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

• (a) $c_{u2}$ — impeller rim par liquid ki exit whirl velocity → centrifugal pump se belong karti hai (yeh Euler head $H=u_2{c}_{u2}/g$ set karta hai).
• (b) ${\text{NPSH}}_A$ — vapor pressure ke upar available suction head → NPSH / cavitation pillar se belong karta hai (dekho Cavitation).
• (c) $\Delta c_u$ — gas jab ek rotor row cross karta hai tab whirl mein change → axial turbine stage se belong karta hai (work $w=u\Delta c_u$, the Euler Turbomachinery Equation).

No inlet swirl ka matlab hai ==$c_{u1}=0$==, toh doosra term zero ho jaata hai aur $H=\frac{u_2{c}_{u2}}{g}$. Designers yeh kyun chahte hain: liquid ko radially/axially feed karne ka matlab hai ki yeh impeller sab whirl add karta hai, ek given rim speed $u_2$ ke liye head maximise karta hai aur formula simple rakhta hai.

rpm → rad/s convert karo (kyun: Euler relations $\omega$ ko rad/s mein use karte hain, rev/min mein nahi): $\omega =\frac{2\pi N}{60}=\frac{2\pi (40000)}{60}=4188.8\text{rad/s}.$ Rim par blade speed (kyun: $u=\omega r$ radius $r$ par ek point ki linear speed hai): $u_2=\omega r_2=4188.8\times 0.08=335.1\text{m/s}.$

Yeh formula kyun: no inlet swirl ke saath Euler head $H=u_2{c}_{u2}/g$ ban jaata hai — yeh exactly liquid ke unit weight par add ki gayi energy hai, matlab whirl work $u_2{c}_{u2}$ (energy per unit mass) ko $g$ se divide karo taaki ise ek column height ki tarah express kar sako. Hum ise yahan use karte hain kyunki $u_2$ (L2.1 se) aur $c_{u2}$ (diya gaya $u_2$ ke fraction ke roop mein) dono known hain. $c_{u2}=0.75\times 335.1=251.3\text{m/s}$. $H=\frac{u_2{c}_{u2}}{g}=\frac{335.1\times 251.3}{9.81}=8585\text{m}.$ (Equivalent liquid column ka lagbhag 8.6 km — dekho Bernoulli's Principle yeh samajhne ke liye ki head aise kyun measure hota hai.)

$\rho g$ se kyun multiply karein: head energy per unit weight hai; $\rho g$ se multiply karo taaki energy per unit volume mile, jo pressure hai. $\Delta p=\rho gH=1140\times 9.81\times 8585=9.60\times 10^7\text{Pa}\approx 960\text{bar}.$ Physical realism: ek stage se 960 bar real pumps jo per stage deliver karte hain usse kaafi zyada hai. Pehle do limits aate hain. (i) Tip speed: $u_2=335\text{m/s}$ impeller-rim stresses ko metal ki strength limit ki taraf push karta hai — zyaatar pumps $u_2$ ko $\sim 400\text{m/s}$ se neeche rakhte hain, aur yeh problem us ceiling ke paas hai kyunki yeh ek chhote radius ko extreme 40000 rpm ke saath combine karta hai. (ii) Cavitation & stage loading: ek impeller mein concentrated 960-bar rise unrealistic hai; real turbopumps large rises ko multiple stages mein split karte hain (SSME LH₂ pump teen use karta hai). Is number ko ek upper-bound sanity figure samjho, buildable single stage nahi — arithmetic sahi hai, single-stage engineering nahi.

