Answer:
C(t) = I/F [1 - e^(-Ft/V) ] + C₀e^(-Ft/V)
as t = 0 ; C(t) = 1/F
Explanation:
dC/dt) = (-F/V)*C+(I/V)
To make it easier to solve, let
Constants: I, V,F
Variables: C, t
workings and solution can be viewed below
(a) The concentration at any time [tex]t[/tex] is determined by [tex]C(t) = \left(C_{o}-\frac{I}{F} \right)\cdot e^{-\frac{F}{V}\cdot t }+\frac{I}{F}[/tex].
(b) The final concentration when [tex]t \to + \infty[/tex] is [tex]C_{\infty} = \frac{I}{V}[/tex].
(c) 6 years are needed to drop pollution to [tex]\frac{C_{o}}{10}[/tex] in the lake Eire.
(d) 431 years are needed to drop pollution to [tex]\frac{C_{o}}{10}[/tex] in the lake Superior.
(a) The differential equation is equivalent to:
[tex]\frac{dC}{dt} + \frac{F}{V} \cdot C = \frac{I}{V}[/tex] (1)
Whose solution is [tex]C(t) = \left(C_{o}-\frac{I}{F} \right)\cdot e^{-\frac{F}{V}\cdot t }+\frac{I}{F}[/tex], where [tex]t[/tex] is time, in minutes.
The concentration at any time [tex]t[/tex] is determined by [tex]C(t) = \left(C_{o}-\frac{I}{F} \right)\cdot e^{-\frac{F}{V}\cdot t }+\frac{I}{F}[/tex]. [tex]\blacksquare[/tex]
(b) [tex]e^{-\frac{F}{V}\cdot t } \to 0[/tex] inasmuch [tex]t \to +\infty[/tex], then we have that [tex]C_{\infty} = \frac{I}{V}[/tex].
The final concentration when [tex]t \to + \infty[/tex] is [tex]C_{\infty} = \frac{I}{V}[/tex]. [tex]\blacksquare[/tex]
(c) In this case, (1) is reduce into this form:
[tex]C(t) = C_{o}\cdot e^{-\frac{F}{V}\cdot t }[/tex] (2)
If we know that [tex]C(t) = \frac{C_{o}}{10}[/tex], then the time required is:
[tex]\frac{C_{o}}{10} = C_{o}\cdot e^{-\frac{F}{V}\cdot t }[/tex] (3)
[tex]\ln \frac{1}{10} = -\frac{F}{V}\cdot t[/tex]
[tex]t = - \frac{V}{F} \cdot \ln \frac{1}{10}[/tex]
[tex]t = -\frac{458}{175}\cdot \ln \frac{1}{10}[/tex]
[tex]t \approx 6.026\,yr[/tex]
6 years are needed to drop pollution to [tex]\frac{C_{o}}{10}[/tex] in the lake Eire. [tex]\blacksquare[/tex]
(d) The final concentration in lake Superior is:
[tex]t = -\frac{12221}{65.2}\cdot \ln \frac{1}{10}[/tex]
[tex]t \approx 431.593\,yr[/tex]
431 years are needed to drop pollution to [tex]\frac{C_{o}}{10}[/tex] in the lake Superior. [tex]\blacksquare[/tex]
The statement present mistakes and is poorly formatted. Correct form is shown below:
This problem models pollution effects in the Great Lakes. We assume pollulants are flowing into a lake at a constant rate of [tex]I[/tex] kilograms per year, and that water in flowing out at a constant rate of [tex]F[/tex] cubic kilometers per year. We also assume that the pollulants are uniformly distributed throughout the lake. If [tex]C(t)[/tex] denotes the concentration (in kilograms per cubic kilometer) of pollulants at time [tex]t[/tex] (in years), then [tex]C(t)[/tex] satisfies the differential equation [tex]\frac{dC}{dt} = -\frac{F}{V}\cdot C + \frac{I}{V}[/tex], where [tex]V[/tex]is the volume of the lake (in cubic kilometers). We assume that (pollutant-free) rain and streams flowing into the lake keep the volume of water in the lake constant.
(a) Suppose that the concentration at time [tex]t= 0[/tex] is [tex]C_{o}[/tex]. Determine the concentration at any time [tex]t[/tex] by solving the differential equation.
(b) Find [tex]\lim_{t \to \infty} C(t)[/tex].
(c) For Lake Eire, [tex]V = 458\,km^{3}[/tex] and [tex]F = 175\,\frac{km^{3}}{yr}[/tex]. Suppose that one day its pollutant concentration is [tex]C_{o}[/tex] and that all incoming pollution suddenly stopped (so, [tex]I = 0\,\frac{kg}{yr}[/tex]). Determine the number of years it would then take the pollution levels to drop to [tex]\frac{C_{o}}{10}[/tex]. Give your answer in decimal form, rounded to the nearest year.
(d) For Lake Superior, [tex]V = 12221\,km^{3}[/tex] and [tex]F = 65.2\,\frac{km^{3}}{yr}[/tex]. Answer the same question a in part (c) for Lake Eire.
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