Combustion thermodynamics — stoichiometry, adiabatic flame temperature: Question bank

3.3.19 · D5 · Physics › Rocket Propulsion › Combustion thermodynamics — stoichiometry, adiabatic flame t

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Koi bhi trap samajhne se pehle, is page par use hone wale har symbol ko pakad lo. Isko skip mat karo — zyaadatar "traps" asal mein symbol confusion ke disguise mein hote hain.

Definition The players (simple words + sign conventions)

Enthalpy $H$ — gas ka total "heat content" bookkeeping variable constant pressure par; joules mein measure hota hai. Hum sirf $H$ ke differences se matlab rakhte hain, kabhi bhi uski absolute value se nahi.
Standard enthalpy of formation $Δ H_{f}^{\circ}$ — reference state ( $298 K$ , $1 atm$ ) par pure elements se 1 mole species banane ka enthalpy change. Elements apni standard state mein convention se $Δ H_{f}^{\circ} = 0$ assign kiye jaate hain. Negative value matlab heat release hoti hai jab species banta hai (woh energy ladder par neeche hota hai).
Enthalpy of reaction $Δ H_{r x n}$ — (products ki formation enthalpies) minus (reactants ki formation enthalpies). Negative $Δ H_{r x n}$ = exothermic (heat released); released heat positive quantity $- Δ H_{r x n}$ hoti hai.
Specific heat $c_{p}$ — 1 mole gas ko $1 K$ raise karne ke liye required joules constant pressure par. Ye temperature ke saath badhti hai jab vibrational modes switch on hoti hain.
Molar mass $M$ — product gas mixture ke ek mole ki mass (kg mein). Halke products = chhota $M$ .
Exhaust velocity $v_{e}$ — nozzle se gas nikalne ki speed; ye $v_{e} \propto T_{c} / M$ ke scale par hoti hai. Dekho Specific Impulse and Exhaust Velocity.

Formula Energy balance jo

$T_{a d}$ ko define karti hai (kyun $Q = 0$ , $p = const$ matter karta hai) Pehle assumption upar se rakho: chamber mein combustion adiabatic ( $Q = 0$ ) hai aur constant pressure par. Constant pressure par first law kehta hai ki exchanged heat enthalpy change ke barabar hoti hai, toh $Q = 0 \Rightarrow Δ H = 0$ poore process ke liye — total enthalpy conserved hoti hai: $H_{r e a c t} (T_{0}) = H_{p r o d} (T_{a d})$ Ab har side ko "species ko $T_{0}$ par form karo" plus "isse $T_{0}$ se uski temperature tak heat karo" mein split karo. Reactants $T_{0}$ par enter karte hain, toh unka heating term zero hai: $\sum_{r e a c t} n_{j} Δ H_{f, j}^{\circ} = \sum_{p r o d} n_{i} Δ H_{f, i}^{\circ} + \sum_{p r o d} n_{i} \int_{T_{0}}^{T_{a d}} c_{p, i} (T) d T$ Rearrange karne par, aur $Δ H_{r x n} = \sum_{p r o d} n_{i} Δ H_{f, i}^{\circ} - \sum_{r e a c t} n_{j} Δ H_{f, j}^{\circ}$ pehchankar: $- Δ H_{r x n} = p r o d \sum n_{i} \int_{T_{0}}^{T_{a d}} c_{p, i} (T) d T$ Reading: chemistry se released heat (left, positive) = product gases ki temperature $T_{a d}$ tak chadne mein absorb hone wali heat (right). Ye woh equation hai jis par neeche har trap quietly rely karta hai. Split ki step-by-step picture neeche hai.

Figure — Combustion thermodynamics — stoichiometry, adiabatic flame temperature

True or false — justify karo

Stoichiometric mixture hamesha sabse hot burn karta hai.

Almost sahi lekin exactly nahi — peak flame temperature kai fuels ke liye

ϕ = 1

se thodi rich side par hoti hai kyunki dissociation aur product composition optimum ko shift karte hain; safe statement hai "near stoichiometric," na ki "exactly at it."

Rocket chamber mein combustion adiabatic hai kyunki walls perfect insulators hain.

False — walls actively cooled hoti hain, insulating nahi; ye adiabatic isliye hai kyunki reaction itni fast hoti hai ki burn ke dauran negligible heat nikalta hai (

Q \approx 0

timescale se, insulation se nahi).

Kyunki

Δ H_{r x n}

negative hai, products reactants se lower enthalpy par end hote hain.

