Real gas effects — dissociation, recombination: Visual walkthrough

3.3.20 · D2 · Physics › Rocket Propulsion › Real gas effects — dissociation, recombination

Step 1 — Molecule aur usse banne wale atoms ko draw karo

KYA. Ek molecule imagine karo jo do identical halves se mila hua hai jo saath stuck hain. Poore molecule ko $A_{2}$ kehte hain ("A-two" = do A's bonded). Agar hum ise alag karo toh do alag $A$ atoms milenge. Hum is do-taraf wale raaste ko likhte hain:

$A_{2} ⇌ 2 A$

$A_{2}$ — ek bonded molecule (do atoms haath pakde hue).
$2 A$ — do free atoms (haath chod diye).
$⇌$ — double arrow: reaction dono taraf chalta hai. Left-to-right dissociation hai (todna); right-to-left recombination hai (wapas banana).

KYUN. Isse pehle ki hum "kitna toota" ki baat karein, hume woh cheez fix karni hogi jo toot rahi hai aur woh tukde jo bante hain. Aage ka sab kuch inhi do boxes ko count karta hai: molecules left mein, atoms right mein.

PICTURE. Left cup mein ek joined dumbbell hai; right cup mein do lone balls hain. Unke beech ki spring chemical bond hai — aur springs energy store karte hain.

Step 2 — "Kitna toota" ko ek number $α$ se measure karo

KYA. Molecules ka ek batch lo. Unka kuch fraction toot jaata hai. Hum us fraction ko $α$ (Greek "alpha") kehte hain:

$α = \frac{moles of A _{2} that broke}{moles of A _{2} we started with}, 0 \leq α \leq 1$

Upar toote hue molecules count hote hain.
Neeche shuruat mein jo tha sab count hota hai.
$α = 0$ → kuch nahi toota (saare molecules intact).
$α = 1$ → sab kuch toot gaya (saare atoms free).

KYUN. Hume ek single dial chahiye jo bole "reaction kitni aage gayi." Ek number se hum energy, temperature, aur average particle kitna bhaari hai — sab ek saath track kar sakte hain, bina alag-alag molecule counts juggle kiye.

PICTURE. $0$ se $1$ tak ek slider. Jaise hum ise right ki taraf kheenchte hain, intact molecules ka bar chhota hota jaata hai aur free atoms ka bar bada — particles ki total count badhti hai, jo hume Step 5 mein chahiye hogi.

Step 3 — Combustion ki heat ko do kamon mein baanto

KYA. Propellant jalane se gas mein ek fixed amount ki heat, $Q_{co mb}$ , jaati hai. Us heat ko exactly do cheezein karni hoti hain:

$Q_{co mb} = Kaam 1: gas ko garam karo n \overset{c}{ˉ}_{p} (T_{c} - T_{0}) + Kaam 2: bonds todo α n Δ H_{d}$

Term by term:

$Q_{co mb}$ — kul chemical heat released, joules mein.
$n$ — shuruat mein gas ke moles ki sankhya.
$\overset{c}{ˉ}_{p}$ — mean specific heat: ek mole ko ek kelvin badhane ke liye joules chahiye.
$T_{c}$ — final chamber temperature jo hum reach karte hain.
$T_{0}$ — shuruat ki temperature.
$(T_{c} - T_{0})$ — humne actually kitne degrees chadhe.
$α n$ — moles jo toote (Step 2 se).
$Δ H_{d}$ — unhe todne ki price per mole (Step 1 se).

KYUN. Energy gayab nahi ho sakti. Ek bond snap karne mein jo joule kharch hua (Kaam 2), woh joule temperature nahi badha paaya (Kaam 1). Yeh single bookkeeping line hi poora reason hai ki dissociation flame ko kyun thanda karta hai.

PICTURE. Heat ki ek incoming pipe do buckets mein split hoti hai. Jab $α$ badhta hai, "break bonds" bucket "heat the gas" bucket se churaata hai — toh thermometer kam padhta hai.

