Combustion thermodynamics — stoichiometry, adiabatic flame temperature: Worked examples

3.3.19 · D3 · Physics › Rocket Propulsion › Combustion thermodynamics — stoichiometry, adiabatic flame t

Shuru karne se pehle, char symbols jo baar baar kaam aayenge (sab parent mein define hain, yahan re-anchor kar rahe hain taaki kuch assumed na ho):

Definition Char quantities jinhe hum juggle karte hain

$(\frac{O}{F})$ = oxidizer-to-fuel mass ratio = kilograms of oxidizer per kilogram of fuel. Bada number = zyada oxidizer per unit fuel.
$ϕ$ = equivalence ratio = actual mixture ko perfect (stoichiometric) se compare karta hai. $ϕ = 1$ perfect, $ϕ > 1$ fuel-rich (extra fuel), $ϕ < 1$ fuel-lean (extra oxidizer).
$- Δ H_{r x n}$ = balanced reaction ke per heat released (kilojoules mein ek positive number), jo phir product gases ko $T_{a d}$ tak heat karta hai.
$M$ = average product molar mass = exhaust gas ke ek "average mole" ka mass, grams per mole mein. Agar exhaust pure water vapour hoti toh yeh $18$ hota; bacha hua halka hydrogen ( $2$ g/mol) is average ko neeche kheechta hai, bhari bacha hui oxygen ( $32$ ) ise upar dhakelti hai. Hum $M$ ki parwah karte hain kyunki exhaust speed $T_{c} / M$ ke scale karti hai — halki gas zyada tezi se nikalti hai.

Scenario matrix

Har combustion problem in cells mein se ek mein land karti hai. Neeche ke examples us cell se label hain jisme woh aate hain, aur saath milke sab cover karte hain.

Cell	Case class	Kya tricky hai	Example
A	Stoichiometric ( $ϕ = 1$ ) balance + O/F	baseline; atoms exact karo	Ex 1
B	Fuel-rich ( $ϕ > 1$ )	bacha hua fuel dead mass hai, heat soakta hai	Ex 2
C	Fuel-lean ( $ϕ < 1$ )	bacha hua oxidizer dead mass hai	Ex 3
D	Non-oxygen oxidizer (oxidizer apna O saath laata hai)	C, H, N aur O ko do compounds mein balance karo	Ex 4
E	Heat of reaction sign / product phase	liquid vs gas water $Δ H$ badal deta hai	Ex 5
F	Constant- $c_{p}$ estimate of $T_{a d}$	naïve upper bound	Ex 6
G	Degenerate / limiting input ( $ϕ \to 0$ , $ϕ \to \infty$ )	edges par kya hota hai	Ex 7
H	Real-world word problem (tank sizing)	story → O/F → mass translate karo	Ex 8
I	Exam twist ("hotter ≠ better")	$T / M$ maximise karo, $T$ nahi	Ex 9

Ex 1 — Cell A: stoichiometric baseline

Forecast: guess karo — kya O/F ratio hydrogen ke liye mile 8 se bada hoga ya chhota? Ek number likho.

Skeleton likho aur carbon conserve karo. Yeh step kyun? Left par har atom right par bhi dikhna chahiye — yahi poora rule hai. $C H_{4} + x O_{2} \to 1 C O_{2} + 2 H_{2} O$ Left par ek C → ek $C O_{2}$ . Left par char H → do $H_{2} O$ (har ek mein 2 H hote hain).
Ab oxygen conserve karo taaki $x$ mile. Yeh step kyun? Oxygen ek hi unknown coefficient bachi hai. Right side O: $2$ ( $C O_{2}$ se) $+ 2$ (do $H_{2} O$ se) $= 4$ . Toh $2 x = 4 \Rightarrow x = 2$ . $C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
Mole ratio → mass ratio convert karo. Yeh step kyun? Engineers tanks mein mass store karte hain, molecules nahi, isliye useful ratio kilogram mein hota hai. $(\frac{O}{F})_{s t} = \frac{2 \cdot M _{O_{2}}}{1 \cdot M _{C H_{4}}} = \frac{2 \cdot 32}{1 \cdot 16} = 4$

