5.3.4 · Chemistry › Combustion Chemistry (Propulsion Bridge)
Intuition Bada picture (YE KYU EXIST KARTA HAI)
Jab ek fuel–oxidizer mixture jalti hai, tab reaction front do fundamentally alag tareekon se unburnt gas mein travel kar sakta hai:
Deflagration — ek subsonic flame jo heat conduction + radicals ke diffusion se aage badhti hai. Pressure almost nahi badlta (thoda sa girta bhi hai).
Detonation — ek supersonic shock wave jo gas ko mechanically compress karke ignite karti hai. Pressure ~15–40 ke factor se jump karta hai.
Dono same chemical energy release karte hain, lekin propagation ka mechanism (diffusion vs shock) bilkul opposite hai. Chapman–Jouguet (CJ) condition woh rule hai jo mathematically allowed infinitely many solutions mein se unique stable detonation speed select karti hai. Yahi propulsion ka bridge hai: rotating detonation engines (RDEs) detonation chahte hain kyunki ye constant-volume (zyada efficient) heat release ke kareeb hoti hai.
Front ko ek patli discontinuity maano. Uss frame mein baitho jo front ke saath move kar raha hai , to gas usme u 1 speed se andar aati hai aur u 2 se bahar jaati hai. Subscript 1 = unburnt (reactants), 2 = burnt (products). Per unit area:
Definition Rankine–Hugoniot jump conditions
Mass, momentum, aur energy — teen conservation laws front ke across conserve hone chahiye:
ρ 1 u 1 = ρ 2 u 2 ( mass )
p 1 + ρ 1 u 1 2 = p 2 + ρ 2 u 2 2 ( momentum )
h 1 + 2 1 u 1 2 + q = h 2 + 2 1 u 2 2 ( energy )
jahan q = chemical heat released per unit mass, h = c p T = specific enthalpy.
Ye teen kyun? Reaction front sirf ek control volume hai. Kuch magic nahi — wahi bookkeeping jaise shock tube mein, bas energy equation mein chemistry ka source term q bhi add ho jaata hai.
Maano v = 1/ ρ specific volume hai. Mass se: u 1 = ρ 2 u 2 / ρ 1 , aur mass flux m ˙ = ρ 1 u 1 = ρ 2 u 2 . To u 1 = m ˙ v 1 , u 2 = m ˙ v 2 .
Momentum p 1 + m ˙ 2 v 1 = p 2 + m ˙ 2 v 2 mein substitute karo:
p 2 − p 1 = − m ˙ 2 ( v 2 − v 1 ) (Rayleigh line)
Intuition Rayleigh line kya bolta hai
( v , p ) plane mein ye initial state ( v 1 , p 1 ) se guzarti ek straight line hai jiska slope − m ˙ 2 ≤ 0 hai. Kyunki slope negative hai, har real solution mein ya to:
p 2 > p 1 aur v 2 < v 1 → detonation (compression, top-left), ya
p 2 < p 1 aur v 2 > v 1 → deflagration (expansion, bottom-right).
Steeper line (bada m ˙ , yani faster front) detonation se correspond karti hai.
Energy equation lo aur velocities eliminate karo. Use karo 2 1 u 1 2 − 2 1 u 2 2 = 2 1 m ˙ 2 ( v 1 2 − v 2 2 ) . Momentum relation m ˙ 2 = − ( p 2 − p 1 ) / ( v 2 − v 1 ) ke saath combine karo:
2 1 u 1 2 − 2 1 u 2 2 = 2 1 ( p 2 − p 1 ) ( v 1 + v 2 )
Energy mein plug karo (h 2 − h 1 = q + 2 1 u 1 2 − 2 1 u 2 2 ):
Actual final state DONO lines satisfy karti hai: ye Rayleigh line aur Hugoniot curve ka intersection hai.
Ek given m ˙ ke liye Rayleigh line Hugoniot ko do points par kaat sakti hai (strong & weak), ek point par (tangent), ya bilkul nahi . CJ point special hai:
Definition Chapman–Jouguet condition
CJ state woh hai jahan Rayleigh line Hugoniot curve ko tangent ho. Physically ye equivalent hai:
u 2 = a 2
Burnt gas front se exactly local speed of sound par bahar nikalta hai (Mach number M 2 = 1 , "sonic" / "choked").