Step 4 — Equilibrium se $α$ ki value chunwao

KYA. $α$ free nahi hai — nature ise settle karta hai. Temperature $T$ par $A_{2} ⇌ 2 A$ ke liye, balance point ek equilibrium constant $K_{p} (T)$ follow karta hai:

$K_{p} (T) = \frac{( p _{A} / p ^{\circ} ) ^{2}}{p _{A_{2}} / p ^{\circ}}$

$p_{A}$ — free atoms ki partial pressure (atoms walls par kitna push karte hain).
$p_{A_{2}}$ — intact molecules ki partial pressure.
$p^{\circ}$ — ek fixed reference pressure (standard pressure, usually 1 bar). Har pressure ko isse divide kiya jaata hai taaki $K_{p}$ ek pure number bane jisme koi units na hon — yahi hume neeche uska logarithm lene deta hai.
Atoms squared hain kyunki reaction ke right side par unke do hain — chemistry har molecule ko ek factor maanti hai.

$K_{p}$ khud thermodynamics se aata hai: $ln K_{p} = - Δ G^{\circ} / (R_{u} T)$ , jahan $Δ G^{\circ}$ measure karta hai ki reaction kitna "jaana chahta hai." Hum $K_{p}$ use karte hain (na ki " $α$ guess karo") kyunki yeh woh ek quantity hai jo ek given temperature par balance point fix karti hai.

KYUN. Akele chhod do toh todna aur banana ek standoff par pahunchte hain: jitne molecules har second snap hote hain, utne hi re-form hote hain. $K_{p}$ us standoff ka mathematical naam hai. $K_{p} = (expression in α)$ solve karne se hume actual $α (T, p)$ milta hai jo Step 3 mein dalna hai.

PICTURE. Do arrows ek loop mein ek doosre ka peecha karte hain — forward (todna) aur backward (banana). Equilibrium par dono equal length hain; loop steady hai. $K_{p}$ ki value batati hai ki standoff kitna ek-taraf wala hai.

Recall Kyun atom pressure squared hai

Kyunki reaction do A atoms per A $_{2}$ banata hai ::: $p_{A}$ ke do identical factors multiply hote hain, jo $p_{A}^{2}$ deta hai; exponent sirf balanced equation mein us species ki count hai.

Step 5 — $K_{p}$ ko $α$ aur $p$ ki equation mein badlo

KYA. Ab particles count karo. 1 mole $A_{2}$ se shuru karo. $α$ fraction tootne ke baad:

intact molecules bache: $1 - α$
free atoms bane: $2 α$
kul particles: $(1 - α) + 2 α = 1 + α$

Har species ki partial pressure hai (crowd mein uska share) × (total pressure $p$ ). In shares ko Step 4 ke $K_{p}$ mein substitute karne par (har pressure $p^{\circ}$ se divide hokar) milta hai:

$K_{p} (T) = \frac{( 2 α ) ^{2}}{( 1 - α ) ( 1 + α )} \frac{p}{p ^{\circ}} = \frac{4 α ^{2}}{1 - α ^{2}} \frac{p}{p ^{\circ}}$

$\frac{2 α}{1 + α}$ — crowd mein atom ka share → squared ho jaata hai.
$\frac{1 - α}{1 + α}$ — molecule ka share → neeche baithta hai.
$p^{\circ}$ — Step 4 wala wahi reference pressure, ab explicitly carry-through kiya gaya.

KYUN. Yeh bridge hai: yeh abstract $K_{p}$ ko concrete dial $α$ se jodata hai aur exactly dikhata hai ki pressure $p$ kahan enter karta hai. Right side par woh $p$ hi hai jo squeezing ko matter banata hai.

PICTURE. Step 2 ka crowd bar, ab apne shares se labelled, $K_{p}$ fraction mein jaata hai. Dekho $p$ upar kaise ride kar raha hai.

Step 6 — Pressure lever padho (Le Chatelier ek picture mein)

KYA. Temperature fixed rakho, toh $K_{p}$ ek constant hai. Mein

$K_{p} = \frac{4 α ^{2}}{1 - α ^{2}} \frac{p}{p ^{\circ}}$

agar hum $p$ double karein, toh $K_{p}$ same rakhne ke liye fraction $\frac{α ^{2}}{1 - α ^{2}}$ ko half hona padega — jo $α$ ko neeche force karta hai.

