Turbopump design — centrifugal pump, axial turbine stages, NPSH: Question bank

3.3.27 · D5 · Physics › Rocket Propulsion › Turbopump design — centrifugal pump, axial turbine stages, N

Shuru karne se pehle, is page mein use hone wale har symbol ka simple shabdon mein matlab diya gaya hai, taaki kuch bhi seedha na aaye. Neeche ke figures do geometric objects draw karte hain jis par sab kuch tika hai: impeller velocity triangle aur NPSH head stack.

Definition Symbol glossary (ek baar padho, baad mein refer karo)

$H$ = head, fluid ko dii gayi energy per unit weight, us fluid ke apne column ke metres mein measure ki gayi.
$u_{2} = ω r_{2}$ = impeller rim par blade speed (rim radius $r_{2}$ times spin rate $ω$ ).
$c_{u 2}$ = exit whirl, fluid ki exit speed ka sideways (tangential) hissa; $c_{u 1}$ wohi cheez inlet par hai.
Sign convention: tangential speed $c_{u}$ ko positive count karte hain jab wo blade ke movement ki same direction mein ho (co-rotating swirl) aur negative jab blade ke against ho. Toh added, forward swirl positive $c_{u}$ hai; blade se ladte hue pre-swirl negative hai.
$Δ c_{u} = c_{u, in} - c_{u, out}$ = ek row ke across fluid mein hone wala whirl ka change. Positive $Δ c_{u}$ ka matlab hai whirl remove hua (turbine ne work nikala).
$Δ p = ρ g H$ = pressure rise, energy per unit volume; $ρ$ = fluid density.
$\overset{m}{˙}$ = mass flow rate, kilograms of fluid passing per second (subscript $g$ turbine gas ke liye, $prop$ pumped propellant ke liye).
$η_{pump}, η_{t}, η_{mech}$ = pump, turbine, aur shaft/bearings ki efficiencies (0 aur 1 ke beech ek fraction) — ideal energy ka kitna hissa actually andar se guzarta hai.
$c_{p}$ = turbine gas ki specific heat: constant pressure par uske 1 kg ko 1 K garam karne ke liye joules. Yeh gas ke temperature ko stored energy per kg mein convert karta hai.
$T_{0, in}$ = turbine gas ki total (stagnation) inlet temperature — wo temperature jo gas ko rest par laaने par hoti, yaani entry par uski poori thermal energy content.
$π_{t}$ = turbine pressure ratio, turbine ke across inlet pressure divided by outlet pressure — gas ko kitna expand karne diya jaata hai.
$γ$ = gas ka specific-heat ratio ( $c_{p} / c_{v}$ ), wo number jo govern karta hai ki gas expand hote waqt kaise thanda hota hai.
$p_{v}$ = vapor pressure, wo pressure jis par liquid apne current temperature par boil karta hai.
$NPSH_{A}$ / $NPSH_{R}$ = plumbing se suction-head available vs. impeller geometry ke liye required.

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

Figure ka left side impeller hai: rim $u_{2}$ par move karta hai, fluid $c_{u 2}$ whirl ke saath nikalta hai, aur unka product (per $g$ ) head hai. Right side NPSH stack hai: tank-pressure margin plus gravity height minus friction, pump eye par available head deta hai.

Yeh doosra figure dikhata hai ki kyun $c_{u 2}$ , $u_{2}$ ko track karta hai: fixed blade exit angle $β_{2}$ ke liye, exit velocity triangle simply scale up hoti hai jab rim tezi se spins karta hai — whirl side $c_{u 2}$ , $u_{2}$ ke proportion mein badhti hai (near-constant "slip" gap ko allow karte hue). Yahi assumption $H \sim u_{2}^{2} / g$ likhne ke peeche hai.

True or false — justify karo

True or false: Fixed RPM par chal raha ek centrifugal pump water ke liye aur liquid oxygen ke liye same head (metres mein) produce karta hai.

True — Euler head

H = u_{2} c_{u 2} / g

mein koi density term nahi hai, toh column ke metres identical hain; sirf pressure rise

Δ p = ρ g H

alag hota hai kyunki woh

ρ

ke saath scale karta hai.

True or false: Impeller rim speed double karne se pressure rise roughly double ho jaata hai.

False — kyunki exit blade angle fixed hai,

c_{u 2}

roughly

u_{2}

ke saath linearly scale karta hai, toh

H = u_{2} c_{u 2} / g \sim u_{2}^{2} / g

; rim speed double karne se rise quadruple hoti hai. Yahi quadratic reason hai ki turbopumps tens of thousands of RPM par chalte hain.

True or false: Turbine aur pump ko independently size kiya ja sakta hai.

False — woh ek shaft share karte hain, toh turbine ko exactly

P_{turb} = P_{pump} / η_{mech}

produce karna padta hai; yeh power-match constraint unhe couple karti hai aur gas generator flow ko size karta hai.

True or false: High

NPSH_{A}

value guarantee karta hai ki koi cavitation nahi hogi.