Verify: left atoms = {1 C, 4 H, 4 O}; right atoms = {1 C (CO₂), 4 H (2×H₂O), 2+2=4 O}. Match. Mass check: reactant mass $16 + 64 = 80$ g = product mass $44 + 2 \times 18 = 80$ g. Conserved. $(O / F)_{s t} = 4$ — hydrogen ke 8 se chhota, kyunki methane pehle se hi apna hydrogen sasta uthati hai.

Ex 2 — Cell B: fuel-rich mixture

Forecast: fuel-rich matlab extra fuel. Kya actual O/F 8 se upar hoga ya neeche? Padhne se pehle decide karo.

Bilkul samjho $ϕ$ kya measure karta hai, phir rearrange karo. Yeh step kyun? Parent ne define kiya $ϕ = \frac{( O / F ) _{s t}}{( O / F ) _{a c t u a l}}$ — ek fraction jiska top fixed hai (perfect ratio, 8) aur jiska bottom woh number hai jis par hum actually run karte hain. Toh $ϕ$ actually yeh sawaal hai "meri real mixture perfect wali se kitni zyada oxidizer-starved hai?" Real ratio paane ke liye, hum us fraction ko invert karte hain: dono sides ko $(O / F)_{a c t u a l}$ se multiply karo aur $ϕ$ se divide karo. Yeh ek unknown isolate karta hai. $ϕ = \frac{( O / F ) _{s t}}{( O / F ) _{a c t u a l}} ⟹ ϕ \cdot (O / F)_{a c t u a l} = (O / F)_{s t} ⟹ (O / F)_{a c t u a l} = \frac{( O / F ) _{s t}}{ϕ} = \frac{8}{1.5} = 5.33$ Algebra ki sanity: $ϕ > 1$ hone se hum 8 ko 1 se bade number se divide kar rahe hain, toh answer 8 se neeche aana chahiye — bilkul wahi jo "extra fuel, kam oxidizer" matlab hona chahiye.
Interpret karo. Yeh step kyun? Meaning ke bina number bekar hai. Hum abhi sirf $5.33$ kg O₂ per kg H₂ de rahe hain, 8 se kam jo zaroori tha — toh hydrogen surplus mein hai. Fuel-rich confirm.
Kaunsa leftover? Excess hydrogen burn se bachta hai. Kyun matter karta hai: bacha hua H₂ bahut halka hai ( $M = 2$ g/mol), jo average product molar mass $M$ ko neeche kheechta hai — exhaust speed ke liye acha hai (dekho Specific Impulse and Exhaust Velocity).

Verify: $ϕ > 1$ ko actual O/F stoichiometric 8 se neeche dena chahiye, aur deta hai ( $5.33 < 8$ ). Fuel/oxidizer form se cross-check: $(F / O)_{a c t u a l} = 1/5.33 = 0.1875$ , $(F / O)_{s t} = 1/8 = 0.125$ , ratio $0.1875/0.125 = 1.5 = ϕ$ . ✓

Ex 3 — Cell C: fuel-lean mixture

Forecast: lean = extra oxidizer. 4 se upar ya neeche?

