4 Modeling with Tanks

1. Introduction   The energy diagrams of systems ecologist H.T. Odum can be used to do environmental modeling in a way that encourages a visual, qualitative, and computational approach. Two basic diagrams will suffice for most of our work.  One is an energy storage tank, the other is an energy interaction box or workgate.  We will begin with the simpler of the two, the storage tank.

Fig. 1   Two fundamental energy diagrams

2. Tanks   The tank symbol is used to represent energy storage.  Some examples of stored, or potential energy are the biochemical energy stored in an apple (or body fat!), the electrical energy stored in a capacitor, and the geothermal energy stored in molten rocks, or magma.  Other examples of stored biochemical energy are a fish pond, a redwood tree, or soil.  Resource biologists often refer to the content of a tank as stock.   

Fig. 2   Various forms of stored, or potential energy

The tank is used to model processes whenever outflows are proportional to the content Q.  Sometimes Q is referred to as a state variable. The flow of energy from the tank is indicated by DQ and is read "Q dot".  For a tank,  DQ  prprtnl  Q.  In other words, the rate at which energy is transferred is large for large stores of energy and small for small stores of energy. The state variable Q can stand for energy, the energy present in material, or even the energy represented by money.

The letters b, c and k are constants of proportionality and are known in various applied fields as transfer coefficients, path parameters or turnover factors. They measure the rate at which energy escapes from storage and have a value between 0 and 1.  Since outflow is proportional to Q, the change in Q per unit time DQ, is given by the equations above the diagrams in Fig. 3.  The negative sign indicates that these are outflows -- the tanks are losing energy.

Fig. 3   Energy Outflow from a Tank                   

   Suppose b =.20 and c =.05 . Then the right path turns over 20% of the energy in the tank per unit time, while the bottom path turns over 5% of the energy in the tank per unit time. For convenience the paths can be combined to yield a single path coefficient k --

    - bQ - cQ  =  - (b+c)Q  =  - kQ  =  -.25Q       ( k  =  b + c  =  .20 + .05  =  .25 ).

Energy might be measured in British Thermal Units and time in minutes. A BTU is the energy needed to raise the temperature of a pound of water 1°F.)  In this case, the units for Q and DQ could be BTU and BTU/min.  Likewise, if energy is measured in Calories and time in hours, then the units of Q and DQ are Cal and Cal/hr.  Suppose the tank started with Q(0) = 5120 BTU, the time units are minutes, and k is still .25.  At the first click of the clock, 25% of 5120 BTU, or 1280 BTU are lost, and 3840 BTU are left in the tank. At the next click, 25% of the remaining 3840 BTU or 960 BTU are lost, and 2880 BTU are left in the tank.  Notice that the percent of energy lost (25%) is constant but the actual amount of energy lost gets smaller at every click of the clock.

Exercise 4.1   In the above "5120 BTU" example, carry out the calculations for three more minutes.

       Another way to get at turnover and turnover factor is to look at a tank that is maintained, on the average, at a steady level. Imagine a large cylindrical tank with 120 gallons of blue liquid as in Figure 4 (below).  A lighter, red-colored liquid enters the top of the tank at the rate of 30 gallons per minute. The blue liquid exits along a pathway with coefficient  k = .25.  So the blue liquid escapes at the same rate as the red liquid enters, 30 gallons per minute. The color of the liquid is changing but the tank keeps a steady level of 120 gallons. At the end of the first minute the 25% of the red liquid is replaced, and each minute thereafter another 25% is replaced or turned over. If the value of k is .10, then 10% of the tank, or 12 gallons would be turned over every minute.

Fig. 4   A steady state situation

      Below are three fundamental properties of tanks --

3. Five-stage Modeling   Let's expand on the five stages outlined in the first chapter.

4. Model #1 -- Tank Fed by Constant-flow Source   Look at the following three situations --

The following table will probably convince you that each of the three situations can be modeled by the same energy diagram, a tank fed by a constant flow source.  The table also starts us on the Five-stage modeling process.

Stage 1  The Energy Diagram  (Model #1)

Figure 5-S1

Stage 2  The Qualitative Graph  (Model #1)

Figure 5-S2

Stage 3  The Flow Equation  (Model #1)

Begin interpreting the diagram . . .		Fig. 5-S3
Jo = constant		Reflects "steady flow, deposits, ... "
Net Flow  =  Inflow - Outflow		Our mantra
DQ  =  Jo  -  bQ  -  cQ		Outflows are proportional to Q
=  Jo - (b+c)Q  =  Jo - kQ    ( k = b+c )		Algebra    ( Notation)
The Flow Equation    DQ  =  Jo - kQ		Algebraic recap
Start value               Q(0)  =  Qo		Tank's content at the beginning

      Suppose the tank starts with 3000 BTU, the outflow turnover factor is 20%, and the stored energy is replenished at the rate of 1200 BTU/sec.  That is,

Stage 4  Step-by-Step Solution  (Model #1)

The change in stored energy Q will be computed every second. The change during each unit step will be indicated by DQ, read "delta-Q".

