Optimum expansion — P_e = P_a for maximum thrust: Exercises

3.3.13 · D4 · Physics › Rocket Propulsion › Optimum expansion — P_e = P_a for maximum thrust

Yeh page ek self-test hai. Har problem pehle padho, paper pe try karo, PHIR collapsible solution kholo. Yahan jo bhi symbols use hue hain woh parent topic mein explain hain; agar kuch unfamiliar lage, toh pehle Thrust equation derivation aur Isentropic flow relations revisit karo.

Level 1 — Recognition

L1·Q1 — Regime ka naam batao

Ek nozzle $P_{e} = 0.4 bar$ par gas exit kar raha hai. Rocket sea level par hai jahan $P_{a} = 1.0 bar$ . Kya yeh nozzle under-expanded, optimally expanded, ya over-expanded hai?

Recall Solution

Dono pressures compare karo. Yahan $P_{e} = 0.4 bar < P_{a} = 1.0 bar$ .

$P_{e} > P_{a}$ → under-expanded
$P_{e} = P_{a}$ → optimal
$P_{e} < P_{a}$ → over-expanded ✔

Gas pehle se hi baahr ke pressure se neeche ja chuki hai, toh atmosphere exhaust ko waapas daba raha hai. Answer: over-expanded.

L1·Q2 — Kaun sa term zero ho jaata hai?

Thrust equation likho, phir batao ki perfectly (optimally) expanded nozzle mein kaunsa ek term zero ho jaata hai, aur equation kya reh jaati hai.

Recall Solution

$F = \overset{m}{˙} v_{e} + (P_{e} - P_{a}) A_{e}$ Optimal matlab $P_{e} = P_{a}$ , toh $(P_{e} - P_{a}) = 0$ . Pressure term $(P_{e} - P_{a}) A_{e}$ khatam ho jaata hai. $F_{opt} = \overset{m}{˙} v_{e}$ Ab saari thrust purely exhaust ke momentum se aa rahi hai — sabse clean possible conversion.

Level 2 — Application

L2·Q1 — Pressure term ke saath thrust calculate karo

Ek rocket mein $\overset{m}{˙} = 250 kg/s$ , $v_{e} = 3000 m/s$ , $A_{e} = 1.5 m^{2}$ , aur $P_{e} = 0.5 bar$ hai. Us altitude par thrust nikalo jahan $P_{a} = 0.3 bar$ hai.

Recall Solution

Pressures convert karo: $P_{e} = 0.5 \times 1 0^{5} = 50, 000 Pa$ , $P_{a} = 0.3 \times 1 0^{5} = 30, 000 Pa$ . $F = \overset{m}{˙} v_{e} + (P_{e} - P_{a}) A_{e}$ $F = (250) (3000) + (50, 000 - 30, 000) (1.5)$ $F = 750, 000 + (20, 000) (1.5) = 750, 000 + 30, 000$ $F = 780, 000 N = 780 kN$ Yahan $P_{e} > P_{a}$ : nozzle under-expanded hai, toh pressure term positive hai aur thodi extra thrust deta hai.

L2·Q2 — Same engine, sea level par

L2·Q1 wala exactly wahi engine ( $\overset{m}{˙} = 250$ , $v_{e} = 3000$ , $A_{e} = 1.5$ , $P_{e} = 0.5 bar$ ) ab sea level par $P_{a} = 1.0 bar$ mein fire karo. Thrust nikalo aur batao kya hua.

Recall Solution

$P_{e} = 50, 000 Pa$ , $P_{a} = 1.0 \times 1 0^{5} = 100, 000 Pa$ . $F = (250) (3000) + (50, 000 - 100, 000) (1.5)$ $F = 750, 000 + (- 50, 000) (1.5) = 750, 000 - 75, 000$ $F = 675, 000 N = 675 kN$ Ab $P_{e} < P_{a}$ : nozzle over-expanded hai, pressure term negative hai, aur thrust momentum-only value se $75 kN$ kam ho gayi. Dekho Altitude compensation jisse pata chale ek fixed nozzle har jagah kyun nahi jeet sakta.

Level 3 — Analysis

L3·Q1 — Thrust loss ka percentage

Vacuum ke liye design kiya gaya engine: $\overset{m}{˙} = 400 kg/s$ , $v_{e} = 3200 m/s$ , $A_{e} = 3.0 m^{2}$ , $P_{e} = 0.08 bar$ . Iska vacuum thrust aur sea-level thrust ( $P_{a} = 1.01 bar$ ) nikalo, phir sea level par kitna thrust lost hua ka percentage batao.