Actual O/F solve karo. Yeh step kyun? Ex 2 jaisi hi inversion — $ϕ = (O / F)_{s t} / (O / F)_{a c t u a l}$ ⇒ $(O / F)_{a c t u a l} = (O / F)_{s t} / ϕ$ — par ab $ϕ < 1$ hai, toh hum 4 ko 1 se chhote number se divide karte hain aur answer 4 se upar jaana chahiye. $(O / F)_{a c t u a l} = \frac{( O / F ) _{s t}}{ϕ} = \frac{4}{0.8} = 5$
Interpret karo. Yeh step kyun? Ab hum 5 kg O₂ per kg fuel supply kar rahe hain jahan sirf 4 chahiye → excess oxygen bachti hai. Lean ka yahi matlab hai.
Consequence. Bacha hua O₂ ( $M = 32$ g/mol) bhari aur unreacted hai — average $M$ badhata hai aur bina koi heat release kiye heat soakta hai, $T$ girata hai. Rocket ke liye dono count par bura, isliye lean launchers mein rare hoti hai.

Verify: $ϕ < 1$ ⇒ actual O/F 4 se upar ⇒ humne 5 paaya. ✓ Consistency: $(F / O)_{a c t u a l} = 0.2$ , $(F / O)_{s t} = 0.25$ , ratio $0.2/0.25 = 0.8 = ϕ$ . ✓

Ex 4 — Cell D: ek non-oxygen oxidizer (oxidizer apni oxygen saath laata hai)

Forecast: itne bhaari oxidizer ke saath (92 vs O₂ ka 32), kya O/F ratio H₂/O₂ ke 8 se bada hoga ya chhota?

Skeleton likho fuel aur oxidizer dono par unknown coefficients ke saath. Yeh step kyun? O₂ (ek single element) ke khilaf, yahan do compounds milte hain, toh dono ko coefficients chahiye. Hamare paas conserve karne ke liye teen elements hain — N, H, O — do nahi. $a N_{2} H_{4} + b N_{2} O_{4} \to c N_{2} + d H_{2} O$
Hydrogen conserve karo taaki $d$ ko $a$ se tie karo. Yeh step kyun? Sirf fuel H carry karta hai, toh hydrogen water coefficient turant fix kar deta hai. Left H $= 4 a$ ; right H $= 2 d$ ⇒ $d = 2 a$ .
Oxygen conserve karo taaki $b$ ko $d$ se (isliye $a$ se) tie karo. Yeh step kyun? Sirf $N_{2} O_{4}$ O carry karta hai, aur sirf water ise bahar le jaata hai — yeh woh step hai jo sirf-O₂ problems mein kabhi nahi hota. Left O $= 4 b$ ; right O $= d = 2 a$ ⇒ $4 b = 2 a \Rightarrow b = \frac{a}{2}$ .
$a = 1$ set karo aur baaki read karo, phir fraction clear karo. Yeh step kyun? Hum simplest fuel coefficient choose karte hain aur scale karte hain. $a = 1$ ke saath: $d = 2$ , $b = \frac{1}{2}$ . $c$ ke liye nitrogen conserve karo: left N $= 2 a + 2 b = 2 + 1 = 3$ , right N $= 2 c$ ⇒ $c = \frac{3}{2}$ . Whole numbers paane ke liye sab 2 se multiply karo: $2 N_{2} H_{4} + N_{2} O_{4} \to 3 N_{2} + 4 H_{2} O$
Mass ratio mein convert karo. Yeh step kyun? Tanks mass store karte hain; yahan $a = 2$ fuel, $b = 1$ oxidizer. $(\frac{O}{F})_{s t} = \frac{b M _{N_{2} O_{4}}}{a M _{N_{2} H_{4}}} = \frac{1 \cdot 92}{2 \cdot 32} = \frac{92}{64} = 1.44$

Verify: atoms — left {N: $2 \times 2 + 1 \times 2 = 6$ , H: $2 \times 4 = 8$ , O: $1 \times 4 = 4$ }; right {N: $3 \times 2 = 6$ , H: $4 \times 2 = 8$ , O: $4 \times 1 = 4$ }. Sab match. ✓ Mass: reactants $2 \times 32 + 1 \times 92 = 156$ g $=$ products $3 \times 28 + 4 \times 18 = 84 + 72 = 156$ g. ✓ $(O / F)_{s t} \approx 1.44$ — H₂/O₂ ke 8 se kaafi neeche, kyunki hydrazine bahut oxygen-hungry-cheap hai aur $N_{2} O_{4}$ har molecule mein char O atoms deliver karta hai.

