Turbopump design — centrifugal pump, axial turbine stages, NPSH: Worked examples

3.3.27 · D3 · Physics › Rocket Propulsion › Turbopump design — centrifugal pump, axial turbine stages, N

Har symbol, use se pehle define kiya gaya

Neeche kuch bhi assume nahi kiya gaya. Inhe ek baar padh lo aur poori page self-contained ho jayegi.

Definition Blade speed, whirl, aur head

Blade speed $u = ω r$ = wheel ke kisi point ki speed, metres per second mein. Yahan $ω$ (omega) spin rate hai radians per second mein (ek poora chakkar $= 2 π$ radians), aur $r$ us point ki radius hai. Hum $u_{1} = ω r_{1}$ eye pe aur $u_{2} = ω r_{2}$ rim pe likhte hain.
Whirl $c_{u}$ = liquid ki absolute velocity ka sideways (tangential) hissa — woh hissa jo axis ke around ghoomta hai. Sirf yahi hissa angular momentum carry karta hai, isliye sirf yahi head change karta hai. $c_{u 1}$ hai woh whirl jo eye pe andar jaata hai; $c_{u 2}$ hai woh whirl jo rim pe bahar aata hai.
Gravitational acceleration $g = 9.81 m/s^{2}$ — woh constant jo "energy per unit weight" ko height mein convert karta hai. Is page ke har head aur NPSH formula mein isi $g$ se divide kiya jaata hai.
Head $H$ = har kilogram-weight liquid ko diya gaya energy, us liquid ki height ke roop mein express kiya gaya (metres): $H = \frac{u _{2} c _{u 2} - u _{1} c _{u 1}}{g}$

Definition NPSH vocabulary (poori tarah spelled out)

NPSH = Net Positive Suction Head — pump inlet pe spare pressure margin, us point se upar jahan liquid boil ho jaati, liquid ki height ke roop mein (metres). Ye do flavours mein aata hai jinke subscripts bilkul mix nahi karne chahiye:

$NPSH_{A}$ — A hai Available ke liye: jo plumbing inlet tak actually deliver karti hai.
$NPSH_{R}$ — R hai Required ke liye: jo impeller boiling rokne ke liye demand karta hai, uski geometry se set hota hai.
Safe operation ke liye chahiye $NPSH_{A} > NPSH_{R}$ . Available margin teen named pieces se banta hai: $NPSH_{A} = pressure margin \frac{p _{tank} - p _{v}}{ρ g} + elevation z - friction h_{loss}$
$p_{tank}$ = propellant tank mein pressure; $p_{v}$ = vapor pressure, woh pressure jis par liquid apne temperature pe boil karti hai. Margin ye hai ki tank boiling se kitna upar baitha hai.
$z$ = tank elevation difference, liquid surface ki height pump se upar. Sign convention: $z > 0$ jab tank pump se upar ho (gravity help kare), $z < 0$ jab tank neeche ho (suction lift, gravity fight kare).
$h_{loss}$ = head loss, woh friction jo flow feed line se neeche jaate waqt khota hai; hamesha subtract hota hai kyunki real pipes margin sirf khaate hain.

The scenario matrix

Is topic mein jo bhi problem aa sakti hai woh in cells mein se ek mein padti hai. Har example neeche tagged hai us cell(s) se jo woh cover karta hai.

Cell	Kya special hai	Covered by
A. Standard pump	normal spin, no inlet swirl ( $c_{u 1} = 0$ )	Ex 1
B. Pre-swirl (sign of $c_{u 1}$ )	inlet whirl present, positive vs negative	Ex 2
C. Degenerate: $ω = 0$ or $r_{1} = r_{2}$	wheel stopped / no radius change → zero head	Ex 3
D. Density swap (invariance trap)	same pump, LOX vs LH₂ — head vs pressure	Ex 4
E. Turbine sign flip	Euler run "backwards", work out	Ex 5
F. Power-match coupling	turbine must equal pump demand	Ex 6
G. NPSH normal + limiting	positive margin, aur boiling edge	Ex 7
H. NPSH degenerate	$p_{tank} = p_{v}$ , suction lift ( $z < 0$ )	Ex 8
I. Word problem (real mission)	story se tool choose karo	Ex 9
J. Exam twist	"doubling speed" scaling / hidden unit trap	Ex 10

Ex 1 — Cell A: plain-vanilla pump

Forecast: Andaaza lagao — kya $H$ $1, 000$ m ke qareeb hoga ya $5, 000$ m ke? (Yaad karo $H \sim u_{2}^{2} / g$ , aur $u_{2}$ kuch sau m/s hogi.)

