Turbopump design — centrifugal pump, axial turbine stages, NPSH: Visual walkthrough

3.3.27 · D2 · Physics › Rocket Propulsion › Turbopump design — centrifugal pump, axial turbine stages, N

Step 1 — Ek spinning disk par liquid ka ek blob

KYA HAI. Ek flat disk imagine karo jo ek central axle ke around spin kar rahi hai. Us disk par kuch distance $r$ (iska radius) par axle se ek chota sa liquid ka blob chipka do. Jab disk ghoomti hai, blob circle mein khicha jaata hai.

KYUN. Ek centrifugal pump bilkul yehi hai: ek impeller (curved blades wali ek spinning disk) liquid ko ghuma kar bahar ki taraf fenkta hai. Kisi bhi formula se pehle, hume ek clean idea chahiye: radius $r$ par koi point kitni tezi se move karta hai?

PICTURE. Figure dekho. Axle kaala dot hai. Blob radius $r$ par baitha hai (yellow line). Blue arrow uski velocity hai — yeh sideways point karta hai, circle ke tangent par, kyunki blob ghoom raha hai, andar ya bahar nahi.

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

$u = turns/s in rad ω \times kitna bahar hai r$

Step 2 — Fluid ki velocity ko do honest arrows mein split karna

KYA HAI. Real liquid do kaam ek saath karta hai: yeh axis ke around swirl karta hai aur rim ki taraf bahar creep karta hai. Hum uski velocity ko do perpendicular arrows mein split karte hain.

KYUN. Un do motions mein se sirf ek hi angular momentum (axle ke around spin) carry karta hai. Bookkeeping cleanly karne ke liye hume us piece ko isolate karna hoga — isliye hum decompose karte hain.

PICTURE. Pink arrow $c$ liquid ki actual velocity hai. Hum ise do directions par drop karte hain: tangential (whirl) part $c_{u}$ circle ke along, aur radial part $c_{r}$ seedha bahar ki taraf. Yeh ek right angle banate hain.

Step 3 — Angular momentum: woh quantity jo blade actually change karta hai

KYA HAI. Ek spinning object ke liye, "spin ki quantity" bookkeeping quantity hai angular momentum = (mass) × (radius) × (tangential speed). Hum track karte hain ki fluid kitna carry karta hai andar aur bahar.

KYUN. Rotation ke liye Newton's law kehta hai: kisi cheez ka angular momentum change karne ke liye tumhe torque (ek twisting force) apply karna hoga. Impeller hi akela cheez hai jo fluid ko twist kar raha hai, isliye uska diya hua torque fluid ke angular momentum ke change per second ke barabar hai.

PICTURE. Do dashed circles: andar ki eye radius $r_{1}$ par jahan fluid enter karta hai, aur bahar ki rim radius $r_{2}$ par jahan yeh nikalti hai. Har ek apna whirl arrow carry karta hai, $c_{u 1}$ aur $c_{u 2}$ .

Steady flow ke liye, mass rate $\overset{m}{˙}$ (kg per second) se pass hota hai. Angular momentum enter karta hai per second $\overset{m}{˙} r_{1} c_{u 1}$ ; nikal ta hai $\overset{m}{˙} r_{2} c_{u 2}$ . Supplied torque difference hai:

$T = \overset{m}{˙} (rim par OUT carry hoti spin r_{2} c_{u 2} - eye par IN carry hoti spin r_{1} c_{u 1})$

Step 4 — Torque se power: $ω$ se multiply karo

KYA HAI. $ω$ par turn karne wale shaft ke through lagaya gaya torque $T$ power deliver karta hai $P = T ω$ (energy per second).

KYUN. Hum ultimately fluid ko di gayi energy chahte hain, torque nahi. Power bridge hai: yeh torque times kitni tezi se tum ghoomte ho, bilkul jaise straight-line power force times speed hai.

PICTURE. $ω$ par turn karta shaft, fluid par torque arrows, aur ek chota sa energy meter jo per second fill ho raha hai. Notice karo $ω r_{1} = u_{1}$ aur $ω r_{2} = u_{2}$ — radii blade speeds mein fold ho jaate hain.

$P = T ω = \overset{m}{˙} ω (r_{2} c_{u 2} - r_{1} c_{u 1}) = \overset{m}{˙} (ω r_{2} = u_{2} u_{2} c_{u 2} - ω r_{1} = u_{1} u_{1} c_{u 1})$

Step 5 — Power se head: weight-flow se divide karo

WHAT. Power lo, ise fluid stream par split karo, aur ise head $H$ kahi jaane wali liquid column ki height mein convert karo.

KYUN. Energy per unit mass $P / \overset{m}{˙}$ hai. Ek baar aur $g$ (gravity ki strength, $9.81 m/s^{2}$ ) se divide karne par "energy per kilogram" se "metres of liquid" ban jaata hai — kyunki 1 kg ko 1 m upar uthane mein exactly $g$ joules lagte hain. Head ek friendly, density-free tarika hai pump ko size karne ka.

PICTURE. Height $H$ ki liquid ki ek tall column; pump jo energy per kilogram add karta hai woh us column ke energy ke barabar hai jo per kilogram store karta hai. Side par ruler $H$ metres mein read karta hai.

$\frac{P}{m ˙} = u_{2} c_{u 2} - u_{1} c_{u 1} ⟹ H = \frac{1}{g} \cdot \frac{P}{m ˙}$

Step 6 — Clean design choice: inlet swirl nahi

KYA HAI. Liquid ko seedha eye mein feed karo — axis ya radially inward ke along pointed, bina kisi pre-spin ke. Tab $c_{u 1} = 0$ .