Ex 5 — Cell E: heat of reaction, aur phase trap

Forecast: kya $Δ H_{r x n}$ negative hoga (heat out) ya positive? Fuel ke per mole kitna roughly?

Hess's law apply karo: products minus reactants. Yeh step kyun? Enthalpy ek state function hai (path-independent), toh reaction ka $Δ H$ sirf tabulated formation enthalpies ka difference hai — dekho First Law of Thermodynamics. $Δ H_{r x n} = [1 (- 393.5) + 2 (- 241.8)] - [1 (- 74.8) + 2 (0)]$
Calculate karo. Yeh step kyun? Sirf arithmetic, par signs watch karo — ek chhuta hua minus exothermic↔endothermic flip kar deta hai. $= (- 393.5 - 483.6) - (- 74.8) = - 877.1 + 74.8 = - 802.3 kJ$
Sign padho. Yeh step kyun? Negative $Δ H_{r x n}$ = exothermic = heat chemistry se nikalta hai aur (chamber mein) gases ko heat karta hai. Toh $- Δ H_{r x n} = + 802.3$ kJ released per mole of CH₄.

Verify: magnitude sensible hai — methane ki textbook lower heating value ≈ 802 kJ/mol hai. ✓ Phase trap: agar water liquid mein condense ho jaati ( $Δ H_{f}^{\circ} (H_{2} O, l) = - 285.8$ ) toh zyada heat milti ( $- 890$ kJ, higher heating value). Rocket mein water vapour rehti hai, toh gas value use karo.

Ex 6 — Cell F: naïve constant- $c_{p}$ estimate of $T_{a d}$

Forecast: kya yeh crude number real ~3000 K se upar ya neeche hoga? Kitna?

Energy balance likho. Yeh step kyun? Adiabatic + constant pressure ⇒ saari released heat product gases ko upar uthane mein jaati hai (parent §3). $- Δ H_{r x n} = (\sum_{i} n_{i} c_{p, i}) Δ T$
Products ki total heat capacity. Yeh step kyun? Har product mole heat soakta hai; unki capacities add karo. $\sum_{i} n_{i} c_{p, i} = 1 (57) + 2 (45) = 57 + 90 = 147 J/K$
$Δ T$ solve karo aur $T_{0}$ add karo. Yeh step kyun? Energy ko heat capacity se divide karo — yahi temperature rise ki definition hai. $Δ T = \frac{802300}{147} = 5458 K, T_{a d} \approx 298 + 5458 = 5756 K$

Verify: units — J ÷ (J/K) = K. ✓ Number $\approx 5760$ K real ~3000 K ka roughly double hai, bilkul wahi overshoot jiske baare mein parent ne warn kiya tha: yeh dissociation aur temperature ke saath $c_{p}$ ka badhna ignore karta hai. Yeh ek upper bound hai, prediction nahi.

Figure 1 — energy-into-one-bucket picture (abhi padho): Neeche figure mein do coloured bars products ki heat capacities hain: violet bar ek mole CO₂ ka hai ( $57$ J/K), orange bar do moles H₂O ka hai ( $90$ J/K). Unki heights dashed magenta line mein add hoti hain $147$ J/K par — woh combined bucket. Pura $802300$ J released chemical energy (magenta arrow) us ek bucket mein daala jaata hai, aur temperature rise simply energy ÷ bucket size hai. Pedagogical punch: bada bucket — zyada product moles, ya dissociation se naye species — chhota rise deta hai, aur yahi reason hai ki real flame is crude number se cooler nikalti hai.

