Adiabatic flame temperature — calculation with enthalpies of formation

5.3.2 · Chemistry › Combustion Chemistry (Propulsion Bridge)

1. Core idea — ek energy bookkeeping problem

Hum kya conserve karte hain: constant pressure par enthalpy. Ek constant-pressure, adiabatic, no-work process ke liye First Law deta hai:

$Q = Δ H = 0 \Rightarrow H_{products} (T_{ad}) = H_{reactants} (T_{in})$

2. First principles se derivation

Har species ki enthalpy ko do hisson mein tod do (yeh key trick hai):

$H_{i} (T) = chemical, at 298 K Δ H_{f, i}^{\circ} + sensible heat \int_{298}^{T} C_{p, i} d T$

Yeh split kyun? Hum sirf $Δ H_{f}^{\circ}$ ko ek reference (298 K) par tabulate karte hain. Baaki sab "yeh molecule 298 K se kitna zyada hot hai" hai, jo sensible heat hai.

$H_{prod} = H_{react}$ apply karo. Reactants ko $T_{in} = 298$ K par enter karte maan lo, toh unka sensible term zero ho jaata hai:

= \sum_{\text{react}} n_i\,\Delta H^\circ_{f,i}$$ Rearrange karo — formation enthalpies ko ek taraf group karo: $$\boxed{\;\sum_{\text{prod}} n_j\!\int_{298}^{T_{\text{ad}}} C_{p,j}\,dT \;=\; -\,\Delta H^\circ_{\text{rxn}}\;}$$ jahan $\displaystyle \Delta H^\circ_{\text{rxn}} = \sum_{\text{prod}} n_j\Delta H^\circ_{f,j} - \sum_{\text{react}} n_i\Delta H^\circ_{f,i}$. > [!formula] Master equation > $$\sum_{\text{prod}} n_j \int_{298}^{T_{\text{ad}}} C_{p,j}\,dT = -\Delta H^\circ_{\text{rxn}}$$ > **Isko aise padho:** "Products dwara soakup ki gayi sensible heat (left) = woh heat jo reaction release karti agar 298 K tak cool kiya jaata (right)." > Constant average $C_p$ ke saath: $\;\sum_j n_j \bar C_{p,j}\,(T_{\text{ad}} - 298) = -\Delta H^\circ_{\text{rxn}}$. ![[5.3.02-Adiabatic-flame-temperature-—-calculation-with-enthalpies-of-formation.png]] --- ## 3. HOW — ek recipe 1. **Balance** karo combustion equation (koi bhi excess air/inert $N_2$ include karo — woh heat absorb karte hain!). 2. $\Delta H^\circ_f$ values se $\Delta H^\circ_{\text{rxn}}$ compute karo. 3. Products ki sensible heat = $-\Delta H^\circ_{\text{rxn}}$ set karo. 4. **$T_{\text{ad}}$ ke liye solve karo** (aksar iterate karna padta hai kyunki $C_p$ temperature $T$ ke saath badhta hai). --- ## 4. Worked Example A — Hydrogen stoichiometric O₂ mein $$\mathrm{H_2(g) + \tfrac12 O_2(g) \to H_2O(g)}, \quad \Delta H^\circ_f(\mathrm{H_2O,g}) = -241.8\ \text{kJ/mol}$$ - **Gas-phase water kyun?** Flame temperatures par paani vapour hota hai; liquid value use karne se latent heat galat tarike se add ho jaati. $\Delta H^\circ_{\text{rxn}} = -241.8 - 0 - 0 = -241.8$ kJ (elements ka $\Delta H^\circ_f = 0$ hota hai). Sirf ek product: 1 mol H₂O. Maan lo $\bar C_p(\mathrm{H_2O,g}) \approx 0.045$ kJ/mol·K. **Yeh step kyun?** Saari released energy sirf us 1 mol steam ko hi heat karti hai. $$1(0.045)(T_{\text{ad}} - 298) = 241.8 \Rightarrow T_{\text{ad}} \approx 298 + 5373 \approx 5670\ \text{K (unrealistically high)}$$ > [!mistake] Pure-O₂ number *bahut zyada high* hai > **Kyun sahi lagta hai:** algebra clean hai. **Fix:** >2500 K par, products **dissociate** ($\mathrm{H_2O \rightleftharpoons OH + H...}$), energy absorb karte hain aur real AFT ko ~3000 K ke paas cap kar dete hain. Constant-$C_p$ + no-dissociation hamesha *overestimate* karta hai. --- ## 5. Worked Example B — Methane in air (zyada realistic) $$\mathrm{CH_4 + 2O_2 + 7.52\,N_2 \to CO_2 + 2H_2O(g) + 7.52\,N_2}$$ - **$7.52\,N_2$ kyun?** Air mein ~79% N₂ / 21% O₂ hota hai, isliye har mol O₂ ke saath $79/21 = 3.76$ mol N₂ aata hai; 2 O₂ ke liye woh $7.52$ hai. $\Delta H^\circ_f$: $\mathrm{CH_4}=-74.8$, $\mathrm{CO_2}=-393.5$, $\mathrm{H_2O(g)}=-241.8$ kJ/mol. $$\Delta H^\circ_{\text{rxn}} = [-393.5 + 2(-241.8)] - [-74.8] = -877.1 - (-74.8) = -802.3\ \text{kJ}$$ Product heat capacities (approx, kJ/mol·K): $C_p(\mathrm{CO_2})\approx0.055$, $C_p(\mathrm{H_2O})\approx0.045$, $C_p(\mathrm{N_2})\approx0.033$. $$\sum n_j \bar C_{p,j} = 1(0.055) + 2(0.045) + 7.52(0.033) = 0.055 + 0.090 + 0.248 = 0.393\ \text{kJ/K}$$ **N₂ kyun include karo?