3.3.19 · Physics › Rocket Propulsion
Ek rocket engine basically ek chemical furnace with a nozzle hai. Fuel + oxidizer react karte hain, aur chemical bond energy ko heat ke roop mein release karte hain. Woh heat kahin ja nahi sakti (combustion bahut fast hoti hai → bahar leak hone ka time nahi → adiabatic ), toh woh poori ki poori product gases mein jaati hai aur unhe ek jalti hui flame temperature T a d tak le jaati hai.
YEH KYUN MATTER KARTA HAI: exhaust speed v e ∝ T c / M . Zyada T a d aur kam product molar mass M = tez exhaust = zyada specific impulse. Aage sab kuch (thrust, I s p ) yahin se shuru hota hai.
Stoichiometry batata hai ki kitna oxidizer chahiye ek unit fuel ke liye; adiabatic flame temperature batata hai ki result kitna garam hoga.
Stoichiometry ek chemical reaction ko balance karna hai taaki left side ka har atom right side par bhi dikhe (mass aur atoms conserve hote hain). Stoichiometric mixture woh exact fuel-to-oxidizer ratio hai jisme dono poori tarah consume ho jaate hain, koi leftover fuel ya oxidizer nahi bachta.
HUMEIN KYA CHAHIYE: coefficients jo atom conservation satisfy karein.
KAISE MILENGE: skeletal reaction likho, har element conserve karo, solve karo.
Worked balance — hydrogen/oxygen (ek real rocket propellant, jaise RS-25):
2 H 2 + O 2 → 2 H 2 O
Yeh step kyun? Left mein 4 H, 2 O hain. H balance karne ke liye 2 water molecules chahiye (→ 4 H, 2 O). Ab O bhi balance ho gaya. Ho gaya.
Intuition Bonds springs hain jo energy store karte hain
Bonds todna energy kharch karta hai; bonds banana energy release karta hai. Agar products ke bonds reactants se zyada strong hain (lower energy), toh surplus heat ke roop mein release hoti hai. Wahi surplus enthalpy of reaction hai.
Definition Standard enthalpy of formation
Δ H f ∘
1 mole of a species banane ki enthalpy uske elements in standard state se (jinki Δ H f ∘ = 0 hoti hai) 298 K , 1 a t m par.
Example: Δ H f ∘ ( H 2 O , g ) = − 241.8 kJ/mol (negative = banate waqt heat release hoti hai).
Intuition Released heat kahan jaati hai?
Adiabatic = koi heat bahar nahi jaati (Q = 0 ). Chamber mein combustion constant pressure par hoti hai (W sirf flow se hota hai). Toh reaction se release honi wali enthalpy 100% products ko heat karne mein jaati hai , T 0 se T a d tak.
Derivation (constant pressure, Q = 0 ):
Constant P par energy balance: total enthalpy conserve hoti hai,
H r e a c t an t s ( T 0 ) = H p r o d u c t s ( T a d ) .
Kyun? Adiabatic + constant-P ⇒ poore process ke liye Δ H = Q = 0 .
Har side ko "formation at T 0 " + "T 0 se T tak heating" mein split karo. Kyunki reactants T 0 par enter karte hain:
H r e a c t ( T 0 ) r e a c t ∑ n j Δ H f , j ∘ = products T 0 par banao p r o d ∑ n i Δ H f , i ∘ + products ko heat karo p r o d ∑ n i ∫ T 0 T a d c p , i d T
Rearrange karo (§2 se Δ H r x n use karke):
Worked example H₂/O₂ → 2 H₂O(g) ka quick estimate
Released: − Δ H r x n = 483.6 kJ 2 mol H₂O ke liye.
Lete hain c p ( H 2 O , g ) ≈ 45 J/mol⋅K (average, garam condition mein).
Δ T = 2 × 45 483600 = 5373 K ⇒ T a d ≈ 298 + 5373 ≈ 5670 K
Yeh step kyun? Saari released energy 2 mol water ko heat karti hai; energy ko total heat capacity se divide karo.
Reality check: real T a d ≈ 3000 –3500 K hoti hai, bahut kam. Kyun? Neeche dekho — dissociation aur badhta hua c p yeh difference kha jaate hain.
Common mistake "Bas constant
c p use karo aur dissociation ignore karo" (Steel-man)
Kyun sahi lagta hai: simple formula clean hai aur low-energy fuels ke liye theek kaam karta hai (methane in air ≈ 2200 K).
Rockets ke liye kyun galat hai: T > 2500 K par molecules dissociate ho jaate hain (H 2 O → OH + H , etc.). Dissociation endothermic hai — woh woh energy absorb karta hai jo otherwise T badhati. Saath mein c p bhi badhta hai T ke saath (vibrational modes activate ho jaate hain). Dono milke T a d ko naive estimate se bahut neeche cap kar dete hain.
Fix: temperature-dependent c p ( T ) use karo aur chemical equilibrium solve karo (saare possible product species par Gibbs energy minimize karo). Tools: NASA CEA. Naive number ek upper bound hai.
