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31 July, 21:41

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.

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  1. 1 August, 01:30
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    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
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