Ch 2.5 Lake Pollution Models

• Pollution in our lakes and rivers has become a major problem, particularly over the past 50 years.

• In order to improve this situation in the future, it is necessary to gain a good understanding of the processes involved.

• Some way of predicting how the situation might improve (or decline) as a result of current management practices is vital.

• To this end we need to be able to predict how pollutant amounts or concentrations vary over time and under different management strategies.

• Before developing the differential equation describing this process for a lake, let's consider the simpler case of a salt solution in a tank.
• We are interested in modeling amount \( S(t) \) kg of salt (the solute) dissolved in a tank of water (the solution).
  

• Suppose that the tank contains 100 liters of salt water.
  
• Initially \( s_0 \) kg of salt are dissolved in tank.
• Salt water flows into the tank at a rate of 10 liters per minute.
• The concentration \( c_{in}(t) \) kg/liter of salt in this incoming flow of salt water varies with time.
• The tank is well-mixed and salt solution flows out at same rate that it flows in (tank volume remains constant).

• Shown below is the input-output compartmental diagram for salt solution in tank.

• From the compartment diagram, we can write the word equation as

• The rate at which salt enters tank is the product of flow rate in and concentration of salt in the incoming mixture:

\[ \left(10 \frac{L}{min}\right) \left(c_{in}(t) \frac{kg}{L}\right) = 10 c_{in}(t) \frac{kg}{min} \]

• The rate at which salt leaves tank is the product of flow rate out and concentration of salt in the outgoing mixture:

\[ \left( 10 \frac{L}{min}\right) \left(\frac{S(t)}{100} \frac{kg}{L} \right) = \frac{S(t)}{10} \frac{kg}{min} \]

• We can now write the word equation as a differential equation (IVP):

\[ \frac{dS}{dt}= 10 c_{in}(t) - \frac{S(t)}{10}, \,\, S(0) = s_0 \]

• Linear first order ODE

\[ \frac{dS}{dt} = 10 c_{in}(t) - \frac{1}{10}S(t), \,\, S(0)= s_0 \]

• Normal form

\[ \frac{dS}{dt} + \frac{1}{10}S(t) = 10 c_{in}(t) , \,\, S(0)= s_0 \]

• Solve using method of integrating factors.

• Normal form

\[ \frac{dS}{dt} + \frac{1}{10}S(t) = 10 c_{in}(t) , \,\, S(0)= s_0 \]

• Solve using method of integrating factors

\[ \begin{align*} \mu(t) & = e^{\int \frac{1}{10} dt } = e^{\frac{t}{10} }\\ S(t) & = e^{-\frac{t}{10} } \int 10 e^{\frac{s}{10} }c_{in}(s) ds + s_0 e^{-\frac{t}{10} } \end{align*} \]

• Response due to initial condition goes to zero as t gets large.

\[ \lim_{t \rightarrow \infty} s_0 e^{-\frac{t}{10}} = 0 \]

• Normal form for IVP

\[ \frac{dS}{dt} + \frac{1}{10}S(t) = 10 c_{in}(t) , \,\, S(0)= s_0 \]

• Solution for \( c_{in}(t) = c_1 \):

\[ S(t) = s_0 e^{-t/10} + 100c_1(1-e^{-t/10}) \]

• Solution for \( c_{in}(t) = 0.2 - 0.1\sin(t) \):

\[ S(t) = s_0 e^{-t/10} + 20 + \frac{10}{101}(\sin(t) - 10\cos(t) - 192e^{-t/10}) \]

• \( c_0 \) = initial concentration of pollution in lake.
• Concentration \( c_{in} \) of pollutant in the flow entering lake is constant:

\[ c_{in}\, \frac{g}{m^3} = \mathrm{constant} \]

• Rate \( F \) at which water flows in and out of the lake is constant:

\[ F \, \frac{m^3}{day} = \mathrm{constant} \]

• For the mass \( M(t) \), we have:

\[ M'(t) = F c_{in} - F \frac{M(t)}{V} \]

• Since \( C(t) = M(t)/V \) and \( C'(t) = M'(t)/V \), we have:

\[ C'(t) = \frac{F}{V} c_{in} - \frac{F}{V}C \]

\[ \begin{align*} \frac{dC}{dt} & = \frac{F}{V}c_{in} - \frac{F}{V}C(t) , \,\, C(0)= c_0 \\ & = \frac{F}{V}\left(c_{in} - C(t)\right) \end{align*} \]

• Solve using separation of variables (see text)

\[ \begin{align*} C(t) & = c_{in} - (c_{in} - c_0) e^{-Ft/V } \\ & = c_{in}\left(1 - e^{-Ft/V }\right) + c_0e^{-Ft/V } \end{align*} \]

• Solution

\[ \begin{align*} C(t) & = c_{in} - (c_{in} - c_0) e^{-Ft/V } \\ & = c_{in}\left(1 - e^{-Ft/V }\right) + c_0e^{-Ft/V } \end{align*} \]

• Concentration of \( C(t) \) as \( t \) gets large.

\[ \lim_{t \rightarrow \infty} \left(c_{in} - (c_{in} - c_0) e^{-Ft/V }\right) = c_{in} \]

• How long will it take for the lake's pollution level to reach 5% of its initial level if only fresh water flows into the lake?
• From previous slide, with \( c_{in} = 0 \):

\[ C(t) = c_{in} - (c_{in} - c_0) e^{-Ft/V } = c_0 e^{-Ft/V } \]

• Solve for \( t \), and then use \( C = 0.05c_0 \):

\[ t = \frac{V}{F} \ln \left( \frac{C}{c_0} \right) = - \frac{V}{F}\ln(0.05) \cong \frac{3F}{V} \]

• From previous slide with \( C = 0.05c_0, \,c_{in} = 0 \):

\[ t \cong \frac{3F}{V} \]

• For Lake Erie, we can plug in values for \( F \) and \( V \)

\[ t \cong 7.8 \, \mathrm{years} \]

• For Lake Ontario:

\[ t \cong 23.5 \, \mathrm{years} \]