G a tank contains 300 liters of fluid in which 40 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.

To solve this problem, we make a component mass balance. Let us say that the salt is called A. Therefore the mass of A in the tank over time is given by the differential equation:

d (mA) / dt = ΦmA (in) - ΦmA (out)

where mA is mass of A, ΦmA mass flow rate of A, and t is time

Since the fluid being pumped into the tank is pure water, therefore:

ΦmA (in) = 0

so,

d (mA) / dt = - ΦmA (out)

We also know that:

mass = concentration * volume

m = A * V

Φm = A * ΦV

Therefore:

d (A * V) / dt = - A ΦV

Since V is constant, we take it out of the derivative:

V dA / dt = - A ΦV

Rearranging:

dA / A = - ΦV dt / V

Integrating with limits of:

A = 40 to A

t = 0 to t

ln (A / 40) = ( - ΦV/V) t

since it is given that ΦV = 5 L/min and V = 300 L therefore:

ln (A / 40) = - (5/300) t

Taking the e on both sides:

A / 40 = e^ - (5t/300)

A = 40 e^ - (5t/300)