Figure — Combustion thermodynamics — stoichiometry, adiabatic flame temperature

Alt text: bar chart of product heat capacities (CO₂ 57 J/K, H₂O 90 J/K) summing to a dashed line at 147 J/K, with an arrow labelling the 802 300 J of released energy divided by that capacity to give a 5458 K rise.

Ex 7 — Cell G: degenerate limits

Forecast: kis extreme par bilkul aag nahi lagti?

Limit $ϕ \to 0^{+}$ (infinitely lean). Yeh step kyun? Definition padho $(O / F)_{a c t u a l} = (O / F)_{s t} / ϕ$ . Jaise $ϕ \to 0$ , $(O / F)_{a c t u a l} \to \infty$ : oxidizer ke samandar, fuel ki ek trace. $lim_{ϕ \to 0^{+}} (O / F)_{a c t u a l} = + \infty$ Thoda sa fuel thodi si heat release karta hai, huge thande excess oxidizer ke mass par spread hoke ⇒ $T_{a d} \to T_{0}$ . Essentially koi temperature rise nahi.
Limit $ϕ \to \infty$ (infinitely rich). Yeh step kyun? Same formula: $(O / F)_{a c t u a l} \to 0$ , almost koi oxidizer nahi. $lim_{ϕ \to \infty} (O / F)_{a c t u a l} = 0$ Bina oxidizer ke, almost kuch nahi jalta ⇒ phir $T_{a d} \to T_{0}$ . Koi aag nahi.
Peak kahan hai? Yeh step kyun? Dono edges thandi hain, toh maximum $T_{a d}$ in dono ke beech baithta hai, $ϕ = 1$ ke paas. Yeh woh shape hai jo tumhe yaad rakhni hai: ek hump jo stoichiometric ke around peak karta hai.

Verify: dono limits $T_{a d} \to T_{0}$ dete hain, consistent with physical fact ki heat release ke liye dono reactants present chahiye. Middle-mein-maximum behaviour parent ke mistake callout se match karta hai ("more fuel ≠ hotter"). ✓

Figure 2 — temperature hump (abhi padho): Neeche figure mein magenta curve $T_{a d}$ hai jo equivalence ratio $ϕ$ (horizontal axis) ke against kelvin (vertical axis) mein plot ki gayi hai. Ise left se trace karo: $ϕ \to 0$ par (left orange arrow, "lean") curve dotted floor " $T_{0}$ , no fire" ko hug karta hai. Ab right ki taraf chalo: $ϕ \to \infty$ par (right orange arrow, "rich") curve usi floor par wapas dab jaata hai. Sirf uoonchi zameen navy dot hai violet vertical line $ϕ = 1$ ke paas. Woh akela hump hi "more fuel ≠ hotter" ka visual proof hai: $ϕ$ ko peak se aage push karo aur tum seedha $T_{0}$ ki taraf wapas slope par aa jaate ho.

Alt text: a hump-shaped magenta curve of adiabatic flame temperature versus equivalence ratio, hugging a dotted "no fire" floor at both extremes and peaking with a navy dot near the violet phi=1 line, with orange arrows labelling the lean and rich cold limits.

Ex 8 — Cell H: real-world tank sizing

Forecast: LOX per unit bhaari hai — kya oxidizer tank mein fuel tank se zyada ya kam mass hogi?

Ratio ko mass mein translate karo. Yeh step kyun? $(O / F) = 8$ literally matlab hai 8 kg O₂ per 1 kg H₂ — multiply karo. $m_{O_{2}} = 8 \times m_{H_{2}} = 8 \times 12000 = 96000 kg$
Total propellant. Yeh step kyun? Dono tanks lift-off mass mein contribute karte hain. $m_{t o t a l} = m_{H_{2}} + m_{O_{2}} = 12000 + 96000 = 108000 kg$
Split par sanity. Yeh step kyun? Fuel sirf $12000/108000 = 11.1%$ propellant mass hai, chahe energetically woh heavy lifting karta hai. Isliye LOX tanks mass se LH₂ tanks se bade hote hain (halaalki LH₂ tanks volume se bade hote hain — hydrogen fluffy hai). Dekho Propellant Selection.

