Combustion thermodynamics — stoichiometry, adiabatic flame temperature: Exercises

3.3.19 · D4 · Physics › Rocket Propulsion › Combustion thermodynamics — stoichiometry, adiabatic flame t

Level 1 — Recognition

Kya tum tool ka naam bata sakte ho aur numbers plug in kar sakte ho? Abhi koi analysis nahi.

L1.1 — Oxidizer-rich pair ko balance karo

Methane aur oxygen ki reaction ko $C O_{2}$ aur water vapour tak balance karo: $C H_{4} +_O_{2} \to_C O_{2} +_H_{2} O$

Recall Solution

KYA karte hain: ek ek karke har atom (C, H, O) ko conserve karo.

Carbon: left side par 1 C hai ⇒ 1 $C O_{2}$ .
Hydrogen: left side par 4 H hain ⇒ 2 $H_{2} O$ (4 H deta hai).
Oxygen: right side par ab $2$ (from $C O_{2}$ ) $+ 2$ (from $2 H_{2} O$ ) $= 4$ O atoms hain $= 2 O_{2}$ . $C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$ Check: left $= 4$ H, $1$ C, $4$ O; right $= 4$ H, $1$ C, $4$ O. ✓

L1.2 — Stoichiometric O/F by mass

Upar wali methane reaction ke liye, stoichiometric oxidizer-to-fuel mass ratio compute karo.

Recall Solution

Kyun mass, moles nahi: tanks mass se bhare jaate hain. Symbol == $O$ matlab oxidizer ki total mass hai aur $F$ matlab fuel ki total mass== hai, isliye ratio $O / F$ literally "kilograms of oxidizer per kilogram of fuel" hai. Hum mole ratio ko is mass ratio mein convert karte hain: $(\frac{O}{F})_{s t} = \frac{b M _{o x}}{a M _{f u e l}}$ jahan $a$ fuel ka stoichiometric coefficient hai (balanced equation mein fuel ke aage likha hua number) aur $b$ oxidizer ka stoichiometric coefficient hai. Balanced reaction $C H_{4} + 2 O_{2} \to \dots$ se, fuel $C H_{4}$ ka coefficient $a = 1$ hai aur oxidizer $O_{2}$ ka coefficient $b = 2$ hai. $M_{o x}$ aur $M_{f u e l}$ unke molar masses hain. $(\frac{O}{F})_{s t} = \frac{2 \times 32}{1 \times 16} = \frac{64}{16} = 4$ Toh 4 kg O₂ per 1 kg CH₄.

L1.3 — Enthalpy ka sign

$Δ H_{f}^{\circ} (C O_{2}, g) = - 393.5 kJ/mol$ negative hai. Physically negative sign ka matlab kya hai?

Recall Solution

Negative $Δ H_{f}^{\circ}$ ka matlab hai ki us molecule ko uske elements ( $C + O_{2}$ ) se banane par heat release hoti hai — product ke bonds energy mein lower (zyada stable) hote hain loose elements se. Positive hota toh matlab hota ki banane mein heat lagti hai. Toh $C O_{2}$ ek stable, heat-releasing "downhill" product hai — exactly wahi jo ek fuel chahta hai ki end tak pahunche.

Level 2 — Application

Ab do ya teen steps chain karo.

L2.1 — Methane ke liye heat of reaction

Hess's law use karke, $C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O (g)$ ke liye $Δ H_{r x n}$ nikalo.

Recall Solution

Kyun Hess's law kaam karta hai: enthalpy ek state function hai — total change sirf start aur end par depend karta hai, isliye $Δ H_{r x n} = \sum_{p r o d} n_{i} Δ H_{f, i}^{\circ} - \sum_{r e a c t} n_{j} Δ H_{f, j}^{\circ}$ . Yaad karo $Δ H_{f}^{\circ} (O_{2}) = 0$ (element in standard state). $Δ H_{r x n} = [1 (- 393.5) + 2 (- 241.8)] - [1 (- 74.9) + 2 (0)]$ $= (- 393.5 - 483.6) - (- 74.9) = - 877.1 + 74.9 = - 802.2 kJ$ $Δ H_{r x n} \approx - 802.2 kJ per mol CH_{4}$ Negative ⇒ exothermic, jaisa ek fuel ko hona chahiye.