Stage 5  Energy versus Time Graph  (Model #1)

Exercise 4.2   Complete the table of Stage 4 for Model#1.

Exercise 4.3   Complete the graph of Stage 5 for Model#1.

5. A Small aetour -- Steady State   Joe decides to get serious about regular saving and has $20/wk deposited directly into his bank account. His long-time spending habit is to make weekly withdrawals that reflect the bank account - the more on deposit, the more he withdraws to spend on entertainment. Joe promises to withdraw just 10% of whatever is in the bank account for entertainment. If he starts with $300, what will happen to his account in the long run?  Numerous bank charges offset the little bit of interest the bank pays so we will put in just one output path.

Suppose he takes out 10% of the $300 just ahead of the $20 deposit.  So $30 comes out, $20 goes in, and the account drops to $290. The next week it will drop again, but not quite as much. The account keeps dropping, but not to zero because of the $20 weekly direct deposit. Here's the key point -- It will settle down to some steady value, say Qss. Can this be predicted without going through a long step-by-step calculation . . . ?   Well, in this special case it is easy.        We are given that                                          Withdrawals = .10Qss $/wk,  Deposits = 20 $/wk.                                                 We want withdrawals.to balance deposits so let's set them equal and solve for the Qss --
.10Qss = 20  or  Qss =  20/.10  =  $200.   This is the steady state value.		Fig. 6

Searching for Qss     The steady state in a general case.	Fig. 7
Jo = constant	Steady deposits or inflow
Net Flow  =  Inflow - Outflow	Our mantra
DQ  =  Jo  -  kQss	Right from the energy diagram
DQ  =  0	Since the contents don't change
kQss    =  Jo	Previous two steps & algebra
Therefore,   Qss  =  Jo / k	The steadty state value

Notice that this result is independent of the start value  Q(0) = Qo.  The steady state value is given by  Qss  =  Jo / k.  It doesn't matter how much is in the tank initially.   To get back to Joe, he could have started with $30 in the account or $3000 in the account -- in the long run he would still end up with $200. The steady state relation  Qss = Jo/k  can be used for any situation that can be modeled with the above energy diagram.

Exercise 4.4   Melissa's father puts $750 in a savings account at the beginning of the Fall semester and promises to put in $50 per week, providing she uses "judgement" in her spending habit. She believes that life is too short to let so much money sit in the bank, a bank whose interest payments just offset its many nitpicking charges. She decides that weekly withdrawals of 40% of whatever is in the account pushes the envelope, but that it will get by OK. Use the Step-by-step approach to construct a chart for the amount in Melissa's account over the first ten weeks of the semester. Predict approximately how much will be left at the semester's end (15 weeks).  Hint: Compute the steady state value.

Exercise 4.5   Three-quarters of a 300-pound pile of leaves rots each year. How long will it take the pile to disappear? (For mathematical leaves, the answer is Never.  However, we are supposed to be representing actual leaves. So decide in advance on the weight of a residue that you consider negligible. This is a judgement call -- you might decide 15 lb, 5 lb, or as little as one pound.)

Exercise 4.6   Suppose we analyze Melissa's situation in Ex. 4.2 a little differently.  Each week the new amount in the account is 60% of the old amount, plus the $50 that is deposited.  So Qnew = .60Qold + 50 . At the end of the first week the new amount is 60% of $750 plus the $50.  This comes to $450 plus $50, or $500.  Carry out this same procedure for the next nine weeks.  Notice how much simpler the calculations are than they were in Ex. 4.2.

6. Model #2   Small Storage Fed from a Large Reserve Storage

A fuel dump delivers a fixed percentage of its store to a field station each week. The field station draws a fixed percentage of its gasoline each week to supply vehicles.

Stage 1  Energy Diagram  (Model #2)

Figure 8-S1

Stage 2  Qualitative Graph  (Model #2)

Figure 8-S2

Stage 3  Flow Equations  (Model #2)

      Suppose the fuel dump starts with 800 gallons of gasoline and delivers a=30% of its store to a field station every week. The station starts with 250 gallons and consumes k = 25% of its stored fuel each week. (The path coefficient k combines vehicle fuel b and loss c, so k = b+c.)  Describe the situation in the dump and the station over a time period of three months.

Begin interpreting the diagram . . .	Figure 8-S3
Net Flow  =  Inflow - Outflow      E-tank	Our mantra for the Fuel Dump
DE  =   0   -   .30E  gal/wk	a = .30, a Given
E(0) = 800  gal	"fuel dump starts with 800 gals"
Net Flow  =  Inflow - Outflow      Q-tank	Our mantra for the Field Station
DQ  =  .30E  -  .25Q   gal/wk	What leaves E goes to Q We're that given  k = .25 = b+c
Q(0)  =  250   gal	"The station starts with 250 gal's"

Stage 4  Step-by-Step Solution  (Model #2)

      Our time units will be weeks.  The total time period is three months, a quarter or 13 weeks. Because of the start-up values, our table will have 14 time entries, from 0 to 13.  As in Model #1, the change during each unit step will be indicated by DQ, read "delta-Q".

