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)