Is example ka exit velocity triangle yahan draw kiya gaya hai — amber whirl arrow $c_{u 2}$ hai, cyan arrow spin direction mein blade speed $u_{2}$ hai:

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

$ω = 2 π N /60 = 2 π (25000) /60 = 2618 rad/s$ . Ye step kyun? rpm poore turns per minute count karta hai; radians per second woh hai jo $u = ω r$ ko chahiye (ek chakkar $= 2 π$ radians, ek minute $= 60$ s).
$u_{2} = ω r_{2} = 2618 \times 0.08 = 209.4 m/s$ . Ye step kyun? Blade speed = rim jo actual metres-per-second travel karta hai; yahi whirl ke liye ceiling set karta hai.
$c_{u 2} = 0.75 \times 209.4 = 157.1 m/s$ . Ye step kyun? Blades $u_{2}$ se zyada whirl force nahi kar sakte; $u_{2}$ se chhota fraction realistic hai.
$H = \frac{u _{2} c _{u 2}}{g} = \frac{209.4 \times 157.1}{9.81} = 3353 m$ . Ye step kyun? $c_{u 1} = 0$ ke saath Euler head — is stage mein saari whirl add hoti hai, aur hum $g = 9.81$ se divide karke energy-per-weight ko metres mein convert karte hain.

Verify: Units: $(m/s) (m/s) / (m/s^{2}) = m$ . ✓ Magnitude dono guesses ke beech hai, isliye $u_{2}^{2} / g$ scaling sahi hai. ✓

Ex 2 — Cell B: inlet pre-swirl, dono signs

Forecast: Kaun sa case zyada head deta hai — wheel ke saath swirl, ya uske against?

Neeche ki figure dikhati hai ki wahi $c_{u 1}$ term head ko no-swirl baseline se upar ya neeche kaise shift karta hai:

Full Euler: $H = \frac{u _{2} c _{u 2} - u _{1} c _{u 1}}{g}$ . Ye step kyun? $c_{u 1} \neq = 0$ ke saath hum doosra term drop nahi kar sakte; yeh us angular momentum ko subtract karta hai jo fluid pehle se andar carry karta tha.
Case (i): $H = \frac{209.4 ( 157.1 ) - 78.5 ( + 40 )}{9.81} = \frac{32897 - 3140}{9.81} = 3033 m$ . Ye step kyun? Positive $c_{u 1}$ matlab fluid pehle se wheel ke saath whirl kar raha tha, isliye impeller kam naya kaam karta hai — head girta hai.
Case (ii): $H = \frac{32897 - 78.5 ( - 40 )}{9.81} = \frac{32897 + 3140}{9.81} = 3673 m$ . Ye step kyun? Counter-swirl matlab impeller ko pehle fluid ki whirl reverse karni padti hai, phir aage drive karna — zyada kaam, zyada head.

Verify: No-swirl answer (Ex 1) $3353$ m tha; case (i) usse neeche hai aur case (ii) usse upar, symmetrically straddling ( $\pm 320$ m). ✓ Yeh symmetry exactly $- u_{1} c_{u 1}$ term ka sign flip hai. ✓ Isi liye designers $c_{u 1} = 0$ choose karte hain — yeh clean middle hai jo counter-swirl hardware ke bina $u_{2}$ per unit head maximize karta hai.

Ex 3 — Cell C: degenerate inputs (zero head)

Forecast: Kya ye chhota head dete hain ya exactly zero?

(a) $u_{2} = ω r_{2} = 0$ , isliye $H = \frac{0 \cdot c _{u 2} - 0 \cdot c _{u 1}}{g} = 0$ . Ye step kyun? Koi blade motion nahi = fluid pe koi kaam nahi. Head added energy hai; ruka hua wheel kuch nahi jodta.
(b) $u_{2} = u_{1}$ aur $c_{u 2} = c_{u 1}$ , isliye $H = \frac{u ( c _{u} ) - u ( c _{u} )}{g} = 0$ . Ye step kyun? Euler head angular momentum flux ka difference hai. Agar in aur out ke beech whirl ya radius ke baare mein kuch nahi badla, toh difference zero hai — pipe hai, pump nahi.

