox pump: $W_{o x} = \frac{22 \times 14 \times 1 0 ^{6}}{1140 \times 0.7} = \frac{3.08 \times 1 0 ^{8}}{798} \approx 3.86 \times 1 0^{5} W \approx 0.386 MW$ Total pump demand: $W_{p u m p} = 1.13 + 0.386 = 1.52 MW$ Compare to turbine $1.76 MW$ : $1.76 MW > 1.52 MW ✓$ The cycle sustains itself with $\approx 0.24 MW$ of margin (used for startup, throttle, control). Notice: the ox pump is cheap ( $0.39 MW$ ) despite moving far more mass, because oxygen is $\sim 16 \times$ denser than hydrogen → tiny volume flow. All the pump cost lives in the light, fluffy hydrogen.

Figure — Expander cycle — hydrogen-cooled nozzle drives turbine

Level 4 — Synthesis

Build a chain of reasoning, include limiting/degenerate cases.

L4.1

An engineer proposes raising the turbine pressure ratio from $p_{2} / p_{1} = 9/13$ to $p_{2} / p_{1} = 5/13$ (a much deeper drop) to squeeze out more turbine power. Compute the new $W_{t u r b}$ (same $\overset{m}{˙} = 4$ , $T_{1} = 420$ , $c_{p} = 14000$ , $γ = 1.4$ , $η_{t} = 0.75$ ). Then state the hidden penalty.

Recall Solution

New ratio $5/13 = 0.3846$ . $ln (0.3846) = - 0.9555$ , times $0.2857 = - 0.2730$ , $e^{- 0.2730} = 0.7611$ . $W_{t u r b} = 4 \times 14000 \times 420 \times (1 - 0.7611) \times 0.75$ $= 23520000 \times 0.2389 \times 0.75 \approx 4.214 \times 1 0^{6} W \approx 4.21 MW$ That is far more power. The hidden penalty: after the turbine, the hydrogen must still be injected into the combustion chamber, which sits at high pressure. If you drop the gas to $5 MPa$ but the chamber needs $\sim 9$ – $10 MPa$ , the gas can no longer get in — you would need the pumps to pre-pressurise even higher, eating the gain. Limiting case: as $p_{2} \to 0$ the bracket $\to 1$ and $W_{t u r b} \to \overset{m}{˙} c_{p} T_{1} η_{t}$ (a hard ceiling — you cannot extract more than all the enthalpy), but $p_{2} = 0$ means the gas has zero pressure and cannot enter the chamber at all. The closed expander therefore deliberately keeps only a modest drop.

L4.2 (degenerate case)

What is $W_{t u r b}$ if $p_{2} = p_{1}$ (no pressure drop at all)? Interpret physically.

Recall Solution

$p_{2} / p_{1} = 1$ , so $(1)^{0.2857} = 1$ , and the bracket $1 - 1 = 0$ : $W_{t u r b} = \overset{m}{˙} c_{p} T_{1} \times 0 \times η_{t} = 0 W$ Physical meaning: a turbine extracts work only from a pressure drop. No drop → no expansion → no work, no matter how hot the gas is. Heat alone is useless to a turbine; you need a pressure difference for the gas to fall through. This is why the fuel pump must first build pressure the turbine can later spend.

Level 5 — Mastery

Full synthesis, scaling law, design judgement.

L5.1 — The square-cube ceiling

A baseline closed expander produces $Q_{ba se} = 21 MW$ of jacket heat and needs $W_{ba se} = 1.52 MW$ of pump power (from L2/L3). You scale every linear dimension of the engine by $k = 2.5$ . Heat scales as surface area $\propto k^{2}$ ; pump demand scales with propellant/volume flow $\propto k^{3}$ . Find the scaled heat and demand, and the ratio (heat available):(pump demand relative to baseline).

Recall Solution

Scaled heat supply factor: $k^{2} = 2. 5^{2} = 6.25$ . Scaled pump demand factor: $k^{3} = 2. 5^{3} = 15.625$ . $Q_{sc a l e d} = 6.25 \times 21 = 131.25 MW, W_{sc a l e d} = 15.625 \times 1.52 = 23.75 MW$ The heat-to-demand advantage relative to baseline: $\frac{k ^{2}}{k ^{3}} = \frac{1}{k} = \frac{1}{2.5} = 0.4$ So a $2.5 \times$ -larger engine has only $40%$ as much relative heat headroom. Interpretation: demand outgrows supply as you scale up. Eventually $k$ is large enough that the walls physically cannot pass enough heat to run the pumps — the cycle stops closing. This is the Square-cube law and it caps closed expanders near a few hundred kN of thrust (e.g. the RL10). It is geometry, not engineering laziness.

L5.2 — Design choice under the ceiling

You need a big engine ( $\sim 2 MN$ ) but love the "burn nothing extra" idea. Given the L5.1 result, which cycle do you pick, and what one trick could keep an expander-like idea alive at larger size?

Recall Solution

At $\sim 2 MN$ the closed expander cannot supply enough wall heat (square-cube). You would choose a Staged combustion cycle (highest performance) or Gas generator cycle (simpler, dumps a little gas). The trick to stretch the expander idea: go to an open (bleed) expander — route only a fraction of the hydrogen through the turbine and dump it overboard. This lets you run more turbine flow at higher drop (more power) at the cost of some Specific impulse, sidestepping the closed-cycle power cap while still using free wall heat. See Turbopump design for how the shaft power sizes the pumps.

Recall Quick self-check (cloze)

The turbine's energy in a closed expander comes from heat conducted through the nozzle walls, not from burning anything extra. Pump power depends on ==volume flow ( $\overset{m}{˙} / ρ$ )==, so the dense propellant (oxygen) is cheap to pump. A turbine with zero pressure drop produces zero work regardless of temperature. Heat scales as $k^{== 2 ==}$ but pump demand scales as $k^{== 3 ==}$ , which caps closed expander thrust.

Turbine power depends on the pressure ratio, not just the temperature — why? ::: Because for an isentropic expansion $T_{2} / T_{1} = (p_{2} / p_{1})^{(γ - 1) / γ}$ , so the enthalpy drop is set by the pressure ratio. Why can't a closed expander scale to 2 MN? ::: Heat supply grows as surface area ( $k^{2}$ ) but pump demand grows as flow ( $k^{3}$ ); the ratio falls as $1/ k$ until the walls can't feed the pumps.