Combustion thermodynamics — stoichiometry, adiabatic flame temperature: Visual walkthrough

3.3.19 · D2 · Physics › Rocket Propulsion › Combustion thermodynamics — stoichiometry, adiabatic flame t

Step 1 — "Temperature" aur "heat" yahan kya hote hain

KYA HAI. Koi bhi equation se pehle, do everyday words fix karo.

Heat = energy jo isliye flow karti hai kyunki ek cheez doosri se zyada hot hai. Hum ise joules ( $J$ ) mein measure karte hain, jo kisi bhi energy ki same unit hai.
Temperature = gas particles kitni tezi se jiggle karte hain. Zyada hot gas = zyada fast jiggling. Hum ise kelvin ( $K$ ) mein measure karte hain, ek scale jahan $0 K$ matlab "bilkul bhi jiggle nahi" aur room temperature lagbhag $298 K$ hai.

YAHAN SE KYUN SHURU KAREIN. Poori derivation ek sentence hai — "reaction se release hone wali energy product gas ki jiggling energy ban jaati hai" — isliye humein bilkul clear rehna hoga ki "released energy" (heat) aur "kitna hot hai" (temperature) do alag cheezein hain jo gas ke through link hoti hain.

PICTURE. Left pe, slow blue dots = cold gas. Right pe, fast orange dots = hot gas. "heat $Q$ add karo" label wala arrow ek ko doosre mein convert karta hai.

Step 2 — Gas pratek joule mein kitna warm hoti hai: quantity $c_{p}$

KYA HAI. Gas ke $n$ moles mein $Q$ joules heat push karo aur woh $Δ T$ amount se warm ho jaata hai. Proportionality yeh hai: $Q = n c_{p} Δ T$

Term by term, exactly wahan jahan woh rehte hain:

$n$ — moles ki sankhya, yaani kitni gas hai (ek mole $= 6.02 \times 1 0^{23}$ particles).
$c_{p}$ — molar heat capacity at constant pressure: ek mole ko ek kelvin warm karne ke liye zaroori joules, units $J\cdotmol^{- 1} K^{- 1}$ .
$Δ T = T_{final} - T_{start}$ — yeh kitna zyada hot hua.

YEH TOOL KYUN, DOOSRA NAHI. Hum energy ko temperature rise se connect karna chahte hain. Jo single number exactly yeh kaam karta hai woh $c_{p}$ hai — yeh joules aur kelvin ke beech "exchange rate" hai. Hum constant-pressure version use karte hain ( $c_{p}$ , $c_{v}$ nahi) kyunki rocket chamber roughly constant pressure par burn karta hai jab gas nozzle ki taraf stream karti hai.

PICTURE. Ek tall thin gas (small $c_{p}$ ) same heat ke liye steeply shoot karta hai; ek wide gas (big $c_{p}$ ) muskil se warm hoti hai. Har line ki slope hai $1/ (n c_{p})$ .

Step 3 — Energy kahaan se aati hai: reaction ki heat

KYA HAI. Fuel + oxidizer jalane se atoms naye molecules mein rearrange hote hain jinke bonds lower energy par hote hain. Energy ka yeh difference heat ke roop mein release hota hai. Hum ise $Δ H_{r x n}$ kehte hain (reaction ki enthalpy mein change; enthalpy $H$ = constant pressure par cheezoon ka heat content). Ise kaise tabulate kiya jaata hai iske liye Hess's Law and Enthalpy of Formation dekho.

$Δ H_{r x n} = \sum_{prod} n_{i} Δ H_{f, i}^{\circ} - \sum_{react} n_{j} Δ H_{f, j}^{\circ}$

$Δ H_{f}^{\circ}$ — enthalpy of formation: raw elements se ek mole species build karne ki energy. Apni natural state mein elements ka score $0$ hota hai.
$\sum_{prod}$ — sabhi product molecules par add karo, har ek apne mole count $n_{i}$ se weighted hoke.
Minus sign — products minus reactants, kyunki woh difference hi release hua tha.

KYUN. Enthalpy ek state function hai — yeh sirf start aur end ki parwah karta hai, path ki nahi (yeh hai Hess's Law). Isliye hum release ko "final bond energy − initial bond energy" ke roop mein compute kar sakte hain aur messy intermediate flames ko ignore kar sakte hain.

