Problem Session 4 [10/7/96]

We will now start serious atmospheric chemistry. The first chemical problem focuses on the Chapman Chemistry because you can solve it analytically and numerically. Lori will have discussed this in some detail in class lectures.

1. First solve for the steady-state values of O(Z) and O3(Z) [dO/dt = dO3/dt = 0.0] analytically, using the Quadratic Formula, for Z = 50, 40, 30, and 20km.

2. Next, solve analytically for the steady-state values of O(Z) and O3(Z) for Z = 50, 40, 30, and 20km using the "family approximation".

3. Thirdly, numerically integrate the 2 coupled non-linear differential equations, dO(Z,t)/dt and dO3(Z,t)/dt, until they have reached steady state, for Z = 50, 40, 30, and 20km. Either use the O3(Z) value that you were given in PS2 for O3(Z,0), your initial value, or start with 0.0. You can start with O(Z,0) = 0.0. Numerical integration is the approach that you will use for the rest of the course as we add species and reactions to your chemical model. We start with the Chapman Chemistry so you can check your own numerical work.

There are a wide variety of numerical integration schemes, which Lori will briefly discuss. The simplest, though also the slowest numerically, is the Euler, but you are free to try more sophisticated appoaches. You must take relatively small time-steps with the Euler, as you will find out, and it may take many of them to converge to steady-state, as you will also find out. DO NOT start your integration with the analytic solutions.

  1. Compare and discuss the analytic and "family approximation" analytic solutions for O and O3. How do they compare as Z changes?

  2. Use the analytic solutions for O(Z) and O3(Z) to check your numerical integration. You should agree to the 3rd significant figure. You should also try to numerically integrate the "family approximation" version of the 2 coupled non-linear differential equations, since this approximation will be used for the more complex reaction schemes that you will be building.

  3. How long did you have to integrate to reach steady-state, and how does the time change with Z? Why does it change with Z?