Common mistake "Zyada fuel = zyada garam" wali confusion
Sahi lagta hai: zyada fuel = zyada energy. Galat: peak T a d ϕ = 1 ke paas hoti hai. Excess fuel/oxidizer dead mass hai jo heat absorb karta hai bina react kiye, T ko kam karta hai . Rockets deliberately thoda off-peak T accept karte hain M kam karne aur walls bachane ke liye.
Recall Khud test karo (answers chhupao)
Combustion ko chamber mein adiabatic kyun maana jaata hai?
Stoichiometric balance mein kya conserve hota hai?
Rockets fuel-rich kyun chalte hain?
Naive T a d reality se zyada kyun hoti hai?
Exhaust speed ke liye asal mein kaun si quantity matter karti hai, aur sirf T kyun nahi?
Recall Feynman: 12-saal ke bachche ko samjhao
Socho ek super-fast campfire ek sealed metal box mein. Tum exactly utna "jalane wala stuff" (fuel) aur "jalane mein help karne wala stuff" (oxygen) daalo ki kuch bhi waste na ho — yahi stoichiometry hai, jaise har bachche ke liye exactly ek kursi match karna. Aag itni fast jalti hai ki heat box se bahar nahi ja sakti, toh SAARI heat andar reh jaati hai aur dhuaan ko paaglon jitna garam kar deti hai — yahi flame temperature hai. Lekin agar woh bahut zyada garam ho jaata hai, toh dhuaan ke molecules literally khud ko tod dete hain, aur todne mein energy lagti hai, toh woh thoda thanda ho jaata hai. Isliye real aag tumhari pehli guess jitni garam nahi hoti. Phir rockets is garam dhuaan ko ek nozzle se bahar phenkte hain — garam aur halka dhuaan = tez phenkna = zyada push.
"BALANCE karo atoms, BURN karo bina leak ke, BLOW karo halka."
Balance → stoichiometry (atoms conserve hote hain).
Burn bina leak → adiabatic, Q = 0 , heat saari products mein jaati hai.
Blow halka → low M ke liye fuel-rich chalo, tez exhaust.
Aur: "Hot ÷ Heavy" — T / M dhundo, sirf T nahi.
"Stoichiometric mixture" ka kya matlab hai? Woh exact fuel:oxidizer ratio jo dono ko poori tarah consume kare, koi excess reactant na bacha.
H₂ + O₂ → water balance karo. 2 H 2 + O 2 → 2 H 2 O .
H₂/O₂ ke liye stoichiometric O/F mass ratio? 2 ⋅ 2 1 ⋅ 32 = 8 (8 kg O₂ per kg H₂).
Equivalence ratio ϕ define karo. ϕ = ( F / O ) a c t u a l / ( F / O ) s t ; ϕ > 1 fuel-rich, ϕ < 1 fuel-lean, ϕ = 1 stoichiometric.
Rockets aksar fuel-rich kyun chalte hain? Excess light fuel (jaise H₂) product molar mass
M kam karta hai,
v e ∝ T / M badhata hai, aur
T ko material limits se neeche rakhta hai.
Heat of reaction ke liye Hess's law? Δ H r x n = ∑ n i Δ H f , p r o d − ∑ n j Δ H f , r e a c t .
Standard state mein element ki standard enthalpy of formation? Definition se zero.
Adiabatic flame temperature define karne wali physical condition? Constant pressure, Q = 0 ⇒ saari reaction enthalpy products ko heat karti hai; H r e a c t ( T 0 ) = H p r o d ( T a d ) .
T a d ke liye governing equation?− Δ H r x n = ∑ i n i ∫ T 0 T a d c p , i d T .
Constant-c p estimate of T a d ? T a d ≈ T 0 + ∑ i n i c p , i − Δ H r x n .
Real T a d naive estimate se kam kyun hoti hai? Products ka endothermic dissociation aur high temperature par badhta hua c p ( T ) energy absorb kar lete hain.
Maximum T a d max fuel par kyun nahi hoti? Excess reactant unreacted mass hai jo heat absorb karta hai; peak T ϕ = 1 ke paas hoti hai.
Exhaust velocity kaunse combustion outputs set karte hain? Chamber temperature
T c ≈ T a d aur product molar mass
M ,
v e ∝ T c / M ke zariye.
Specific Impulse and Exhaust Velocity — v e ∝ T c / M directly T a d use karta hai.
Nozzle Theory and Isentropic Expansion — garam high-P gas ko kinetic energy mein convert karta hai.
Chemical Equilibrium and Dissociation — Gibbs minimization se T a d correct karta hai.
First Law of Thermodynamics — T a d ke neeche enthalpy balance.
Hess's Law and Enthalpy of Formation — Δ H r x n ka source.
Propellant Selection — kyun H₂/O₂, RP-1/LOX T aur M mein alag hain.
O/F mass ratio 8 for H2/O2
Heat of reaction Hess law