5.3.1Combustion Chemistry (Propulsion Bridge)
Stoichiometric vs fuel-rich vs fuel-lean combustion
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1. The master quantity: Air–Fuel Ratio (AFR)
WHY we need a reference — derive AFR_st from scratch
Take a general hydrocarbon . WHAT must be true for complete combustion? Every C must become CO₂, every H must become H₂O. Balance step by step:
Why this step? Carbon balance forces CO₂; hydrogen balance forces H₂O (since each H₂O holds 2 H atoms).
Now balance oxygen. Right side O atoms . Left side . So:
Why this step? Oxygen is conserved; equate atoms on both sides and solve for the moles of O₂.
Air is 21% O₂ and 79% N₂ by mole, so per mole O₂ we drag along mol N₂:
=\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}}$$ > ($28.96$ g/mol = mean molar mass of air, $4.76 = 1+3.76$ total mol air per mol O₂.) --- ## 2. Equivalence ratio $\phi$ — the dimensionless dial > [!definition] Equivalence ratio > $$\phi = \frac{\text{AFR}_{\text{st}}}{\text{AFR}_{\text{actual}}} > = \frac{(F/A)_{\text{actual}}}{(F/A)_{\text{st}}}$$ > where $F/A$ is the fuel–air ratio (inverse of AFR). WHY define $\phi$? Because raw AFR depends on the fuel; $\phi$ normalizes so that **one number describes every fuel**. | Regime | $\phi$ | AFR vs AFR_st | Meaning | |---|---|---|---| | **Lean** | $\phi<1$ | AFR **larger** | excess O₂ | | **Stoichiometric** | $\phi=1$ | AFR = AFR_st | perfect | | **Rich** | $\phi>1$ | AFR **smaller** | excess fuel | > [!intuition] Reading the dial > $\phi$ literally means "==how many times the stoichiometric *fuel* I poured in==". > $\phi=2$ → twice the fuel that the air can handle → very rich. ![[5.3.01-Stoichiometric-vs-fuel-rich-vs-fuel-lean-combustion.png]] --- ## 3. What comes out of the tailpipe (HOW products shift) > [!intuition] Why rich makes CO & soot, lean makes "clean but cool" > Oxygen is the *limiting reagent* when rich. There isn't enough O to finish oxidizing > carbon to CO₂, so it stalls at **CO** (and unburnt fuel/soot). When lean there's *spare* > O₂ that just passes through, diluting and cooling the flame. - **Lean ($\phi<1$):** products = CO₂, H₂O, **leftover O₂**, N₂. Cooler flame, low CO/soot, but high O₂+high T near $\phi\approx0.9$ breeds **NOₓ**. - **Stoichiometric ($\phi=1$):** *hottest* flame (all chemical energy released, no diluent excess) — peak adiabatic flame temperature. - **Rich ($\phi>1$):** products include **CO, H₂, soot, unburnt $C_xH_y$**; flame cooler again because incomplete oxidation releases less heat. > [!example] Rich methane combustion (mass-balanced idea) > For $\phi>1$ you supply only $a/\phi$ moles of O₂. Carbon is shared between CO₂ and CO so > that available O is exactly consumed — that's WHY CO appears: the O ran out. --- ## 4. Worked examples > [!example] Example 1 — AFR_st of methane $CH_4$ > $x=1,\,y=4 \Rightarrow a = 1+4/4 = 2$. > **Why?** Plug into $a=x+y/4$. > $M_{CH_4}=16$ g/mol. > $$\text{AFR}_{\text{st}}=\frac{4.76\times 2\times 28.96}{16}=\frac{275.7}{16}\approx 17.2$$ > **Why this step?** Total air moles $=4.76a$, times air molar mass, divided by fuel mass. > ✅ Real gasoline is ≈14.7; methane is a touch higher — consistent (H-rich fuels need more air per kg). > [!example] Example 2 — Is AFR = 20 for methane lean or rich? > $\phi = \text{AFR}_{\text{st}}/\text{AFR} = 17.2/20 = 0.86 < 1$. > **Why?** More air than needed ⇒ excess O₂ ⇒ lean. Engine runs cooler, lower CO. ✅ > [!example] Example 3 — Octane $C_8H_{18}$, find $a$ and AFR_st > $a = 8 + 18/4 = 8 + 4.5 = 12.5$. **Why?** direct formula. > $M=114$ g/mol. $\text{AFR}_{\text{st}}=\dfrac{4.76\times 12.5\times 28.96}{114}=\dfrac{1723}{114}\approx 15.1$. > **Why this step?** Same machinery; close to the famous 14.7 for petrol. ✅ > [!example] Example 4 — Percent excess air at $\phi=0.8$ > Excess air factor $\lambda = 1/\phi = 1.25$, i.e. **25% excess air**. > **Why?** $\lambda = \text{AFR}/\text{AFR}_{\text{st}} = 1/\phi$; 1.25 means 125% of needed air. --- ## 5. Common mistakes (Steel-man + fix) > [!mistake] "Rich means more oxygen because it's fuel-*rich*... rich = lots of everything?" > **Why it feels right:** "rich" sounds like "abundant", so people imagine abundant air too. > **Fix:** *Rich* refers only to the **fuel**. Rich = excess **fuel**, *deficient* oxygen. > Hence rich = $\phi>1$. > [!mistake] "Stoichiometric gives best fuel economy." > **Why it feels right:** "perfect ratio" sounds optimal. > **Fix:** Stoichiometric gives **hottest flame & max power**, but engines often run slightly > **lean** for *economy* (less unburnt fuel) — at the cost of NOₓ. Different goals, different $\phi$. > [!mistake] "$\phi = \text{AFR}/\text{AFR}_{\text{st}}$." > **Why it feels