Verify: Dono exactly $0$ m dete hain, aur dono physically sense banate hain: pump sirf $u c_{u}$ ko impeller ke across change karke kaam karta hai. ✓ Yeh har Euler problem ki degenerate boundary hai.

Ex 4 — Cell D: density-swap invariance trap

Yahan $ρ$ (rho) hai density — liquid ka mass per cubic metre.

Forecast: Same pump, same speed — kya head liquids ke beech badlega? Kya pressure badlega?

Head unchanged hai: $H = 3353 m$ dono ke liye. Ye step kyun? $H = u_{2} c_{u 2} / g$ mein koi $ρ$ nahi hai. Head energy per unit weight hai, isliye column height wahi rehta hai chahe kuch bhi flow ho.
LOX: $Δ p = ρ g H = 1140 \times 9.81 \times 3353 = 3.75 \times 1 0^{7} Pa = 375 bar$ . Ye step kyun? Head (energy/weight) ko $ρ g$ (weight/volume) se multiply karo to get energy/volume = pressure.
LH₂: $Δ p = 71 \times 9.81 \times 3353 = 2.34 \times 1 0^{6} Pa = 23.4 bar$ . Ye step kyun? Same head, lekin halka hydrogen per litre almost kuch nahi weighs, isliye same column chhota pressure hai.

Verify: Pressures ka ratio $= ρ_{LOX} / ρ_{LH_{2}} = 1140/71 = 16.1$ , aur $375/23.4 = 16.0$ . ✓ Head identical, pressure density ke saath scale karta hai. Yahan bubbles banna hoga Cavitation — heat se nahi balki low pressure se boiling, exactly woh cheez jo NPSH guard karta hai.

Ex 5 — Cell E: Euler ulta chalao (turbine work)

Forecast: Pump ke liye humne whirl add kiya (work in). Yahan whirl rotor ke across girta hai — kya $w$ positive aayega (extracted work)?

$Δ c_{u} = c_{u, in} - c_{u, out} = 520 - 120 = 400 m/s$ . Ye step kyun? Turbine whirl ke loss ko harvest karta hai; woh drop shaft work mein convert hota hai.
$w = u Δ c_{u} = 350 \times 400 = 1.40 \times 1 0^{5} J/kg$ . Ye step kyun? Pump wali hi $u Δ c_{u}$ Euler relation — sirf sign convention flip hota hai kyunki fluid momentum blade ko deta hai receive karne ki jagah. (Yeh $w = u Δ c_{u}$ exactly woh Euler turbomachinery relation hai jo parent note ne derive ki: work per kg = blade speed times change in whirl.)

Verify: Units $(m/s)^{2} = m^{2} / s^{2} = J/kg$ . ✓ Positive → kaam genuinely extract hua, turbine ke liye sahi. ✓

Ex 6 — Cell F: power-match coupling

Yahan $\overset{m}{˙}$ (m with a dot) matlab mass flow rate — kilograms passing per second; $η$ (eta) 0 aur 1 ke beech efficiency hai.

Forecast: Pump power kuch MW hogi. Kya gas flow $150$ kg/s ka bada fraction hoga, ya chhoti sip?

$P_{pump} = \frac{m ˙ _{prop} Δ p}{ρ η _{p}} = \frac{150 \times 1.8 \times 1 0 ^{7}}{1140 \times 0.68} = 3.48 \times 1 0^{6} W = 3.48 MW$ . Ye step kyun? $Δ p / ρ$ ideal energy per kg hai; $η_{p}$ se divide karna real pump losses account karta hai.
Shaft ko supply karni hogi $P_{turb} = P_{pump} / η_{mech} = 3.48/0.98 = 3.55 MW$ . Ye step kyun? Shared shaft pe bearing/seal friction matlab turbine ko thoda over-produce karna padta hai.
$\overset{m}{˙}_{g} = \frac{P _{turb}}{w} = \frac{3.55 \times 1 0 ^{6}}{1.40 \times 1 0 ^{5}} = 25.4 kg/s$ . Ye step kyun? Gas ka har kg $w$ joules yield karta hai; kg/s pane ke liye power demand ko us se divide karo.