Recall Solution

Momentum term (har jagah same): $\overset{m}{˙} v_{e} = (400) (3200) = 1, 280, 000 N$ .

Vacuum ( $P_{a} = 0$ ), $P_{e} = 0.08 \times 1 0^{5} = 8000 Pa$ : $F_{vac} = 1, 280, 000 + (8000 - 0) (3.0) = 1, 280, 000 + 24, 000 = 1, 304, 000 N$

Sea level, $P_{a} = 1.01 \times 1 0^{5} = 101, 000 Pa$ : $F_{SL} = 1, 280, 000 + (8000 - 101, 000) (3.0) = 1, 280, 000 + (- 93, 000) (3.0)$ $F_{SL} = 1, 280, 000 - 279, 000 = 1, 001, 000 N$

Loss: $\frac{F _{vac} - F _{SL}}{F _{vac}} \times 100% = \frac{303 , 000}{1 , 304 , 000} \times 100% \approx 23.2%$ Sea level par lagbhag ek chauthai thrust gayab ho jaati hai — yeh classic penalty hai jab bade-area vacuum nozzle ko atmosphere mein neeche use karo.

L3·Q2 — Optimum operation ka altitude dhundho

L3·Q1 ka engine $P_{e} = 0.08 bar$ fixed hai (geometry se). Kis ambient pressure $P_{a}$ par yeh nozzle perfectly expanded hoga, aur kya woh low altitude hai ya high altitude?

Recall Solution

Optimum ke liye $P_{e} = P_{a}$ chahiye, toh $P_{a} = 0.08 bar = 8000 Pa$ . Sea level $\approx 1.01 bar$ hai; upar jaane se ambient pressure girta hai, toh $0.08 bar$ sirf high altitude par milega (roughly $17$ – $18 km$ upar, jahan hawa patli hoti hai). Exactly isliye yeh nozzle sea level par over-expanded hai: woh us patli hawa ka wait kar raha hai jiske liye design hua tha.

Level 4 — Synthesis

Is level ko shuru karne se pehle Figure 1 neeche dhyan se dekho. Isme thrust $F$ (vertical axis) ko exit pressure $P_{e}$ (horizontal axis) ke against plot kiya gaya hai, jaise hum nozzle ko dheere dheere kholte hain. Horizontal axis exit pressure dikhata hai left = low $P_{e}$ (bada area, over-expanded) se right = high $P_{e}$ (chota area, under-expanded) tak. Yellow dashed line $P_{e} = P_{a}$ mark karta hai; blue curve par yellow dot ek akela peak hai. Dhyaan do ki curve left se upar jaati hai, dashed line par top out karti hai, aur phir right ki taraf girne lagti hai — yahi is level ki poori kahani hai.

Figure — Optimum expansion — P_e = P_a for maximum thrust

L4·Q1 — Slope se prove karo ki peak $P_{e} = P_{a}$ par hai

Thrust equation aur steady flow ke Euler's relation se dikhao ki thrust ka exit pressure ke saath slope $\frac{d F}{d P _{e}} = (P_{e} - P_{a}) \frac{d A _{e}}{d P _{e}}$ hai, phir explain karo ki akela interior maximum sirf $P_{e} = P_{a}$ par kyun hai, aur Figure 1 par under-expanded regime locate karo.

Recall Solution

Cancellation ka recap (taaki yeh self-contained rahe). $F = \overset{m}{˙} v_{e} + (P_{e} - P_{a}) A_{e}$ ko $P_{e}$ ke saath differentiate karo (sab kuch nozzle-opening family ke saath vary karta hai): $\frac{d F}{d P _{e}} = \overset{m}{˙} \frac{d v _{e}}{d P _{e}} + A_{e} + (P_{e} - P_{a}) \frac{d A _{e}}{d P _{e}} .$ Steady 1-D flow ke liye Euler's momentum relation deta hai $\overset{m}{˙} d v_{e} = - A_{e} d P_{e}$ , toh $\overset{m}{˙} \frac{d v _{e}}{d P _{e}} = - A_{e}$ . Substitute karne par: $\frac{d F}{d P _{e}} = - A_{e} + A_{e} + (P_{e} - P_{a}) \frac{d A _{e}}{d P _{e}} = (P_{e} - P_{a}) \frac{d A _{e}}{d P _{e}} .$ Dono $A_{e}$ terms exactly cancel ho jaate hain — yahi result ka core hai.

