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Question

A $500$-gal aquarium is cleansed by the recirculating filter system schematically. Water containing impurities is pumped out at a rate of $15 \mathrm{gal} / \mathrm{min}$, filtered, and returned to the aquarium at the same rate. Assume that passing through the filter reduces the concentration of impurities by a fractional amount $\alpha$, as shown in the figure. In other words, if the impurity concentration upon entering the filter is $c(t)$, the exit concentration is $\alpha c(t)$, where $0<\alpha<1$. (a) Apply the basic conservation principle (rate of change = rate in - rate out) to obtain a differential equation for the amount of impurities present in the aquarium at time t. Assume that filtering occurs instantaneously. If the outflow concentration at any time is $c(t)$, assume that the inflow concentration at that same instant is $\alphac(t)$. (b) What value of filtering constant a will reduce impurity levels to $1%$ of their original values in a period of $3$ hr?

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First, for familiarity with previous problems let us denote the amount of impurities in the aquarium at a given time by Q(t)Q(t). As the rate at which water is pumped in and out of the aquarium is constant and given by r=15galminr= 15 \frac{gal}{min}, the volume of water in the aquarium is constant and is V=500galV = 500 gal. The concentration of the impurities is calculated by c(t)=Q(t)Vc(t) = \frac{Q(t)}{V} so we can model the change in the amount of impurities by:

dQ(t)dt=αc(t)rc(t)r\boxed{ \frac{dQ(t)}{dt} = \alpha c(t) r - c(t)r }

since at each moment we have water flowing in at the rate rr with concentration of impurities αc(t)\alpha c(t) and water flowing our at the same rate with the concentration c(t)c(t). Using the fact that c(t)=Q(t)Vc(t) = \frac{Q(t)}{V} we can rewrite this equation as \boldmath

dQ(t)dt=αrVQ(t)rVQ(t)==(α1)rVQ(t)=15(α1)500Q(t)\boxed{\begin{aligned} & \frac{dQ(t)}{dt} = \frac{\alpha r }{V} Q(t) - \frac{r}{V}Q(t)= \\ & = \frac{(\alpha -1)r}{V}Q(t) = \frac{15 (\alpha -1)}{500} Q(t)\end{aligned}}

We have completed the a)a) part of the exercise.

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