Over-expanded nozzle — oblique shocks in plume, efficiency loss: Exercises

3.3.14 · D4 · Physics › Rocket Propulsion › Over-expanded nozzle — oblique shocks in plume, efficiency l

Yeh page parent topic ke liye ek self-test ladder hai. Har problem apna poora worked solution chhupaati hai — pehle khud try karo, phir reveal karo. Hum L1 (sirf words pehchaano) se L5 (design-level synthesis) tak chadhte hain.

Shuru karne se pehle, yeh symbols apne karo — neeche koi bhi equation inhe use nahi karti jab tak tumne yeh nahi padha:

Thrust equation jo hum baar baar use karenge (parent mein build ki gayi):

Hum nozzle ki quality ko ek velocity efficiency $C_{v}$ se bhi grade karenge. Yahan kyun hai ki yeh square-root form leti hai — L3 problems mein use hone se pehle yeh padho:

Formula Why

$C_{v} = p_{0, final} / p_{0, initial}$ $C_{v} = \frac{v _{e, actual}}{v _{e, ideal}} = \frac{p _{0, final}}{p _{0, initial}}$ Square root kyun, ratio khud kyun nahi? Ek nozzle jo exit velocity reach kar sakta hai woh us kinetic energy se set hoti hai jo woh extract karta hai, aur kinetic energy $v^{2}$ ke saath scale karti hai. Ek isentropic expansion mein woh usable energy jo flow abhi bhi convert kar sakta hai, uske surviving stagnation pressure $p_{0}$ (the "budget") ke proportional hoti hai. Toh $v_{e}^{2} \propto p_{0} ⟹ \frac{v _{e, actual}^{2}}{v _{e, ideal}^{2}} = \frac{p _{0, final}}{p _{0, initial}} .$ Dono sides ka square root lo aur tumhe $C_{v}$ milta hai. Words mein: velocity energy budget ke square-root ki tarah jaati hai, isliye budget (say) 0.62 tak cut hone par sirf $0.62 \approx 0.79$ velocity milti hai — square root blow ko soften karta hai, lekin yeh blow hai zaroor.

Neeche wala figure har problem ka physical picture hai is page par — jab bhi koi problem "plume," "lip," ya "oblique shocks" mention kare toh isko refer karo:

Figure — Over-expanded nozzle — oblique shocks in plume, efficiency loss

Isse left se right trace karo: laal arrow hai supersonic flow jo grey diverging nozzle se nikal rahi hai; exit plane par $p_{e} < p_{a}$ , isliye orange arrows (ambient air) andar ki taraf squeeze karte hain; woh squeeze flow ko angle $β$ par nozzle lips se nikalne wali do blue oblique shocks mein fold kar deta hai; woh centreline par cross karti hain aur green lines dikhate hain ki flow agli diamond mein re-expand ho raha hai. Problems 2.3, 3.x aur 5.1 sab un blue shock lines par rehte hain.

Level 1 — Recognition

Problem 1.1 (L1)

Ek nozzle $p_{e} = 40 kPa$ par gas exit karti hai $p_{a} = 101 kPa$ wali hawa mein. Kya yeh over-expanded hai, under-expanded hai, ya perfectly matched?

Worked example Solution 1.1

$p_{e}$ ko $p_{a}$ se compare karo. Yahan $p_{e} = 40 < p_{a} = 101$ , toh jet pressure ambient se neeche hai. Physically iska matlab: bahar ki hawa jeet rahi hai, plume ko andar ki taraf squeeze kar rahi hai. Answer: Over-expanded ( $p_{e} < p_{a}$ ).

Problem 1.2 (L1)

Thrust equation $F = \overset{m}{˙} v_{e} + (p_{e} - p_{a}) A_{e}$ mein, ek over-expanded nozzle ke liye kaun sa term negative ho jaata hai, aur ek negative thrust term ka matlab kya hota hai?

Worked example Solution 1.2

Pressure thrust term $(p_{e} - p_{a}) A_{e}$ . Kyunki $p_{e} < p_{a}$ , bracket negative hai, aur $A_{e} > 0$ , isliye poora term negative hai. Physical meaning: yeh ek retarding force hai — ambient air exit disc par flight ki direction ke against push karta hai, tumhare total thrust se subtract karta hai.

Problem 1.3 (L1)

Jab ek over-expanded plume ambient pressure tak wapas compress hoti hai, toh kya woh shock waves ke zariye karta hai ya expansion fans ke zariye? Ek word.