L2.2 — Naïve adiabatic flame temperature

Methane burn ke liye $T_{a d}$ estimate karo, $c_{p}$ ko constant maanke. Average hot-gas $c_{p} (C O_{2}) \approx 57 J/mol\cdotK$ aur $c_{p} (H_{2} O) \approx 45 J/mol\cdotK$ lo. Products exactly $1 C O_{2} + 2 H_{2} O$ hain.

Recall Solution

KYA governing equation kehti hai: saari released chemical heat product gases ko garam karne mein jaati hai, $- Δ H_{r x n} = \sum_{i} n_{i} c_{p, i} Δ T$ . Products ki total heat capacity (per mol of fuel burned): $\sum_{i} n_{i} c_{p, i} = 1 (57) + 2 (45) = 57 + 90 = 147 J/K$ $Δ T = \frac{- Δ H _{r x n}}{\sum _{i} n _{i} c _{p, i}} = \frac{802200 J}{147 J/K} = 5457 K$ $T_{a d} \approx 298 + 5457 \approx 5755 K$ Reality check: methane/oxygen actually $3500 K$ ke paas jalta hai — naïve number ek wild over-estimate hai exactly L5.1 ke reasons ki wajah se.

L2.3 — Equivalence ratio from a filled tank

Ek engine mein $3 kg$ of $C H_{4}$ aur $9 kg$ of $O_{2}$ load hai. Kya yeh fuel-rich, lean, ya stoichiometric hai? Equivalence ratio $ϕ$ compute karo.

Recall Solution

Pehle, definition. ==Equivalence ratio $ϕ$ == compare karta hai ki tumne kitna fuel actually supply kiya versus kitna ek perfect (stoichiometric) burn ko chahiye: $ϕ = \frac{( F / O ) _{a c t u a l}}{( F / O ) _{s t}} = \frac{( O / F ) _{s t}}{( O / F ) _{a c t u a l}}$ By construction $ϕ = 1$ stoichiometric hai, $ϕ > 1$ fuel-rich hai (ek perfect burn se zyada fuel), aur $ϕ < 1$ fuel-lean hai (excess oxidizer).

Ab numbers. Actual mass ratio: $(\frac{O}{F})_{a c t} = \frac{9}{3} = 3$ . L1.2 se, $(\frac{O}{F})_{s t} = 4$ . $ϕ = \frac{( O / F ) _{s t}}{( O / F ) _{a c t}} = \frac{4}{3} \approx 1.33$ $ϕ > 1$ ⇒ fuel-rich (stoichiometry se kam oxidizer hai, isliye fuel excess mein hai). Intuition check: humne fuel unit per oxidizer ke sirf $3$ "units" supply kiye, $4$ chahiye the, isliye kuch methane puri tarah nahi jal sakta — rich ke saath consistent hai.

Level 3 — Analysis

Ab competing effects weighing karo aur trends explain karo.

L3.1 — Peak temperature $ϕ = 1$ par kyun hoti hai

Yahan $T_{a d}$ adiabatic flame temperature hai — woh temperature jo product gases tab reach karti hain jab saari released chemical heat unke andar hi rehti hai ( $Q = 0$ , koi heat bahar nahi jaati); yeh "burn kitni hot hoti hai" wala number hai jo parent note mein build kiya gaya hai. Ek student kehta hai "zyada fuel daalna hamesha flame ko hotter banata hai." Sketch/argue karo ki $T_{a d}$ ka kya hota hai jab $ϕ$ lean ( $ϕ < 1$ ) se $ϕ = 1$ se rich ( $ϕ > 1$ ) tak sweep karta hai, aur shape explain karo. Figure dekho.