$2 H_{2} + O_{2} \to 2 H_{2} O (g)$ ke liye: $Δ H_{r x n} = 2 (- 241.8) - [2 (0) + 0] = - 483.6 kJ$

Result negative hai — matlab energy chemistry se nikli aur ab gas ko heat karne ke liye free hai.

PICTURE. Ek energy staircase: reactants upar perched hain, products neeche baithey hain, drop labelled $- Δ H_{r x n} = 483.6 kJ$ = deliver ki gayi heat.

Step 4 — Woh ek law jo sab kuch joddta hai: koi heat nahi nikalti

KYA HAI. "Adiabatic" ka matlab hai $Q_{to surroundings} = 0$ : combustion itni fast hoti hai ki heat walls ke through time par leak nahi kar sakti. Constant pressure par, First Law of Thermodynamics tab kehta hai ki pehle ki total enthalpy = baad ki total enthalpy: $H_{reactants} (T_{0}) = H_{products} (T_{a d})$

Left side — sab kuch jo andar aa raha hai, cold start temperature $T_{0}$ par.
Right side — sab kuch jo bahar ja raha hai, hot end temperature $T_{a d}$ par.
Woh equal hain kyunki koi energy boundary cross nahi ki.

YEH TOOL KYUN. First law simply "energy conserved hoti hai" hai. $Q = 0$ aur koi shaft work nahi hone par, ledger mein sirf do columns hain — in aur out — aur unhe match karna zaroori hai. Yeh single equality poore result ka seed hai.

PICTURE. Ek sealed box. Cold reactants left se $T_{0}$ par enter karte hain; hot products right se $T_{a d}$ par nikalte hain. Ek dashed boundary ke saath ek bada red "✗" escaping-heat arrow ke upar: kuch bhi leak nahi hota.

Step 5 — Journey ko do legs mein split karna

KYA HAI. Hum cold reactants se hot products tak ek move mein jump nahi kar sakte, isliye hum ek imaginary two-leg path se route karte hain (legal hai, kyunki enthalpy path-independent hai):

Cold temperature $T_{0}$ par react karo — atoms rearrange hote hain, $- Δ H_{r x n}$ release karte hain.
Products ko heat karo $T_{0}$ se $T_{a d}$ tak, uss released energy ka use karke.

Released = absorbed set karte hue: $reaction se free hui heat - Δ H_{r x n} = product gases dwara absorb ki gayi heat prod \sum n_{i} \int_{T_{0}}^{T_{a d}} c_{p, i} (T) d T$

Left — Step 3 se positive joules ka dher.
$\int_{T_{0}}^{T_{a d}} c_{p, i} d T$ — product species $i$ ke ek mole ko $T_{0}$ se $T_{a d}$ tak warm karne ki heat. Integral isliye aata hai kyunki $c_{p}$ temperature ke saath change ho sakta hai; integral sirf raaste mein kelvin-by-kelvin heat add karta hai.
$\sum n_{i}$ — yeh har product species ke liye karo aur add karo.

INTEGRAL KYUN. Agar $c_{p}$ ek fixed number hota, hum multiply karte. Lekin bucket size badhti hai jab gas heat hoti hai (Step 7), isliye hum thin slices $c_{p} d T$ ka stack karte hain — woh stacking-up-of-slices exactly wahi hai jo integral hai.

PICTURE. Ek two-arrow detour: energy staircase se neeche (cold react karo), phir temperature ramp se upar (products ko heat karo). Direct diagonal jaisi same destination.

Step 6 — $T_{a d}$ solve karna (simple constant- $c_{p}$ case)

KYA HAI. Maan lo $c_{p}$ constant hai (pehla estimate). Tab integral ek plain multiplication mein collapse ho jaata hai: $- Δ H_{r x n} = (\sum_{i} n_{i} c_{p, i}) (T_{a d} - T_{0})$ Unknown ke liye solve karo: $T_{a d} = T_{0} + \frac{- Δ H _{r x n}}{\sum _{i} n _{i} c _{p, i}}$

$T_{0}$ — jahan se shuru kiya.
$- Δ H_{r x n}$ — free hui energy (top, push).
$\sum_{i} n_{i} c_{p, i}$ — sabhi products ka total bucket size (bottom, resistance).