Verify: $96000/12000 = 8 = (O / F)_{s t}$ . ✓ Fuel fraction $12000/108000 = 0.1111$ . ✓ Units sab kilograms.

Ex 9 — Cell I: exam twist — hotter hamesha better nahi hota

Forecast: Mix P zyada hot hai. Kya hotter jeet jaata hai?

Figure of merit likho. Yeh step kyun? Exhaust speed $T_{c} / M$ ke scale karta hai, $T_{c}$ akele nahi — toh us quantity ko compare karo, temperatures ko nahi. $score \propto \frac{T _{c}}{M}$
Mix P score karo. Yeh step kyun? Hotter, heavier pair plug in karo. $\frac{T _{c}}{M}_{P} = \frac{3600}{18} = 200 = 14.14$
Mix Q score karo. Yeh step kyun? Same formula, cooler par lighter fuel-rich pair. $\frac{T _{c}}{M}_{Q} = \frac{3300}{13} = 253.8 = 15.93$
Compare karo aur conclude karo. Yeh step kyun? Zyada score = zyada exhaust speed. Kyunki $15.93 > 14.14$ , cooler, fuel-rich Mix Q jeetta hai, chahe woh 300 K thanda ho. Gain $15.93/14.14 = 1.127$ hai, matlab roughly 13% zyada exhaust speed, kyunki $M$ ko 18 se 13 tak giraa dena temperature ke loss se zyada outweigh karta hai. Isliye real engines peak temperature par nahi, thoda rich run karte hain.

Figure 3 — Hot ÷ Heavy trade-off (abhi padho): Neeche figure mein har bar ek mixture ka exhaust-speed score $T_{c} / M$ (vertical axis) hai. Violet bar (Mix P) hotter, heavier stoichiometric case hai; magenta bar (Mix Q) cooler, lighter fuel-rich case hai. 300 K thanda hone ke bawajood, magenta bar taller hai — orange arrow mark karta hai ki Q roughly $12.7%$ se jeetta hai. Yaad karne ka lesson: tum $T / M$ chase karte ho, $T$ akele nahi.

Alt text: two bars comparing exhaust-speed scores, violet Mix P at 14.14 and magenta Mix Q at 15.93, with an orange arrow noting the cooler-but-lighter mixture wins by 12.7 percent.

Verify: scores ka ratio $15.93/14.14 = 1.127$ , Q ke liye 12.7% gain. ✓ Parent ke "chase $T / M$ , not $T$ " mnemonic ko confirm karta hai. Full $v_{e}$ formula ke liye dekho Nozzle Theory and Isentropic Expansion.

Active Recall

Recall Kaun sa cell kaun sa tha?

Fuel-rich matlab actual O/F ::: stoichiometric se neeche hota hai (kam oxidizer per unit fuel), excess halka fuel bachta hai.
$ϕ \to 0$ aur $ϕ \to \infty$ dono dete hain ::: $T_{a d} \to T_{0}$ — koi complete reaction nahi, toh peak $T$ $ϕ = 1$ ke paas baithta hai.
Naïve constant- $c_{p}$ flame temperature ::: ek upper bound hai; dissociation aur rising $c_{p}$ real value ko kaafi neeche le jaate hain.
Fuel-rich H₂/O₂ stoichiometric ko beat kar sakta hai kyunki ::: average molar mass $M$ kam hone se $T_{c} / M$ itna badhta hai ki temperature drop ka loss poora ho jaata hai.
$N_{2} O_{4}$ jaisa non-O₂ oxidizer tumhe force karta hai ::: oxygen (aur nitrogen) ko balance karne ke liye jo ek compound ke andar rehta hai, free O₂ ke roop mein nahi.