Verify: Gas flow $25.4$ kg/s propellant flow ka roughly $17%$ hai — plausibly high (yeh work-heavy stage hai). ✓ Units: $W / (J/kg) = kg/s$ . ✓ Ek Gas Generator Cycle mein yeh gas ek chhote side-chamber mein burn hoti hai aur turbine spin karne ke baad bahar dump ho jaati hai; ek Staged Combustion Cycle mein ise main chamber mein feed kiya jaata hai taki koi mass waste na ho.

Ex 7 — Cell G: NPSH normal, phir boiling edge

Forecast: Kya $+ z - h_{loss}$ terms zyada matter karenge, ya pressure term dominate karega?

Margin picture (available bars vs required amber line) yahan hai:

$\frac{p _{tank} - p _{v}}{ρ g} = \frac{( 4 - 1.8 ) \times 1 0 ^{5}}{1140 \times 9.81} = \frac{2.2 \times 1 0 ^{5}}{11183} = 19.67 m$ . Ye step kyun? Yeh driving pressure margin hai — tank pressure boiling se kitni upar hai, LOX ke metres mein ( $g = 9.81$ use karke).
$NPSH_{A} = 19.67 + 1.5 - 0.8 = 20.37 m$ . Ye step kyun? Elevated tank se gravity head $z$ add karo, friction loss $h_{loss}$ subtract karo (tank surface se inlet tak energy balance).
Safety: $20.37 > 18$ , margin $= 2.4$ m. Safe hai, par tight. Ye step kyun? Rule hai $NPSH_{A} > NPSH_{R}$ ; neeche eye pressure $p_{v}$ ke neeche jaata hai aur bubbles bante hain.
(b) Edge: $NPSH_{A} = 18$ set karo. Tab $\frac{p _{tank} - p _{v}}{ρ g} = 18 - 1.5 + 0.8 = 17.3 m$ , isliye $p_{tank} - p_{v} = 17.3 \times 11183 = 1.935 \times 1 0^{5}$ , giving $p_{tank} = 1.8 \times 1 0^{5} + 1.935 \times 1 0^{5} = 3.73 \times 1 0^{5} Pa = 3.73 bar$ . Ye step kyun? Hum safety equality ko ulta solve karte hain woh lowest tank pressure nikalne ke liye jo $NPSH_{R}$ clear karta ho.

Verify: $p_{tank} = 3.73$ bar pe, recompute karo: $\frac{( 3.73 - 1.8 ) \times 1 0 ^{5}}{11183} = 17.26$ , $+ 1.5 - 0.8 = 17.96 \approx 18$ m. ✓ Koi bhi tank pressure $3.73$ bar se neeche → inlet pe boiling. ✓

Ex 8 — Cell H: NPSH degenerate cases

Forecast: (a) mein koi margin bacha hai kya? (b) mein, kya neeche-pump tank kabhi safe ho sakta hai?

(a) Pressure term $= \frac{p _{v} - p _{v}}{ρ g} = 0$ , isliye $NPSH_{A} = 0 + 1.5 - 0.8 = 0.7 m$ . Ye step kyun? Jab $p_{tank} = p_{v}$ tab liquid pehle se apni boiling edge pe hai; single margin jo bacha hai woh gravity head minus friction hai. Almost koi bhi $NPSH_{R}$ fail karega — isi liye tanks ko $p_{v}$ se upar pressurize kiya jaata hai.
(b) $NPSH_{A} = 3.0 + (- 1.2) - 0.8 = 1.0 m$ . Ye step kyun? Pump se neeche tank $z$ ko negative banata hai (hamari sign convention) — gravity ab flow ke against karti hai aur margin khaati hai. Sirf bada pressure term NPSH ko positive rakhta hai.

Verify: (a) $0.7$ m aur (b) $1.0$ m dono chhote aur positive hain — exactly woh marginal region jahan inducer (impeller ke aage ek patla axial screw jo flow ko gently pre-pressurize karta hai, $NPSH_{R}$ kam karta hai) add kiya jaata hai. ✓ $z$ ka sign sahi handle hua: tank upar uthane se help ( $+$ ), neeche karne se hurt ( $-$ ). ✓

Ex 9 — Cell I: real-mission word problem

Forecast: Same shaft speed, same target pressure — kya dono pumps ko same head chahiye, ya wildly different?

