A 1000 L tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flows into the tank at a rate of 80 L/min. The fluid mixes instantaneously and is pumped out at a specified rate Rout. Let y (t) denote the quantity of salt in the tank at time t. Find the salt concentration when the tank overflows assuming that Rout=40 L/min.

Salt Concentration = 40g/L

Explanation:

let v (t) is the volume of tank which is written as v (t) = 200 + 40t, considering v (t) ≤ 1000

dy/dt = Rate in - Rate out

y' (t) = 50 x 80 - 40y (t) / v (t)

y' (t) = 4000 - 40y (t) / 40t + 500

y' (t + 12.5) + y (t) = 4000 (t + 12.5)

∫d/dt. [t + 12.5]. y (t). dt = ∫4000 (t + 12.5). dt

(t + 12.5). y (t) = 2000 (t + 12.5). dt

(t + 12.5). y (t) = 2000 (t + 12.5) ² + c

y (t) = 2000 (t + 12.5) + c / (t + 12.5)

at y (0) = 5000, as tank has 500L x 10g/L = 5000

c = (5000) (12.5) - 2000 (12.5) (12.5) = 62500 - 312500 = - 250000

y (t) = 2000 (t + 12.5) - 250000 / (t + 12.5)

v (t) = 500 + 40t, at tank full v (t) = 1000, solving for t, we get t = (1000 - 500) / 40 = 12.5

Concentration at t = 12.5 is y (t) / v (t)

Concentration = [{2000 (12.5 + 12.5) ² - 250000}/12.5 + 12.5]/1000

Concentation (12.5) = 40g/L