KYUN. Dissociation se zyada particles bante hain ( $1 \to 1 + α$ ). Gas ko squeeze karo aur woh kam particles wali side — molecules — ko favour karke react karta hai. Yeh Le Chatelier's Principle quantitative form mein hai, aur isliye real engines high chamber pressure par run hote hain: kam dissociation → zyada garam, zyada complete combustion.

PICTURE. $α$ ka $p$ ke against curve (temperature fixed): yeh neeche aur right ki taraf slide karta hai — zyada pressure, kam tootna.

Step 7 — Nozzle: kya energy wapas aati hai? (frozen vs. equilibrium)

KYA. Gas nozzle se neeche dodte hai aur thandi hoti hai. Thanda hona reaction ko backward chalana chahta hai ( $2 A \to A_{2}$ ), $Δ H_{d}$ ko heat ke roop mein wapas karta hai. Kya iske paas waqt hai yeh Damköhler number decide karta hai:

$D a = \frac{τ _{f l o w}}{τ _{c h e m}} = \frac{time gas spends in the nozzle}{time the reaction needs}$

$τ_{f l o w}$ — gas ka ek parcel nozzle mein kitni der rehta hai (residence time).
$τ_{c h e m}$ — recombination ko finish hone mein kitna waqt lagta hai.
$D a ≫ 1$ → reaction fast hai, gas equilibrium mein rehta hai, energy wapas aati hai → zyada exhaust velocity.
$D a ≪ 1$ → reaction bahut slow hai, composition frozen hai, energy lock rahti hai → kam exhaust velocity.

KYUN. Yeh poore page ka payoff hai. Step 3 ne jo bond energy churai thi, woh tabhi hamesha ke liye kho jaati hai jab recombination apna window miss kar deta hai. Real nozzles dono limits ke beech hote hain, equilibrium se kuch percent neeche. Dekho Damköhler Number aur Nozzle Flow and Exhaust Velocity.

PICTURE. Thande hote nozzle se do paths: equilibrium path atoms ko wapas jodta hai aur uski energy line wapas upar jaati hai; frozen path apne atoms rakhta hai aur uski energy line flat aur neech rehti hai.

Ek-picture summary

Is page ka sab kuch causes ki ek chain hai. Final figure saaton steps ko arrows ke ek single flow ke roop mein redraw karta hai — wohi heat-split bucket Step 3 se, wohi equilibrium loop Step 4 se, wohi pressure curve Step 6 se, aur wohi thanda hota nozzle Step 7 se, sab jode hue hain taaki tum combustion heat ka ek joule flame se exhaust tak ek hi nazar mein trace kar sako.

Recall Feynman: poora walkthrough ek 12-saal ke bachche ko batao

Ek bade crowd ki picture karo jahan har do log haath pakde khade hain — yeh molecules hain. Jab hum patakhe jalate hain (combustion), us patakhe ki energy ka kuch hissa sabko garam karta hai, lekin kuch hissa logon ke haath alag karne mein kharcha hota hai, aur jinhone haath chod diye (atoms) woh akele ghoomte hain. Letter $α$ bas "pairs ka kaunsa fraction alag hua" hai. Kitne alag hue yeh random nahi hai — yeh ek tug-of-war hai jo ek balance point par settle hota hai, aur $K_{p}$ us balance ka naam hai. Agar tum crowd ko saath crush karo (pressure badhao), log phir se haath pakadne par majboor ho jaate hain, kyunki haath pakadne mein kam jagah lagti hai — yeh pressure lever hai. Finally, jab sab log exit tunnel (nozzle) se bahar daud rahe hote hain toh thanda ho jaata hai, aur thande log phir se haath pakadna chahte hain, energy ko darwaze se ek dhakke ke roop mein wapas dete hain. Lekin sirf tabhi jab tunnel itni lambi ho ki unhe time par partner mil sake. Lambi tunnel (bada Damköhler) → energy wapas aati hai → tez rocket. Chhoti tunnel → energy kho jaati hai. Yeh poora story hai, pehle toote bond se lekar aakhri dhakke tak.

Connections

Chemical Equilibrium and Kp
Adiabatic Flame Temperature
Nozzle Flow and Exhaust Velocity
Specific Impulse
Le Chatelier's Principle
Damköhler Number