Yeh formula kyun: fluid mein add ki gayi hydraulic power $\dot{m}\Delta p/\rho$ hai (energy/mass $=\Delta p/\rho$); efficiency se divide karna pump ke andar real losses account karta hai. $P_{\text{pump}}=\frac{\dot{m}\Delta p}{\rho {\eta }_p}=\frac{150\times 180\times 10^5}{810\times 0.72}=4.63\times 10^6\text{W}\approx 4.63\text{MW}.$

Yeh formula kyun: turbine gas ki thermal energy ka ek fraction extract karta hai jo isentropic temperature drop se set hoti hai; bracket woh ideal drop fraction hai, aur ${\eta }_t$ use real machine tak scale karta hai. $P_{\text{turb}}={\dot{m}}_g{c}_p{T}_0{\eta }_t\left(1-\frac{1}{{\pi }_t^{(\gamma -1)/\gamma }}\right).$ Exponent: $(\gamma -1)/\gamma =0.3/1.3=0.2308$. Toh ${\pi }_t^{0.2308}=8^{0.2308}=1.622$, aur $1-1/1.622=0.3833$. $P_{\text{turb}}=4.0\times 2100\times 950\times 0.62\times 0.3833=1.90\times 10^6\text{W}\approx 1.90\text{MW}.$ Check: pump ko 4.63 MW chahiye lekin turbine sirf 1.90 MW banata hai ⇒ kaafi nahi. Design power match fail karta hai; tumhe ${\dot{m}}_g$, $T_0$, ya ${\pi }_t$ badhana hoga. Exactly aise hi Gas Generator Cycle size hota hai.

$P_{\text{turb}}\propto {\dot{m}}_g$, toh linearly scale karo: $\frac{{\dot{m}}_{g,\text{new}}}{{\dot{m}}_{g,\text{old}}}=\frac{4.63}{1.90}=\mathrm{2.44.}$ Naya gas flow ${\dot{m}}_g=2.44\times 4.0=9.8\text{kg/s}$. Yeh matter kyun karta hai: gas generator mein jitna zyada gas burn hoga woh propellant main chamber mein nahi jaayega — yeh Specific Impulse par seedha asar dalta hai.

Full Bernoulli, tank surface → pump inlet (dekho Bernoulli's Principle). Har head term likhte hue: $\frac{p_{\text{tank}}}{\rho g}+z-h_{\text{loss}}=\frac{p_{\text{inlet}}}{\rho g}+\frac{v^2}{2g}.$ ${\text{NPSH}}_A$ vapor pressure ke upar total (stagnation) inlet head hai, isliye yeh velocity head rakhta hai: rearrange karne par, $\frac{p_{\text{inlet}}}{\rho g}+\frac{v^2}{2g}-\frac{p_v}{\rho g}=\frac{p_{\text{tank}}-p_v}{\rho g}+z-h_{\text{loss}}$. $\frac{v^2}{2g}$ term gayab nahi hota — yeh left side mein absorb ho jaata hai aur exactly yahi reason hai ki NPSH static pressure ki jagah stagnation pressure par define hota hai. Right side par jo bachta hai woh driving margin hai: ${\text{NPSH}}_A=\frac{p_{\text{tank}}-p_v}{\rho g}+z-h_{\text{loss}}.$ Inlet velocity head final expression se kyun gayab ho jaata hai: yeh Bernoulli ke dono sides ke beech cancel ho jaata hai — tezi se move karta hua inlet fluid static pressure ko velocity head ke liye trade karta hai, lekin unka sum (stagnation head) wohi hai jo NPSH track karta hai, isliye alag $v^2/2g$ term nahi bachta. Pressure-margin head: $\frac{(2.5-1.0)\times 10^5}{71\times 9.81}=\frac{1.5\times 10^5}{696.5}=215.4\text{m}$. ${\text{NPSH}}_A=215.4+1.5-2.0=214.9\text{m}.$ Bahut bada, kyunki LH₂ ka $\rho$ bahut chhota hai — ek chhota pressure margin ek giant column ban jaata hai.

