Pollutants — NOₓ, soot, unburned hydrocarbons: Exercises

This page is a self-test. Every problem has a collapsible solution — try first, then reveal. Problems climb from recognising facts to synthesising a whole combustor design. Numeric answers all check out in the verify block. Parent: 5.3.09 Pollutants — NOₓ, soot, unburned hydrocarbons (Hinglish).

Before we start, we need to be crystal clear on the symbols that appear on this page. None will be used before it is defined here.

Definition Notation used on this page

Square brackets $[X]$ mean "the concentration of species X"—how many moles of X sit in each unit of volume (mol/m³). So $[N_{2}]$ is "how much N₂ is present", $[O]$ is "how much atomic oxygen is present", and $[NO]$ is "how much NO has built up". A rate written $\frac{d [ NO ]}{d t}$ then reads "how fast the NO concentration grows per second".
$A$ is the Arrhenius pre-exponential factor—it counts how often the right molecules collide with the right geometry, before we ask whether they are energetic enough. It sets the overall scale of the reaction speed. On this page $A$ is a fixed constant: because every exercise compares the same reaction at two temperatures, $A$ (and $[N_{2}], [O]$ ) cancels out and never needs a number. It only reminds us "there is a collision-frequency prefactor sitting there."
$φ$ (equivalence ratio) will be defined at Level 5 where it is first needed. See Equivalence Ratio and Flammability Limits for the full story.

Formula The toolbox for this page

Gas constant $R = 8.314 J mol^{- 1} K^{- 1}$ (the number that converts "energy per mole" into "how likely a collision is at temperature $T$ ").
Thermal-NOₓ activation energy $E_{a} \approx 319 kJ/mol$ (the energy needed to crack the $N \equiv N$ bond in the slow step). See Arrhenius Equation and Activation Energy.
Master rate: $\frac{d [ NO ]}{d t} = 2 A [N_{2}] [O] e^{- E_{a} / R T}$ , where every symbol is now defined above.

The ratio of NOₓ rate at two temperatures $T_{2}$ vs $T_{1}$ (same $A, [N_{2}], [O]$ —which is why they drop out) is $\frac{rate ( T _{2} )}{rate ( T _{1} )} = exp [- \frac{E _{a}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})] .$ Read it slowly: if $T_{2} > T_{1}$ , then $1/ T_{2} < 1/ T_{1}$ , the bracket is negative, the minus makes the exponent positive, so the rate rises. Hotter = more NOₓ. This one formula powers half the exercises.

Level 1 — Recognition

L1.1

Which two chemical species make up the label "NOₓ", and where does the nitrogen atom in thermal NOₓ originate?

Recall Solution

NOₓ $=$ ==NO $+$ NO₂. In the thermal (Zeldovich) route the nitrogen comes from the N₂ molecules already in the air== (air is ~78% N₂), not from the fuel. The air's normally-inert N₂ gets oxidised only because the flame is very hot.

L1.2

Sort these three zones by which pollutant they produce: (a) a hot but slightly-lean pocket, (b) a hot, oxygen-starved rich pocket, (c) a cold pocket near a wall. Choose from NOₓ, soot, UHC.

Recall Solution

(a) hot + lean + plenty of O $\to$ NOₓ (thermal route loves heat and spare oxygen atoms).
(b) hot + rich (no O to finish carbon) $\to$ soot.
(c) cold (reactions freeze) $\to$ UHC (fuel never finishes burning).

This is exactly the "too hot / too rich / too cold" trilogy from the parent note.

Level 2 — Application

L2.1

A gas turbine burns with peak temperature $T = 2100 K$ . A redesign cools the peak to $T = 1950 K$ . By what factor does the thermal NOₓ rate change? (Use $E_{a} / R = 319000/8.314$ .)

Recall Solution

What we do: plug into the ratio formula, $T_{1} = 2100$ (old), $T_{2} = 1950$ (new).

First, $\frac{E _{a}}{R} = \frac{319000}{8.314} = 38369 K$ .

Bracket: $\frac{1}{1950} - \frac{1}{2100} = 5.1282 \times 1 0^{- 4} - 4.7619 \times 1 0^{- 4} = 3.663 \times 1 0^{- 5} K^{- 1}$ .