Apne aap mein False — safety ke liye

NPSH_{A} > NPSH_{R}

chahiye; ek bahut bada

NPSH_{A}

bhi agar geometry-driven

NPSH_{R}

aur bada ho toh eye par boil karta hai.

True or false: Ek inducer

NPSH_{A}

raise karke kaam karta hai.

False — inducer flow ko gently pre-pressurize karke

NPSH_{R}

lower karta hai, taaki same available head requirement ko clear kar sake; plumbing-side

NPSH_{A}

untouched rehta hai.

True or false: Euler relation

u Δ c_{u}

pump aur turbine dono ko describe karta hai.

True — yeh same angular-momentum bookkeeping hai (Euler Turbomachinery Equation); pump whirl add karta hai (work in,

u_{2} c_{u 2} - u_{1} c_{u 1}

) jabki turbine ise remove karta hai (work out,

u Δ c_{u}

), sirf sign/direction mein alag.

True or false: Cavitation ka ek maatra harmful effect yeh hai ki yeh head reduce karta hai.

False — head collapse performance zaroor hurt karta hai, lekin zyada buri damage yeh hai ki bubbles blades par collapse hote hain, pitting erosion aur vibration cause karte hain jo impeller ko physically destroy karte hain. Toh "only" statement ko false banata hai.

True or false: LH₂ pumps bahut bade NPSH values (hundreds of metres) dikhate hain kyunki unke inlets unusually safe hain.

False — bade metres ek modest pressure margin ko tiny density se divide karne se aate hain (

ρ_{L H_{2}} \approx 71

); actual pressure margin chota hota hai, aur yahi reason hai ki LH₂ cavitation-prone hai aur inducers ki zaroorat hoti hai.

Error dhundho

"Head

H = u_{2} c_{u 2} / g

, toh denser propellant ke liye pressure raise karne ke liye hume kuch change karne ki zaroorat nahi."

Head same rehta hai, lekin pressure rise

Δ p = ρ g H

density ke saath badhta hai, toh pump ko same head par denser fluid ke liye zyada shaft power absorb karni padti hai (

P = \overset{m}{˙} g H / η

"Hum pump ke liye axial stages choose karte hain kyunki axial rows series mein stack hoti hain aur zyada energy extract karti hain."

Reasoning sahi hai lekin machine swap ho gayi — turbines axial hoti hain (low-density, high-velocity gas, stack karna easy). Pump centrifugal hota hai kyunki ek single impeller dense liquid ko compactly high pressure tak fling karta hai.

"Head maximize karne ke liye humein bahut saara inlet swirl add karna chahiye, toh

c_{u 1}

bada ho."

Intuition ka sign galat hai — inlet whirl

c_{u 1}

H = (u_{2} c_{u 2} - u_{1} c_{u 1}) / g

mein subtract hota hai. Designers

c_{u 1} = 0

set karte hain (radial/axial entry) taaki saari whirl is stage dwara add ho, head maximize ho.

"Bernoulli kehta hai eye par tez liquid ka high pressure hota hai, toh eye cavitation se sabse safe spot hai."

Ulta — Bernoulli's Principle tez flow ko low static pressure deta hai, toh eye hi sabse zyada likely jagah hai jahan

p_{v}

se neeche drop ho aur boiling ho. Yahi poora reason hai ki NPSH inlet par define hota hai.

"NPSH

_{A}

mein sirf tank pressure aur vapor pressure involve hote hain."

Incomplete —

NPSH_{A} = \frac{p _{tank} - p _{v}}{ρ g} + z - h_{loss}

mein pump ke upar tank ki static height

z

bhi add hoti hai aur feedline friction

h_{loss}

subtract hoti hai.

"Turbine power sirf gas mass flow par depend karta hai."

Nahi —

P_{turb} = \overset{m}{˙}_{g} c_{p} T_{0, in} η_{t} (1 - π_{t}^{- (γ - 1) / γ})

inlet temperature aur pressure ratio ke saath bhi scale karta hai. $T_{0, in}$ kyun matter karta hai:

c_{p} T_{0}

hot gas ke per kg mein stored energy hai, toh zyada garam gas simply zyada extractable energy carry karta hai. $π_{t}$ kyun matter karta hai: bada pressure ratio gas ko zyada expand karne deta hai, aur gas sirf expand hote waqt hi work surrender kar sakta hai — zyada expansion, per kg zyada work.

"Kyunki head mein density nahi hai, ek water test LOX pump ki shaft power perfectly predict karta hai."

Head transfer hota hai, lekin power

P = \overset{m}{˙} g H / η = \overset{m}{˙} Δ p / (ρ η)

mass flow aur

Δ p

par depend karta hai, dono

ρ

ke saath change karte hain; tumhe power measurement scale karni hogi.

Why questions

Turbine work

Δ c_{u} = c_{u, in} - c_{u, out}

ke saath kyun likha jaata hai, single velocity se nahi?

Kyunki work fluid ke whirl ke change se aata hai jab woh rotor cross karta hai; rotor sirf woh angular momentum harvest karta hai jo gas de deta hai, jo tangential speed mein difference hai.

