Exercises — Turbopump design — centrifugal pump, axial turbine stages, NPSH

• (a) $c_{u2}$ — exit whirl velocity of the liquid at the impeller rim → belongs to the centrifugal pump (it sets the Euler head $H=u_2{c}_{u2}/g$).
• (b) ${\text{NPSH}}_A$ — available suction head above vapor pressure → belongs to the NPSH / cavitation pillar (see Cavitation).
• (c) $\Delta c_u$ — change in whirl as gas crosses a rotor row → belongs to the axial turbine stage (work $w=u\Delta c_u$, the Euler Turbomachinery Equation).

No inlet swirl means ==$c_{u1}=0$==, so the second term vanishes and $H=\frac{u_2{c}_{u2}}{g}$. Why designers want it: feeding the liquid in radially/axially means this impeller adds all the whirl, maximising head for a given rim speed $u_2$ and keeping the formula simple.

Convert rpm → rad/s (why: the Euler relations use $\omega$ in rad/s, not rev/min): $\omega =\frac{2\pi N}{60}=\frac{2\pi (40000)}{60}=4188.8\text{rad/s}.$ Blade speed at the rim (why: $u=\omega r$ is the linear speed of a point at radius $r$): $u_2=\omega r_2=4188.8\times 0.08=335.1\text{m/s}.$

Why this formula: with no inlet swirl the Euler head collapses to $H=u_2{c}_{u2}/g$ — this is exactly energy added per unit weight of liquid, i.e. the whirl work $u_2{c}_{u2}$ (energy per unit mass) divided by $g$ to express it as a column height. We use it here because both $u_2$ (from L2.1) and $c_{u2}$ (given as a fraction of $u_2$) are known. $c_{u2}=0.75\times 335.1=251.3\text{m/s}$. $H=\frac{u_2{c}_{u2}}{g}=\frac{335.1\times 251.3}{9.81}=8585\text{m}.$ (About 8.6 km of equivalent liquid column — see Bernoulli's Principle for why head is measured this way.)

Why multiply by $\rho g$: head is energy per unit weight; multiply by $\rho g$ to get energy per unit volume, which is pressure. $\Delta p=\rho gH=1140\times 9.81\times 8585=9.60\times 10^7\text{Pa}\approx 960\text{bar}.$ Physical realism: 960 bar from one stage is far above what real pumps deliver per stage. Two limits bite first. (i) Tip speed: $u_2=335\text{m/s}$ pushes impeller-rim stresses toward the strength limit of the metal — most pumps keep $u_2$ below $\sim 400\text{m/s}$, and this problem sits near that ceiling because it combines a small radius with an extreme 40000 rpm. (ii) Cavitation & stage loading: a 960-bar rise concentrated in one impeller is unrealistic; real turbopumps split large rises across multiple stages (the SSME LH₂ pump uses three). Treat this number as an upper-bound sanity figure, not a buildable single stage — the arithmetic is right, the single-stage engineering is not.

Why this formula: hydraulic power added to the fluid is $\dot{m}\Delta p/\rho$ (energy/mass $=\Delta p/\rho$); dividing by efficiency accounts for real losses inside the pump. $P_{\text{pump}}=\frac{\dot{m}\Delta p}{\rho {\eta }_p}=\frac{150\times 180\times 10^5}{810\times 0.72}=4.63\times 10^6\text{W}\approx 4.63\text{MW}.$

Why this formula: the turbine extracts a fraction of the gas's thermal energy set by the isentropic temperature drop; the bracket is that ideal drop fraction, and ${\eta }_t$ scales it to the real machine. $P_{\text{turb}}={\dot{m}}_g{c}_p{T}_0{\eta }_t\left(1-\frac{1}{{\pi }_t^{(\gamma -1)/\gamma }}\right).$ Exponent: $(\gamma -1)/\gamma =0.3/1.3=0.2308$. So ${\pi }_t^{0.2308}=8^{0.2308}=1.622$, and $1-1/1.622=0.3833$. $P_{\text{turb}}=4.0\times 2100\times 950\times 0.62\times 0.3833=1.90\times 10^6\text{W}\approx 1.90\text{MW}.$ Check: the pump needs 4.63 MW but the turbine only makes 1.90 MW ⇒ not enough. The design fails the power match; you must raise ${\dot{m}}_g$, $T_0$, or ${\pi }_t$. This is exactly how the Gas Generator Cycle gets sized.

$P_{\text{turb}}\propto {\dot{m}}_g$, so scale linearly: $\frac{{\dot{m}}_{g,\text{new}}}{{\dot{m}}_{g,\text{old}}}=\frac{4.63}{1.90}=\mathrm{2.44.}$ New gas flow ${\dot{m}}_g=2.44\times 4.0=9.8\text{kg/s}$. Why this matters: more gas burned in the gas generator is propellant not going to the main chamber — a direct hit to Specific Impulse.