Worked example Solution 1.3

Shock waves. Supersonic flow mein rising pressure compression hai, aur supersonic flow mein compression shocks ke zariye hoti hai. (Expansion fans iska ulta karte hain — woh pressure smoothly giraaate hain.)

Level 2 — Application

Problem 2.1 (L2)

Ek nozzle ka $p_{e} = 30 kPa$ , $p_{a} = 101 kPa$ , $A_{e} = 1.5 m^{2}$ hai. Pressure thrust term compute karo. Kya yeh help kar raha hai ya hurt?

Worked example Solution 2.1

Hum kya karte hain: $(p_{e} - p_{a}) A_{e}$ mein plug karo. $(30 - 101) \times 1 0^{3} \times 1.5 = (- 71 \times 1 0^{3}) (1.5) = - 106.5 \times 1 0^{3} N = - 106.5 kN$ Kyun: hum pressure part ko akele isolate karte hain taaki penalty akele dekh saken. Answer: $- 106.5 kN$ — yeh hurt karta hai, momentum thrust se 106.5 kN subtract karta hai.

Problem 2.2 (L2)

2.1 wale nozzle ke liye, momentum thrust hai $\overset{m}{˙} v_{e} = 480 kN$ . Actual total thrust nikalo, aur over-expansion ki wajah se fractional thrust loss nikalo.

Worked example Solution 2.2

Total thrust: $F = 480 + (- 106.5) = 373.5 kN$ Fractional loss momentum thrust ke relative: $\frac{106.5}{480} = 0.2219 \approx 22.2%$ Answer: $F = 373.5 kN$ ; tum pressure mismatch ki wajah se akele jet ke thrust ka lagbhag $22.2%$ khote ho.

Problem 2.3 (L2)

Ek shock mein upstream Mach number $M_{1} = 3.0$ aur shock angle $β = 3 0^{\circ}$ hai. Normal Mach component $M_{1 n} = M_{1} sin β$ nikalo. Yeh number, $M_{1}$ khud nahin, shock strength ko kyun govern karta hai?

Worked example Solution 2.3

$M_{1 n} = 3.0 \times sin 3 0^{\circ} = 3.0 \times 0.5 = 1.5$ Normal component kyun rule karta hai: ek shock compression ki ek wall hai jise flow shock face ke perpendicular cross karta hai. Sirf woh velocity component jo us wall mein point karta hai, use slam aur slow kiya jaata hai. Jo component shock ke saath saath slide karta hai woh untouched nikal jaata hai. Toh shock ki "violence" $M_{1 n}$ se set hoti hai, aur yahan $M_{1 n} = 1.5$ (ek modest shock) hai, even though $M_{1} = 3$ fierce lagta hai.

Level 3 — Analysis

Problem 3.1 (L3)

Ek oblique shock ke across static pressure ratio hai $\frac{p _{2}}{p _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{1 n}^{2} - 1) .$ $γ = 1.4$ aur $M_{1 n} = 1.5$ ke liye, $p_{2} / p_{1}$ compute karo.

Worked example Solution 3.1

Hum kya karte hain: $γ = 1.4$ , $M_{1 n} = 1.5$ substitute karo. Pehle front coefficient: $\frac{2 γ}{γ + 1} = \frac{2 ( 1.4 )}{2.4} = \frac{2.8}{2.4} = 1.1667$ . Phir $M_{1 n}^{2} - 1 = 2.25 - 1 = 1.25$ . $\frac{p _{2}}{p _{1}} = 1 + 1.1667 \times 1.25 = 1 + 1.4583 = 2.4583$ Answer: $p_{2} / p_{1} \approx 2.46$ . Ek modest $M_{1 n} = 1.5$ shock pressure ko 2.46 factor se badhata hai, yeh confirm karta hai ki ek shallow oblique shock ek gentle compressor hai.

Problem 3.2 (L3)

Stagnation-pressure ratio (jo "energy quality budget" survive karta hai) ek shock ke across hai $\frac{p _{02}}{p _{01}} = (\frac{( γ + 1 ) M _{1 n}^{2}}{2 + ( γ - 1 ) M _{1 n}^{2}})^{\frac{γ}{γ - 1}} (\frac{γ + 1}{2 γ M _{1 n}^{2} - ( γ - 1 )})^{\frac{1}{γ - 1}} .$ $γ = 1.4$ , $M_{1 n} = 1.5$ ke saath, ise compute karo. Stagnation pressure ka kitna fraction lost hota hai?

