Exercises — Thrust equation F = ṁv_e + (P_e − P_a)A_e — derivation

Momentum thrust hai $\dot{m}{v}_e$ — har second kitna mass throw kiya times kitni fast throw ki. Pressure thrust hai $(P_e-P_a)A_e$ — exit pressure aur bahar ke pressure ka mismatch, exit area par act karta hua. Perfectly expanded ka matlab hai $P_e=P_a$, isliye $(P_e-P_a)=0$ aur pressure term khatam ho jaati hai. Sirf $\dot{m}{v}_e$ bachta hai.

Perfect expansion → pressure term $0$ hai, isliye $F=\dot{m}{v}_e$. $F=300\times 3000=900,000\text{N}=900\text{kN}.$

(a) $P_e>P_a$ → underexpanded → pressure thrust positive. (b) $P_e<P_a$ → overexpanded → pressure thrust negative. (c) $P_e=P_a$ → perfectly expanded → pressure thrust zero.

Momentum thrust: $\dot{m}{v}_e=250\times 2800=700,000\text{N}$. Pressure thrust: $(P_e-P_a)A_e=(95,000-101,000)(1.2)=(-6,000)(1.2)=-7,200\text{N}$ (overexpanded, isliye negative). Total: $F=700,000-7,200=692,800\text{N}=692.8\text{kN}$.

Momentum thrust unchanged: $700,000\text{N}$. Pressure thrust: $(95,000-0)(1.2)=114,000\text{N}$ (ab fully positive — kuch push back nahi karta). Total: $F=700,000+114,000=814,000\text{N}=814\text{kN}$. Sea level par gain: $814,000-692,800=121,200\text{N}=121.2\text{kN}$. Yeh gain exactly $P_a{A}_e=101,000\times 1.2=121,200\text{N}$ ke barabar hai — exactly missing atmospheric back-push.

$F=\dot{m}{v}_e+P_e{A}_e$ se shuru karo (kyunki $P_a=0$). Momentum thrust: $180\times 3100=558,000\text{N}$. Isliye pressure term ko $600,000-558,000=42,000\text{N}$ supply karna hoga. $A_e=\frac{42,000}{P_e}=\frac{42,000}{40,000}=1.05{\text{m}}^2.$

Pressure term ko pehle velocity mein collect kyun karte hain? Thrust ke do visually alag pieces hain — ek hai $(\text{mass/s})\times (\text{speed})$, doosra hai $(\text{pressure})\times (\text{area})$. Yeh baad ke formulas (jaise Tsiolkovsky) mein awkward hota hai jo sirf ek "exhaust velocity" jaante hain. Isliye hum poochhte hain: agar poora thrust sirf mass throw karne se aata, toh kaun si ek speed same force deti? $F=\dot{m}{v}_{eff}$ set karo aur us pretend speed ke liye solve karo. Pressure force $(P_e-P_a)A_e$ ko $\dot{m}$ se divide karne par "extra newtons" ko "extra m/s" mein convert kiya jaata hai, taaki yeh $v_e$ ke saath baith sake. Yahi woh formula hai jo diya gaya hai.

Pressure term tha $-7,200\text{N}$. $v_{eff}=2800+\frac{-7,200}{250}=2800-28.8=2771.2\text{m/s}.$ Check: $\dot{m}{v}_{eff}=250\times 2771.2=692,800\text{N}$ ✓. Overexpansion ne effectively exhaust ko $28.8\text{m/s}$ push ke hisaab se slow kar diya.

Sirf term $-P_a{A}_e$ $P_a$ par depend karti hai. Kyunki isme minus sign hai, $P_a$ kam hone par $F$ badhta hai. Isliye thrust monotonically vacuum tak badhti rehti hai, jahan yeh maximum hai. Sea level se vacuum tak total change: $\Delta F=-A_e(0-101,000)=1.2\times 101,000=121,200\text{N}=121.2\text{kN}.$ Isliye ek booster ki thrust upar jaate waqt badhti hai, engine bilkul unchanged chal raha ho tab bhi.