Figure — Combustion thermodynamics — stoichiometry, adiabatic flame temperature

Figure — $T_{a d}$ versus $ϕ$ (self-contained caption). Horizontal axis equivalence ratio $ϕ$ (dimensionless) hai, $0.4$ (lean, left) se $2.0$ (rich, right) tak sweep karta hai. Vertical axis adiabatic flame temperature $T$ kelvin mein hai. Amber curve lean side par rise karti hai, $ϕ = 1$ ke paas maximum tak pahunchti hai (dashed cyan line), aur phir rich side par wapas girती hai — ek asymmetric hump. White dot peak mark karta hai; LEAN block (left) note karta hai "excess oxidizer soaks heat" aur RICH block (right) note karta hai "excess fuel is dead mass". Agar image render nahi ho rahi: ek single rounded hump imagine karo jo $ϕ = 1$ ke thoda aage peak karta hai, dono ends par low.

Recall Solution

Figure kya dikhata hai. Amber curve $ϕ$ ka function hone par $T_{a d}$ hai: yeh lean side par low values se rise karti hai, $ϕ = 1$ ke paas maximum tak pahunchti hai (dashed cyan line stoichiometric mark karti hai), phir rich side par wapas girती hai — ek asymmetric hump jo thoda sa rich side par peak karta hai. White dot peak mark karta hai; left/right text blocks LEAN aur RICH regions label karte hain aur explain karte hain ki har ek cooler kyun hai.

Lean side ( $ϕ < 1$ ): excess oxidizer. Woh surplus $O_{2}$ react nahi karta but phir bhi chamber mein baith kar heat absorb karta hai (uski heat capacity hoti hai). Released energy limiting fuel se fix hai, lekin ab zyada moles mein share hoti hai ⇒ lower $Δ T$ .
Stoichiometric ( $ϕ = 1$ ): har fuel aur oxidizer molecule react karta hai, koi inert dead mass heat nahi soakता ⇒ maximum energy per unit heat capacity ⇒ peak $T_{a d}$ .
Rich side ( $ϕ > 1$ ): excess fuel. Unburnt fuel dead mass hai jo energy release kiye bina garam hota hai, aur kuch carbon/hydrogen partially oxidised nikaalta hai (kam energy release) ⇒ lower $T_{a d}$ . Conclusion: "zyada fuel = hotter" galat hai. $ϕ = 1$ ke baad, extra fuel ek heat sink hai, heat source nahi. Real peak often thoda rich land karta hai kyunki dissociation (L5.1) exactly stoichiometric par sabse zyada hoti hai jahan $T_{a d}$ highest hoti hai.

L3.2 — Do fuels, same energy: kaun tez udega?

Yahan $T_{c}$ combustion chamber temperature hai — combustion se nikalne wale gas ki temperature, essentially parent note mein build ki gayi adiabatic flame temperature $T_{a d}$ ke barabar. Fuel A ka exhaust $T_{c} = 3600 K$ aur product molar mass $M = 22 g/mol$ produce karta hai. Fuel B ka $T_{c} = 3000 K$ aur $M = 12 g/mol$ hai. $v_{e} \propto T_{c} / M$ use karke, kaun faster exhaust deta hai? Kitne factor se?

Recall Solution

Kyun $T / M$ aur sirf $T$ nahi: nozzle thermal energy ko directed kinetic energy mein convert karta hai; halke molecules ( $M$ small) same energy ke liye zyada accelerate hote hain, isliye dono hot aur light matter karte hain. Ratio compare karo: $\frac{v _{e, B}}{v _{e, A}} = \frac{T _{B} / M _{B}}{T _{A} / M _{A}} = \frac{3000/12}{3600/22} = \frac{250}{163.6} = 1.528 \approx 1.236$ $v_{e, B} \approx 1.24 v_{e, A}$ Fuel B jeet jaata hai despite $600 K$ colder hone ke, kyunki uska exhaust almost aadha weight ka hai. Exactly yahi reason hai ki hydrogen-rich exhaust ( $M$ low) hotter lekin heavier alternatives ko beat karta hai — Specific Impulse and Exhaust Velocity aur Propellant Selection ke peeche ka lesson.

