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Suppose that a large mixing tank initially holds 300 liters of water in which 50 kg of salt have been dissolved. Another brine solution, with a concentration of 250 g/L, is pumped into the tank at a rate of 5 L/min, and the tank is also being emptied at a rate of 5 L/min. The fluid in the tank is being constantly stirred, and we can assume that, at any moment, the concentration is uniform throughout the tank. If A(t) represents the amount of salt (in kg) in the tank at time t, find a first-order differential equation for A(t), and write down the initial condition that goes with this equation.

Sagot :

Manethoidn

Answer:

[tex]\therefore \quad \frac{d A}{d t}=6-\frac{2 A(t)}{300+t}[/tex]

Explanation:

[tex]\because \quad \frac{d A}{d t}=R_{i n}-R_{\text {out }}[/tex]

Thus, we find [tex]R_{i n}$ and $R_{\text {out first }}[/tex]

[tex]\because R_{i n}=( concentration of salt in inflow ) \cdot[/tex] (input rate of brine)

[tex]\therefore \quad R_{i n}=(2 \mathrm{lb} / \mathrm{gal}) \cdot(3 \mathrm{gal} / \mathrm{min})=6 \mathrm{lb} / \mathrm{min}[/tex]

Since the solution is pumped out at a slower rate, thus it is accumulating

at the rate of [tex](3-2)=1 \mathrm{gal} / \mathrm{min}[/tex]

Thus, after t minutes there will be [tex]300+t[/tex] gallons in tank

[tex]\because R_{\text {out }}=[/tex] (concentration of salt in outflow) \cdot (output rate of brine)

[tex]\therefore \quad R_{o u t}=\left(\frac{A(t)}{300+t} \mathrm{lb} / \mathrm{gal}\right) \cdot(2 \mathrm{gal} / \mathrm{min})=\frac{2 A(t)}{300+t} \mathrm{lb} / \mathrm{min}[/tex]

Now, we substitute by these results in the [tex]\mathrm{DE}[/tex] to get

[tex]\therefore \quad \frac{d A}{d t}=6-\frac{2 A(t)}{300+t}[/tex]

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