Worked examples — Optimum expansion — P_e = P_a for maximum thrust

3.3.13 · D3 · Physics › Rocket Propulsion › Optimum expansion — P_e = P_a for maximum thrust

Yeh page optimum-expansion topic ka drill ground hai. Parent note ne yeh idea build kiya tha ki thrust tab peak karti hai jab $P_e=P_a$. Yahan hum har uss case ko dhundhte hain jo yeh idea produce kar sakta hai — under, over, aur perfect expansion; vacuum aur sea level; zero aur degenerate inputs; limiting cases; ek word problem; aur ek exam twist.

Shuru karne se pehle, symbols ke baare mein ek promise. Neeche har cheez sirf wahi reuse karta hai jo parent ne define kiya tha, aur hum har symbol ko pehli baar aate hi restate karte hain taaki tumhe kabhi upar scroll na karna pade.

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Scenario matrix

Is topic mein jo bhi problem aaye, woh in cells mein se kisi ek mein fit hogi. Har row ek case class hai; aakhri column us worked example ka naam deta hai jo isse best cover karta hai.

#	Case class	Sign / condition	Kya test ho raha hai	Example
A	Under-expanded	$P_e>P_a$	pressure term thrust add karta hai	Ex 1
B	Perfectly expanded	$P_e=P_a$	pressure term vanish ho jaata hai, thrust pure momentum hai	Ex 2
C	Over-expanded	$P_e<P_a$	pressure term thrust subtract karta hai	Ex 3
D	Vacuum limit	$P_a\to 0$	maximum possible pressure thrust	Ex 4
E	Degenerate: zero pressure term	$A_e\to 0$ ya $P_e=P_a$	pressure thrust do alag tareekon se khatam ho sakta hai	Ex 5
F	Design problem ($\epsilon$ dhundho)	geometry choose karo taaki $P_e=P_a$	inverse problem via Isentropic flow relations	Ex 6
G	Word problem: altitude sweep	$P_a$ rocket ke chadhne ke saath girta hai	ek fixed nozzle poori flight ke across	Ex 7
H	Exam twist: kaun zyada efficient hai?	do nozzles ko Specific impulse se compare karo	trap — bada thrust ≠ behtar nozzle	Ex 8

Shuru karne se pehle neeche wali figure dekho — yeh woh map hai jis par har example rehta hai.

Red curve thrust versus exit pressure hai. Teen colored dots cells A, B, C hain. Jo kuch bhi aage aata hai woh sirf inhi dots ko numbers ke saath visit karna hai.

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Cell A — Under-expanded ($P_e>P_a$)

Forecast: $P_e>P_a$, isliye pressure term positive hai. Aage padhne se pehle guess karo ki yeh bonus $\dot{m}{v}_e$ ke muqable bada hai ya chhota.

1. Bars ko pascals mein convert karo. Yeh step kyun? Thrust equation $\dot{m}{v}_e$ (already newtons mein) ko pressure-times-area term ke saath mix karta hai, isliye pressures pascals (Pa) mein hone chahiye taaki newtons mile. $P_e=1.4\times 10^5\text{Pa},P_a=0.9\times 10^5\text{Pa}$
2. Momentum thrust. Kyun? Yeh $F$ ka hamesha maujood backbone hai. $\dot{m}{v}_e=250\times 2700=675,000\text{N}$
3. Pressure thrust. Kyun? $P_e>P_a$ matlab gas sky ko outpush kar rahi hai, exit rim par force add ho rahi hai. $(P_e-P_a)A_e=(1.4-0.9)\times 10^5\times 1.5=0.5\times 10^5\times 1.5=75,000\text{N}$
4. Add karo. $F=675,000+75,000=750,000\text{N}=750\text{kN}$

Verify: Units — (kg/s)(m/s) = kg·m/s² = N ✓, aur (Pa)(m²) = (N/m²)(m²) = N ✓. Bonus (75 kN) momentum thrust ka roughly $11\%$ hai — real hai par modest, exactly wahi jo "slightly under-expanded" dikhna chahiye.

