Environmental Data Analysis BC ENV 3017


Indoor - Outdoor PM

Background

The experiment

Conceptual model



 

We are picturing the bedroom as a box in which the air is well mixed (homogeneous). Exchange with the outside occurs through the windows or possibly the doors. There may be sources or sinks or particles in the room (deposition or re-suspension of particles). The following factors might have an effect on the indoor PM concentration:

Our goal is to construct a mathematical model that includes the relevant processes above and allows us to predict the indoor PM time series. Unfortunately, we do not have a good handle on some of the  factors (in red) mentioned above. We can measure the outdoor/indoor exchange rate using SF6 and can make an assumption about the deposition rate inside, and will ignore all the other processes.

That means we envision two processes controlling the indoor PM concentrations:

Mathematical model

We are assuming that the change of PM concentration inside (Cin) is proportional to the difference between outside and inside concentration (Cout - Cin) and that the deposition rate is proportional to the inside PM concentration (Cin).

dCin
-----  = -l (Cin-Cout)-R Cin
  dt

The same equation holds true for the SF6 experiment, except that Cout is practically 0 and R is 0 as well:

dCSF6
-----  = -l CSF6
  dt

This differential equation has an analytical solution (a mathematical function) while the other equation can only be solved numerically:

CSF6 (t) = CSF6(t=0) * exp (-lt)

By fitting an exponential function to the measured SF6 data, we can determine l and CSF6(t=0) and use the same l to solve the differential equation for PM.

The equation for PM can only be solved numerically:

(Cin (t+Dt) - Cin(t)) / Dt = -l (Cin(t) - Cout(t)) - R Cin(t)
or
Cin(t+Dt) = Cin(t) + Dt (-l (Cin(t) - Cout(t)) -R Cin)
or
Cin(t+Dt) = Cin(t) (1-Dt (l + R)) + Cout(t) Dt l

We are basically calculating Cin at the end of the time step Dt from Cin at the beginning of the time step. Our best estimate of Cout during the time step is actually the average of Cout(t) and Cout(t+Dt). The final equation then is:

Cin(t+Dt) = Cin(t) (1-Dt (l + R)) + 0.5(Cout(t)+Cout(t+Dt)) Dt l

In the lab, we will use this equation to calculate Cin as a function of time and compare it to the measured data to find out if our model does describe the experiment reasonably well.