Exponent: $- 38369 \times 3.663 \times 1 0^{- 5} = - 1.405$ .

Why the minus sign is correct: the new temperature is lower, so NOₓ should drop — a factor below 1. Indeed $\frac{rate(new)}{rate(old)} = e^{- 1.405} \approx 0.245.$ Answer: NOₓ rate falls to about $0.245 \times$ the original — roughly a 4-fold reduction from a mere 150 K cut. This is why lean-premixed combustion chases lower peak T.

L2.2

Near $2000 K$ , how large a temperature rise $Δ T$ doubles the thermal NOₓ rate? (Doubling means the ratio equals 2.)

Recall Solution

What we do: set the ratio $= 2$ and solve for the small rise $Δ T$ above $T = 2000$ K.

$2 = exp [- \frac{E _{a}}{R} (\frac{1}{2000 + Δ T} - \frac{1}{2000})] .$ Take $ln$ : $ln 2 = 0.6931 = - \frac{E _{a}}{R} (\frac{1}{2000 + Δ T} - \frac{1}{2000})$ .

For small $Δ T$ , $\frac{1}{2000 + Δ T} - \frac{1}{2000} \approx - \frac{Δ T}{200 0 ^{2}}$ . So $0.6931 \approx \frac{E _{a}}{R} \cdot \frac{Δ T}{200 0 ^{2}} = 38369 \cdot \frac{Δ T}{4 \times 1 0 ^{6}} .$ $Δ T \approx \frac{0.6931 \times 4 \times 1 0 ^{6}}{38369} = 72.3 K .$ Answer: about a 70 K rise doubles NOₓ near 2000 K — matching the parent note's rule of thumb. The physics of doubling is entirely the giant $E_{a}$ .

Level 3 — Analysis

L3.1

Two engineers argue. Engineer A says "increase residence time 3× to burn out CO — that will also make 3× more NOₓ." Engineer B says "at fixed high T, NOₓ formation is not yet at equilibrium, so more time does add NOₓ, but temperature still dominates." Analyse: is NOₓ linear in residence time in the early (kinetic) regime, and why is temperature still the bigger lever?

Recall Solution

The kinetic regime. Early on, $[NO]$ is far below its equilibrium value, so the reverse reactions are negligible and $[NO] \approx (\frac{d [ NO ]}{d t}) t = 2 A [N_{2}] [O] e^{- E_{a} / R T} \cdot t .$ So yes — in this regime NOₓ is linear in residence time $t$ . Tripling $t$ triples NOₓ. Engineer A's time statement is right.

Why T still wins. Compare the two levers:

Tripling time: factor $3$ .
A modest 200 K rise (2000→2200): from L2 logic, factor $exp [- 38369 (\frac{1}{2200} - \frac{1}{2000})]$ .

Watch the signs carefully. Here $T_{2} = 2200 > T_{1} = 2000$ , so $\frac{1}{2200} < \frac{1}{2000}$ and the bracket $\frac{1}{2200} - \frac{1}{2000} = - 4.545 \times 1 0^{- 5}$ is negative. The formula puts a leading minus in front: $- 38369 \times (- 4.545 \times 1 0^{- 5})$ . Two negatives multiply to a positive exponent $+ 1.744$ . That is why heating gives a factor above 1: $e^{+ 1.744} \approx 5.7.$ So a "small" 200 K temperature change (factor 5.7) beats a full 3× time change (factor 3). Engineer B is right that temperature dominates: time enters linearly, temperature enters exponentially.

What the figure shows: it plots two curves against how hard you push each lever. The blue straight line is NOₓ vs residence time (linear—triple the time, triple the NOₓ). The red curve is NOₓ vs a temperature rise (exponential). Notice the red curve shoots past the blue line almost immediately: by the 200 K mark the red curve is already at ~5.7 while the blue line is only at 3. The visual lesson is "exponential outruns linear from the start."

Figure — Pollutants — NOₓ, soot, unburned hydrocarbons

Level 4 — Synthesis

L4.1

A combustor produces soot in a hot rich pocket. The parent note says "net soot = formation − oxidation." Two fixes are proposed: (i) quench the pocket fast (chill it), (ii) add air to the pocket (make it leaner while still hot). Explain, using formation vs oxidation, which fix genuinely reduces soot and which can make it worse.