Head jo har pump ko reach karni hai: $H = \frac{Δ p}{ρ g}$ . Ye step kyun? Hum $Δ p = ρ g H$ invert karte hain kyunki requirement pressure hai, lekin pump ka kaam head mein measure hota hai.
LH₂: $H = \frac{2.7 \times 1 0 ^{7}}{71 \times 9.81} = 38, 760 m$ . Ye step kyun? Feather-light hydrogen ko $270$ bar banane ke liye monstrous column chahiye.
LOX: $H = \frac{2.7 \times 1 0 ^{7}}{1140 \times 9.81} = 2414 m$ . Ye step kyun? Dense oxygen same pressure $16 \times$ chhote column se banata hai.
Ratio $= 38760/2414 = 16.1$ . LH₂ pump ko $16 \times$ head chahiye → bahut zyada stages ya bahut bada rim → yahi tall multistage machine hai. Ye step kyun? Kyunki $H \sim u_{2}^{2} / g$ fixed $ω$ pe, $16 \times$ head ki zaroorat matlab ya $\sim 16$ stages stack karo ya bada radius — physical reason yahi hai ki LH₂ turbopumps giants hote hain.

Verify: Ratio density ratio $1140/71 = 16.1$ ke barabar hai. ✓ Head answers $ρ$ ke inversely scale karte hain, jaise hona chahiye. ✓ Isi liye Chamber Pressure and Thrust mein rise (higher $Δ p$ ki zaroorat) ever-taller hydrogen pumps demand karta hai.

Ex 10 — Cell J: exam twist (scaling + unit trap)

Forecast: RPM double karo — kya head double hoga, ya kuch zyada dramatic?

(a) $H \propto u_{2}^{2} \propto N^{2}$ (kyunki $c_{u 2} \propto u_{2} \propto N$ ). Ye step kyun? Fixed whirl fraction ke saath Euler head $H = \frac{k u _{2}^{2}}{g}$ hai; dono $u_{2}$ aur $c_{u 2}$ $N$ ke saath scale karte hain, isliye $H$ $N^{2}$ ke saath scale karta hai.
$H_{new} = 2500 \times (40000/20000)^{2} = 2500 \times 4 = 10, 000 m$ . Ye step kyun? Speed double karna head chaar guna kar deta hai — centrifugal pumps ki headline non-linearity.
(b) Student rpm → rad/s convert karna bhool gaya. Sach mein $ω = 2 π \times 40000/60 = 4189 rad/s$ , isliye $u_{2} = 4189 \times 0.06 = 251.3 m/s$ — naki $40000 \times 0.06 = 2400$ m/s (impossibly supersonic). Ye step kyun? $u = ω r$ ko radians per second chahiye; raw rpm use karna blade speed $\approx 9.5 \times$ aur head $\approx 90 \times$ inflate karta hai — ek classic exam-killer.

Verify: (a) factor $4$ $N^{2}$ se. ✓ (b) wrong to right $ω$ ka ratio $40000/4189 = 9.55 = 60/ (2 π)$ hai. ✓ Unit slip exactly missing $2 π /60$ hai. ✓

Recall Aage badhne se pehle self-test

Rim speed double karna head kitni baar multiply karta hai ::: chaar (head $\propto u_{2}^{2}$ ) Woh Euler term jo no inlet swirl mein vanish hoti hai ::: $u_{1} c_{u 1}$ Same pump, LOX vs LH₂ — kya same rehta hai? ::: head $H$ metres mein (pressure $ρ$ se differ karta hai) Turbine vs pump Euler: kya flip hota hai? ::: sign — whirl remove hoti hai (work out) naki add (work in) NPSH mein "A" vs "R" kya stand karta hai? ::: Available (plumbing se deliver) vs Required (impeller ki demand) Tank pump se neeche move karna $z$ ko kya banata hai ::: negative, $NPSH_{A}$ shrink karta hai Jab $NPSH_{A}$ marginal ho toh fix kya hai ::: inducer add karo ( $NPSH_{R}$ kam karta hai)

Dekho bhi: Euler Turbomachinery Equation, Specific Impulse.