Find the amount of salt in a tank at any time 𝑑.Β ?

Β A tank initially holds 60π‘”π‘Žπ‘™π‘  of brine solution containing 1 𝑙𝑏 of salt 6

per gallon. At 𝑑 = 0, another brine solution containing 2 𝑙𝑏 of salt per gallon is poured into the tank at the rate of 5π‘”π‘Žπ‘™/π‘šπ‘–π‘›, while the well- stirred mixture leaves the tank at the rate of 7π‘”π‘Žπ‘™/π‘šπ‘–π‘›. Find the amount of salt in the tank at any time 𝑑. Also the time at which the mixture in the tank contains 8 𝑙𝑏 of salt.

2 Answers

Relevance
  • 2 months ago
    Favorite Answer

    Vo = 60 gallons

    a = 1 lb

    b = 2 lb

    e = Β 5gal/min , f = 7gal/min

    Vo is the initial volume in the tank

    a = brine solution in the tank initially

    b = amount of salt present in the other brine solution

    e and f = rate at which new solution is being poured in the tank = rate at which well stirred

    Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β mixture is leaving the tank

    => dQ/dt + f/[Q/(Vo + et - ft)]=be

    dQ/dt +7*[Q/(100 + 5t - 7t)]=2*5Β 

    dQ/dt = 2*5 - 7*[Q/(100 - 2t)]

    => Q(t) = 10(60-2t) + C1(60-2t)Β²

    Applying the initial condition

    at t= 0 , Q = a = 10 we get : 10(60) + C1(60)Β²

    => C1 = -59/360

    Therefore, the amount of salt in the tank at any time t

    => Q(t) Β = 10(60-2t) + -(59/360)(60-2t)Β²

    when Q(t) = 8Β 

    => t = 29.59 minutes

    Your comments is greatly appreciated . Thank you.

  • Bryce
    Lv 7
    1 month ago

    dQ/dt= 2*5 – 7Q/(60 – 2t)

    integrating factor: e^∫ 7/(60 – 2t) dt= (-7/2)ln(60 – 2t)= (60 – 2t)^(-7/2)

    d/dt[Q(60 – 2t)^(-7/2)]= 10(60 – 2t)^(-7/2)

    Q(60 – 2t)^(-7/2)= 10*-2/5*-1/2(60 – 2t)^(-5/2) + c

    Q(t)= 2(60 – 2t) + c(60 – 2t)^(7/2)

    Q(0)= 60, c= -1/60^(5/2)

    Q(t)= 2(60 – 2t) – 1/60^(5/2)(60 – 2t)^(7/2)

    Q(28)β‰ˆ 8 lb salt

Still have questions? Get your answers by asking now.