KYUN. Agar fluid pehle se whirling aata hai, exit whirl ka kuch hissa is impeller ne nahi daala — woh free swirl tha jo hum saath laaye. Ise khatam karna ( $c_{u 1} = 0$ ) matlab hai ki rim par har bit of whirl genuine kaam hai jo is stage ne kiya, jo formula ko simplify bhi karta hai aur ek given rim speed ke liye head maximize bhi karta hai.

PICTURE. Left: swirling inflow (messy, head ka kuch hissa "borrowed" hai). Right: clean radial inflow, $c_{u 1} = 0$ , isliye poora $u_{2} c_{u 2}$ term hamara hai.

$H \approx \frac{u _{2} c _{u 2}}{g} (setting c_{u 1} = 0)$

Step 7 — Edge cases: formula extremes par kya karta hai

KYA HAI. Hum formula ko teen corners par test karte hain taaki reader kabhi koi unshown scenario na dekhe.

KYUN. Woh formula jis par tum trust karte ho woh hai jise tumne apni limits par behave karte dekha ho: kuch bhi spinning nahi, koi whirl add nahi, aur density ka sawaal.

PICTURE. Teen mini-panels: (a) disk ruka hua, (b) fluid bina whirl ke seedha bahar nikal jaata hai, (c) identical pumps par do alag liquids.

(a) $ω = 0$ (disk nahi ghoom raha). Tab $u_{2} = ω r_{2} = 0$ , isliye $H = 0$ . Jo pump spin nahi karta woh koi head add nahi karta. ✔ Samajh aata hai.
(b) $c_{u 2} = 0$ (fluid bina whirl ke nikalti hai). Chahe disk tezi se ghoomta rahe, $u_{2} c_{u 2} = 0 \Rightarrow H = 0$ . Whirl actually impart hona chahiye; ek bald spinning disk jo fluid ko radially slide off hone deta hai kuch nahi karta. Isliye blades curved hote hain — woh $c_{u 2} \neq = 0$ force karte hain.
(c) Do alag liquids. Head $H = u_{2} c_{u 2} / g$ mein koi density $ρ$ nahi hai. LOX ya water ko same speed par chalao aur tumhe metres mein same head milega. Sirf jab tum pressure rise mein convert karte ho tab density enter karta hai, $Δ p = ρ g H$ ke through (bhaari liquid $\Rightarrow$ same metres ke liye bada pressure). Yeh woh subtle point hai jo parent note steel-man karta hai.

Ek-picture summary

Yahan ek single board par poori derivation hai: blade speed $u = ω r$ Step 1 par paida hua → velocity whirl $c_{u}$ mein split hui Step 2 par → angular-momentum torque $T = \overset{m}{˙} Δ (r c_{u})$ Step 3 par → power $P = T ω = \overset{m}{˙} Δ (u c_{u})$ Step 4 par → head $H = Δ (u c_{u}) / g$ Step 5 par → clean $H \approx u_{2} c_{u 2} / g$ Step 6 par.

Recall Feynman retelling — plain words mein wapas bolo

Ek pump ek spinning disk hai jis par curved blades hain. Rim tezi se move karti hai — woh speed $u = ω r$ hai, bas "kitna bahar" times "kitni tezi se ghoomta hai." Liquid do kaam karta hai: yeh swirl karta hai around ( $c_{u}$ ) aur bahar creep karta hai ( $c_{r}$ ), lekin sirf swirl hi axle ke around spin carry karta hai, isliye hum sirf usi ko track karte hain. Fluid ko exit par entry se zyada swirl karwane ke liye, blade ko ise twist karna hoga — aur twisting matlab torque, jo per second $\overset{m}{˙}$ times $r c_{u}$ ka change hai. Torque ko turning speed $ω$ se multiply karo aur tumhe power milti hai; $ω r$ 's blade speeds $u$ ban jaate hain, isliye power per kilogram $u c_{u}$ ka change hai. Gravity $g$ se divide karo aur woh energy-per-kilogram liquid ki ek height ban jaata hai — the head $H$ . Fluid ko bina pre-swirl ke andar feed karo aur sirf rim term bachta hai: $H \approx u_{2} c_{u 2} / g$ . Kyunki exit whirl rim speed ko chase karta hai, head rim speed ke square ki tarah badhta hai, isliye hum inhe insanely tezi se spin karte hain. Aur notice karo: ek bhi density kahin nahi — metres mein head hydrogen ya oxygen ke liye same hai; density sirf tab show up karta hai jab tum pressure maango.

Recall Quick self-test

Head mein sirf $c_{u}$ kyun aata hai ( $c_{r}$ nahi)? ::: Radial part axle ke through point karta hai, isliye axis ke about uska zero angular momentum hai; sirf tangential whirl hi fluid ko twist karta hai. Power equation mein $ω r_{2}$ ki jagah kya aata hai aur kyun? ::: Blade speed $u_{2}$ — yeh exactly $ω r_{2}$ hai, measurable rim-tip speed. $c_{u 1} = 0$ ke saath, head kya hai? ::: $H \approx u_{2} c_{u 2} / g$ . Agar rim speed double ho, head kya karta hai? ::: Roughly chaar guna ho jaata hai, kyunki $H \sim u_{2}^{2} / g$ . Kya head liquid ki density par depend karta hai? ::: Nahi — metres mein head density-free hai; sirf pressure rise $Δ p = ρ g H$ density ke saath scale karta hai.

Prerequisite threads: Bernoulli's Principle (fluids mein energy bookkeeping), Cavitation (inlet ko kya limit karta hai), aur broader Euler Turbomachinery Equation jiska yeh derivation ek special case hai. Downstream yeh head achievable Chamber Pressure and Thrust set karta hai aur isliye Specific Impulse, cycle choices ko feed karta hai jaise Gas Generator Cycle aur Staged Combustion Cycle.