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Cell B — Perfectly expanded ($P_e=P_a$)

Forecast: $P_e=P_a$ ke saath, poora doosra chunk vanish ho jaana chahiye. Predict karo $F=\dot{m}{v}_e$ exactly.

1. Condition check karo. Kyun? Yahi is topic ka poora point hai — tab pehchano jab pressure term zero ho. $P_e-P_a=(1.4-1.4)\times 10^5=0$
2. Pressure thrust. Kyun? Kuch bhi times zero zero hota hai, $A_e$ chahe kuch bhi ho. $(P_e-P_a)A_e=0\times 1.5=0\text{N}$
3. Thrust pure momentum hai. $F=\dot{m}{v}_e+0=675,000\text{N}=675\text{kN}$

Verify: Yeh Ex 1 ke 750 kN se neeche hai — surprising? Nahi: Ex 1 under-expanded tha aur use genuine bonus mila. Lekin yeh iss altitude ke liye design kiye gaye nozzle ka peak hai, kyunki yahan perfectly expanded nozzle ka $A_e$ bada hoga aur $P_e$ kam, jo expansion ko internally capture karta hai. Figure mein dot B curve ka maximum hai, pressure term ka maximum nahi. Units pehle jaisi check hain ✓.

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Cell C — Over-expanded ($P_e<P_a$)

Forecast: $P_e<P_a$, isliye pressure term negative hai — yeh momentum thrust ko khaayega. Bite ka size guess karo.

1. Convert karo. $P_e=0.3\times 10^5$ Pa, $P_a=1.0\times 10^5$ Pa. Kyun? Same units rule.
2. Momentum thrust. $\dot{m}{v}_e=300\times 3000=900,000\text{N}$
3. Pressure thrust (negative). Kyun? Atmosphere exhaust ko outpush kar raha hai, isliye net exit-plane force backward point karti hai. $(P_e-P_a)A_e=(0.3-1.0)\times 10^5\times 3.0=(-0.7\times 10^5)(3.0)=-210,000\text{N}$
4. Add karo. $F=900,000-210,000=690,000\text{N}=690\text{kN}$

Verify: Pressure penalty momentum thrust ka $210/900\approx 23\%$ hai — ek over-expanded nozzle ki classic sea-level penalty. Savdhaan: agar yeh nozzle bahut zyada over-expand kare toh flow physically wall se tear off ho jaata hai (Supersonic flow separation) aur yeh simple formula loss ko overstate karta hai. Units ✓.

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Cell D — Vacuum limit ($P_a\to 0$)

Forecast: Bahar kuch bhi push nahi kar raha, toh pressure term sabse zyada positive hai. Predict karo $F$ 900 kN se upar jaayega.

1. $P_a=0$ ke saath pressure term. Yeh step kyun? $P_a=0$ set karna limiting case hai — bahar kuch nahi, toh exit gas ka push unopposed hai. $(P_e-0)A_e=0.3\times 10^5\times 3.0=90,000\text{N}$
2. Thrust. $F_{\text{vac}}=900,000+90,000=990,000\text{N}=990\text{kN}$
3. Ex 3 (sea level, 690 kN) se compare karo. Kyun? Isliye wahi nozzle space mein bahut zyada worth ka hota hai. $\Delta F=990-690=300\text{kN}$

Verify: Poora swing $300$ kN equals $P_a^{\text{SL}}{A}_e=1.0\times 10^5\times 3.0=300,000$ N ✓ — atmosphere ka constant $P_a{A}_e$ back-push exactly wahi hai jo vacuum mein disappear ho jaata hai. Yahi Altitude compensation ki poori logic hai. Units ✓.

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Cell E — Degenerate: pressure thrust do tareekon se khatam ho sakta hai

Forecast: Formula $(P_e-P_a)A_e$ mein, product zero hota hai agar koi bhi ek factor zero ho. Predict karo X bracket se kill karta hai, Y area se kill karta hai.