Recall Solution

Net soot $S = F - X$ where $F$ = formation, $X$ = oxidation. The pocket is hot and rich, so formation $F$ is already running (pyrolysis $\to$ C₂H₂ $\to$ PAH $\to$ particles).

(i) Quench fast. Chilling stops formation, yes — but it also stops the burnout reactions (soot oxidation by O₂/OH needs high T). You freeze in whatever soot already formed: $X \to 0$ . Net soot can stay high. This is precisely the "hot-rich pocket then quenched" worst case the parent note flags. See Quenching and Wall Heat Transfer.

(ii) Add air (still hot). Now the pocket moves toward stoichiometric/lean. Two wins: fewer spare carbon atoms so $F$ falls, and the O₂/OH from the added air drives $X$ up (soot burns off). Net soot drops. The catch — done clumsily this raises local T and O, feeding NOₓ, so it must be staged carefully (this is why real designs use staged air addition; see Lean Premixed Combustion & Staging).

Answer: Fix (ii) genuinely reduces soot (kills formation and boosts oxidation); Fix (i) can worsen net soot by freezing oxidation while soot is already present.

Level 5 — Mastery

L5.1 — Design a low-emission operating point

First, a definition we finally need.

Problem. You control $φ$ and can estimate peak temperature $T (φ)$ . Reason where each pollutant peaks on the $φ$ axis, then state a single operating strategy (lean-premixed) and justify it against all three pollutants at once. Finally, estimate the NOₓ payoff of running at $T = 1800 K$ instead of $T = 2300 K$ .

Recall Solution

Map each pollutant onto $φ$ (see figure):

CO & UHC: high at $φ ≪ 1$ (too lean, flame-out, slow finish) and at $φ > 1$ (not enough O). Minimum near slightly lean of stoichiometric.
Soot: appears only on the rich side $φ > 1$ and grows with richness (spare carbon clumps).
NOₓ: peaks slightly lean of stoichiometric, where $T (φ)$ is highest and O atoms are plentiful.

What the figure shows: the horizontal axis is $φ$ running from lean (left) through $φ = 1$ (dashed white line) to rich (right). The blue CO/UHC curve is U-shaped—high at both ends, dipping to a minimum just left of $φ = 1$ . The yellow soot curve is flat zero on the lean side and only lifts off once $φ$ passes 1, climbing with richness. The red NOₓ curve is a single hump peaking just left of $φ = 1$ . The shaded green band marks the lean-premixed target: sitting in the CO/UHC dip while safely left of the NOₓ peak. The takeaway is visual: no single $φ$ zeroes all three, but a slightly-lean band is the sweet compromise.

Strategy: lean-premixed. Premix fuel and air before the flame so there are no rich pockets (kills soot) and no local hot stoichiometric spikes (the flame burns uniformly cooler). Run lean, $φ < 1$ , to pull peak T down (kills thermal NOₓ). Keep $φ$ not too lean, and give enough residence time / hot volume, so CO and UHC still finish (avoids the lean CO/UHC rise). That single choice — uniform, lean, controlled T — attacks all three: soot (no rich zones), NOₓ (lower T), CO/UHC (still hot and mixed enough to burn out).

NOₓ payoff, 2300 K → 1800 K. $ratio = exp [- 38369 (\frac{1}{1800} - \frac{1}{2300})] .$ Bracket $= 5.5556 \times 1 0^{- 4} - 4.3478 \times 1 0^{- 4} = 1.2077 \times 1 0^{- 4} K^{- 1}$ (positive, because the new T is lower). Exponent $= - 38369 \times 1.2077 \times 1 0^{- 4} = - 4.634$ (the leading minus keeps it negative $\Rightarrow$ a drop). $ratio = e^{- 4.634} \approx 0.00972.$ Answer: NOₓ rate drops to about 1% of the original — a ~100× reduction. This enormous payoff (from an exponential in $1/ T$ ) is the entire reason lean-premixed combustion exists, even though it must be babysat to avoid the lean CO/UHC penalty.