Ask Question
23 April, 15:02

A brine solution of salt flows at a constant rate of 7 L/min into a large tank that initially held 100 L of brine solution in which was dissolved 0.1 kg of salt. The solution inside the tank is kept well stirred and flows out of the tank at the same rate. If the concentration of salt in the brine entering the tank is 0.02 kg/L, determine the mass of salt in the tank after t min. When will the concentration of salt in the tank reach 0.01 kg/L? Determine the mass of salt in the tank after t min.

+1
Answers (1)
  1. 23 April, 18:43
    0
    Let x (t) represent the amount of salt (in kg) at any time, t (in minutes).

    Step 1

    Write down the input rate and output rate in terms of x (showing how you get units). What is the differential equation for x (simplified) ?

    Input rate: (7L/min) x (0.02kg/L) = 0.14 = 7/50

    Output rate: (7L/min) (x/100 kg/L) = 7x / 100

    Differential Equation: δx/δt =

    => 7/50 - 7x/100

    step 2

    Find the general solution for your differential equation

    DE is linear with μ (t) = e^ (7t/100), so solving we have

    x = x (t) = e^ ( - 7t/100) ∫7/50 e^ ( - 7t/100) dt

    = 50 + C e^ ( - 7t/100)

    The initial condition mass of salt given is 0.1kg, using this we find the constant of integration

    so, when t = 0, x = 0.1, therefore

    0.1 = 50 + C

    C = - 49.9

    Hence x (t) = 50 - 49.9e^ ( - 7t/100)

    When will the concentration of salt in the tank reach 0.01 kg/L?

    concentration = 0.01kg/L

    This is the same as

    x/100 = 0.01

    => x = 1

    Substituting x = 1 into the solution yields

    1 = 50 - 49.9e^ ( - 7t/100)

    t = 9.2486

    Therefore the concentration of salt in the tank reach 0.01kg / L after 9.2486 minutes.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A brine solution of salt flows at a constant rate of 7 L/min into a large tank that initially held 100 L of brine solution in which was ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers