## Related questions with answers

# 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?

Solution

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

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

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

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

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

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