Lake Erie has a volume of roughly 100 cubic miles, and its equal inflow and outf

Question

Lake Erie has a volume of roughly 100 cubic miles, and its equal inflow and outflow rates are 40 cubic miles per year. At year t = 0, a certain pollutant has a volume concentration of 0.05%, but after that the concentration of pollutant flowing into the lake drops to 0.01%. Answer the following questions, assuming that the pollutant leaving the lake is well mixed with lake water.

a) What is the IVP satisfied by the volume V (in cubic miles) of pollutant in the lake?

b) What is the volume V of pollutant in the lake at time t?

c) How long will it take to reduce the pollutant concentration to 0.02% in volume?

Explanation / Answer

The rate at which the amount of pollutant in the late changes with time is given by:

rate of change in amount of pollutant = rate pollutant enters lake - rate pollutant leaves lake

Let V be the total volume of Lake Erie (100mi^3), and let P(t) be the amount (measured in units of mi^3) of pollutant in the lake at time t. We are told that the rate at which water enters and leaves the lake (call this f) is the same (40 mi^3/yr), so the total volume of the lake stays the same over time.

The rate at which pollutant is added to the lake is simply the concentration of pollutant in the incoming water (call this C_in, and we are told that this is 0.01% = 1*10^-4) multiplied by the rate at which water enters the lake:

rate pollutant enters lake = C_in * f

The rate at which pollutant leaves the lake is simply the concentration of pollutant in the lake at a given time, multiplied by the rate at which water leaves the lake. The concentration of pollutant in the lake is given by the amount of pollutant in the lake at that time divided by the total volume of the lake:

rate pollutant leaves lake = (P(t)/V)*f

Putting this all together gives the differential equation that describes the rate of change in the amount of pollutant in the lake at a given time:

dP(t)/dt = C_in*f - (f/V)*P(t)

Plugging in the numbers for this case gives:

dP(t)/dt = (1*10^-4)*(40 mi^3/yr) - ((40mi^3/yr)/(100mi^3))*P(t)

dP(t)/dt = (4*10^-3 mi^3/yr) -( 0.40/yr)*P(t)

For part (b), we have to solve the differential equation. This is a separable equation of the form: dy/dt = a - b*y.

dP(t)/dt = C_in*f - (f/V)*P(t)

dP/(C_in - (1/V)*P) = f dt

(-v)*ln[(C_in - (1/V)*P)/c] = f*t

where c is the constant of integration.

ln[(C_in - (1/V)*P)/c] = (-f/V)*t

(C_in - (1/V)*P) = c*exp((-f/V)*t)

(1/V)*P = C_in - c*exp((-f/V)*t)

P(t) = (V)*[C_in - c*exp((-f/V)*t)]

Plugging in the appropriate constants for this case, we have:

P(t) = (100mi^3)*[1*10^-4 - c*exp(-0.4*t/yr)]

Now use the initial condition to solve for the constant of integration:

At t = 0, P(0)/V = P(0)/(100 mi^3) = 0.05% = 5*10^-4, so P(0) = 0.05 mi^3.

0.05 = 100[(1*10^-4 - c*exp(0)]

-4.0*10^-4 = c

P(t) = (100mi^3)*[10^-4 + 4*10^-4 * exp(-0.4*t/yr)]

P(t) = (0.01 mi^3)*[1 + 4*exp(-0.4*t/yr)]

This last equation gives the amount of pollutant in the lake (in mi^3) as a function of time in years.

For part c, we want to know when P(t)/V = 0.02% = 2*10^-4, which is equivalent to asking when P(t) = 0.02 mi^3. Plug this value of P(t) into the above equation and solve for t:

0.02mi^3 = 0.01mi^3 * [1 + 4*exp(-0.4*t/yr)]

1 = 4*exp(-0.4*t/yr)

ln(1/4)*yr = -0.4*t

t = ln(4)*yr/0.4

t = 3.47 yr

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