Differntial equations Mathmatical model compartmental anaylyis. A simple mathema

Question

Differntial equations Mathmatical model compartmental anaylyis.

A simple mathematical model that can be used to determine the time it would take to clean up the Great Lakes can be developed using a multiple compartmental analysis approach. In particular, we can view each lake as a tank that contains a liquid in which is dissolved a particular pollutant (DDT. phosphorus, mercury). Schematically, we view the lakes as consisting of five tanks connected as indicated in Figure 5.55. For a detailed discussion of this model, see An Introduction to Mathematical Modeling by Edward A. Bender (Krieger, New York. 1991). Chapter 8. For our model, we make the following assumptions: The volume of each lake remains constant. The flow rates are constant throughout the year. When a liquid enters the lake, perfect mixing occurs and the pollutants are uniformly distributed. Pollutants are dissolved in the water and enter or leave by inflow or outflow of solution. Before using this model to obtain estimates on the cleanup times for the lakes, we consider some simpler models: Use the outflow rates given in Figure 5.55 to determine the time it would take to "drain" each lake. This gives a lower bound on how long it would lake to remove all the pollutants. A belter estimate is obtained by assuming that each lake is a separate lank with only clean water flowing in. Use this approach to determine how long it would take the pollution level in each lake to be reduced to 50 percentage of its original level. How long would it take to reduce the pollution lo 5c/c of its original level? Finally, to lake into account the fact that pollution from one lake flows into the next lake in the chain, use the entire multiple compartment model given in Figure 5.55 to determine when the pollution level in each lake has been reduced to 50 percentage of its original

Explanation / Answer

Solution :

dA/dt = rate in-rate out where A is the amount of pollution at time t.

First I wrote equations for each lake. Rather than using A as my variable, I used the first letter of each lake (with n for Ontario) to stand for the amount of pollution in the given lake at time t. This gives...

ds/dt = 15s/ 2900

dm/dt = 38m/1180

dh/dt = 15s/2900 + 38m/1180 68h/850

de/dt = 68h/850 85e/116

dn/dt = 85e/116 99n/393

Rearranging and pulling out the differential operator leads to the following system:

(D + 38/1180)[m] = 0

(D 68/850)[h] + 38m/1180 15s/2900 = 0

(D + 15/2900)[s] = 0

(D + 85/116)[e] 68h/850 = 0

(D + 99/393)[n] 85e/116 = 0

Converted all the fractions into decimal form i.e.

38/1180 = 0.0322

68/850 = 0.0800

15/2900 = 0.0051

85/116 = 0.7327

99/393 = 0.2519

Using the Laplace method for converting these differential equations into algebraic equations , thereby we will solve the equations and get the Laplace transforms of S, H, M, E and N and then take the inverse Laplace transforms of these to get the final equations of pollution for these lakes . Now s(0) = h(0) = m(0) = e(0) = n(0) = p , since the initial conditions are same for all the lakes .

Therfore using Laplace notation we have ,

sS s(0) + 0.0051S = 0

Therefore S = p/(s+0.0051)

Similarly we have M = p/(s+0.0322)

Now for (2) differential equation we have

(D 68/850)[h] + 38m/1180 15s/2900 = 0

Applying Laplace we will have

H = 0.0322p/((s + 0.0322)(s + 0.08)) * 0.0051p/((s + 0.0051)(s+0.08)) + p/(s + 0.08)

Similarly , you can plug in this value of H in the remaining differential equations (along with the value of M and S found previously to find the Laplace expressions for E and N , and then take the Laplace inverse of all and you will get the equations for all lakes .

Now for getting 5% time, just put 0.05p and 0.5p for 50%. Equate it to get t for each case .

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.