# Daily Incremental Exposure

As we have seen, a single cigarette can cause very high pollutant concentrations indoors due to the large quantity of particulate mass and other pollutants it emits. Furthermore, the pollutants emitted by a cigarette tend to linger indoors for many hours due to the relatively low air change rates found in most homes. An EPA measurement study of 178 homes in Riverside, CA, found that homes with smokers had indoor concentrations that averaged 30 μg/m$^3$ higher than homes without smokers $^{5-7}$, an important number that has been confirmed in other measurement studies in U.S. homes with smokers8. Of course, some homes with smokers have lower indoor concentrations than the average and some homes have higher indoor concentrations than the average, but on average smoking in a home adds about 30 μg/m$^3$ to the indoor fine particle concentration.

In the single bedroom example in the figure shown in the previous section, how does each additional cigarette smoked per day change the 24-h average concentration? A useful formula exists for the average Daily Incremental Exposure (DIE) contributed by one cigarette:

$DIE = \frac{x_o}{24D} (1 - e^{-24D})$ ~ $\frac{x_o}{24D}$

where $DIE$ = 24-h Average Concentration Contributed by the Source, $x_o$ = Initial Maximum Concentration (μg/m$^3$), $D$ = Decay Rate (h$^{-1}$)

Notice that the volume of the room does not appear in this formula, because the volume already is reflected by the initial maximum concentration $x_o$ caused by smoking the cigarette in this room. Each 24-hour average concentration predicted by this equation for every source can be added together, a useful property of the incremental daily exposure.

To apply this formula to the single cigarette smoked in the bedroom, we need first to find the decay rate for the main portion of the particle concentration curve that is decreasing with the door closed. The method for finding this important parameter is discussed elsewhere in the literature$^9$, and it requires performing regression analysis on the logarithm of the concentration versus time, which gives the result of $D$ = 0.5 h$^{-1}$. Substituting the initial concentration of $x_o$ = 341 μg/m$^3$ and this decay rate into the above equation for the daily average concentration gives:

$DIE = \frac{x_o}{(24)(0.5)} (1 - e^{-(24)(0.5)})$ ~ $\frac{341}{12}$ = 28.4 μg/m$^3$

Thus, a single cigarette smoked in this bedroom raises the average 24-h concentration by 28.4 μg/m$^3$, which is quite high but is less than the federal 24-h health-based standard of 35 μg/m$^3$. Smoking just one additional cigarette in this bedroom, however, will bring the average for 2 cigarettes in 24 h to (2)(28.4) = 56.8 μg/m$^3$, which is well above the EPA health-based standard. Violating the federal health-based standards happens easily because of the incredibly high particle emissions of each cigarette. In summary, even though a single cigarette lasts for only about 8.5 min, its aftermath of elevated concentrations in the room can persist for several hours, and smoking just 2 cigarettes per day in this bedroom will cause the indoor concentration to exceed the 24 h federal standard of 35 μg/m$^3$ designed to protect health outdoors under the US Clean Air Act.

It is clear that smoking many additional cigarettes per day in this bedroom would raise the 24-h average particle concentration to extremely high levels. At first glance, these high concentrations in the bedroom may seem inconsistent with the known average incremental concentration of 30 μg/m$^3$ for smoking homes reported in the EPA field studies mentioned above. A smoking parent and a small child may spend time together in a bedroom with the door closed, but usually the smoker does not stay in one room but instead smokes in different parts of the house, and the volume of the entire house then becomes important. If a person smokes a half-pack per day (10 cigarettes) in a typical house with a volume of 300 m$^3$ and the house has this same rate of decay for particles, then the 24-h concentration in the home will be 38.8 μg/m$^3$, causing the indoor air of the entire home to violate the federal 24-hour ambient air quality standard for PM$_{2.5}$.

[To be extended by referring to the California house volume data in the literature, illustrating that each cigarette is a potent source of fine particulate matter, pushing 24-h averages close to and often exceeding the NAAQS for PM$_{2.5}$.]