Level 4 — Synthesis

Ek problem mein stoichiometry + enthalpy + $T_{a d}$ + performance combine karo.

L4.1 — H₂/O₂ ke liye full pipeline

Raw elements se shuru karke, $2 H_{2} + O_{2} \to 2 H_{2} O (g)$ ke liye: (a) stoichiometric O/F confirm karo; (b) $Δ H_{r x n}$ nikalo; (c) $c_{p} (H_{2} O) \approx 45 J/mol\cdotK$ ke saath naïve $T_{a d}$ estimate karo; (d) average product molar mass nikalo; (e) performance figure $T_{a d} / M$ banao aur methane case se compare karo (methane ki naïve $T_{a d} = 5755 K$ use karo jo L2.2 se hai, product $M$ same tarike se compute kiya hua).

Recall Solution

(a) O/F — kyun ye numbers: balanced equation se fuel $H_{2}$ ka coefficient $a = 2$ aur oxidizer $O_{2}$ ka coefficient $b = 1$ hai; mole ratio ko mass ratio mein $O / F = b M_{o x} / (a M_{f u e l})$ se convert karne par $(\frac{O}{F})_{s t} = \frac{1 \times 32}{2 \times 2} = 8 kg O_{2} per kg H_{2} . ✓$ Yeh parent note se match karta hai, humaari bookkeeping confirm karta hai.

(b) $Δ H_{r x n}$ — kyun Hess's law: enthalpy ek state function hai, isliye reaction heat formation enthalpies ka (products − reactants) hai, aur elements $H_{2}, O_{2}$ ka $Δ H_{f}^{\circ} = 0$ hota hai: $Δ H_{r x n} = 2 (- 241.8) - [2 (0) + 1 (0)] = - 483.6 kJ per 2 mol H_{2} .$ Negative ⇒ exothermic, $483.6 kJ$ release karta hai.

(c) Naïve $T_{a d}$ : products $2$ mol H₂O hain, total heat capacity $2 \times 45 = 90 J/K$ . $Δ T = \frac{483600}{90} = 5373 K \Rightarrow T_{a d} \approx 298 + 5373 \approx 5671 K$

(d) Product molar mass: saara product water hai, isliye $M = 18 g/mol$ .

(e) Performance figure: $\frac{T _{a d}}{M}_{H_{2} / O_{2}} = \frac{5671}{18} = 315.1 \approx 17.75$ Methane products ${1 C O_{2}, 2 H_{2} O}$ ke liye, total mass $= 44 + 2 (18) = 80 g$ over $3$ mol ⇒ $M = 80/3 \approx 26.7 g/mol$ : $\frac{5755}{26.7} = 215.5 \approx 14.68$ Ratio: $17.75/14.68 \approx 1.21$ . Hydrogen/oxygen lagbhag 21% higher exhaust speed deliver karta hai — similar temperature par bhi — kyunki uska water-only exhaust methane ke $C O_{2}$ -laden exhaust se kaafi halka hai. Yeh Propellant Selection ka quantitative core hai aur seedha Nozzle Theory and Isentropic Expansion mein jaata hai.

Level 5 — Mastery

Ab tumhe judge karna padega ki physically real kya hai, sirf compute nahi.

L5.1 — Kyun naïve number 5000 K se upar nonsense hai

Tumhara L4 answer H₂/O₂ ke liye $T_{a d} \approx 5671 K$ diya, lekin real engines $\approx 3400 K$ measure karte hain. Do independent physical reasons explain karo ki true value itni kam kyun hai, aur batao ki naïve number actually kya represent karta hai.