Stage 5  Energy versus Time Graph  (Model #2)

 Exercise 4.7   Complete the table of Stage 4 for Model #2.  

Exercise 4.8  Use the data from Ex. 4.7 to construct an Energy vs Time graph.  Plot 14 points for each storage facility. Connect each set of dots -- one from the E-tank and one from the Q-tank -- with a smooth curve.
    •   Be sure that the intervals on the Time axis are equal
    •   Be sure that the intervals on the Energy axis are equal (or the chart will be distorted)
    •   Be sure to include physical units
See the graph in Stage 5 of Model #1 to see how it should be done. Notice the ID box for the chart.

Exercise 4.9   Estimate the maximum number of gallons of gas in the field station. When will the dump run out of fuel? When will the station run out of fuel? (The answer for mathematical gasoline might be Never, but this does not address the intention of the question, which is about physical, not mathematical fuel.)



7. The Dynamical System Approach   There is another approach to Stage 3 that leads to numerical solutions.  It is not much different from the way we think of New Year.  For example, suppose you were alive in 1985.  At midnight December 31st the Old Year, 1985 became the New Year, 1986.  This records that the earth made a complete circuit around the sun.  In other words, at each click of the sun-earth clock we get a change in the year,  Ynew = Yold + 1.  Of course 1986 gets the same treatment when 1987 rolls around.  We update at each click of the clock.  Any system that is described by such an equation is called a dynamical system. This is our basic pattern for the new approach Model #2.. 

Figure 9  is the energy diagram for Model #2.  The system has a certain number of gallons of fuel in each tank and is subject to change at each click of the weekly clock.  We can think of it as going from an old state to a new state  So the old E will be tranformed to a new E, and the old Q will be transformed to a new Q.  Briefly, Eold changes to.Enew, and Qold changes to Qnew. .Does the energy diagram suggesta way we can detemine the new values from the old . . . ?  The answer is Yes.        The E-tank loses 30% of its store every week, so the new E has only 70% of the old E.  The new Q keeps only 75% of the old Q but gains 30% of the old E.  Notice that this can, like the floequation, be read directly from the diagram.  Here it is in our notation language --                      Enew  = .70·Eold   and   Qnew  = .75·Qold  + .30·Eold.
		Figure 9  Model #2

Now we can go from the start values E(0) = 800 gal, Q(0) = 250 gal to values at the end of the 1st week, and then use those values to find values at the end of the 2nd wk, and so forth.  Let's actually calculate a few time steps, and some familar numbers will appear.

The new-old notation is very friendly and is excellent for finding relations directly from a diagram.   However, we need to re-express the equations into a more explicit form.  Suppose we are at the week n. Then the next week is n+1.  A notation that captures this information is written alongside our new-old notation --

Here are the first three steps in the new notation --

Since this new notation returns the same answers and bristles with subscripts, why would we use it?  First of all, the subscripts in the calculations tell you exactly what week (or time-step) you are working on.  Also, notice that the pair of equations in the box actually contain all possible pairs of equations since n represents any week.  Finally, this standard dynamical systems notation will allow us to enter the equations in graphers and other computational devices and automatically carry out the calculations.

 Exercise 4.10   Carry out the rest of the dynamical system calculations for three more weeks and verify that they agree with the table that was completed in Ex. 4.7.Use whichever notation you prefer, the new-old or the "porcupine".

Exercise 4.11   Replace the input and output parameters of Model #2 by .20 and .10, resp. (See Fig. 9)  Write down the dynamical system equations that describe this system.  Use whichever notation you prefer, the new-old or the "porcupine".

Exercise 4.12   The equations used In Stage 4 to calculate the E and Q values for Model #2 are   DE = -.30E  and  DQ =.30E - .25Q. These are changes are per unit time-step, so we could write them as  DE = -.30En  and     DQ = .30En - .25Qn.  However, we can also write  DE =  En+1 - En  and   DQ = Qn+1 - Qn, since the right side of each equation is also a change per unit time-step.  Can you derive the dynamical system "porcupine" equations in the box by setting corresponding "D's" to each other . . . ?

Project P-1  (Based on Model #2)

     In the island warfare of the Pacific (WW II) a small fuel dump could be replenished by PT boats.  Suppose a PT boat can sneak in under cover of darkness once a week and deliver two barrels of fuel to the dump (A barrel holds 55 gallons). The dump delivers a percentage of its stored fuel to a field station every week. The field station consumes a certain percentage of its fuel every week.
      Take as your initial conditions and path parameters the same values as in Model #2.  Analyze this situation over three months and predict the long-run situation for the dump and the station.  Write up the project (outlined below), using no more than four pages.