Peak kahan hai. Slope zero set karo: $(P_{e} - P_{a}) \frac{d A _{e}}{d P _{e}} = 0$ . Nozzle kholne se area hamesha change hota hai, toh $\frac{d A _{e}}{d P _{e}} \neq = 0$ . Product zero sirf tab hoga jab $P_{e} - P_{a} = 0$ , yaani $P_{e} = P_{a}$ . Yahi woh ek condition hai jo Figure 1 mein yellow dot (flat top) hai.

Direction check (Figure 1 par map karo). Jaise nozzle kholte hain, $A_{e}$ badhta hai aur $P_{e}$ girta hai — toh nozzle kholna Figure 1 ke horizontal axis par hame left ki taraf le jaata hai, aur $\frac{d A _{e}}{d P _{e}} < 0$ .

Peak ke right ( $P_{e} > P_{a}$ , under-expanded): $(P_{e} - P_{a}) > 0$ times negative $\frac{d A _{e}}{d P _{e}}$ deta hai $\frac{d F}{d P _{e}} < 0$ . Kyunki nozzle kholne par $P_{e}$ ghar jaata hai, left ki taraf move karna (peak ki taraf) thrust badhata hai — curve dashed line ki taraf chadh rahi hai. Tum abhi bhi peak se neeche ho; nozzle kholna help karta hai.
Peak ke left ( $P_{e} < P_{a}$ , over-expanded): $(P_{e} - P_{a}) < 0$ times negative $\frac{d A _{e}}{d P _{e}}$ deta hai $\frac{d F}{d P _{e}} > 0$ ; aur kholne par ( $P_{e}$ aur girna, aur left jaana) ab thrust lose hogi — curve wahan se neeche girti hai. Tum peak ko overshoot kar chuke ho.

Dono slopes $P_{e} = P_{a}$ ki taraf waapas point karte hain: yeh genuine maximum hai, exactly woh yellow dot. Dekho Nozzle expansion ratio jisse samjho $A_{e}$ aur $P_{e}$ saath kaise move karte hain.

L4·Q2 — Isentropic flow se optimum expansion ratio

Ek engine mein $P_{0} = 40 bar$ (chamber), $γ = 1.2$ hai, aur $P_{a} = 0.5 bar$ par optimum hona chahiye. Isentropic exit-Mach relation use karke $\frac{P _{e}}{P _{0}} = (1 + \frac{γ - 1}{2} M_{e}^{2})^{- \frac{γ}{γ - 1}},$ exit Mach number $M_{e}$ nikalo (upar defined: exit speed ÷ local speed of sound). Yeh Nozzle expansion ratio $ε$ ko feed karta hai.

Recall Solution

Optimum ⇒ $P_{e} = P_{a} = 0.5 bar$ , toh pressure ratio hai $\frac{P _{e}}{P _{0}} = \frac{0.5}{40} = 0.0125.$ $γ = 1.2$ ke saath: exponent $- \frac{γ}{γ - 1} = - \frac{1.2}{0.2} = - 6$ , aur $\frac{γ - 1}{2} = 0.1$ . $0.0125 = (1 + 0.1 M_{e}^{2})^{- 6}$ $- 6$ power undo karne ke liye dono sides ko $- \frac{1}{6}$ power se raise karo: $0.012 5^{- 1/6} = 1 + 0.1 M_{e}^{2}$ Left side logarithms se carefully compute karo: $ln (0.0125) = - 4.382$ , toh $- \frac{1}{6} ln (0.0125) = + 0.7304$ , aur $e^{0.7304} = 2.076$ . Isliye $2.076 = 1 + 0.1 M_{e}^{2} \Rightarrow 0.1 M_{e}^{2} = 1.076 \Rightarrow M_{e}^{2} = 10.76$ $M_{e} \approx 3.28$ Ek strongly supersonic exit ( $M_{e} \approx 3.3$ ), exactly woh jo ek achhe se expand kiya hua rocket nozzle produce karta hai.

Level 5 — Mastery

L5·Q1 — Poore climb ke liye do fixed nozzles mein se best kaun sa

Ek booster apna burn do equal-duration conditions mein split karta hai:

Phase A (low altitude): $P_{a} = 0.9 bar$
Phase B (high altitude): $P_{a} = 0.1 bar$

Do candidate nozzles $\overset{m}{˙} = 500 kg/s$ , $v_{e} = 3000 m/s$ share karte hain lekin exit design mein alag hain:

Nozzle X (sea-level design): $A_{e} = 2.0 m^{2}$ , $P_{e} = 0.9 bar$ .
Nozzle Y (vacuum design): $A_{e} = 5.0 m^{2}$ , $P_{e} = 0.1 bar$ .

