Stoichiometric vs fuel-rich vs fuel-lean combustion

5.3.1 · Chemistry › Combustion Chemistry (Propulsion Bridge)

1. Master quantity: Air–Fuel Ratio (AFR)

WHY ek reference chahiye — AFR_st ko scratch se derive karo

Ek general hydrocarbon $C_{x} H_{y}$ lo. Complete combustion ke liye WHAT sach hona chahiye? Har C ko CO₂ banna chahiye, har H ko H₂O banna chahiye. Step by step balance karo:

$C_{x} H_{y} + a O_{2} \to x C O_{2} + \frac{y}{2} H_{2} O$

Yeh step WHY? Carbon balance $x$ CO₂ force karta hai; hydrogen balance $\frac{y}{2}$ H₂O force karta hai (kyunki har H₂O mein 2 H atoms hote hain).

Ab oxygen balance karo. Right side O atoms $= 2 x + \frac{y}{2}$ . Left side $= 2 a$ . Toh:

$2 a = 2 x + \frac{y}{2} \Rightarrow a = x + \frac{y}{4}$

Yeh step WHY? Oxygen conserved hoti hai; dono sides ke atoms barabar karo aur O₂ ke moles ke liye solve karo.

Air 21% O₂ aur 79% N₂ by mole hai, isliye har mole O₂ ke saath $79/21 = 3.76$ mol N₂ bhi aata hai:

=\frac{a\,(1+3.76)\;\times\;28.96}{1\times M_{\text{fuel}}}$$ > [!formula] Stoichiometric coefficient & AFR > $$a = x+\frac{y}{4},\qquad > \text{AFR}_{\text{st}}=\frac{4.76\,a\cdot 28.96}{M_{C_xH_y}}\ \frac{\text{kg air}}{\text{kg fuel}}$$ > ($28.96$ g/mol = air ka mean molar mass, $4.76 = 1+3.76$ total mol air per mol O₂.) --- ## 2. Equivalence ratio $\phi$ — dimensionless dial > [!definition] Equivalence ratio > $$\phi = \frac{\text{AFR}_{\text{st}}}{\text{AFR}_{\text{actual}}} > = \frac{(F/A)_{\text{actual}}}{(F/A)_{\text{st}}}$$ > jahan $F/A$ fuel–air ratio hai (AFR ka inverse). $\phi$ define WHY karte hain? Kyunki raw AFR fuel par depend karta hai; $\phi$ normalize karta hai taaki **ek number har fuel describe kar sake**. | Regime | $\phi$ | AFR vs AFR_st | Matlab | |---|---|---|---| | **Lean** | $\phi<1$ | AFR **bada** | excess O₂ | | **Stoichiometric** | $\phi=1$ | AFR = AFR_st | perfect | | **Rich** | $\phi>1$ | AFR **chhota** | excess fuel | > [!intuition] Dial padhna > $\phi$ ka matlab literally hai "==kitni baar stoichiometric *fuel* daali maine==". > $\phi=2$ → double fuel jo air handle kar sakti hai → bahut rich. ![[5.3.01-Stoichiometric-vs-fuel-rich-vs-fuel-lean-combustion.png]] --- ## 3. Tailpipe se kya niklega (HOW products shift karte hain) > [!intuition] Rich CO & soot kyun banata hai, lean "clean but cool" kyun hota hai > Oxygen *limiting reagent* hoti hai jab rich condition ho. Itna O nahin hota ki carbon ko > CO₂ tak fully oxidize kar sake, toh yeh **CO** par ruk jaata hai (aur unburnt fuel/soot). Jab lean hota hai toh *spare* > O₂ hoti hai jo bas pass-through ho jaati hai, flame ko dilute aur cool karti hai. - **Lean ($\phi<1$):** products = CO₂, H₂O, **leftover O₂**, N₂. Cooler flame, kam CO/soot, lekin high O₂+high T near $\phi\approx0.9$ se **NOₓ** banta hai. - **Stoichiometric ($\phi=1$):** *sabse hot* flame (saari chemical energy release, koi diluent excess nahin) — peak adiabatic flame temperature. - **Rich ($\phi>1$):** products mein **CO, H₂, soot, unburnt $C_xH_y$** shamil hote hain; flame phir se cooler hoti hai kyunki incomplete oxidation kam heat release karti hai. > [!example] Rich methane combustion (mass-balanced idea) > $\phi>1$ ke liye aap sirf $a/\phi$ moles of O₂ supply karte ho. Carbon CO₂ aur CO ke beech share hota hai taaki > available O exactly consume ho jaye — isliye WHY CO appear hota hai: O khatam ho gaya. --- ## 4. Worked examples > [!example] Example 1 — Methane $CH_4$ ka AFR_st > $x=1,\,y=4 \Rightarrow a = 1+4/4 = 2$. > **WHY?** $a=x+y/4$ mein plug karo. > $M_{CH_4}=16$ g/mol. > $$\text{AFR}_{\text{st}}=\frac{4.76\times 2\times 28.96}{16}=\frac{275.7}{16}\approx 17.2$$ > **Yeh step WHY?** Total air moles $=4.76a$, times air molar mass, divided by fuel mass. > ✅ Real gasoline ≈14.7 hai; methane thoda higher hai — consistent (H-rich fuels ko per kg zyada air chahiye). > [!example] Example 2 — Methane ke liye AFR = 20 lean hai ya rich? > $\phi = \text{AFR}_{\text{st}}/\text{AFR} = 17.2/20 = 0.86 < 1$. > **WHY?** Zarorat se zyada air ⇒ excess O₂ ⇒ lean. Engine cooler chalta hai, lower CO. ✅ > [!example] Example 3 — Octane $C_8H_{18}$, $a$ aur AFR_st nikalo > $a = 8 + 18/4 = 8 + 4.5 = 12.5$. **WHY?** Direct formula. > $M=114$ g/mol. $\text{AFR}_{\text{st}}=\dfrac{4.76\times 12.5\times 28.96}{114}=\dfrac{1723}{114}\approx 15.1$. > **Yeh step WHY?** Same machinery; petrol ke famous 14.7 ke kareeb. ✅ > [!example] Example 4 — $\phi=0.8$ par percent excess air > Excess air factor $\lambda = 1/\phi = 1.25$, yaani **25% excess air**. > **WHY?