In class you have seen how to derive rate expressions for elementary chemical reactions. Here you will apply this to an important environmental problem, the ozone dynamics in the lower atmosphere, and how this is impacted by combustion or the burning of fuel.

The ozone model --- problem formulation

The Earth's ozone levels are of great interest as at high concentrations ozone is harmful to humans and animals, and because ozone is also a green-house gas. On the other hand, ozone plays an important role in absorbing the UV radiation coming to the Earth from the Sun.

You will make a model that can be used to investigate the behaviour of ozone in the lower atmosphere. Specifically, you will model how the anthropogenic emissions of NO due to combustion or burning of fuels alter the natural ozone dynamics.

Assumptions

The dynamics of ozone in the lower atmosphere is governed by the following chemical reactions between the molecular oxygen ($O_2$), ozone ($O_3$), atomic oxygen ($O$), nitric oxide ($NO$), and nitrogen dioxide ($NO_2$). These reactions represent the natural processes involved in ozone cycling.

$$ Reaction~1: \qquad NO_2 (+ h\nu) \rightarrow NO+O $$ $$ Reaction~2: \qquad O+O_2 \rightarrow O_3 $$ $$ Reaction~3: \qquad NO+O_3 \rightarrow O_2 + NO_2 $$

The first reaction is the photo-dissociation of $NO_2$ to form $NO$ and $O$. This reaction depends on solar radiation ($h\nu$), and therefore its rate, $R_1$, changes drastically at sunrise and sunset.

The second reaction (rate $R_2$) describes the production of ozone; this is in reality a reversible process, but for our purpose we assume it only occurs in one direction.

In the third reaction, ozone is scavenged by reacting with $NO$ (rate $R_3$).

The reaction rates $R_1$, $R_2$ and $R_3$ are controlled by the rate constants $k_1$, $k_2$ and $k_3$.

In addition to these natural processes, nitric oxide ($NO$) is emitted through combustion or fuel burning processes. Assume that these processes are represented as a constant input of $NO$, at a rate $\sigma = 10^{11}~mol~d^{-1}$.

Tasks

| Name | Value | Unit | | ---------- | ---------- | ------------------ | | $k_1$ | $100$ | $d^{-1}$ | | $k_2$ | $10^{10}$ | $d^{-1}$ | | $k_3$ | $10^{-11}$ | $mol^{-1}~d^{-1}$ |

Here, maxrad is the maximal radiation at midday. Assume a value of $1200~\mu mol~photons~m^{-2}~s^{-1}$. The max statement assures that light intensity does not become negative, but is 0 during nighttime. Output the radiation as an ordinary variable.

Note: the equations used in this model are what mathematicians call a very "stiff" set of differential equations, meaning that processes occur on widely varying time scales (look at rate constants that range from $10^2$ to $10^{10}$). Because of that, the default solver of the deSolve package may have a hard time finding a solution. You can remedy that by using the solver called "vode" as follows:

out <- ode( , method = "vode")


dynamic-R/RTM documentation built on Feb. 28, 2025, 1:23 p.m.