Physics: 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

Answer is d) part as an everday example is when you run fast your body needs more oxygen so your heartbeat increases...

Suppose that a large mixing tank initially holds

Ответ:

ginger87771

03.05.2021

Physics

$\therefore \quad \frac{d A}{d t}=6-\frac{2 A(t)}{300+t}$

Explanation:

$\because \quad \frac{d A}{d t}=R_{i n}-R_{\text {out }}$

Thus, we find $R_{i n}$ and $R_{\text {out first }}$

$\because R_{i n}=( concentration of salt in inflow ) \cdot$ (input rate of brine)

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

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

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

Thus, after t minutes there will be 300+t gallons in tank

$\because R_{\text {out }}=$ (concentration of salt in outflow) \cdot (output rate of brine)

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

Now, we substitute by these results in the $\mathrm{DE}$ to get

$\therefore \quad \frac{d A}{d t}=6-\frac{2 A(t)}{300+t}$

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