Worked example Solution 3.2

Formula kahan se aata hai (taaki yeh black box na rahe): stagnation pressure do cheezoin se set hoti hai jo ek shock change karta hai — density-like squeeze aur temperature-like heating. Formula literally un dono effects ka product hai.

Bracket 1 density ratio hai jo $\frac{γ}{γ - 1}$ tak raise kiya gaya hai. Woh exponent isentropic link hai density aur pressure ke beech: change ke smooth (reversible) part ke liye, pressure density $^{γ / (γ - 1)}$ ke saath scale karta hai jab stagnation terms mein likha jaaye.
Bracket 2 irreversible piece carry karta hai — entropy jump jo sirf ek shock produce karta hai — exponent $\frac{1}{γ - 1}$ ke zariye. Yahi woh bracket hai jo product ko 1 se neeche giraaata hai; shock ki entropy rise ke bina yeh exactly 1 hota. Inhe multiply karne se milta hai "total-pressure budget ka kitna hissa survive karta hai." Dono exponents sirf do tarike hain jisme $γ$ compressible flow mein enter karta hai (reversible scaling vs. entropy accounting).

Ab numbers ( $γ = 1.4$ toh $\frac{γ}{γ - 1} = 3.5$ , $\frac{1}{γ - 1} = 2.5$ ):

$M_{1 n}^{2} = 2.25$ .
Bracket 1: $\frac{( 2.4 ) ( 2.25 )}{2 + ( 0.4 ) ( 2.25 )} = \frac{5.4}{2.9} = 1.8621$ . $3.5$ tak raise karo: $1.862 1^{3.5} = 8.798$ .
Bracket 2: $\frac{2.4}{2 ( 1.4 ) ( 2.25 ) - 0.4} = \frac{2.4}{6.3 - 0.4} = \frac{2.4}{5.9} = 0.40678$ . $2.5$ tak raise karo: $0.4067 8^{2.5} = 0.10565$ .
Product: $8.798 \times 0.10565 = 0.9297$ . Answer: $p_{02} / p_{01} \approx 0.930$ , toh stagnation-pressure budget ka lagbhag $7.0%$ is ek shock se destroy ho jaata hai. Interpretation: ek gentle shock bhi ( $M_{1 n} = 1.5$ ) tumhari usable energy quality ka 7% burn kar deta hai — aur diamonds mein aise kai stacked hote hain.

Problem 3.3 (L3)

Ek plume teen identical shocks se guzarta hai, har ek ka $p_{02} / p_{01} = 0.930$ . Overall stagnation-pressure survival nikalo, phir velocity efficiency $C_{v} = \frac{p _{0, final}}{p _{0, initial}} .$

Worked example Solution 3.3

Hum kya karte hain: losses multiply hote hain, kyunki har shock pichle wale ke survivor par act karta hai. $\frac{p _{0, final}}{p _{0, initial}} = 0.93 0^{3} = 0.8044$ Velocity efficiency (yaad karo: velocity surviving energy budget ke square-root ki tarah jaati hai): $C_{v} = 0.8044 = 0.8969 \approx 0.897$ Answer: stagnation pressure ka lagbhag $80.4%$ survive karta hai; $C_{v} \approx 0.90$ . Tumne exhaust velocity ka ~10% diamond stack se akele khoya hai.

Level 4 — Synthesis

Problem 4.1 (L4)

Ek vacuum-optimised engine sea level par test-fire kiya jaata hai. Data: $p_{e} = 6 kPa$ , $p_{a} = 101 kPa$ , $A_{e} = 0.6 m^{2}$ , $\overset{m}{˙} v_{e} = 520 kN$ . Iske alawa plume shocks se guzarta hai jisme overall $p_{0, final} / p_{0, initial} = 0.62$ hai. (a) Pressure thrust loss. (b) Sirf pressure mismatch ke baad thrust. (c) Velocity efficiency $C_{v}$ . (d) Ek sentence mein: kaun sa loss dominate karta hai aur kyun.

Worked example Solution 4.1

(a) Pressure thrust: $(p_{e} - p_{a}) A_{e} = (6 - 101) \times 1 0^{3} \times 0.6 = (- 95 \times 1 0^{3}) (0.6) = - 57 kN$ (b) Pressure mismatch ke baad thrust: $F = 520 + (- 57) = 463 kN$ (c) Velocity efficiency: $C_{v} = 0.62 = 0.7874 \approx 0.787$ (d) Kaun dominate karta hai: Pressure mismatch momentum thrust ka $57/520 = 11.0%$ kharta hai, lekin shock losses exhaust velocity ko ~ $21%$ cut karte hain ( $1 - 0.787$ ). Shock losses dominate karte hain, kyunki woh energy ki quality (stagnation pressure) destroy karte hain, jo achievable $v_{e}$ set karta hai, jabki pressure term exit disc par ek one-time subtraction hai.

Problem 4.2 (L4)

Same engine, lekin tum ek diffuser add karte ho (ek suppressor jo effective back-pressure ko smoothly raise karta hai) taaki plume ko ab shocks ki zaroorat na ho: $p_{0, final} / p_{0, initial} = 0.97$ aur exit par effective ambient aise low kiya gaya ki $(p_{e} - p_{a}) A_{e} = - 12 kN$ . $F$ aur $C_{v}$ recompute karo, aur 4.1 ke versus total thrust recovered batao.

Worked example Solution 4.2

Naya pressure thrust: $- 12 kN$ (given). Naya total thrust: $F = 520 + (- 12) = 508 kN$ . Nayi efficiency: $C_{v} = 0.97 = 0.9849 \approx 0.985$ . Thrust recovered: $508 - 463 = 45 kN$ zyada usable thrust sirf pressure term se, aur velocity efficiency $0.787$ se $0.985$ tak jump karti hai. Kyun kaam karta hai: diffuser plume ko gradually compress karne deta hai (near-isentropic) violent shocks ke zariye nahin, stagnation-pressure budget bacha leta hai.

Problem 4.3 (L4) — the matched edge case

Same engine ab us altitude par fly kiya jaata hai jahan uski nozzle exactly matched hai: $p_{e} = p_{a} = 40 kPa$ , $A_{e} = 0.6 m^{2}$ , $\overset{m}{˙} v_{e} = 520 kN$ , aur no over-expansion ki wajah se koi plume shocks nahi hain ( $p_{0, final} / p_{0, initial} = 1$ ). (a) Pressure thrust term. (b) Total thrust. (c) Velocity efficiency. (d) Ek line mein, kyun yeh design-optimal condition hai.

Worked example Solution 4.3

(a) Pressure thrust: $(p_{e} - p_{a}) A_{e} = (40 - 40) \times 1 0^{3} \times 0.6 = 0 N$ . Term vanish ho jaata hai. (b) Total thrust: $F = 520 + 0 = 520 kN$ — pure momentum thrust, kuch subtract nahi. (c) Velocity efficiency: $C_{v} = 1 = 1.00$ (koi shock losses nahi, isliye poora stagnation-pressure budget survive karta hai). (d) Kyun optimal hai: $p_{e} = p_{a}$ ke saath pressure term na penalty hai (over-expanded) na missed opportunity (under-expanded), aur koi shocks nahi hain stagnation pressure destroy karne ke liye — thermal energy ka har joule jo ordered jet motion ban sakta tha, bana. Yeh target hai jo har altitude-compensation scheme pursue karti hai; dekho Altitude Compensation.

Level 5 — Mastery

Problem 5.1 (L5)

Oblique formula se ek normal shock ke liye pressure ratio $p_{2} / p_{1}$ derive karo, aur ise sanity check ki tarah use karo. Oblique relation hai $\frac{p _{2}}{p _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{1}^{2} sin^{2} β - 1) .$ (a) $β$ ki kaun si value ek oblique shock ko normal shock mein turn karti hai, aur kyun? (b) $M_{1} = 4.2$ , $γ = 1.2$ ke liye, normal-shock pressure ratio compute karo. (c) Ek weak oblique shock $β = 2 5^{\circ}$ par compare karo ( $p_{2} / p_{1}$ compute karo) aur ek line mein explain karo ki nature oblique wala kyun prefer karta hai.

Worked example Solution 5.1

(a) Ek normal shock flow ke perpendicular khada hota hai, yaani uski face velocity se $9 0^{\circ}$ par hai, matlab $β = 9 0^{\circ}$ . Tab $sin β = 1$ aur $M_{1} sin β = M_{1}$ — poori velocity shock ke normal hai. Yeh maximum-strength, maximum-entropy case hai. (b) $β = 9 0^{\circ}$ ke saath, $M_{1 n} = M_{1} = 4.2$ , $γ = 1.2$ : $\frac{2 γ}{γ + 1} = \frac{2.4}{2.2} = 1.0909, M_{1}^{2} - 1 = 17.64 - 1 = 16.64$ $\frac{p _{2}}{p _{1}} = 1 + 1.0909 \times 16.64 = 1 + 18.153 = 19.153$ Toh $M = 4.2$ par ek normal shock pressure ko ek violent step mein ~19.2× jump karta hai. (c) $β = 2 5^{\circ}$ par oblique: $M_{1 n} = 4.2 sin 2 5^{\circ} = 4.2 \times 0.42262 = 1.775$ , toh $M_{1 n}^{2} = 3.151$ . $\frac{p _{2}}{p _{1}} = 1 + 1.0909 (3.151 - 1) = 1 + 1.0909 \times 2.151 = 1 + 2.346 = 3.346$ Nature oblique kyun prefer karta hai: wahi required pressure rise kai gentle oblique steps mein reach ki ja sakti hai (har ek $3.3 \times$ ) instead of ek $19 \times$ normal shock ke, aur kyunki stagnation-pressure loss $M_{1 n}$ ke saath sharply badhti hai, kai weak shocks ek strong shock se kaafi kam energy destroy karte hain. Nature sabse sasta compression leta hai.

Problem 5.2 (L5)

Design task. Ek first-stage engine sea level ( $p_{a} = 101 kPa$ ) se $30 km$ ( $p_{a} \approx 1.2 kPa$ ) tak fly karna hai. Uski fixed nozzle $10 km$ par matched ( $p_{e} = p_{a}$ ) hai jahan $p_{a} = 26 kPa$ hai, jisse $p_{e} = 26 kPa$ milta hai. Exit area $A_{e} = 1.0 m^{2}$ . (a) Sea level par, kya yeh over- ya under-expanded hai, aur pressure thrust term kya hai? (b) $30 km$ par, same questions. (c) Ek sentence mein, is baat se connect karo ki boosters staged kyun hote hain.

Worked example Solution 5.2

(a) Sea level: $p_{e} = 26 < p_{a} = 101$ , toh over-expanded. $(p_{e} - p_{a}) A_{e} = (26 - 101) \times 1 0^{3} \times 1.0 = - 75 \times 1 0^{3} N = - 75 kN$ Neeche 75 kN ka penalty (plus shock losses upar se). (b) 30 km: $p_{e} = 26 > p_{a} = 1.2$ , toh under-expanded. $(p_{e} - p_{a}) A_{e} = (26 - 1.2) \times 1 0^{3} \times 1.0 = + 24.8 \times 1 0^{3} N = + 24.8 kN$ Ab pressure term 24.8 kN add karta hai — lekin gas finish expand nahi hui, toh yeh ek khoya hua opportunity hai, destroyed nahi (expansion fans near-isentropic hote hain, koi shock losses nahi). (c) Staging link: kyunki koi bhi single fixed nozzle 100:1 ambient swing mein matched nahi ho sakti, hum ascent ko stages mein split karte hain, har ek apna nozzle carry karta hai apne altitude band ke liye sized — dekho Rocket Staging aur Altitude Compensation.

Recap Ladder

Question

Kaun si single inequality ek over-expanded nozzle define karti hai?

Answer

p_{e} < p_{a}

— exit pressure ambient se neeche.

Question

Kaun sa Mach component shock ki strength set karta hai, aur kyun?

Answer

Normal component

M_{1 n} = M_{1} sin β

, kyunki sirf woh velocity jo shock face ko perpendicularly cross karti hai compress hoti hai.

Question

C_{v}

stagnation-pressure ratio ka square-root kyun leta hai?

Answer

Kyunki kinetic energy

v^{2}

ke saath scale karti hai aur extractable energy

p_{0}

ke saath, isliye

v_{e}^{2} \propto p_{0}

aur

C_{v} = v_{e, actual} / v_{e, ideal} = p_{0, final} / p_{0, initial}

Question

Shocks ki chain ke across stagnation-pressure losses add hote hain ya multiply?

Answer

Woh multiply hote hain:

p_{0, final} / p_{0, initial} = \prod (p_{02} / p_{01})_{i}

Question

Perfectly matched case

p_{e} = p_{a}

mein pressure thrust term aur efficiency ka kya hota hai?

Answer

Pressure term exactly zero ho jaata hai aur (koi shocks nahi)

C_{v} = 1

— design-optimal condition.

Question

Over-expansion under-expansion se worse kyun hai?

Answer

Over-expansion irreversible shocks force karta hai (compression, entropy rise); under-expansion reversible expansion fans allow karta hai (koi quality lost nahi).

Prerequisite webs touched here: Shock Wave Fundamentals, Isentropic Flow, Gas Dynamics, Thrust Equation, Nozzle Flow Separation, Nozzle Area Ratio.