Safe operation ke liye ${\text{NPSH}}_A>{\text{NPSH}}_R$ chahiye. Yahan $214.9<240$ ⇒ safe nahi; impeller eye vapor pressure se neeche drop ho jaayegi aur cavitate karegi. Shortfall: $240-214.9=25.1\text{m}$. Fix: ek inducer add karo — impeller ke aage ek slender axial screw jo dhire se flow ko pre-pressurise karta hai, required ${\text{NPSH}}_R$ ko kam karta hai taaki pump ek bhaari, zyada-pressurised tank ke bina low inlet head tolerate kar sake. (Poora point Cavitation se bachna hai.)

${\text{NPSH}}_A$ ko $25.1\text{m}$ badhana hai, matlab pressure-margin head $215.4$ se $240.5\text{m}$ tak jaanaa chahiye. Invert karo: $p_{\text{tank}}-p_v=(\text{margin head})\times \rho g=240.5\times 71\times 9.81=1.675\times 10^5\text{Pa}$. $p_{\text{tank}}=1.675\times 10^5+1.0\times 10^5=2.675\times 10^5\text{Pa}\approx 2.68\text{bar}.$ Sirf $\sim 0.18\text{bar}$ zyada tank pressure — sunne mein sasta lagta hai, lekin zyada pressure wale tank ko har jagah thick aur heavy hona padega, mass fraction hurt hogi. Inducer usually jeeetta hai; yahi trade reason hai ki turbopumps exist karte hain.

$u_2=\omega r_2\propto N$. $c_{u2}=\kappa u_2$ ke saath: $\Delta p=\rho gH=\rho u_2{c}_{u2}=\rho \kappa u_2^2\propto N^2.$ Yeh design ko dominate kyun karta hai: pressure rise speed ke square ke saath badhta hai — RPM double karne par $\Delta p$ chaar guna ho jaata hai, isliye impellers hazaaron RPM par scream karte hain. $\Delta p$ double karne ke liye $N^2$ double chahiye, matlab $N_{\text{new}}=N_{\text{old}}\sqrt{2}$: $N_{\text{new}}=40000\times \sqrt{2}=56569\text{rpm}.$

$H=\frac{u_2\cdot 0}{g}=0.$ Physical meaning: tezi se spin karna useless hai jab tak blades actually flow ko tangential motion mein turn na karein. Whirl $c_{u2}$ wohi hai jo angular momentum bahar le jaata hai; koi deliver na ho toh angular-momentum change nahi ⇒ koi work nahi ⇒ koi head nahi. Ek radial-tipped blade jo fluid ko rim se seedha slip karne deta hai kuch nahi karta.

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 ko supply karna hoga (mechanical losses account karte hue): $P_{\text{turb}}=\frac{P_{\text{pump}}}{{\eta }_{\text{mech}}}=\frac{6.67\times 10^6}{0.98}=6.80\times 10^6\text{W}.$ 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: $\frac{13.7}{180}=0.0762=7.6\%$ propellant pump drive karne mein kharach hota hai — ek Gas Generator Cycle mein low $I_{sp}$ par overboard dump hota hai. Us gas ko chamber mein wapas route karna exactly wohi hai jo Staged Combustion Cycle karta hai use reclaim karne ke liye, overall Specific Impulse aur Chamber Pressure and Thrust protect karta hai.

Kin pillar ke head formula mein koi $\rho$ nahi hai? ::: Centrifugal pump — $H=u_2{c}_{u2}/g$ (density sirf $\Delta p=\rho gH$ mein aati hai). Kaunsa single Euler relation pump aur turbine dono ke work govern karta hai? ::: $w=u\Delta c_u$ — pump whirl add karta hai, turbine use remove karta hai. ${\text{NPSH}}_A$ mein kaun sa term LH₂ ke liye bahut bada hai aur kyun? ::: Pressure-margin head $\frac{p_{\text{tank}}-p_v}{\rho g}$, kyunki $\rho$ chhota hai isliye ek chhota pressure ek giant column ban jaata hai. Sabse sasta safe fix jab ${\text{NPSH}}_A<{\text{NPSH}}_R$ ho? ::: Ek inducer add karo ${\text{NPSH}}_R$ kam karne ke liye, tank light rakhte hue.