Question

tank contains 250 liters of fluid in which 20 grams of salt is dissolved. Pure water is then pumped into the tank at a rate of 5 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t.

Answer 1

Solution :

Given data :

$c_{in}$ = 1 g/L

$r_{in}$ = 5 L/min

$r_{out}$ = 5 L/min

$v_0$ = 250 L

$A_0$ = 20 g

∴ $r_{net} = r_{in}- r_{out}$

         = 5 - 5

          = 0

$c_{out} = \frac{A}{250} \ g/L$

Now, $\frac{dA}{dt}=(r_{in} \times c_{in}) - (r_{out} \times c_{out})$

$\frac{dA}{dt} = 5-5\left(\frac{A}{250}\right)$

$\frac{dA}{dt}+5 \left(\frac{A}{250}\right) = 5$

$\frac{dA}{dt}+5 \left(\frac{A}{250}\right) = 5 \text{ with} \ A_0 = 20$

Integrating factor = exp(5 t/250)

Therefore,

$A \times \exp (5t \ /250) = \text{integral of}\ 5 \times \exp (5t / 250) + C$

Put $A_0=250+C$

C = -230

$A \times \exp(5t/250) = 250 \exp(5t/250) + (-230)$

$A(t) = 250-230 \exp(-5t/250)$

$A(t) = 250-230e^{\left(\frac{-t}{50}\right)} \ g$