File last modified 5 November 1998
In this section we will look at a multi-box model of the world ocean phosphorus to examine the steady state distribution of phosphorus as a result of a crude approximation of the world ocean general circulation.
As an example let us consider the following box model of the world distribution of oceanic phosphorus. The volumes are in 1017 m3, the fluxes of water are in Sv (106 m3/s), the surface areas of the three surface boxes are in 1014 m2, and the concentrations of phosphorus (P) are in mmol P/m3.
We are largely concerned with the distribution between the upper and lower oceans, but have a suspicion that the Antarctic waters act as some sort of mixing bowl/redistribution center and so we represent it with its own box. Figure 16.4.1 shows the reservoirs and their linkages.
If we assume that 95% of the phosphorus in the surface waters is converted by biological activity into sinking detritus and that the remaining 5% stays in solution and is transported by the water flows. Of the 95% of the phosphorus that is transported to the bottom by sinking all but 1% remineralizes back into solution, the 1% is removed permanently into the sediments and this flux of phosphorus is balanced by the input of phosphorus from the river inputs (0.25 and 0.75 Sv in the surface Atlantic and Indo-pacific respectively).
So the fluxes are given simply by the concentration of the box it is leaving times the water flux (in Sv). For example let's consider the surface Atlantic box (SAT).
In equation 16.4.1 the first term id the amount of phosphorus delivered from the deep Atlantic (DAT), the next term the amount of P from the Antarctic box (ANT), and the third term the amount of P delivered from the rivers. These terms all have to be multiplied by (1-0.95) to represent the phosphorus lost as particles. The we subtract off the phosphorus sent back (as dissolved P) to the deep Atlantic. All of that is then divided by the volume of the surface Atlantic to put it back into concentration units.
But wait, what are the units in the equation 16.4.1? The flows are in terms of Sverdrups (106 m3 s-1), the concentrations in mmol-P m-3, and the volumes in 1017 m3. This yields the differential in units of 10-11 mmol-P m-3 s-1, multiplying each equation by the number of seconds in a year (roughly 3.16 x 107) will put everything on a per year footing (this may not seem terribly important now, but see below under the Runge-Kutta solution).
Now we can write the coupled equations that represent this box model:
I have coded these equations up in MatLab as an m-function for ODE45 (see phos.m). When coding them up I combined and simplified where possible so the equations in the m-file won't be exactly the same as these.
You can download the MatLab file phos.m and use it in MatLab with one of MatLab's ODE integrators (my favorite is ODE45, but it will work the others such as their "stiff" ODE integrator ODE15). To make a run of this model and display the evolution of phosphorus to steady state use the following MatLabese:
[t,x]=ode45('phos',[0 1500],[1 1 1 1 1]);
plot(t,x);
it's that simple.
Now you may be wondering what all this unit business is about. If you look at the MatLab statement that calls ODE45 you'll see that we have to give it start and stop times. If we had not added that timcon variable we would have had to specify those bounds in seconds. That's not a very convenient unit of time to use if you have to integrate out 1500 years! It makes no difference to MatLab's ODE45, it takes the same number of time steps whether you tell it to integrate from 0 to 1500 years or from 0 to 4.7336 x 1010 seconds. Figure 16.4.2 shows this model integrated out 2000 years.
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