Intuition KYU tangency = sonic = chosen speed
Do intersections ka matlab hota ki front kaafi saari speeds par chal sakta hai. Nature woh speed chunti hai jo self-sustaining aur stable ho.
Agar u 2 < a 2 (strong detonation), to peeche se rarefaction waves pakad leti hain aur front ko kamzor karti hain jab tak u 2 = a 2 nahi ho jaata.
u 2 = a 2 par flow choked ho jaati hai: peeche ki disturbances front se aage nahi ja sakti, to ye steadily aur autonomously propagate karta hai. Ye minimum-speed self-sustaining detonation hai.
Mathematically, Rayleigh line ka Hugoniot ko sirf graze karna precisely sabse chhota m ˙ (aur sabse chhoti detonation velocity) hai jiska detonation branch par solution exist karta hai.
Typical numbers: stoichiometric H₂–O₂ ke liye D C J ≈ 2840 m/s, p 2 / p 1 ≈ 18 .
Property
Deflagration
Detonation
Propagation mechanism
heat conduction + radical diffusion
shock compression
Speed
subsonic (∼ 0.1–10 m/s lab flames)
supersonic (∼ 1500–3000 m/s)
p 2 vs p 1
thoda lower (p 2 < p 1 )
strongly higher (p 2 ≫ p 1 )
v 2 vs v 1
gas expands (v 2 > v 1 )
gas compressed (v 2 < v 1 )
Front se nikalna wali flow
accelerate hoti hai, M 2 < 1
CJ par M 2 = 1
Hugoniot branch
lower-right
upper-left
Propulsion relevance
normal flames, ramjets
RDE, PDE — near constant-volume, higher efficiency
q = 3.0 × 1 0 6 J/kg, γ = 1.3 ke liye D C J estimate karo
D C J = 2 ( γ 2 − 1 ) q = 2 ( 1.69 − 1 ) ( 3.0 × 1 0 6 )
Ye step kyun? Hum strong-detonation estimate use karte hain kyunki question sirf q aur γ deta hai.
= 2 ( 0.69 ) ( 3.0 × 1 0 6 ) = 4.14 × 1 0 6 ≈ 2035 m/s
Sanity check: supersonic, order of km/s — real detonation ke saath consistent hai. ✓
Worked example (b) Kaun sa branch? Ek front
p 2 / p 1 = 12 , v 2 / v 1 = 0.55 deta hai.
Ye step kyun? Pressure badha aur volume gira — dono compression ke signatures hain.
→ Ye upper-left Hugoniot branch par hai ⇒ detonation . ✓
Worked example (c) Rayleigh slope sanity check
Diya gaya p 1 = 1 atm, v 1 = 1 (units), aur detonation state p 2 = 15 , v 2 = 0.6 .
Slope = v 2 − v 1 p 2 − p 1 = 0.6 − 1 15 − 1 = − 0.4 14 = − 35 .
Ye step kyun? Rayleigh slope − m ˙ 2 hai, to m ˙ 2 = 35 ⇒ m ˙ = 5.9 .
Negative slope physically valid front confirm karta hai; bada magnitude ek fast (detonation) front confirm karta hai. ✓
Common mistake "Detonation sirf ek bahut fast deflagration hai."
Kyun sahi lagta hai: dono same fuel jalate hain aur same chemical energy release karte hain, to lagta hai jaise speeds ka ek continuum ho.
Fix: mechanism categorically different hai . Deflagration diffusion of heat/radicals se drive hoti hai (subsonic, pressure girta hai); detonation ek shock se drive hoti hai jo gas ko compression se ignite karti hai (supersonic, pressure spike karta hai). Ye Hugoniot ke alag branches par hain — ek se doosre par smoothly slide nahi ho sakta.
Common mistake "CJ point woh hai jahan energy release maximum hai."
Kyun sahi lagta hai: "special point" ek energy extremum jaisa lagta hai.
Fix: CJ ek tangency / sonic (M 2 = 1 ) condition hai, minimum self-sustaining detonation speed , energy max nahi. Heat q chemistry se fixed hai; CJ unique speed select karta hai jiske liye downstream disturbances front ko disrupt nahi kar sakti.
Common mistake "Pressure hamesha reacting front ke across badhta hai."
Kyun sahi lagta hai: explosions = high pressure.
Fix: sirf detonations pressure badhate hain. Deflagrations mein p 2 ≲ p 1 hota hai kyunki gas expand aur accelerate hoti hai — yahi reason hai ki open flames cheezein nahi todtein jabki detonations todti hain.
Recall Feynman: 12-year-old ko samjhao
Socho ek corridor mein log ek line mein khade hain aur "burn" pass kar rahe hain.
Deflagration: har person agale ko shoulder tap karta hai burn start karne ke liye. Slow, gentle, corridor shant rehta hai.
Detonation: koi corridor mein itni tez cannonball chodta hai ki har person smash hoke jal jaata hai. Ye sound se faster hai, aur peeche high pressure ki wall chhodta hai.
Chapman–Jouguet woh rule hai jo kehta hai: cannonball ek exact speed par settle ho jaata hai — woh speed jis par peeche ke log aage instructions shout nahi kar sakte (unki awaaz pakad nahi sakti). Woh self-running speed detonation speed hai.
"DEFLAGRATION = DIFFUSE & DEFLATE; DETONATION = SHOCK & SPIKE."
Aur CJ ke liye: "CJ = Choked Junction" — burnt gas Mach 1 par bahar nikalta hai.
Deflagration aur detonation mein mechanism ke hisaab se kya farq hai? Deflagration heat conduction + radical diffusion se propagate hoti hai (subsonic); detonation shock compression se propagate hoti hai (supersonic).
Deflagration vs detonation mein pressure ke saath kya hota hai? Deflagration: p 2 thoda lower than p 1 (gas expands). Detonation: p 2 ≫ p 1 (gas compressed).
Teen Rankine–Hugoniot jump conditions batao. Mass ρ 1 u 1 = ρ 2 u 2 ; momentum p 1 + ρ 1 u 1 2 = p 2 + ρ 2 u 2 2 ; energy h 1 + 2 1 u 1 2 + q = h 2 + 2 1 u 2 2 .
Rayleigh line kya hai aur uska slope kya hai? p 2 − p 1 = − m ˙ 2 ( v 2 − v 1 ) ; ( v 1 , p 1 ) se guzarti straight line jiska slope − m ˙ 2 hai (hamesha negative).
Chapman–Jouguet condition define karo. Rayleigh line Hugoniot curve ko tangent hoti hai; equivalently burnt gas front se exactly local sound speed par bahar nikalta hai, u 2 = a 2 (M 2 = 1 ).
Detonation CJ speed par kyun settle hoti hai? M 2 = 1 par flow choked hoti hai, to rearward rarefaction waves front ko overtake karke kamzor nahi kar sakti — ye minimum detonation speed par self-sustaining ban jaati hai.
CJ velocity ka ideal-gas estimate do. D C J ≈ 2 ( γ 2 − 1 ) q (strong-detonation limit); zyada heat release
q → faster detonation.
Detonation kaun sa Hugoniot branch hai? Deflagration kaun sa? Detonation = upper-left (p 2 > p 1 , v 2 < v 1 ); deflagration = lower-right (p 2 < p 1 , v 2 > v 1 ).
Detonation propulsion (RDE/PDE) ke liye attractive kyun hai? Ye constant-volume heat release approximate karta hai, jo constant-pressure deflagration cycles se zyada thermodynamic efficiency deta hai.
Steeper Rayleigh line physically kya mean karti hai? Bada m ˙ , yani faster front — deflagration ki bajaye detonation ki characteristic.
approaches constant volume
Rankine-Hugoniot jump conditions
Chapman-Jouguet condition
Detonation supersonic shock
Deflagration subsonic flame
Rotating Detonation Engines