Ask Question
9 April, 04:56

A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into the tank at a rate of 5 gal/minute. The well mixed solution is then pumped out at a rate of 10 gal/minute. Find the number A (t) of pounds of salt in the tank at time t. When is the tank empty?

+5
Answers (1)
  1. 9 April, 08:45
    0
    A) A (t) = 10 (100 - t) + c (100 - t) ²

    B) Tank will be empty after 100 minutes.

    Explanation:

    A) The differential equation of this problem is;

    dA/dt = R_in - R_out

    Where;

    R_in is the rate at which salt enters

    R_out is the rate at which salt exits

    R_in = (concentration of salt in inflow) * (input rate of brine)

    We are given;

    Concentration of salt in inflow = 2 lb/gal

    Input rate of brine = 5 gal/min

    Thus;

    R_in = 2 * 5 = 10 lb/min

    Due to the fact that the solution is pumped out at a faster rate, thus it is reducing at the rate of (5 - 10) gal/min = - 5 gal/min

    So, after t minutes, there will be (500 - 5t) gallons in the tank

    Therefore;

    R_out = (concentration of salt in outflow) * (output rate of brine)

    R_out = [A (t) / (500 - 5t) ]lb/gal * 10 gal/min

    R_out = 10A (t) / (500 - 5t) lb/min

    So, we substitute the values of R_in and R_out into the Differential equation to get;

    dA/dt = 10 - 10A (t) / (500 - 5t)

    This simplifies to;

    dA/dt = 10 - 2A (t) / (100 - t)

    Rearranging, we have;

    dA/dt + 2A (t) / (100 - t) = 10

    This is a linear differential equation in standard form.

    Thus, the integrating factor is;

    e^ (∫2 / (100 - t)) = e^ (In (100 - t) ^ (-2)) = 1 / (100 - t) ²

    Now, let's multiply the differential equation by the integrating factor 1 / (100 - t) ².

    We have;

    So, we;

    (1 / (100 - t) ²) (dA/dt) + 2A (t) / (100 - t) ³ = 10 / (100 - t) ²

    Integrating this, we now have;

    A (t) / (100 - t) ² = ∫10 / (100 - t) ²

    This gives;

    A (t) / (100 - t) ² = (10 / (100 - t)) + c

    Multiplying through by (100 - t) ², we have;

    A (t) = 10 (100 - t) + c (100 - t) ²

    B) At initial condition, A (0) = 0.

    So, 0 = 10 (100 - 0) + c (100 - 0) ²

    1000 + 10000c = 0

    10000c = - 1000

    c = - 1000/10000

    c = - 0.1

    Thus;

    A (t) = 10 (100 - t) + - 0.1 (100 - t) ²

    A (t) = 1000 - 10t - 0.1 (10000 - 200t + t²)

    A (t) = 1000 - 10t - 1000 + 20t - 0.1t²

    A (t) = 10t - 0.1t²

    Tank will be empty when A (t) = 0

    So, 0 = 10t - 0.1t²

    0.1t² = 10t

    Divide both sides by 0.1t to give;

    t = 10/0.1

    t = 100 minutes
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into the tank at ...” in 📙 Engineering if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers