Worked examples — Nozzle area ratio ε = A_e - A - — choosing for optimal performance

3.3.18 · D3 · Physics › Rocket Propulsion › Nozzle area ratio ε = A_e - A - — choosing for optimal perf

Kuch bhi calculate karne se pehle, hum alphabet pe agree karte hain. Neeche har symbol earn kiya hua hai — agar tum isse pehle mile nahi ho, yeh box padho.

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Scenario matrix

Har rocket-nozzle question inhi cells mein se ek hai. Neeche ke examples har ek ko cover karte hain — last column batata hai kaun sa example kiska cover hai.

Cell	Case class	Kya special hai	Covered by
A	$p_e=p_a$ (matched)	optimum, pressure term $=0$	Ex 1
B	$p_e>p_a$ (under-expanded)	ε bahut chhota, positive pressure thrust baaki hai	Ex 2
C	$p_e<p_a$ (over-expanded)	ε bahut bada, negative pressure thrust, separation risk	Ex 3
D	$p_a\to 0$ (vacuum limit)	limiting behaviour, bahut bada ε	Ex 4
E	$\epsilon =1$ (degenerate)	throat = exit, sonic ($M_e=1$) only	Ex 5
F	ek ε ke do Mach roots	sign/branch choice (sub- vs supersonic)	Ex 6
G	real-world word problem	flight ke dauran ε choose karna	Ex 7
H	exam twist (thrust comparison)	do nozzles rank karo, trap pakdo	Ex 8
I	sub-critical: kabhi nahi chokes	$p_0/p_a<1.772$, flow subsonic rehti hai	Ex 9

Poore page mein, $\gamma =1.20$ lo taaki $\frac{\gamma }{\gamma -1}=6$ aur $\frac{\gamma +1}{2(\gamma -1)}=5.5$ ho, parent note se match karte hue. Sea-level $p_a=1.0$ bar; $1\text{bar}=10^5\text{Pa}$.

Figure 1 neeche teeno physical cases side by side dikhata hai — jab hum sky sweep karein tab bar bar isko dekhte rehna.

Figure 1 mein left panel (blue plume, seedhe edges) Cell A ka matched case hai; middle (yellow, ballooning plume) under-expanded Cell B hai; right (pink, pinched plume) over-expanded Cell C hai. Yellow dashed line throat $A^{\ast }$ hai; blue vertical line exit plane $A_e$ hai.

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Cell A — matched (optimum) expansion

Forecast: calculate karne se pehle andaza lagao — kya $\epsilon$ 3, 15, ya 150 ke kareeb hoga? Pressure thrust $+$, $0$, ya $-$? Figure 1 ka left panel dekho — uska plume seedha hai kyunki $p_e=p_a$.

1. Pressure ratio. $p_0/p_e=70/0.5=140$. Yeh step kyun? Pressure–Mach equation $p_0/p_e$ ke terms mein likhi hai; pehle woh number chahiye.
2. $M_e$ solve karo. Invert: $1+0.1M_e^2=140^{1/6}=2.276$, toh $M_e^2=12.76$, $M_e=3.573$. Yeh step kyun? Mach number bridge hai — area equation sirf Mach bolti hai, pressure nahi.
3. Area ratio. $M_e=3.573$ ke saath: $\epsilon =\frac{1}{3.573}{\left[\frac{2}{2.2}(2.276)\right]}^{5.5}=\frac{1}{3.573}(2.069{)}^{5.5}\approx \frac{54.8}{3.573}\approx \mathrm{15.3.}$ Yeh step kyun? Chosen pressure ko uss geometry mein convert karta hai jo hum machine kar sakte hain.
4. Pressure thrust. Design altitude par $p_e=p_a=0.5$ bar, toh $(p_e-p_a)A_e=0$. Yeh step kyun? Yahi Cell A ki definition hai — matched matlab doosra thrust term zero ho jaata hai.

Verify: $M_e=3.573$ ko pressure–Mach mein wapas plug karo: $(1+0.1(12.77){)}^6=2.277^6=140.0$ ✓. Magnitude sanity: upper-stage/vacuumish design jo $\epsilon \approx 15$ deta hai, exactly "tens" range mein hai jaise parent ne predict kiya tha. Units: ε dimensionless hai ✓.

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Cell B — under-expanded ($p_e>p_a$)

Forecast: chhota ε matlab kam expansion. Kya $p_e$ matched case se zyada ya kam hogi? $(p_e-p_a)$ ka sign? Figure 1 ka middle panel (ballooning plume) wahi hai jo hum test kar rahe hain.

1. $\epsilon =8$ se $M_e$ nikalo. Supersonic root ke liye area equation root-find karo (bracket $M\in [1,10]$, upar di gayi recipe se bisection): $M_e\approx 3.117$. Yeh step kyun? Is baar geometry di gayi hai, pressure nahi — toh hum area → Mach → pressure jaate hain, aur area → Mach ke liye root-finding chahiye kyunki ise algebra se solve nahi kar sakte.
2. $p_e$ nikalo. $p_0/p_e=(1+0.1(3.117{)}^2{)}^6=(1.9716{)}^6=58.9$, toh $p_e=70/58.9=1.188$ bar. Yeh step kyun? $p_e$ ko $p_a$ se compare karna hai — woh comparison cell ka naam batata hai.
3. Classify karo. $p_e=1.188>p_a=0.20$ → under-expanded (Cell B). Gas abhi bhi pressure ke saath baahar phootati hai. Yeh step kyun? $p_e-p_a$ ka sign hi classification hai.
4. Pressure thrust. $(p_e-p_a)A_e=(1.188-0.20)\times 10^5\times 0.16=+1.58\times 10^4\text{N}=+15.8\text{kN}$. Yeh step kyun? Positive term ⇒ chhota karna "safe" tha — koi separation nahi, aur yeh abhi bhi push add karta hai. Lekin agar hum ise is altitude ke matched ε ki taraf badhate toh aur bhi zyada $v_e$ milti.

Verify: $M_e$ cross-check karo: area formula $\frac{1}{3.117}[\frac{2}{2.2}(1.9716){]}^{5.5}=\frac{1}{3.117}(1.7924{)}^{5.5}=\frac{24.9}{3.117}=8.0$ ✓. Pressure term positive "under-expanded still pushes" se match karta hai ✓. Units: $\text{Pa}\cdot {\text{m}}^2=\text{N}$ ✓.

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Cell C — over-expanded ($p_e<p_a$)

Forecast: pressure term ab kaun sa sign lega? Cell A se bada ya chhota ε? (Same nozzle — trick: ε fixed geometry hai; sirf sky badli.) Figure 1 ka right panel (pinched plume) yahi cell hai.

1. Nozzle ke baare mein kuch nahi badla. $p_e$ abhi bhi $0.5$ bar hai kyunki ε fixed geometry hai. Yeh step kyun? Core insight: ek rigid nozzle apni $p_e$ rakhta hai; sirf $p_a$ $0.5$ se $1.0$ bar ho gayi.
2. Classify karo. $p_e=0.5<p_a=1.0$ → over-expanded (Cell C). Bahari hawa bell ke andar dhakka deti hai. Yeh step kyun? $p_e-p_a$ ab negative hai — danger cell.
3. Momentum thrust. $\dot{m}{v}_e=250\times 2650=6.625\times 10^5\text{N}=662.5\text{kN}$. Yeh step kyun? Dominant, hamesha-positive term — dekhna hai ki penalty kitna khaati hai.
4. Pressure thrust. $(0.5-1.0)\times 10^5\times 0.30=-1.5\times 10^4\text{N}=-15\text{kN}$. Yeh step kyun? Yahi penalty hai — yeh subtract karti hai, exactly parent ke number se match karta hai.
5. Total. $F=662.5-15=647.5\text{kN}$; pressure term $\approx 2.3\%$ cost karta hai. Yeh step kyun? Loss ko perspective mein rakhta hai, aur possible Flow Separation in Over-expanded Nozzles ki warning deta hai jab $p_e$ $p_a$ se bahut neeche gir jaati hai.

Verify: $\dot{m}{v}_e=250\cdot 2650=662500$ N ✓. Pressure term $=(0.5-1.0)\cdot 10^5\cdot 0.30=-15000$ N ✓. Fraction $15/662.5=0.0226\approx 2.3\%$ ✓.

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Cell D — vacuum limit ($p_a\to 0$)

Forecast: kya $\epsilon$ kisi fixed number ki taraf jaata hai, ya $p_e\to 0$ hone par without bound badhta rehta hai?

1. Pressure ratio. $p_0/p_e=70/0.02=3500$. Yeh step kyun? Chhoti $p_e$ ek bada ratio banati hai — yahi bade ε ka source hai.
2. $M_e$ solve karo. $1+0.1M_e^2=3500^{1/6}=3.911$, $M_e^2=29.11$, $M_e=5.396$. Yeh step kyun? Zyada pressure ratio ⇒ zyada exit Mach (faster, thanda exhaust).
3. Area ratio. $\epsilon =\frac{1}{5.396}[\frac{2}{2.2}(3.911){]}^{5.5}=\frac{1}{5.396}(3.556{)}^{5.5}\approx \frac{1069}{5.396}\approx 198$. Yeh step kyun? Geometry mein convert karta hai — ε do sau ke paas, exactly deep-vacuum regime.
4. Limit reasoning. Jaise $p_e\to 0$, $M_e\to \infty$ aur $\epsilon \to \infty$. Perfect vacuum expansion ke liye infinitely bada bell chahiye — impossible — toh real vacuum engines sirf ε "mass/packaging ke hisaab se jitna bada ho sake" banate hain. Is trap se bachne ke liye Altitude Compensation — Aerospike Nozzles dekho. Yeh step kyun? Limiting behaviour cell ko honestly cover karta hai: optimum unreachable hai, sirf approached hota hai.

Verify: $(1+0.1\cdot 29.11{)}^6=3.911^6=3500$ ✓. Area: $\frac{1}{5.396}(3.556{)}^{5.5}\approx 198$ ✓. Trend: bada $p_0/p_e\Rightarrow$ bada ε ✓.

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Cell E — degenerate throat ($\epsilon =1$)

Is example se pehle, master curve se milo. Figure 2 $\gamma =1.20$ ke liye area ratio $\epsilon$ ko exit Mach $M_e$ ke against plot karta hai. Abhi ise study karo — Cells E aur F dono isme rehte hain.

Figure 2 mein bilkul neeche yellow dot Cell E ka throat hai ($\epsilon =1$ at $M=1$); dashed $\epsilon =2$ line par pink aur blue dots Cell F ke do roots hain. Curve $M$ ke 1 ki taraf badhne par gir ta hai, apna single lowest point $\epsilon =1$ exactly $M=1$ par hit karta hai, phir $M>1$ ke liye wapas utha ta hai. Yeh single-minimum shape dono cells ki key hai.

Forecast: bilkul bhi flare nahi hone par, kya exhaust subsonic, sonic, ya supersonic hogi?

1. $\epsilon =1$ area equation mein daalo. Figure 2 mein $\epsilon (M)$ curve ka unique minimum $\epsilon =1$ hai, aur woh minimum exactly $M=1$ par aata hai. Toh $\epsilon =1\iff M_e=1$ — curve par aur koi jagah utni neeche nahi. Yeh step kyun? $\epsilon =1$ curve ka bilkul bottom hai; kyunki minimum unique hai, yeh sirf single Mach $M=1$ par hi reach ho sakta hai (throat choke point hai, Choked Flow and the Throat Condition).
2. Check karo. $M_e=1$ par: $\frac{1}{1}[\frac{2}{2.2}(1+0.1){]}^{5.5}=(1.0{)}^{5.5}=1$ ✓. Yeh step kyun? Confirm karta hai ki degenerate case self-consistent hai — koi diverging section nahi, koi supersonic exit nahi.
3. Exit par pressure. $p_0/p_e=(1+0.1{)}^6=1.1^6=1.772$, toh $p_e=70/1.772=39.5$ bar. Yeh step kyun? Dikhata hai ki exhaust abhi bhi bahut high pressure par nikalta hai — kisi bhi altitude par hugely under-expanded; almost sari expansion waste ho jaati hai. Isliye tumhe diverging bell chahiye.

Verify: $\frac{2}{2.2}(1.1)=1.0$ exactly, toh $\epsilon =1.0$ at $M_e=1$ ✓. $1.1^6=1.7716$, $70/1.7716=39.5$ bar ✓.

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Cell F — ek ε ke do Mach roots (branch choice)

Forecast: throat ke downstream kaun sa root rehta hai? Figure 2 mein horizontal $\epsilon =2$ line trace karo — yeh curve ko do baar stab karti hai.

1. Do roots kyun exist karte hain. Area–Mach curve apne minimum ($\epsilon =1$) tak $M=1$ par dip karti hai, phir dono taraf utha ti hai. Toh koi bhi $\epsilon >1$ do baar hit hota hai — ek baar subsonic jaate hue, ek baar supersonic. Yeh step kyun? Yahi is cell ka poora point hai: $\epsilon$ akela flow pin nahi karta; sign/branch choice karta hai. Figure 2 mein do marked dots dekho.
2. Subsonic root. Bracket $M\in [0.01,1]$ par root-find karo (bisection): $g(M)$ chhote $M$ par large-positive se zero se neeche jaata hai. Solution $M_e\approx 0.306$ hai (pink dot). Yeh step kyun? Yeh branch throat se pehle converging hisse mein hai — exhaust nahi. Hum teen decimals quote karte hain kyunki bisection ki kuch halvings wahi deliver karti hain.
3. Supersonic root. Bracket $M\in [1,10]$ par root-find karo: solution $M_e\approx 2.197$ hai (blue dot). Yeh step kyun? Throat ke baad flow $M=1$ par choke hoti hai aur diverging bell mein accelerate hoti rehti hai — toh yahi root hai.
4. Choose karo. Nozzle exit ke liye (throat ke downstream), $M_e=2.197$ lo. Yeh step kyun? Physical placement, algebra nahi, branch select karti hai — classic exam trap jiske baare mein parent ne warning di thi.

Verify: Supersonic: $\frac{1}{2.197}[\frac{2}{2.2}(1+0.1\cdot 4.827){]}^{5.5}=\frac{1}{2.197}(1.348{)}^{5.5}=\frac{1}{2.197}(4.41)=2.01$ ✓. Subsonic: $\frac{1}{0.306}[\frac{2}{2.2}(1+0.1\cdot 0.0936){]}^{5.5}=\frac{1}{0.306}(0.9176{)}^{5.5}=\frac{0.614}{0.306}=2.01$ ✓.

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Cell G — real-world word problem

Forecast: first stage ka bell bada hona chahiye ya chhota?

1. First-stage strategy. Yeh apna zyada kaam moti hawa mein karta hai; $p_e$ apni range ke mid ke paas choose karo taaki liftoff par heavy over-expansion se bacha ja sake. Design $p_e\approx 0.7$ bar lo. Yeh step kyun? Exactly $1.0$ bar (liftoff) ke liye design karne par upar jaake badly over-expand hoga; thoda compromise liftoff par $t=0$ separation se bachata hai. Dekho Flow Separation in Over-expanded Nozzles.
2. First-stage ε. $p_0/p_e=100$; $1+0.1M_e^2=100^{1/6}=2.154$, $M_e^2=11.54$, $M_e=3.397$; $\epsilon =\frac{1}{3.397}[\frac{2}{2.2}(2.154){]}^{5.5}=\frac{1}{3.397}(1.958{)}^{5.5}\approx \frac{40.7}{3.397}\approx 12.0$. Yeh step kyun? Chosen $p_e$ ko buildable geometry mein turn karta hai — ek modest ε ≈ 12, booster ke liye bilkul sahi.
3. Second-stage strategy. Yeh thin air mein kaam karta hai, toh low $p_e$ match karo. $p_e\approx 0.05$ bar lo. Yeh step kyun? Low ambient ⇒ low optimum $p_e$ ⇒ bada expansion, bada bell.
4. Second-stage ε. $p_0/p_e=1400$; $1+0.1M_e^2=1400^{1/6}=3.386$, $M_e^2=23.86$, $M_e=4.885$; $\epsilon =\frac{1}{4.885}[\frac{2}{2.2}(3.386){]}^{5.5}=\frac{1}{4.885}(3.078{)}^{5.5}\approx \frac{484}{4.885}\approx 99$. Yeh step kyun? Confirm karta hai ki second stage ko bahut bada bell chahiye (ε ≈ 99) — "size ε to the sky" mnemonic action mein; bada ε ⇒ vacuum mein zyada Specific Impulse Isp.

Verify: Stage 1: $2.154^6=100.0$ ✓, $\epsilon \approx 12.0$. Stage 2: $3.386^6=1400$ ✓, $\epsilon \approx 99$. Ordering ${\epsilon }_1<{\epsilon }_2$ (12 < 99) "higher altitude ⇒ bigger ε" se match karta hai ✓.

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Cell H — exam twist (do nozzles rank karo)

Forecast: N1 ka exhaust faster hai — kya yahi settle karta hai?

1. N1 thrust. Momentum $=200\times 2650=530,000$ N. Pressure $=(0.5-0.4)\times 10^5\times 0.30=+3000$ N. Total $=533.0\text{kN}$. ($p_e>p_a$, yahan thoda under-expanded — achha.) Yeh step kyun? $p_a=0.4$ par, bada nozzle near matched hai, toh uska pressure term chhota aur positive hai.
2. N2 thrust. Momentum $=200\times 2500=500,000$ N. Pressure $=(1.188-0.4)\times 10^5\times 0.16=+12,608$ N. Total $=512.6\text{kN}$. Yeh step kyun? Chhota nozzle zyada under-expanded hai (bada positive pressure term) lekin uski slower $v_e$ zyada cost karti hai.
3. Compare karo. $533.0>512.6\text{kN}$ ⇒ N1 yahan jeeta hai — lekin kyunki yeh near-matched hai, sirf "bada" hone ki wajah se nahi. Yeh step kyun? Trap ko defuse karta hai: agar hum sea level par fire karte ($p_a=1.0$), N1 ka term $(0.5-1.0)\cdot 10^5\cdot 0.30=-15$ kN hota aur N2 ka $(1.188-1.0)\cdot 10^5\cdot 0.16=+3008$ N — gap chhota hota aur ranking flip ho sakta hai. Optimality matching ke baare mein hai, size ke nahi.

Verify: N1 $=530000+3000=533000$ N ✓. N2 $=500000+12608=512608$ N ✓. N1 > N2 ✓. Sea-level check: N1 $=530000-15000=515000$; N2 $=500000+3008=503008$ — N1 wahan bhi jeetta hai lekin kam se; general warning stand karta hai (zyada over-expanded N1 haar sakta tha).

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Cell I — sub-critical: flow kabhi choke nahi hoti

Forecast: sirf $1.5\times$ pressure ratio ke saath, kya tumhare khyaal se gas kahin bhi Mach 1 reach kar sakti hai?

1. Choke threshold nikalo. Ex 5 se, throat par $M=1$ reach karne ke liye kam se kam $(1+\frac{\gamma -1}{2}{)}^{\gamma /(\gamma -1)}=1.1^6=1.772$ ka pressure ratio chahiye. Yeh step kyun? Yeh single number gatekeeper hai: iske neeche throat sonic nahi ja sakta.
2. Compare karo. $p_0/p_a=1.5<1.772$ → flow sub-critical hai. Throat Mach 1 se neeche rehti hai; nozzle mein kahin bhi gas sound ki speed tak nahi pohunchti. Yeh step kyun? $1.5<1.772$ poora case decide karta hai — koi choking nahi.
3. Nozzle ki jagah kya karta hai. Koi sonic throat nahi hone se, diverging "bell" flow ko ab accelerate nahi karta — subsonic gas ke liye ek widening duct use slow down karta hai (jaise ek river phail kar slow hoti hai). Toh diverging section ek diffuser ki tarah act karta hai, exit subsonic rehta hai, aur $p_e=p_a=1.0$ bar automatically hota hai (ek subsonic jet hamesha ambient pressure match karta hai). Yeh step kyun? Yahi crucial reversal hai: same geometry jo choked supersonic flow ko accelerate karta hai woh un-choked subsonic flow ko decelerate karta hai. Figure 2 mein area–Mach curve ki subsonic branch (yellow dot ke left) yahan govern karti hai.
4. Thrust consequence. Kyunki $M_e<1$ aur $p_e=p_a$, exit velocity modest hai aur pressure term zero hai — yeh rig feeble thrust produce karta hai. Real rockets hamesha $p_0/p_a$ ko $1.772$ se bahut upar run karte hain exactly is dud regime se bachne ke liye. Yeh step kyun? Case close karta hai: sub-critical operation ek failure mode hai jisse design mein bachna chahiye, koi useful cell nahi.

Verify: Threshold $1.1^6=1.7716$ ✓. Kyunki $1.5<1.7716$, sub-critical (no choke) ✓. Ek choked run ke liye $p_0/p_a\ge 1.772$ chahiye hoga, e.g. upar ke $p_0=70$ bar engines sab ise wide margin se clear karte hain ✓.

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Recall Har cell ka one-line recap

Matched ::: pressure term $=0$, optimum (Ex 1). Under-expanded ::: $p_e>p_a$, term $+$, ε bahut chhota (Ex 2). Over-expanded ::: $p_e<p_a$, term $-$, ε bahut bada, separation risk (Ex 3). Vacuum limit ::: $p_a\to 0\Rightarrow \epsilon \to \infty$, unreachable (Ex 4). Degenerate $\epsilon =1$ ::: sirf $M_e=1$, sonic, hugely under-expanded (Ex 5). Two roots ::: throat ke downstream supersonic choose karo (Ex 6). Multi-stage ::: chhota ε neeche, bada ε upar (Ex 7). Ranking trap ::: bada ε $\ne$ hamesha zyada thrust; matching jeetta hai (Ex 8). Sub-critical ::: $p_0/p_a<1.772$, kabhi choke nahi hota, diverging section diffuser ki tarah act karta hai (Ex 9).

Connections

• Parent topic
• Thrust Equation and Effective Exhaust Velocity
• De Laval Converging-Diverging Nozzle
• Isentropic Flow Relations
• Choked Flow and the Throat Condition
• Flow Separation in Over-expanded Nozzles
• Altitude Compensation — Aerospike Nozzles
• Specific Impulse Isp