Full Bernoulli, tank surface → pump inlet (see Bernoulli's Principle). Writing every head term: $\frac{p_{\text{tank}}}{\rho g}+z-h_{\text{loss}}=\frac{p_{\text{inlet}}}{\rho g}+\frac{v^2}{2g}.$ ${\text{NPSH}}_A$ is the total (stagnation) inlet head above vapor pressure, so it keeps the velocity head: rearranging, $\frac{p_{\text{inlet}}}{\rho g}+\frac{v^2}{2g}-\frac{p_v}{\rho g}=\frac{p_{\text{tank}}-p_v}{\rho g}+z-h_{\text{loss}}$. The $\frac{v^2}{2g}$ term does not vanish — it is absorbed into the left side and is exactly why NPSH is defined on stagnation rather than static pressure. What survives on the right is the driving margin: ${\text{NPSH}}_A=\frac{p_{\text{tank}}-p_v}{\rho g}+z-h_{\text{loss}}.$ Why the inlet velocity head disappears from the final expression: it cancels between the two sides of Bernoulli — the fast-moving inlet fluid trades static pressure for velocity head, but their sum (the stagnation head) is what NPSH tracks, so no separate $v^2/2g$ term remains. Pressure-margin head: $\frac{(2.5-1.0)\times 10^5}{71\times 9.81}=\frac{1.5\times 10^5}{696.5}=215.4\text{m}$. ${\text{NPSH}}_A=215.4+1.5-2.0=214.9\text{m}.$ Huge, because $\rho$ for LH₂ is tiny — a small pressure margin becomes a giant column.

Safe operation needs ${\text{NPSH}}_A>{\text{NPSH}}_R$. Here $214.9<240$ ⇒ not safe; the impeller eye would drop below vapor pressure and cavitate. Shortfall: $240-214.9=25.1\text{m}$. Fix: add an inducer — a slender axial screw ahead of the impeller that gently pre-pressurises the flow, lowering the required ${\text{NPSH}}_R$ so the pump tolerates the low inlet head without a heavier, more-pressurised tank. (Avoiding Cavitation is the whole point.)

Need ${\text{NPSH}}_A$ to rise by $25.1\text{m}$, i.e. the pressure-margin head must go from $215.4$ to $240.5\text{m}$. Invert: $p_{\text{tank}}-p_v=(\text{margin head})\times \rho g=240.5\times 71\times 9.81=1.675\times 10^5\text{Pa}$. $p_{\text{tank}}=1.675\times 10^5+1.0\times 10^5=2.675\times 10^5\text{Pa}\approx 2.68\text{bar}.$ Only $\sim 0.18\text{bar}$ more tank pressure — sounds cheap, but a higher-pressure tank must be thicker and heavier everywhere, hurting mass fraction. The inducer usually wins; that trade is why turbopumps exist at all.

$u_2=\omega r_2\propto N$. With $c_{u2}=\kappa u_2$: $\Delta p=\rho gH=\rho u_2{c}_{u2}=\rho \kappa u_2^2\propto N^2.$ Why this dominates design: pressure rise grows with the square of speed — doubling RPM quadruples $\Delta p$, which is why impellers scream at tens of thousands of RPM. To double $\Delta p$ we need $N^2$ to double, i.e. $N_{\text{new}}=N_{\text{old}}\sqrt{2}$: $N_{\text{new}}=40000\times \sqrt{2}=56569\text{rpm}.$

$H=\frac{u_2\cdot 0}{g}=0.$ Physical meaning: spinning fast is useless unless the blades actually turn the flow into tangential motion. Whirl $c_{u2}$ is what carries angular momentum out; with none delivered, no angular-momentum change ⇒ no work ⇒ no head. A radial-tipped blade that lets fluid slip straight through the rim does nothing.

Pump power: $P_{\text{pump}}=\frac{{\dot{m}}_{\text{prop}}\Delta p}{\rho {\eta }_p}=\frac{180\times 210\times 10^5}{810\times 0.70}=6.67\times 10^6\text{W}.$ Turbine must supply (accounting for mechanical losses): $P_{\text{turb}}=\frac{P_{\text{pump}}}{{\eta }_{\text{mech}}}=\frac{6.67\times 10^6}{0.98}=6.80\times 10^6\text{W}.$ Per-unit-mass turbine work: exponent $(\gamma -1)/\gamma =0.2308$; ${\pi }_t^{0.2308}=10^{0.2308}=1.702$; $1-1/1.702=0.4125$. $w=c_p{T}_0{\eta }_t\left(1-\frac{1}{{\pi }_t^{0.2308}}\right)=2000\times 1000\times 0.60\times 0.4125=4.95\times 10^5\text{J/kg}.$ Gas flow: ${\dot{m}}_g=\frac{P_{\text{turb}}}{w}=\frac{6.80\times 10^6}{4.95\times 10^5}=13.7\text{kg/s}.$ Turbine tax: $\frac{13.7}{180}=0.0762=7.6\%$ of propellant is spent driving the pump — dumped overboard at low $I_{sp}$ in a Gas Generator Cycle. Routing that gas back into the chamber instead is precisely what the Staged Combustion Cycle does to reclaim it, protecting overall Specific Impulse and Chamber Pressure and Thrust.

Which pillar has no $\rho$ in its head formula? ::: The centrifugal pump — $H=u_2{c}_{u2}/g$ (density only enters $\Delta p=\rho gH$). What single Euler relation governs both pump and turbine work? ::: $w=u\Delta c_u$ — the pump adds whirl, the turbine removes it. Which term in ${\text{NPSH}}_A$ is huge for LH₂ and why? ::: The pressure-margin head $\frac{p_{\text{tank}}-p_v}{\rho g}$, because $\rho$ is tiny so a small pressure becomes a giant column. Cheapest safe fix when ${\text{NPSH}}_A<{\text{NPSH}}_R$? ::: Add an inducer to lower ${\text{NPSH}}_R$, keeping the tank light.