Flux and steady state mass of P in the global ocean. Units of mass (Tg) are taragram for the global oceans. Fluxes are in units of Tg/year.
Homework: (due 2/25) Consider the following global phosphorus cycle:
Calculate the evolution of atmospheric oxygen assuming zero oxygen in the atmosphere at time zero. Assume that the flux of oxygen TO the atmosphere is 138 times (mole/mole) the flux of P from the surface to deep ocean and the flux of oxygen FROM the atmosphere is 138 times the flux of P from the deep to surface ocean. Modeling the figure above using simple finite difference approximations is tricky because the surface box needs short time steps (<100 years) and the evolution of oxygen in the atmosphere takes at least 0.6 billion years. It would be a very long spreadsheet indeed!
An alternative is to assume that the entire ocean rapidly reaches steady state between the surface and deep reservoirs.
The net rate of oxygen production is F2-F3, where F2 is a small fraction of the total phosphorus added to the ocean via weathering. Over time, net photosynthesis will increase atmospheric oxygen with 138 O2 produced per mole of P consumed. A problem is that the ocean can’t hold enough organic material [(CH2O)106(NH3)16PO4] to produce significant amounts of oxygen. Significant amounts of organic material must be buried through subduction or precipitation (Prec. Flux).
Assuming that water that leaves the lake is equal to rainwater that falls into the drainage area of the lake, compute the residence time of water in each lake.
Why is the residence time calculated in question 1 too short?
The water level on each lake can be controlled up or down by about 0.5 meters by adjusting the gates on the outflow dam. Estimate how long it would take reach a new steady state water level if the dam was raised 0.5 meters on May 1 and dropped 0.5 meters on October 15.
2) Predicting the water level in a bathtub – an example of Steady State systems: due 2/11/2016.
Consider a simple bathtub with a faucet that adds water to the tub at a constant rate and a drain where water flows out at a rate proportional to the height of the water level. In this exercise you will calculate the water level in the tub as a function of time, determine the steady state water level, and calculate the residence time of water in the tub. This bathtub model is a good analogy for many dynamic environmental systems typical of the atmosphere, ocean, lakes, and living organisms. Assume that water is added to the tub from a faucet with a constant flow rate of 10 liters/minute. The inflow rate expression is a simple zero-order process where the inflow is constant and independent of water level.
The outflow is proportional to the height of the water in the tub. This is a first-order process where the higher the water level the greater the outflow rate.
Notice that the sign of the change in volume with time (dV/dt) is negative since water is leaving the tub. The rate constant k will have units of (liters/minute cm) so that the product of the rate constant and water height is outflow rate (liters/minute). From geometry we also know that the height of the water in the tub is the volume of the tub divided by the area of the tub.
When you start filling a tub the initial volume is usually zero. Adding water increases the volume and outflow decreases the volume.
Volume = initial volume + water added – water lost down the drain (3)
Equation 3 describes the conservation of mass (water) for the tub. We will write similar conservation equations for may environmental systems.
If the amount of water added is greater than the water lost then the volume will increase. As the volume increases the depth of the water in the tub will increase which will increase the outflow rate. Eventually the outflow rate will equal the inflow rate and the bathtub fill level will reach steady state.
Steady State is reached when INFLOW = OUTFLOW. Notice that you can set equation 1 equal to equation 2 to calculate the water height at steady state. It is interesting to think about what steady state means. For the bathtub, steady state water level is only “steady” when the water is running constantly into the tub. Adding water is like adding energy to the system. Steady state systems need energy inputs to maintain their positions. This can be compared to the equilibrium water level of the tub which will be zero after the faucet is turned off.
The residence time of the water is the average time that a water molecule will spend in the tub. This can be defined as the volume of water in the tub divided by the inflow or outflow rate when the system is at steady state (volume/(volume/time) = residence time).
Water inflow rate = 10 l/minute
Water outflow rate constant (k) = 1 (l/min cm)
Area of the tub = 10,000 cm^2
Starting water volume in tub = 0 liters
time step for the model = 0.5 minutes
1) Build and Excel model of bathtub levels as a function of time. Use the attached YouTube video as a model for your program.
2) Try running the model with an initial volume of 0 liters, 100, and 200 liters. How do the simulations compare? What does this tell you about Steady State conditions?
3) Manipulate equations one and two to calculate the expected Steady State level of the bathtub and compare to the results obtained in question two.
4) Calculate the residence time of water in the bathtub at steady state.
Turn in the answers to the questions 1-4 on one piece of paper showing your model on the front of the page and your results on the back side of the page. Don’t waste paper!