** $\lambda = \text{AFR}/\text{AFR}_{\text{st}} = 1/\phi$; 1.25 matlab needed air ka 125%. --- ## 5. Common mistakes (Steel-man + fix) > [!mistake] "Rich matlab zyada oxygen kyunki yeh fuel-*rich* hai... rich = sab kuch zyada?" > **WHY yeh sahi lagta hai:** "Rich" sunne mein "abundant" lagta hai, toh log abundant air bhi imagine karte hain. > **Fix:** *Rich* sirf **fuel** ke baare mein hai. Rich = excess **fuel**, *deficient* oxygen. > Isliye rich = $\phi>1$. > [!mistake] "Stoichiometric best fuel economy deta hai." > **WHY yeh sahi lagta hai:** "Perfect ratio" optimal lagta hai. > **Fix:** Stoichiometric **hottest flame & max power** deta hai, lekin engines aksar thoda > **lean** run karte hain *economy* ke liye (kam unburnt fuel) — NOₓ ki cost par. Alag goals, alag $\phi$. > [!mistake] "$\phi = \text{AFR}/\text{AFR}_{\text{st}}$." > **WHY yeh sahi lagta hai:** AFR headline number hai, toh seedha divide karo. > **Fix:** $\phi$ **fuel/air** use karta hai, toh yeh *inverse* hai: > $\phi = \text{AFR}_{\text{st}}/\text{AFR}_{\text{actual}}$. Bada AFR ⇒ leaner ⇒ chhota $\phi$. > [!mistake] "Air balance mein nitrogen bhool jaana." > **WHY yeh sahi lagta hai:** N₂ react nahin karta, toh ignorable lagta hai. > **Fix:** N₂ mass carry karta hai (3.76 mol per O₂) aur flame ko **dilute/cool** karta hai; yeh > AFR mass mein aur flame-temperature calculations mein hona chahiye. --- ## 6. Forecast-then-Verify > [!recall] Compute karne se pehle predict karo > Example 3 padhne se pehle: **Forecast** karo — kya octane ka AFR_st methane (17.2) se > zyada hoga ya kam? Methane bahut H-rich hai (4 H per C); octane kam. Forecast: octane > *lower*. **Verify karo:** 15.1 < 17.2. ✅ Teri reasoning (zyada H per C ⇒ per kg zyada air chahiye) > sahi hai. --- ## 7. Flashcards #flashcards/chemistry AFR ke terms mein equivalence ratio $\phi$ kya hota hai? ::: $\phi=\text{AFR}_{\text{st}}/\text{AFR}_{\text{actual}}$ $\phi<1$ kaun sa regime describe karta hai? ::: Fuel-lean (excess oxygen). $\phi>1$ kaun sa regime describe karta hai? ::: Fuel-rich (excess fuel, deficient O₂). $C_xH_y$ ke liye stoichiometric O₂ coefficient? ::: $a=x+y/4$. Air mein 1 mol O₂ ke saath kitne moles N₂ aate hain? ::: 3.76 (kyunki 79/21). Kaun sa regime sabse hot (peak adiabatic) flame deta hai? ::: Stoichiometric, $\phi=1$. Fuel-rich combustion CO aur soot kyun produce karta hai? ::: Oxygen limiting hai, isliye carbon CO₂ tak fully oxidize nahin ho sakta. Methane ka AFR_st (approx)? ::: ≈17.2 kg air/kg fuel. $\phi$ ke terms mein excess-air factor $\lambda$? ::: $\lambda=1/\phi$. N₂ inert hone ke bawajood balance mein kyun rakhte hain? ::: Yeh mass carry karta hai aur flame ko dilute/cool karta hai (aur high T par NOₓ banata hai). --- > [!recall]- Feynman: 12-year-old ko explain karo > Ek campfire imagine karo. Wood **fuel** hai, air **oxygen** deta hai. Agar bahut saari > wood daalo lekin fire saans na le sake, toh kala dhuan aur orange smoky flames banenge — yeh > **rich** hai (bahut zyada fuel). Agar ek chhoti si twig par bahut zyada air phunko, toh flame chhoti aur > cool hogi — yeh **lean** hai (bahut zyada air). Jab wood aur air *bilkul* match ho jaayein, toh fire > sabse hot aur clean roar karta hai — yeh **stoichiometric** hai. Number $\phi$ bas yahi hai "perfect amount ke comparison mein kitna wood": 1 perfect hai, 1 se zyada smoky-rich hai, 1 se kam breezy-lean hai. > [!mnemonic] Direction yaad rakho > **"RICH = fuel king hai, oxygen garib."** Aur **"$\phi$ Fat → Fuel-rich"**: > bada $\phi$ matlab bada fuel. Aur: **L**ean mein e**L**evated air hai (aur **L**ow φ). ## Connections - [[Adiabatic Flame Temperature]] — $\phi=1$ par peak karta hai, energy balance se derived. - [[Combustion Stoichiometry & Balancing]] — $a=x+y/4$ ka source. - [[NOx Formation (Zeldovich)]] — thoda lean & hot par sabse bura. - [[Soot & Incomplete Combustion]] — rich-side products. - [[Limiting Reagent]] — kyun rich oxygen par starve karta hai. - [[Rocket Propulsion — Why Engines Run Fuel-Rich]] — lighter exhaust, cooler nozzle. - [[Heat of Combustion & Calorific Value]] — AFR ko released energy se link karta hai. ## 🖼️ Concept Map ```mermaid flowchart TD AFR[Air-Fuel Ratio AFR] -->|reference value| AFRst[AFR stoichiometric] Balance[Balance CxHy combustion] -->|solve O2 moles| Coeff[a = x + y/4] Coeff -->|feeds| AFRst Air[Air 21% O2 79% N2] -->|3.76 mol N2 per O2| AFRst AFR -->|ratio with AFRst| Phi[Equivalence ratio phi] AFRst -->|normalizes fuel| Phi Phi -->|phi < 1| Lean[Lean: excess O2] Phi -->|phi = 1| Stoich[Stoichiometric: perfect] Phi -->|phi > 1| Rich[Rich: excess fuel] Stoich -->|hottest flame| Propulsion[Engine temp, thrust, emissions] Lean -->|risk flameout| Propulsion Rich -->|CO and soot| Propulsion ```