Exercises — Stoichiometric vs fuel-rich vs fuel-lean combustion

WHAT: compare $\varphi$ to $1$. WHY: $\varphi$ literally counts fuel relative to the perfect amount. $\varphi =1.4>1$ ⇒ we poured in $1.4\times$ the fuel the air can finish ⇒ fuel-rich. Consequence: oxygen is the limiting reagent, so carbon stalls at CO (plus $H_2$ and soot). ✅

$\varphi =\frac{{\text{AFR}}_{\text{st}}}{\text{AFR}}=\frac{17.2}{22}=0.78$. Since $\varphi <1$ there is excess air ⇒ fuel-lean. Bigger AFR always means leaner, hence smaller $\varphi$. ✅

Step 1 (WHAT/WHY): every C→CO₂, every H→H₂O forces $a=x+y/4$. $a=2+\frac{6}{4}=2+1.5=3.5$ Step 2: feed Machine A.

Actual AFR: rearrange Machine B, ${\text{AFR}}_{\text{actual}}={\text{AFR}}_{\text{st}}/\varphi$. ${\text{AFR}}_{\text{actual}}=\frac{15.1}{0.75}=20.1\frac{\text{kg air}}{\text{kg fuel}}$ Excess air: $\lambda =1/\varphi =1/0.75=1.333$, so we supplied $133.3\%$ of the needed air ⇒ 33.3% excess air. ✅

Step 1 — oxygen available. Stoich $a=2$, so O₂ supplied $=a/\varphi =2/1.25=1.6$ mol, i.e. $2\times 1.6=3.2$ mol of O atoms. Step 2 — spend O on hydrogen first. All $4$ H → $2H_{2}O$, costing $2$ O atoms. Remaining O atoms $=3.2-2=1.2$. Step 3 — split the 1 carbon. Let $p$ = mol CO₂, $q$ = mol CO. Carbon balance $p+q=1$; oxygen balance $2p+q=1.2$. Subtract: $p=0.2$, so $q=0.8$. Result: ${\text{CO}}_2=0.2$ mol, $\text{CO}=0.8$ mol per mole $C{H}_4$. WHAT IT MEANS: running $25\%$ rich, most carbon exits as toxic, un-finished CO — the O simply ran out. See Soot & Incomplete Combustion. ✅

Lean side ($\varphi <1$): every bit of fuel does burn fully, but leftover cold O₂ (and its N₂) sits in the products soaking up heat — extra mass, same energy ⇒ lower temperature per kg. Rich side ($\varphi >1$): now fuel is left un-oxidised (CO, H₂, soot). Chemical energy stays locked in those molecules and is never released, so there's less heat to begin with. At $\varphi =1$: all energy is released and there is no cold excess to dilute it — both penalties vanish, giving the peak. The curve rises to a maximum at $\varphi =1$ and falls on both sides (the rich peak is often nudged slightly to $\varphi \approx 1.05$ because a little dissociation matters, but $\varphi =1$ is the textbook answer). ✅

(a) Propane: $a=3+8/4=3+2=5$. (b) Per mole of mixture ($0.5C{H}_4+0.5C_3{H}_8$): $a_{\text{mix}}=0.5(2)+0.5(5)=1+2.5=3.5{\text{mol O}}_2.$ (c) Why a mole ratio is still the same $\varphi$. Fix the fuel and vary only the air. Then ${\text{AFR}}_{\text{actual}}=k{n}_{O_2,\text{supplied}}$ and ${\text{AFR}}_{\text{st}}=k{a}_{\text{mix}}$ with the same proportionality constant $k$ (mass of air per mole of O₂, divided by fuel mass — identical top and bottom). That constant cancels:

Step 1 — carbon. $p+q=1$ with $p:q=1:3\Rightarrow p=0.25$ (CO₂), $q=0.75$ (CO). Step 2 — count O atoms in products. H₂O: $2$ O. CO₂: $2p=0.5$. CO: $q=0.75$. Total O atoms $=2+0.5+0.75=3.25$, so O₂ supplied $=3.25/2=1.625$ mol. Step 3 — φ. Stoich O₂ is $a=2$; $\varphi =a/n_{O_2}=2/1.625=1.231$. So the flame runs about $\varphi \approx 1.23$ — rich, exactly the kind of window that feeds Soot & Incomplete Combustion. ✅

(a) ${\text{AFR}}_{\text{actual}}={\text{AFR}}_{\text{st}}/\varphi =14.7/1.6=9.19$ kg air/kg fuel. (b) Reason 1 — thermal survival: $\varphi =1$ gives peak Adiabatic Flame Temperature, which would melt the throat and turbine. Running rich pulls the temperature off the peak into a survivable band. It also starves the Zeldovich mechanism (needs O₂ and heat), cutting NOx. Reason 2 — light exhaust = more thrust: incomplete combustion leaves lots of low-mass species ($H_2$, CO, unburnt $C{H}_x$). Exhaust velocity $\propto \sqrt{T/M}$, so a lighter average molecular mass $M$ can raise exhaust speed even at lower $T$ — a net specific-impulse win. Rich is a deliberate optimisation, not a mistake. ✅

(a) ${\text{AFR}}_{\text{actual}}=15.1/0.70=21.6$ kg air/kg fuel. $\lambda =1/0.70=1.4286\Rightarrow$ 42.9% excess air. (b) At $\varphi \approx 0.9$ you still have near-peak flame temperature plus spare O₂ — the perfect recipe for thermal NOx. Dropping to $\varphi =0.70$ dumps in so much cold excess air that the flame temperature collapses; NOx production, which depends exponentially on temperature (Zeldovich), falls off a cliff even though O₂ is more abundant. Temperature beats availability. ✅

Regime at $\varphi =1.4$? ::: Fuel-rich (extra fuel, excess CO). $a$ for ethane $C_2{H}_6$? ::: $2+6/4=3.5$. ${\text{AFR}}_{\text{st}}$ of ethane ($M=30$)? ::: $\approx 16.1$ kg air/kg fuel. Actual AFR for octane at $\varphi =0.75$? ::: $15.1/0.75\approx 20.1$. Rich $C{H}_4$ at $\varphi =1.25$: mol CO per mol fuel? ::: $0.8$ (with $0.2$ CO₂). Reverse-engineer $\varphi$ from $C{O}_2:CO=1:3$ (methane)? ::: $\varphi \approx 1.23$. $\varphi$ of the 50/50 methane+propane mix at $6$ mol O₂? ::: $3.5/6.0\approx 0.58$ (lean). Why run a rocket rich? ::: Survivable temperature + lighter exhaust ($v\propto \sqrt{T/M}$) + less NOx.