Pressure thrust tab khatam hoti hai jab $P_e-P_a=0$, yani $P_a=P_e=60\text{kPa}$ (perfect expansion). Wahan: $F=\dot{m}{v}_e=220\times 2600=572,000\text{N}=572\text{kN}.$ Sea level par $P_a=101\text{kPa}$ (ab overexpanded, $P_e<P_a$): $F=572,000+(60,000-101,000)(1.5)=572,000-61,500=510,500\text{N}=510.5\text{kN}.$ Sea level par overexpansion $61.5$ kN ki cost lagaati hai.

$I_{sp}=\frac{814,000}{250\times 9.81}=\frac{814,000}{2452.5}=331.9\text{s}.$ Kyunki vacuum mein thrust zyada hai, isliye $I_{sp}$ bhi zyada hai — yahi reason hai ki engines ko alag (bada) vacuum $I_{sp}$ ke saath quote kiya jaata hai.

Vacuum thrust (L2·Q2 se) $F=814,000\text{N}$, yahan constant hai. Impulse $J=Ft=814,000\times 2=1,628,000\text{N⋅s}$. Exhaust mass ejected $=\dot{m}t=250\times 2=500\text{kg}$. Newton's third law aur momentum conservation se, yeh total impulse $J$ us forward momentum ke barabar hai jo rocket gain karta hai. Note karo ki $J$ pehle se dono thrust pieces ko bundle karta hai: momentum-thrust part exhaust ka backward momentum hai, aur pressure-thrust part exit-pressure push hai — single number $J=Ft$ dono count karta hai, kyunki $F$ khud hi full thrust hai.

Pehle $v_{eff}=F/\dot{m}=814,000/250=3256\text{m/s}$. Mass ratio $m_0/m_f=20,000/8,000=2.5$, isliye $\ln 2.5=0.91629\dots$ $\Delta v=3256\times 0.91629=2983.4\text{m/s}.$ Pressure thrust ne quietly $v_{eff}$ ko raw $v_e=2800$ se upar uthaya, extra $\Delta v$ khareed ke.

Momentum thrust (fixed): $\dot{m}{v}_e=400\times 3200=1,280,000\text{N}$. (i) Design altitude, $P_e=P_a$ → pressure term $0$: $F=1,280,000\text{N}=1280\text{kN}.$ (ii) Sea level, $P_e=26\text{kPa}<P_a=101\text{kPa}$ (overexpanded): $F=1,280,000+(26,000-101,000)(2.0)=1,280,000-150,000=1,130,000\text{N}=1130\text{kN}.$ (iii) Vacuum, $P_a=0$ (underexpanded): $F=1,280,000+(26,000)(2.0)=1,280,000+52,000=1,332,000\text{N}=1332\text{kN}.$ Wahi engine, teen altitudes: $1130$ → $1280$ → $1332$ kN jaise hawa patli hoti hai. Notice karo ki thrust "perfect" point ke baad bhi badhti rehti hai — perfection ek design target tha us altitude ke liye, raw force ka peak nahi (L3 ki echo).

Momentum thrust: $50\times 200=10,000\text{N}$. Pressure thrust: $(500,000-0)(0.8)=400,000\text{N}$. Total: $F=10,000+400,000=410,000\text{N}=410\text{kN}$. Pressure fraction: $400,000/410,000=0.9756=97.56\%$. Yeh degenerate regime hai jahan pressure term dominate karta hai — physically ek thanda, barely-moving, highly pressurised jet. Real rocket nozzles isse avoid karti hain gas ko expand karke taaki momentum thrust dominate kare.

$F=\dot{m}{v}_e+(P_e-P_a)A_e$ ko $v_e$ ke liye rearrange karo: $v_e=\frac{F-(P_e-P_a)A_e}{\dot{m}}.$ Pressure term: $(80,000-101,000)(1.0)=-21,000\text{N}$. $v_e=\frac{445,000-(-21,000)}{200}=\frac{445,000+21,000}{200}=\frac{466,000}{200}=2330\text{m/s}.$ Double-negative note karo: ek negative pressure term subtract karne par $21$ kN wapas add hota hai jo atmosphere ne chura liya tha, isliye true $v_e$ ek naive $445,000/200=2225$ se bada hai.

Ek fixed engine ke liye, thrust badhti hai jaise ambient pressure girta hai, vacuum mein sabse zyada hoti hai, aur exactly pure $\dot{m}{v}_e$ ke barabar hoti hai us altitude par jahan $P_e=P_a$ ::: Sach — is ek sentence mein L1–L5 sab samaya hua hai.