Question

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.

Answer 1

Answer:

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