Har nozzle ka average thrust $\frac{1}{2} (F_{A} + F_{B})$ compute karo aur decide karo poore climb mein kaun sa booster better hai.

Recall Solution

Dono ke liye momentum term: $\overset{m}{˙} v_{e} = (500) (3000) = 1, 500, 000 N$ . Pa mein pressures: $P_{a}^{A} = 90, 000$ , $P_{a}^{B} = 10, 000$ ; $P_{e}^{X} = 90, 000$ , $P_{e}^{Y} = 10, 000$ .

Nozzle X ( $A_{e} = 2.0$ ):

Phase A: $F = 1, 500, 000 + (90, 000 - 90, 000) (2.0) = 1, 500, 000 N$ (yahan optimum!)
Phase B: $F = 1, 500, 000 + (90, 000 - 10, 000) (2.0) = 1, 500, 000 + 160, 000 = 1, 660, 000 N$
Average: $\frac{1}{2} (1, 500, 000 + 1, 660, 000) = 1, 580, 000 N$

Nozzle Y ( $A_{e} = 5.0$ ):

Phase A: $F = 1, 500, 000 + (10, 000 - 90, 000) (5.0) = 1, 500, 000 - 400, 000 = 1, 100, 000 N$ (buri tarah over-expanded)
Phase B: $F = 1, 500, 000 + (10, 000 - 10, 000) (5.0) = 1, 500, 000 N$ (yahan optimum!)
Average: $\frac{1}{2} (1, 100, 000 + 1, 500, 000) = 1, 300, 000 N$

Nozzle X jeet jaata hai, $1, 580 kN$ vs $1, 300 kN$ average. Bada vacuum nozzle Y neeche ek bhaari $400 kN$ over-expansion penalty suffer karta hai (bada $A_{e}$ negative pressure gap ko amplify karta hai). Exactly yahi trade-off Altitude compensation aur Rocket staging ko motivate karta hai: neeche compact nozzle use karo, upar bada nozzle.

L5·Q2 — Jahan flow separation vacuum nozzle ko khatam karta hai

Nozzle Y ke Phase A mein ( $P_{e} = 0.1 bar$ , $P_{a} = 0.9 bar$ ), pressure ratio $P_{e} / P_{a} = 0.111$ hai. Ek common rule of thumb (Summerfield criterion) kehta hai ki exhaust wall se tab separate hota hai jab $P_{e} / P_{a} ≲ 0.35$ – $0.4$ . Kya Nozzle Y low altitude par separate hota hai, aur iska physically kya matlab hai? (Link: Supersonic flow separation.)

Recall Solution

Ratio compute karo: $\frac{P _{e}}{P _{a}} = \frac{0.1}{0.9} = 0.111$ . Kyunki $0.111 < 0.35$ , flow nozzle wall se exit se pehle hi separate ho jaata hai.

Physically: ambient pressure exit pressure se itna zyada hai ki woh diverging section mein waapas andar dhakelta hai, supersonic jet ko wall se peel kar deta hai. Gas phir effectively ek chote area par higher pressure ke saath exit karti hai ideal $0.1 bar$ ki jagah, aur flow unsteady ho jaata hai (side-loads, buffeting). L5·Q1 mein compute kiya hua clean thrust number actually ek optimistic bound hai — real over-expanded losses aur bure ho sakte hain aur nozzle ko damage bhi kar sakte hain. Exactly isliye bade vacuum nozzles ko physically sea level par fire nahi kiya ja sakta.

Recall Quick self-quiz (answers cover kar lo)

Optimum thrust condition ::: $P_{e} = P_{a}$ $P_{e} > P_{a}$ wala nozzle kehlata hai ::: under-expanded $P_{e} < P_{a}$ wala nozzle kehlata hai ::: over-expanded Optimum par jo term vanish hoti hai ::: pressure term $(P_{e} - P_{a}) A_{e}$ Exit Mach $M_{e}$ ka matlab ::: exit speed divided by the local speed of sound Sea level par vacuum nozzle kyun bura hai ::: $P_{e} ≪ P_{a}$ ⇒ large negative pressure thrust aur flow separation