Recall Solution

Reason 1 — Dissociation (endothermic sink). $\sim 2500 K$ se upar product molecules toot jaate hain: $H_{2} O ⇌ OH + H$ , $H_{2} ⇌ 2 H$ , $O_{2} ⇌ 2 O$ . In bonds ko todne mein energy absorb hoti hai jo otherwise $T$ raise karti. Jitna zyada hot hota hai, utna zyada dissociate hota hai — ek self-limiting brake (dekho Chemical Equilibrium and Dissociation). Reason 2 — Rising heat capacity. Real $c_{p} (T)$ temperature ke saath badhta hai jab molecular vibration aur (eventually) electronic modes activate hoti hain. Same released energy ab smaller $Δ T$ khareedti hai kyunki har degree zyada joules maangti hai. Humara constant $c_{p} \approx 45 J/mol\cdotK$ true hot-gas capacity ko badly under-count karta hai, isliye real $Δ T$ humari computation se kaafi chhoti hai. Naïve number actually kya represent karta hai. Yeh $T_{a d}$ ka ek strict upper bound hai — woh temperature jo tum tab reach karte agar koi energy dissociation mein leak nahi hoti aur $c_{p}$ temperature ke saath kabhi nahi badhti. Reality hamesha isse neeche hoti hai. Sahi route (temperature-dependent $c_{p} (T)$ plus Gibbs-energy minimisation over all possible product species, e.g. NASA CEA) First Law of Thermodynamics aur Hess's Law and Enthalpy of Formation par tikaa hai. Toh humara $5671 K$ "wrong physics" nahi hai — yeh woh ceiling hai jise true $\approx 3400 K$ kabhi cross nahi kar sakta.

L5.2 — Design judgement: mixture ratio choose karo

H₂/O₂ ke liye stoichiometric O/F 8 hai, phir bhi RS-25 engine O/F ≈ 6 (fuel-rich) par run karta hai. Dikhao ki yeh rich hai, aur ek paragraph mein argue karo ki ek designer temperature peak se deliberately door kyun rehta hai.

Recall Solution

Kya yeh rich hai? Actual $(O / F)_{a c t} = 6 < 8 = (O / F)_{s t}$ , isliye $ϕ = (O / F)_{s t} / (O / F)_{a c t} = 8/6 \approx 1.33 > 1$ ⇒ fuel-rich. Leftover H₂ hai. Designer iska choose kyun karta hai? Teen reasons stack hote hain:

Lower molar mass. Excess H₂ ( $M = 2$ ) exhaust mein mix hota hai, average $M$ ko neeche kheenchta hai. Kyunki $v_{e} \propto T_{c} / M$ , lower $M$ $v_{e}$ ko raise kar sakta hai even though $T$ thoda girta hai — Specific Impulse and Exhaust Velocity trade.
Material survival. $ϕ = 1$ ke paas peak $T_{a d}$ us level se zyada ho sakta hai jo chamber walls aur turbine blades tolerate kar sakti hain. $ϕ \approx 1.33$ par back off karna gas ko us level tak cool karta hai jo hardware survive kar sake, throat aur injector face protect karta hai jahan heat flux worst hoti hai.
Kam dissociation loss. Thoda cooler gas kam dissociate karta hai (dekho L5.1), isliye zyada released chemical energy usable directed flow ke roop mein rehti hai instead of tode hue molecules mein lock hone ke. Verdict: designer $T_{c} / M$ ko wall-temperature limit ke subject mein optimise karta hai, raw $T$ ko nahi — yahi reason hai ki real operating point deliberately, sensibly, temperature peak se door hai.

Active Recall

Recall Rapid self-check (hide answers)

Tum kaun sa atom pehle balance karte ho, aur kyun? ::: Woh jo sirf ek product mein appear karta ho (jaise C phir H); oxygen last kyunki woh kai products mein spread hai. $ϕ = 1.33$ ka matlab hai mixture… ::: Fuel-rich hai (excess fuel, stoichiometry se kam oxidizer). Hydrogen/oxygen methane/oxygen ko similar $T$ par bhi kyun beat karta hai? ::: Uske water-only exhaust ka molar mass $M$ kam hai, aur $v_{e} \propto T_{c} / M$ . Naïve $T_{a d}$ ko best describe kiya jaata hai… ::: Upper bound ke roop mein; real $T$ dissociation aur rising $c_{p}$ ki wajah se lower hoti hai. Engine ko temperature peak se door kyun run karate hain? ::: $M$ lower karne ke liye (faster exhaust) aur wall-temperature limit ke andar rehne ke liye.