1. Nozzle X — bracket zero hai. Kyun? Perfect expansion: $(P_e-P_a)=0$. $(0.8-0.8)\times 10^5\times 2.0=0\text{N}\Rightarrow F_X=500,000\text{N}$
2. Nozzle Y — area zero hai. Kyun? Vanishing exit area ko mismatch push karne ke liye kuch nahi milta. Yeh ek degenerate (unphysical, lekin instructive) limit hai. $(0.8-0.5)\times 10^5\times 0=0\text{N}\Rightarrow F_Y=500,000\text{N}$

Verify: Dono $F=500$ kN dete hain, lekin meaning alag hai. X real optimum hai (thrust curve ka peak). Y ek mathematical corner hai: $A_e$ ko zero tak shrink karna us flow ko bhi strangle kar deta hai jo pehli jagah $\dot{m}{v}_e$ produce karta hai — toh Y ek warning hai ki "$A_e=0$ pressure term vanish karta hai" ek formula quirk hai, design goal nahi. Units ✓.

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Cell F — Design problem: expansion ratio nikalo

Forecast: Kam $P_e$ ke liye bada nozzle munh chahiye, isliye $\epsilon$ 1 se kaafi upar hona chahiye. "Kai tens" guess karo.

1. Optimum set karo. Kyun? Hum chahte hain $P_e=P_a$, isliye hum isse impose karte hain. $\frac{P_e}{P_0}=\frac{0.5}{50}=0.01$
2. Isentropic flow relations se exit Mach number $M_e$ nikalo. Yeh tool kyun? Pressure aur Mach number ek isentropic (constant-entropy) expansion ke along locked hote hain; yahi ek relation hai jo us pressure ko connect karta hai jo hum chahte hain us geometry se jo hum bana sakte hain. $\frac{P_e}{P_0}={\left(1+\frac{\gamma -1}{2}{M}_e^2\right)}^{-\frac{\gamma }{\gamma -1}}$ $\gamma =1.20$ ke saath: exponent $-\gamma /(\gamma -1)=-6$, aur $(\gamma -1)/2=0.1$. $0.01=(1+0.1M_e^2{)}^{-6}$ Dono sides ko $-1/6$ tak raise karo: $0.01^{-1/6}=1+0.1M_e^2\Rightarrow 2.1544=1+0.1M_e^2\Rightarrow M_e^2=11.544\Rightarrow M_e=3.398$
3. $M_e$ ko area-ratio relation mein daalo. Kyun? Isentropic area–Mach relation ek Mach number ko geometric ratio $\epsilon$ mein convert karta hai jo hum machine kar sakte hain. $\epsilon =\frac{A_e}{A^{\ast }}=\frac{1}{M_e}{\left[\frac{2}{\gamma +1}\left(1+\frac{\gamma -1}{2}{M}_e^2\right)\right]}^{\frac{\gamma +1}{2(\gamma -1)}}$ Yahan $\frac{2}{\gamma +1}=0.9091$, $1+0.1M_e^2=2.1544$, aur exponent $\frac{\gamma +1}{2(\gamma -1)}=\frac{2.2}{0.4}=5.5$. $\epsilon =\frac{1}{3.398}[0.9091\times 2.1544{]}^{5.5}=\frac{1}{3.398}(1.9585{)}^{5.5}$ $(1.9585{)}^{5.5}\approx 40.35\Rightarrow \epsilon \approx \frac{40.35}{3.398}\approx 11.9$

Verify: $\epsilon \approx 11.9$ — throat area ka roughly baara guna nozzle munh, hamare forecast ke "kai tens" ballpark mein, aur ek mid-altitude engine ka typical. $M_e=3.398$ ko step 2 mein wapas feed karne par $P_e/P_0=0.01$ reproduce hota hai (Verify mein check kiya). Har step sirf Isentropic flow relations use karta hai — koi free-floating assumptions nahi.

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Cell G — Word problem: ek nozzle poori climb ke across

Forecast: Jaise $P_a$, $P_e=0.7$ bar se neeche girta hai, engine over-expanded → perfect → under-expanded walk karta hai, aur thrust poori tarah upar chahiye.