Adiabatic case ke liye False — chemical enthalpy drop hoti hai, lekin woh energy products ki temperature raise karne mein jaati hai, toh total enthalpy

H_{r e a c t} (T_{0}) = H_{p r o d} (T_{a d})

conserved hoti hai.

Stoichiometric se zyada fuel add karne par hamesha released heat badhti hai, toh

T_{a d}

bhi badhna chahiye.

False — extra fuel react nahi karta (uske liye koi oxidizer nahi bacha); ye inert mass ban jaata hai jo heat absorb karta hai, toh

T_{a d}

actually peak ke baad girta hai.

Elements apni standard state mein

Δ H_{f}^{\circ} = 0

measurement se hain.

False — ye ek convention hai (enthalpy scale ka chosen zero), measured value nahi; sirf enthalpy ke differences physical hain.

Naïve constant-

c_{p}

estimate of

T_{a d}

real value ka lower bound hai.

False — ye upper bound hai; dissociation aur

c_{p}

ka temperature ke saath badhna ignore karna dono milke true

T_{a d}

ko estimate se neeche le jaate hain.

Dissociation rocket ke liye har tarah se buri hai.

False — dissociation

T_{a d}

ko lower karti hai (ek loss), lekin ye halki species bhi produce karti hai jo mean molar mass

M

reduce karti hai, aur nozzle mein ye recombine ho sakti hain energy release karke; net effect subtle hai. Dekho Chemical Equilibrium and Dissociation.

Fuel-rich chalaane se flame temperature girta hai, toh exhaust velocity bhi hamesha girni chahiye.

False —

v_{e} \propto T_{c} / M

; fuel-rich

M

bhi lower karta hai, aur

M

ka drop

T

ke drop ko outweigh kar sakta hai, jisse higher

v_{e}

milta hai.

Error dhundo

2 H_{2} + O_{2} \to 2 H_{2} O

, toh O/F mass ratio

1/2

hai kyunki do

H_{2}

par ek

O_{2}

hai."

Error ye hai ki mole ratio ko mass ratio ki jagah use kiya ja raha hai; molar mass se weight karna padega:

(1 \times 32) / (2 \times 2) = 8

, na ki

1/2

"Combustion exothermic hai, toh

Δ H_{r x n} > 0

Sign error — exothermic matlab heat release hoti hai, jo

Δ H_{r x n} < 0

banata hai; released heat

- Δ H_{r x n} > 0

hai.

T_{a d}

nikaalне ke liye, product enthalpy se reactant enthalpy subtract karo aur zero set karo."

Incomplete — har species ko formation enthalpy plus sensible-heat integral

\int_{T_{0}}^{T_{a d}} c_{p} d T

mein split karna hoga; heating term ke bina solve karne ke liye koi temperature hi nahi hogi.

"Reader ne reactants ke liye bhi

c_{p}

T_{0}

T_{a d}

tak integrate kar diya."

Error — reactants

T_{0}

par enter karte hain, toh unka sensible-heat term zero hai; sirf products

T_{a d}

tak heat hote hain.

T_{a d}

5670 K

nikla, toh H₂/O₂ engine actually itna hot run karta hai."

Constant-

c_{p}

, no-dissociation wala number ek naïve ceiling hai; dissociation aur rising

c_{p}

include karne par real value

\approx 3000

–

3500 K

hai.

"Equivalence ratio

ϕ = 2

matlab stoichiometric se double oxidizer hai."

Ulta hai —

ϕ = (F / O)_{a c t u a l} / (F / O)_{s t}

, toh

ϕ > 1

matlab excess fuel (fuel-rich) hai, yaani stoichiometric se kam oxidizer, zyada nahi.

"Hum yahan Hess's law use kar sakte hain sirf tabhi jab hum flame ke andar exact reaction path jaante hain."

Hess's law ka poora point yahi hai ki enthalpy ek state function hai — ye path-independent hai, toh beech ke messy intermediate steps matter nahi karte.

Why questions

High

T_{a d}

alone design goal kyun nahi hai?

Kyunki exhaust speed

T_{c} / M

ke scale par hoti hai — tum

M

(molar mass) lower karke temperature ke last few hundred kelvin chase karne se zyada gain kar sakte ho. Details mein Specific Impulse and Exhaust Velocity.

Combustion products ke liye

c_{p}

temperature ke saath kyun badhta hai?

Higher

T

par molecules ke vibrational modes thermally activate ho jaate hain aur energy store karne lagte hain, toh temperature ke har degree ke liye zyada heat chahiye — extra internal degrees of freedom ka direct consequence.