Turbopump bilkul exist kyun karta hai, tanks ko seedha pressurize karne ki bajay?

Pressure-fed tanks ko chamber-level pressure hold karne ke liye thick, heavy walls chahiye hongi; turbopump tanks ko light aur low-pressure rehne deta hai jabki ek chhoti turbine thoda propellant kharach karke pump par pressure 100× raise karti hai.

Euler head mein sirf tangential fluid velocity use kyun hoti hai, radial part nahi?

Shaft axis ke baare mein angular momentum sirf axis ke around motion se carry hota hai; radial component axis ke through point karta hai aur zero moment contribute karta hai, toh yeh impeller ke saath koi torque exchange nahi kar sakta.

Power-match constraint gas generator ko kyun size karta hai?

Turbine ko exactly pump ki demanded power supply karni hai; pump power fix karne se required

\overset{m}{˙}_{g} c_{p} T_{0, in} η_{t} (1 - π_{t}^{- (γ - 1) / γ})

fix hota hai, jo set karta hai ki gas generator ko kitna propellant burn karna hai.

Staged-combustion engines gas-generator engines se zyada chamber pressures kyun tolerate karte hain?

Staged combustion turbine exhaust ko chamber mein route karta hai instead of dump karne ke, toh koi propellant waste nahi hota aur zyada pump power justify ki ja sakti hai — pump

Δ p

higher enable karta hai aur better Specific Impulse ke liye higher chamber pressure.

Power ko

g

se divide karne par (sirf mass se nahi) "head" metres mein kyun milta hai?

Power/

\overset{m}{˙}

energy per unit mass hai; aage

g

se divide karne par energy per unit weight mein convert hota hai, jiska unit length hai — us same potential energy wale fluid column ki height.

Δ p = ρ g H

head ko

ρ g

se multiply karke kyun milta hai?

Head energy per unit weight hai; weight-per-unit-volume (

ρ g

) se multiply karne par energy per unit volume mein convert hota hai, aur energy per volume ka unit pressure hota hai.

Edge cases

Agar impeller spin nahi kar raha (

ω = 0

) toh head kya hoga?

Tab

u_{2} = ω r_{2} = 0

, toh

H = u_{2} c_{u 2} / g = 0

— ek stationary impeller koi energy add nahi karta; "pump" sirf ek stationary passage hai.

Agar tank pressure exactly vapor pressure ke equal ho toh

NPSH_{A}

ka kya hoga?

Pressure margin term vanish ho jaata hai,

NPSH_{A} = z - h_{loss}

reh jaata hai; sirf gravity head minus friction bachi rehti hai, toh jab tak tank pump se kaafi upar nahi, inlet boiling ke edge par hai.

Agar friction loss

h_{loss}

tank pressure margin plus

z

se zyada ho jaaye?

NPSH_{A}

negative ho jaata hai, matlab inlet pressure already vapor pressure se neeche hai — liquid impeller tak pahunchne se pehle hi boil karta hai aur pump fail ho jaata hai.

Agar exit whirl blade speed ke exactly equal ho (

c_{u 2} = u_{2}

), toh head kya hoga?

H = u_{2}^{2} / g

, us rim speed ke liye theoretical maximum; real blades

c_{u 2} < u_{2}

dete hain (slip aur finite blade angle), toh achievable head thoda kam hota hai.

Practice mein

u_{2}

(aur thus head) kitna grow kar sakta hai, kya limit karta hai?

Material strength — impeller mein centrifugal stress

u_{2}^{2}

ke saath badhta hai, toh disc burst ho jaayegi pehle head binaa bound ke grow kare; blade-tip speed metal dwara cap hoti hai, fluid se nahi.

Pressure ratio

π_{t} \to 1

(no expansion) par turbine work kya approach karta hai?

Factor

(1 - π_{t}^{- (γ - 1) / γ}) \to 0

, toh extracted work zero ho jaata hai — uske across koi pressure drop nahi, gas koi energy surrender nahi karta aur shaft ko koi power nahi milti.

Agar ek rotor row whirl remove karne ki bajay add kare toh

Δ c_{u}

ka sign kya hoga?

Convention ke saath ki

c_{u}

blade motion ke saath positive hai, whirl add karne par

c_{u, out} > c_{u, in}

hota hai, toh

Δ c_{u} = c_{u, in} - c_{u, out}

negative ho jaata hai — machine ab fluid par work kar rahi hai (pump/compressor), extract nahi kar rahi.

Recall Poore trap set ka ek-line summary

Head density-free hai lekin pressure aur power nahi hain; Euler ( $u Δ c_{u}$ ) dono machines ko opposite sign ke saath chalata hai; aur NPSH safety $NPSH_{A} > NPSH_{R}$ hai — inducers ise required side lower karke fix karte hain. Teen traps confirm karo ::: (1) $H$ mein koi $ρ$ nahi; (2) pump whirl add karta hai, turbine remove karta hai; (3) inducer NPSH $_{R}$ lower karta hai, NPSH $_{A}$ raise nahi karta.