1. Momentum thrust (constant). Kyun? $\dot{m}$ aur $v_e$ baahri hawa par depend nahi karte. $\dot{m}{v}_e=400\times 2900=1,160,000\text{N}$
2. Sea level, $P_a=1.0$ bar — over-expanded (Cell C). $(0.7-1.0)\times 10^5\times 2.5=-75,000\text{N}\Rightarrow F=1,085,000\text{N}=1085\text{kN}$
3. Mid-altitude, $P_a=0.7$ bar — perfect (Cell B). $(0.7-0.7)\times 10^5\times 2.5=0\Rightarrow F=1,160,000\text{N}=1160\text{kN}$
4. Near-vacuum, $P_a=0.05$ bar — under-expanded (Cell A). $(0.7-0.05)\times 10^5\times 2.5=162,500\text{N}\Rightarrow F=1,322,500\text{N}=1322.5\text{kN}$

Verify: Thrust monotonically badhti hai $1085\to 1160\to 1322.5$ kN jaise rocket chadhta hai — exactly woh kahani jo Altitude compensation ke peeche hai aur isliye designers $P_e$ kisi single altitude ke liye nahi balki ek average flight condition ke liye choose karte hain. Har pressure term ka sign uske regime label se match karta hai ✓. Yahi wajah bhi hai ki Rocket staging ek bade-$\epsilon$ nozzle mein switch karta hai jab atmosphere patla ho jaata hai.

Upar wala plot exactly yahi sweep dikhata hai: thrust $P_a$ girne ke saath chadhti hai, perfect-expansion point mark kiya gaya hai jahan curve $P_a=P_e$ cross karti hai.

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Cell H — Exam twist: bada thrust automatically better nahi hota

Forecast: $I_{sp}$ unit weight-flow per thrust measure karta hai — true efficiency. Q ka thrust zyada hai; kya uska $I_{sp}$ bhi zyada hai? (Trap: haan yahan, lekin sirf isliye kyunki $\dot{m}$ equal hai. Dekho yeh metric actually kya reward karta hai.)

1. P ke liye $I_{sp}$ compute karo. Yeh tool kyun? Specific impulse strip out karta hai ki aap kitna propellant phenk rahe ho aur puchta hai ki har kilogram kitna achha hai — fair comparison metric. $I_{sp,P}=\frac{560,000}{200\times 9.81}=\frac{560,000}{1962}\approx 285.4\text{s}$
2. Q ke liye $I_{sp}$ compute karo. $I_{sp,Q}=\frac{590,000}{1962}\approx 300.7\text{s}$
3. Rank karo. Q yahan jeet jaata hai kyunki dono ka $\dot{m}$ same hai, isliye zyada thrust genuinely zyada $I_{sp}$ matlab hai.

Verify: $I_{sp,Q}>I_{sp,P}$ roughly $15$ s se. Twist: agar Q ne apna 590 kN simply zyada propellant jala kar (higher $\dot{m}$) achieve kiya hota, toh uska $I_{sp}$ P se kam ho sakta tha bade thrust ke bawajood — wahi exam trap hai. Jab $\dot{m}$ fixed ho, thrust aur $I_{sp}$ saath chalte hain; sirf tabhi "zyada thrust = better" safe hai. Units: N / (kg/s · m/s²) = N / (N/s) = s ✓.

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Recall

Recall Kaun sa regime

positive pressure thrust deta hai? Under-expanded ($P_e>P_a$) ::: exhaust atmosphere ko outpush karta hai, isliye $(P_e-P_a)A_e>0$.

Recall Perfect expansion peak thrust kyun deta hai aur sabse bada pressure term kyun nahi?

Kyunki perfectly expanded nozzle saari expansion internally capture karta hai (bada $A_e$, high $v_e$); pressure term ka zero hona optimum ka ek symptom hai, uska cause nahi ::: momentum thrust $\dot{m}{v}_e$ us geometry ke liye maximise hota hai.

Recall Do alag tarike jinse pressure thrust

$(P_e-P_a)A_e$ zero ho sakta hai? Ya toh $P_e=P_a$ (real optimum) ya $A_e=0$ (degenerate, non-physical corner) ::: sirf pehla ek design goal hai.

Recall Fixed nozzle atmosphere mein chadhte hue — regimes ka kya sequence hoga?

Over-expanded → perfectly expanded → under-expanded jaise $P_a$, $P_e$ se neeche girta hai ::: thrust poori tarah chadhti rehti hai.