Stack test to fence line

Source testing · from the analyser to the receptor

The stack tester's van is parked at the base of a natural-gas boiler stack, sample line run up to the port. The analyser settles at 120 ppm of NOx with the flue oxygen reading 5.2 % — but the permit talks in concentrations corrected to 3 % O₂, so the raw number means nothing to a regulator yet. The stack moves 18 m³/s of exhaust. Downwind, the nearest neighbour sits where the dispersion tables say σy = 200 m and σz = 100 m, the wind at stack height is 5 m/s, and the effective stack height works out to 120 m. Correct the reading, convert it to mass, find what the stack actually emits, and find what the neighbour breathes.

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given

C_meas = 120 ppm — Analyser NOx reading
O₂,meas = 5.2 % — Flue oxygen at the probe
O₂,ref = 3 % — Permit reference oxygen
Q_v = 18 m³/s — Stack volumetric flow
u = 5 m/s — Wind speed at stack height
H = 120 m — Effective stack height

Determine

(a)the concentration corrected to 3 % O₂
(b)that concentration as a mass per cubic metre
(c)the emission rate the stack actually puts up
(d)the ground-level concentration at the neighbour's fence

Step 1 of 4(a) · solve for Corrected concentration

The correction exists because dilution is free: pull extra air through a boiler and every ppm reading drops while the actual emission does not change at all. Normalising to 3 % O₂ takes the tramp air back out — 120 ppm at 5.2 % oxygen is really 136.8 ppm at the permit's reference, a 14 % difference a regulator will absolutely notice.

Rearranged for C_corr

C_{corr} = C_{meas} \, \frac{20.9 - O_{2,ref}}{20.9 - O_{2,meas}}

C_meas

O₂,ref

O₂,meas

Your values, in your units

C_{corr} = \left( 120\ \text{ppm} \right) \cdot \frac{20.9 - \left( 3\ \text{\%} \right)}{20.9 - \left( 5.2\ \text{\%} \right)}

Answer

C_{corr} = 136.82\ \text{ppm}

in

Carried onward at full precision, not this rounded figure.

Open the Emission Correction to Reference Oxygen solver →

Step 2 of 4(b) · solve for Mass concentration

A ppm is a count of molecules; a limit is written in mass. The bridge is the molar mass — and for NOx the convention is to report AS NO₂ at 46.01 g/mol, whatever mix of NO and NO₂ actually left the flame. 136.8 ppm × 46.01 / 24.45 lands at 257.5 mg/m³.

Rearranged for C

C = \frac{ppm \cdot M}{24.45}

ppm136.82 ppmcarried from step 1

M

Your values, in your units

C = \frac{\left( 0.0136815\ \text{\%} \right) \cdot \left( 46.01\ \text{g/mol} \right)}{24.45}

Converted to base units

C = \frac{\left( 136.815\ \text{ppm} \right) \cdot \left( 46.01\ \text{g/mol} \right)}{24.45}

Answer

C = 257.46\ \mu\text{g/L}

in

Carried onward at full precision, not this rounded figure.

Open the ppm to mg/m³ Conversion solver →

Step 3 of 4(c) · solve for Emission rate

Concentration says how dirty the exhaust is; the emission rate says how much is actually reaching the sky, and only the second is a quantity of anything. 257.5 mg/m³ leaving at 18 m³/s is 4.63 g/s — about 400 kg of NO₂ equivalent per day.

Rearranged for E

E = C \, Q_v

C257.46 μg/Lcarried from step 2

Q_v

Your values, in your units

E = \left( 0.000257459\ \text{kg/m}^{3} \right) \cdot \left( 18\ \text{m}^{3}\text{/s} \right)

Converted to base units

E = \left( 257.459\ \text{mg/m}^{3} \right) \cdot \left( 18\ \text{m}^{3}\text{/s} \right)

Answer

E = 16.683\ \text{kg/h}

in

Carried onward at full precision, not this rounded figure.

Open the Emission Rate from Stack Concentration solver →

Step 4 of 4(d) · solve for Ground-level concentration

Now the atmosphere does what the stack cannot: spread it. Across those dispersion coefficients and under that wind the 4.63 g/s arrives at ground level as 7.2 µg/m³ — the exponential on H² is doing most of the protecting, which is exactly why stacks are tall.

Rearranged for C

C = \frac{Q}{\pi \sigma_y \sigma_z u} \, e^{-H^{2}/(2\sigma_z^{2})}

Q16.683 kg/hcarried from step 3

σ_y

σ_z

u

H

Your values, in your units

C = \frac{\left( 0.00463426\ \text{kg/s} \right)}{\pi \cdot \left( 200\ \text{m} \right) \cdot \left( 100\ \text{m} \right) \cdot \left( 5\ \text{m/s} \right)} \, e^{-\left( 120\ \text{m} \right)^{2}/(2 \cdot \left( 100\ \text{m} \right)^{2})}

Converted to base units

C = \frac{\left( 4.63426\ \text{g/s} \right)}{\pi \cdot \left( 200\ \text{m} \right) \cdot \left( 100\ \text{m} \right) \cdot \left( 5\ \text{m/s} \right)} \, e^{-\left( 120\ \text{m} \right)^{2}/(2 \cdot \left( 100\ \text{m} \right)^{2})}

Answer

C = 7.1802e-09\ \text{kg/m}^{3}

in

Carried onward at full precision, not this rounded figure.

Open the Gaussian Plume Ground-Level Concentration solver →

Why this order

This is the whole logic of source testing in one run: the analyser reads a ratio, the permit reads a corrected ratio, the inventory reads a mass rate, and the neighbour breathes a concentration — four different quantities that get called “the emissions” interchangeably in conversation and never are. The oxygen correction comes first because it is the anti-cheating step: without it, leaning the burner or cracking a dilution damper would “comply” any stack in an afternoon. The molar-mass conversion carries the one convention worth memorising — NOx is reported as NO₂, so 46.01 g/mol goes in even though most of what leaves the flame is NO at 30.01. Use 30 and every downstream number is 35 % flattering.

The place people go wrong is at the end, reading the 7.2 µg/m³ as “the” concentration at the fence. It is the concentration on the plume centreline, under one stability class, at one wind speed — move the wind, change the sun, and σy and σz change by factors, not percent. A screening number tells you whether to do the real modelling; it never replaces it.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.