Issue 014 - Air quality - Exposure dose
How much wildfire smoke particulate does a region inhale in a day?
Reuters reported wildfire smoke blanketing the eastern U.S. from the Great Lakes to Washington, D.C., with 68 large fires burning in 15 states and more than 100 million Americans under some level of air-quality alert. Convert that abstract alert into inhaled PM2.5 mass.
The problem
Estimate the total extra mass of PM2.5 inhaled by people in the affected U.S. region during one smoky day.
For the main estimate, use 100 ug/m3 as a round smoky-day outdoor PM2.5 concentration for affected areas. If you want to estimate only the extra smoke burden above a normal day, use a lower effective concentration such as 50 to 100 ug/m3 after subtracting baseline pollution and accounting for indoor sheltering.
Is the total inhaled fine-particle mass closer to a few pounds, a few hundred pounds, or several tons?
Because Fermi problems target an order of magnitude, I normally use no more than two significant digits and write most calculations in scientific notation; the Fermi reference explains both conventions.
Before checking sources
Matt's first pass
I started with a few assumptions. I took microgram to mean 10^-6 grams. I assumed that most affected people would be in "unhealthy" air rather than "very unhealthy" air, and I used 100 ug/m3, or 1 x 10^-4 g/m3, for the particulate concentration.
I estimated daily inhaled air from breathing rate and tidal volume. I used 14 respirations per minute, 0.6 L per breath, and 1,000 L per cubic meter.
breaths per day ~= 14 breaths/minute x 1,440 minutes/day
~= 2.0 x 10^4 breaths/day
air breathed ~= (2.0 x 10^4 breaths/day) x (0.6 L/breath)
~= 1.2 x 10^4 L/day
~= 12 m3/dayThen I assumed people spend about 2 hours outdoors and 22 hours indoors or in partially filtered air. That meant about 1 m3 of air at full outdoor concentration and 11 m3 at about 20% of outdoor concentration.
outdoor PM2.5 inhaled ~= 1 m3 x 1 x 10^-4 g/m3
~= 1 x 10^-4 g/person/day
indoor PM2.5 inhaled ~= 11 m3 x 20% x 1 x 10^-4 g/m3
~= 2.2 x 10^-4 g/person/day
total ~= 3.2 x 10^-4 g/person/dayIf about 100 million people are exposed:
population total ~= (3.2 x 10^-4 g/person/day) x (1 x 10^8 people)
~= 3.2 x 10^4 g/day
~= 32 kg/day
~= 70 lb/daySo my first answer was nearly on the order of 100 lb of particulate inhaled in one day across the exposed U.S. population.
Calibration Score
Matt's Calibration Score: 90 / 100
Higher is better: earn points for accurate pegs, sound models, correct math, and a result close to the sourced answer. The image shows percent full of it: 100 minus the Calibration Score.
Pegs: 20/30. The concentration, breathing-rate, and indoor-exposure pegs were reliable enough.
Model: 30/30. Concentration times air breathed times people was exactly the right dose model.
Math: 10/10. The arithmetic was clean.
Result: 30/30. The final mass estimate landed in the corrected range.
Grounding facts
One hundred pounds of PM2.5 spread across 100 million people is only about 0.45 milligrams per person. That is a tiny mass per person, but PM2.5 is defined by particle size, not by bulk weight. Fine particles can penetrate deep into the lungs, and health effects depend on concentration, duration, particle chemistry, vulnerability, and activity level.
This is why air-quality warnings can matter even when the total mass sounds almost ordinary at the population scale.
After checking sources
Check and recalibrate
Matt's structure was strong. The unit relationship is the heart of the problem:
ug/m3 x m3 = ug
1 ug = 1 x 10^-6 g
1 kg = 1 x 10^3 g
1 lb ~= 0.45 kgEPA's AQI breakpoints put 24-hour PM2.5 in the "Unhealthy" range at 55.5 to 125.4 ug/m3 and "Very Unhealthy" at 125.5 to 225.4 ug/m3. EPA's Exposure Factors Handbook gives adult daily inhalation values in the rough range of 10 to 20 m3/day, so Matt's 12 m3/day is a good Fermi value.
The main adjustment is to call the answer an extra smoke exposure rather than total particulate from all sources. Use 50 to 100 ug/m3 as a plausible extra outdoor smoke burden, then apply an effective exposure fraction for indoor sheltering and filtration. A broad useful range is about 25% to 75% of outdoor exposure.
low scenario:
1 x 10^8 people x 50 ug/m3 x 12 m3/day x 25%
~= 1.5 x 10^10 ug/day
~= 1.5 x 10^4 g/day
~= 15 kg/day
~= 30 lb/day
middle scenario:
1 x 10^8 people x 75 ug/m3 x 15 m3/day x 50%
~= 5.6 x 10^10 ug/day
~= 5.6 x 10^4 g/day
~= 56 kg/day
~= 120 lb/day
higher scenario:
1 x 10^8 people x 100 ug/m3 x 16 m3/day x 75%
~= 1.2 x 10^11 ug/day
~= 1.2 x 10^5 g/day
~= 120 kg/day
~= 260 lb/dayA good Fermi answer is therefore tens to a few hundred pounds of extra PM2.5 inhaled in one smoky day across 100 million exposed people. That is much closer to a few hundred pounds than to several tons.
The counterintuitive part is that the total physical mass is not huge. The public-health problem comes from the mass being divided into microscopic particles, spread through breathing air, and delivered into bodies across an enormous population, with children, older adults, outdoor workers, and people with heart or lung disease at higher risk.
Post-check reflection
Matt's reflection
I did not really make any major errors on this one. I just did not look at excess particulate; I looked at total particulate from this source, but still ended up well within the right order of magnitude. My mental maps were reliable.
The corrected number is not especially difficult to believe or grasp. Spread across 100 million people, that is really not a lot of mass.
Recommended memory peg
Remember 1 ug = 1 x 10^-6 g, adult breathing is roughly 10 to 20 m3/day, and exposure dose is concentration x air volume breathed.
Reader results
Bars show how submitted estimates sort into the answer choices from the gut-check prompt.