H₂/O₂ mein plug karo → 2 mol water, $c_{p} \approx 45 J\cdotmol^{- 1} K^{- 1}$ : $T_{a d} = 298 + \frac{483600}{2 \times 45} = 298 + 5373 \approx 5670 K$

YEH AISA KYUN LAGTA HAI. Upar zyada energy → zyada hot. Neeche zyada gas ya bade buckets → same energy thinly spread hoti hai → cooler. Isliye parent note "Hot ÷ Heavy" preaches karta hai: yehi maths Specific Impulse and Exhaust Velocity govern karta hai.

PICTURE. Fraction ek see-saw ke roop mein draw kiya gaya: left pan par energy temperature ko lift karti hai; right pan par total heat capacity use hold karta hai.

Step 7 — Reality check: $5670 K$ ek jhooth kyun hai

KYA HAI. Real H₂/O₂ chambers sirf $\approx 3000$ – $3500 K$ tak pahunchte hain, $5670 K$ nahi. Do effects difference chura lete hain:

$c_{p}$ $T$ ke saath badhta hai. Hot molecules energy store karne ke extra ways unlock kar lete hain (vibration). Bada bucket → same joules mein kam temperature rise.
Dissociation. $\sim 2500 K$ se upar water phatt jaata hai: $H_{2} O \to OH + H$ . Bonds todna energy absorb karta hai — woh energy jo warna jiggling kar rahi hoti. Dekho Chemical Equilibrium and Dissociation.

Dono effects $T_{a d}$ ko cap karte hain. Naive number sirf ek upper bound hai.

KYUN MATTER KARTA HAI. Ek rocket designer jo $5670 K$ trust karta hai galat wall material aur galat nozzle pick karta hai. Honest tool hai temperature-dependent $c_{p} (T)$ plus equilibrium chemistry (NASA CEA).

PICTURE. "Energy absorbed vs temperature" ke do curves: seedhi naive line $5670 K$ tak shoot karti hai; real curve upar bend karti hai (rising $c_{p}$ ) aur jump karti hai (dissociation), fixed released-energy line ko bahut pehle cross karti hai, $3200 K$ ke paas.

Ek-picture summary

KYA HAI. Ek diagram saaton steps compress karta hai: reactants cold enter karte hain, chemistry ek fixed energy ka dher release karti hai, woh dher product gases ke bucket mein pour hota hai, level $T_{a d}$ tak rise karta hai — aur real level growing $c_{p}$ aur dissociation se neeche khicha jaata hai.

Recall Feynman retelling — plain words mein walkthrough

Ek sealed box imagine karo jisme ek fixed pile of "heat coins" hain jo fire fuel aur oxygen ke milte hi de deti hai (Step 3, energy staircase). Kyunki box sealed hai aur ek flash mein burn hota hai, ek bhi coin nahi nikalti (Step 4). Toh saari coins andar ke smoke ki "hotness" kharidne mein kharch honi chahiyein. Ek degree per kitni coins? Yahi bucket size $c_{p}$ hai — chota bucket matlab har coin bahut saare degrees kharidti hai, bada bucket matlab degrees dheere aate hain (Step 2). Paise ko do steps se route karo: pehle chemistry ko cold hote hue hone do, phir har coin smoke ko heat karne mein kharch karo (Step 5). Coins ko bucket size se divide karo aur temperature mil jaati hai (Step 6) — yahi see-saw hai. Lekin ek catch hai: jab smoke scorching hot hoti hai, bucket phool jaata hai (molecules energy chhupane ke naye tarike dhundh lete hain) aur kuch molecules khud ko phad dete hain, jisme coins kharch hoti hain. Toh real temperature tumhare pehle greedy guess se kaafi cooler hoti hai (Step 7). Aur agar tum kuch nahi jalate, ya fire ko extra gas mein duba dete ho, toh important stuff ka zero coins per degree milta hai — no heat, no rise (degenerate cases). Yahi poori flame-temperature story hai: fixed coins ÷ growing bucket = flame temperature.

Recall Khud test karo

Step 5 mein plain product ki jagah integral kyun aata hai?
$c_{p}$ ek sentence mein kya hai, aur "constant pressure" kyun?
$5670 K$ upper bound kyun hai, answer kyun nahi?
Jab tum bahut zyada excess oxidizer add karte ho toh $T_{a d}$ ka kya hota hai, aur kyun?