Governing equation "chemistry se released heat" ek side aur "gases dwara absorbed heat" doosri side par kyun rakhti hai?

Kyunki constant pressure par

Q = 0

ke saath total enthalpy conserved hoti hai, toh bonds rearrange hone se har joule jo free hoti hai woh product gas mein sensible heat ke roop mein wapas aani chahiye — First Law of Thermodynamics se energy-conservation bookkeeping.

Hum formation enthalpies ko fixed reference (

298 K

1 atm

) par kyun evaluate karte hain, flame temperature par kyun nahi?

Kyunki tabulated

Δ H_{f}^{\circ}

us reference par defined hain; hum phir species ko sensible-heat integral use karke

T_{a d}

tak transport karte hain, chemistry aur heating ko cleanly alag rakhte hue.

Naïve estimate methane-in-air (

\approx 2200 K

) ke liye acceptable kyun hai lekin H₂/O₂ ke liye buri kyun hai?

2500 K

se neeche dissociation negligible hai aur

c_{p}

thoda vary karta hai, toh constant-

c_{p}

theek hai; H₂/O₂ us threshold se kaafi aage jaata hai jahan dono effects hard bite karte hain.

Real

T_{a d}

solve karne ke liye enthalpy balance se zyada Gibbs free energy minimize karna kyun zaroori hai?

Kyunki product composition khud unknown hai (kitna dissociate hota hai ye

T

par depend karta hai), toh chemistry aur temperature ko simultaneously equilibrium via solve karna padta hai — Chemical Equilibrium and Dissociation ka subject.

Nozzle mein recombination overall performance ke liye kyun matter karta hai?

Dissociated fragments stable molecules dobara form kar sakte hain jab gas expand hote waqt cool hoti hai, energy release karke jo exhaust ki kinetic energy mein add hoti hai — ek bonus jo frozen-flow assumptions miss kar dete hain. Dekho Nozzle Theory and Isentropic Expansion.

Edge cases

ϕ \to 0

par

T_{a d}

ka kya hota hai (extremely lean, almost pure oxidizer)?

Ye initial temperature ki taraf plummet karta hai — energy release karne ke liye almost koi fuel nahi hai, aur vast excess oxidizer ek cold heat sink ki tarah kaam karta hai.

ϕ \to \infty

par

T_{a d}

ka kya hota hai (almost pure fuel)?

Ye bhi

T_{0}

ki taraf collapse karta hai — negligible oxidizer matlab negligible reaction, aur excess fuel sirf thodi si release hone wali heat absorb kar leta hai.

Agar reactants

298 K

se pre-heated hoke enter karein, toh

T_{a d}

ka kya hota hai?

Ye badhta hai — reactants already extra sensible heat carry karte hain, toh wahi chemical release unhe higher final temperature par land karti hai (LHS gain karta hai, products warmer start karte hain).

Ek reaction ke liye jo water ko liquid produce karti hai gas ki jagah, temperature raise karne ke liye zyada ya kam heat available hai?

Nominally zyada heat release hoti hai (condensation latent enthalpy add karta hai), lekin flame temperatures par water hamesha vapour hoti hai, toh gas-phase

Δ H_{f}^{\circ}

physically relevant hai — liquid value use karna

T_{a d}

ko overstate karta.

Exact stoichiometric point par, heat absorb karne ke liye koi leftover reactant hota hai?

Nahi — by definition dono fuel aur oxidizer fully consumed ho jaate hain, yehi wajah hai ki temperature waahan apne maximum ke qareeb hoti hai. Dekho Propellant Selection.

Agar tum

T_{c}

raise karte raho aur

M

fixed rahe, toh

v_{e}

ka limiting behaviour kya hai?

T_{c}

ki tarah badhta hai — diminishing returns, kyunki temperature double karne se exhaust speed sirf

2

se multiply hoti hai; isliye

M

lower karna often better lever hota hai.

Active Recall

Recall One-line self-check

Exothermic burn mein $Δ H_{r x n}$ ka sign? ::: Negative; released heat $- Δ H_{r x n}$ hai.
Naïve $T_{a d}$ upper bound kyun hai? ::: Ye endothermic dissociation aur $c_{p}$ ka temperature ke saath badhna ignore karta hai, dono products ko estimate se neeche cool karte hain.
$ϕ > 1$ rich hai ya lean? ::: Fuel-rich — excess fuel, stoichiometric se kam oxidizer.
Kaunse do assumptions $T_{a d}$ derivation mein total enthalpy conserved banate hain? ::: Adiabatic ( $Q = 0$ ) aur constant pressure, toh $Δ H = Q = 0$ .