right:** AFR is the headline number, so you divide it directly. > **Fix:** $\phi$ uses **fuel/air**, so it's the *inverse*: > $\phi = \text{AFR}_{\text{st}}/\text{AFR}_{\text{actual}}$. Bigger AFR ⇒ leaner ⇒ smaller $\phi$. > [!mistake] "Forgetting nitrogen in the air balance." > **Why it feels right:** N₂ doesn't react, so it seems ignorable. > **Fix:** N₂ carries mass (3.76 mol per O₂) and **dilutes/cools** the flame; it must be in > AFR mass and in flame-temperature calculations. --- ## 6. Forecast-then-Verify > [!recall] Predict before you compute > Before reading Example 3: **Forecast** — will octane's AFR_st be higher or lower than > methane's (17.2)? Methane is very H-rich (4 H per C); octane less so. Forecast: octane > *lower*. **Verify:** 15.1 < 17.2. ✅ Your reasoning (more H per C ⇒ needs more air per kg) > holds. --- ## 7. Flashcards #flashcards/chemistry What does the equivalence ratio $\phi$ equal in terms of AFR? ::: $\phi=\text{AFR}_{\text{st}}/\text{AFR}_{\text{actual}}$ $\phi<1$ describes which regime? ::: Fuel-lean (excess oxygen). $\phi>1$ describes which regime? ::: Fuel-rich (excess fuel, deficient O₂). Stoichiometric O₂ coefficient for $C_xH_y$? ::: $a=x+y/4$. How many moles of N₂ accompany 1 mol O₂ in air? ::: 3.76 (since 79/21). Which regime gives the hottest (peak adiabatic) flame? ::: Stoichiometric, $\phi=1$. Why does fuel-rich combustion produce CO and soot? ::: Oxygen is limiting, so carbon can't fully oxidize to CO₂. AFR_st of methane (approx)? ::: ≈17.2 kg air/kg fuel. Excess-air factor $\lambda$ in terms of $\phi$? ::: $\lambda=1/\phi$. Why is N₂ kept in the balance though it's inert? ::: It carries mass and dilutes/cools the flame (and forms NOₓ at high T). --- > [!recall]- Feynman: explain to a 12-year-old > Imagine a campfire. Wood is the **fuel**, air gives the **oxygen**. If you pile on tons of > wood but the fire can't breathe, it makes black smoke and orange smoky flames — that's > **rich** (too much fuel). If you blow lots of air on a tiny twig, the flame is small and > cool — that's **lean** (too much air). When the wood and air are *just* matched, the fire > roars hottest and cleanest — that's **stoichiometric**. The number $\phi$ is just "how much > wood compared to the perfect amount": 1 is perfect, more than 1 is smoky-rich, less than 1 > is breezy-lean. > [!mnemonic] Remember the direction > **"RICH = fuel is the king, oxygen the pauper."** And **"$\phi$ Fat → Fuel-rich"**: > a *big* $\phi$ means *big* fuel. Also: **L**ean has e**L**evated air (and **L**ow φ). ## Connections - [[Adiabatic Flame Temperature]] — peaks at $\phi=1$, derived from energy balance. - [[Combustion Stoichiometry & Balancing]] — source of $a=x+y/4$. - [[NOx Formation (Zeldovich)]] — worst slightly lean & hot. - [[Soot & Incomplete Combustion]] — the rich-side products. - [[Limiting Reagent]] — why rich starves on oxygen. - [[Rocket Propulsion — Why Engines Run Fuel-Rich]] — lighter exhaust, cooler nozzle. - [[Heat of Combustion & Calorific Value]] — links AFR to energy released. ## 🖼️ 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 ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, combustion ka pura khel ek hi cheez pe tika hai: **kitni air laaye vs kitni > chahiye thi**. Agar bilkul exact air laaye (jitni fuel ko poori tarah jalne ke liye chahiye) > to wo hai **stoichiometric** — yahan flame sabse zyada garam hoti hai kyunki saari chemical > energy release ho jaati hai. Agar fuel zyada aur air kam → **fuel-rich**, aur agar air zyada > fuel kam → **fuel-lean**. Bas itni si baat. > > Isko measure karne ke liye hum **equivalence ratio $\phi$** use karte hain: > $\phi = \text{AFR}_{st}/\text{AFR}_{actual}$. $\phi=1$ matlab perfect, $\phi>1$ matlab rich > (fuel jyada, oxygen kam — isliye CO aur kaala soot banta hai), $\phi<1$ matlab lean (extra > oxygen bachta hai, flame thandi). Yaad rakho **"rich" ka matlab sirf fuel rich**, oxygen nahi > — ye sabse common galti hai. > > Stoichiometric coefficient nikalna easy hai: $C_xH_y$ ke liye O₂ chahiye $a = x + y/4$. Phir > air mein har 1 mol O₂ ke saath 3.76 mol N₂ aata hai (kyunki air 21% O₂, 79% N₂), to total air > $= 4.76a$. Methane ka AFR ≈17.2, petrol ka ≈14.7 nikalta hai — ye numbers exam aur engines > dono mein kaam aate hain. > > Propulsion mein ye crucial kyun? Engine ki **thrust, temperature aur emissions (CO, soot, > NOₓ)** sab is ratio se decide hote hain. Rockets aksar thoda **fuel-rich** chalte hain taaki > exhaust halka aur nozzle thanda rahe, jabki cars economy ke liye thoda **lean** ki taraf > jaati hain. To $\phi$ ek chhota sa dial hai jo poore engine ka behaviour control karta hai. ![[audio/5.3.01-Stoichiometric-vs-fuel-rich-vs-fuel-lean-combustion.mp3]]