** Yeh ek bada thermal sponge hai — isi liye **air flames, O₂ flames se bahut zyada cooler hoti hain**. $$T_{\text{ad}} = 298 + \frac{802.3}{0.393} \approx 298 + 2042 \approx 2340\ \text{K}$$ (Real value ~2230 K jab $T$-dependent $C_p$ aur minor dissociation include karo — kaafi close!) > [!example] Quick sanity contrast > N₂ hata do (CH₄ ko pure O₂ mein jalao): denominator gir ke $0.145$ ho jaata hai → $T_{\text{ad}} \approx 298 + 5530 \approx 5800$ K. Sirf N₂ ne AFT ko hazaron kelvin kam kar diya. **Yahi dual-coding takeaway hai: diagram mein "N₂ sensible heat" ka bar sabse tall hai.** --- ## 6. Forecast-then-Verify > [!recall] Compute karne se pehle predict karo > Q: Agar main methane ko **excess air** (lean mixture) ke saath jalata hoon, toh AFT upar jaayegi ya neeche? > > Forecast → **neeche**: extra cold air/N₂ product moles add karta hai jinhein heat karna padta hai lekin koi extra energy release nahi hoti → bada denominator, same numerator → lower $T_{\text{ad}}$. --- ## 7. Common mistakes (Steel-manned) > [!mistake] Liquid-water $\Delta H^\circ_f$ use karna > **Sahi lagta hai:** $-285.8$ kJ/mol "famous" number hai (HHV). **Fix:** flame products vapour hote hain → $-241.8$ use karo (LHV basis). Liquid value released heat ko overstate karta hai. > [!mistake] Nitrogen ko product maan ke bhool jaana > **Sahi lagta hai:** N₂ "react nahi karta." **Fix:** woh react nahi karta lekin *heat up* hota hai, energy soak karta hai. Isse omit karo toh $T_{\text{ad}}$ massively overestimate ho jaata hai. > [!mistake] $C_p$ ko constant maanna > **Sahi lagta hai:** simple algebra. **Fix:** 300→2500 K se $C_p$ ~30–50% badhta hai. Constant-$C_p$ se $T_{\text{ad}}$ overestimate hota hai; precision ke liye $C_p(T)$ ya enthalpy tables use karke iterate karo. --- ## 8. Flashcards #flashcards/chemistry Constant-pressure adiabatic flame ke liye kaunsi thermodynamic quantity conserve hoti hai? ::: Total enthalpy: $H_{\text{prod}}(T_{ad}) = H_{\text{react}}(T_{in})$, i.e. $\Delta H = 0$. Har species ki enthalpy ko hum kitne hisson mein split karte hain? ::: Formation enthalpy at 298 K + sensible heat $\int_{298}^T C_p\,dT$. Real AFT, constant-$C_p$ estimate se lower kyun hoti hai? ::: Dissociation of products energy absorb karta hai aur $C_p$ temperature ke saath badhta hai. Air mein 1 mol O₂ ke saath kitne mol N₂ hote hain? ::: 3.76 mol (79/21). Gas-phase (liquid nahi) water ka $\Delta H^\circ_f$ kyun use karte hain? ::: Flame temperature par products vapour hote hain; liquid value unphysical latent heat add kar deti. Kya excess air AFT badhata ya ghataata hai? ::: Ghataata hai — inert N₂/air heat absorb karta hai bina koi chemical energy add kiye. AFT ke liye master equation words mein? ::: Sensible heat soaked by products = $-\Delta H^\circ_{rxn}$. Standard state mein kisi bhi element ka $\Delta H^\circ_f$? ::: Zero. --- > [!recall]- Feynman: ek 12-saal ke bachche ko samjhao > Ek patakha socho. Gunpowder ke bonds mein energy lock hai. Jab woh darwaza band karke (koi heat bahar nahi jaati) jalta hai, tab saari energy ko kahin nahi jaana hota siwaye smoke aur gas ko *super hot* banane ke. Jis temperature par woh pahunchte hain woh "adiabatic flame temperature" hai. Agar tum usme bahut saari thandi hawa milao, woh hawa bhi garam karni padegi — jaise hot soup mein thanda paani daalte ho — toh sab kuch zyada thanda ho jaata hai. Isliye pure oxygen use karne wale rockets, aam hawa use karne walon se kaafi zyada hot burn karte hain. > [!mnemonic] Steps yaad karo > **"BRSS"** — **B**alance, **R**eaction-enthalpy, **S**ensible-heat-balance, **S**olve. > Aur: *"Nitrogen ek silent sponge hai."* ## Connections - [[Enthalpy of formation]] - [[Hess's Law]] - [[Heat capacity Cp and its temperature dependence]] - [[Chemical equilibrium and dissociation at high temperature]] - [[Stoichiometric vs lean vs rich mixtures]] - [[Rocket propulsion — specific impulse]] - [[First Law of Thermodynamics]] ## 🖼️ Concept Map ```mermaid flowchart TD A[Fuel burns, bonds release energy] B[Adiabatic Q=0, no work] C[First Law dH=0] D[H products = H reactants] E[Split Hi into parts] F[Formation enthalpy at 298K] G[Sensible heat integral Cp dT] H[Master equation] I[Delta H rxn from formation enthalpies] J[Adiabatic flame temperature Tad] K[Propulsion thrust ceiling] A -->|energy stays inside| B B -->|constant P, no work| C C -->|implies| D D -->|apply per species| E E -->|chemical part| F E -->|thermal part| G F -->|combine reactants minus products| I G -->|sum over products| H I -->|equals negative of| H H -->|solve for| J J -->|sets limit for| K ```