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Help ?

  1. Jan 14, 2012 #1
    Help.... ???

    A 200 liter tank initially contains 100 liters of water with a salt concentration of 0.1 grams per liter.
    Water with a salt concentration of 0.5 grams per liter flows into the tank at a rate of 20 liters per
    minute. Assume that the fluid is mixed instantaneously and that this well-mixed fluid is pumped out
    at a rate of 10 liters per minute. Let c (t) and
    v(t), be the concentration of salt and the volume of
    water in the tank at time t (in minutes), respectively. Then,
    v`(t)=10
    v(t) c`(t) +20c(t)=10

    a) Solve these differential equations to find the particular solutions for v(t) and c(t).
    b) What is the concentration of salt in the tank when the tank first overflows?



    ---------
    v(0)= 100 L , c(0) = 0.1 gm/L , c(t)=0.5 gm/L

    v`= 10 ==> v (t) = 10t + k
    ==> v(0) = 0 + k =100 ==> k=100
    so : v(t) = 10 t + 100
    (mass) m= c(t) * v(t)
    m`=c`(t)v(t) + c(t) v`(t)
    10 = c`(t)(10t+100)+ c(t) *10
    c`(t) + c(t) (10/(10t+100)) = 10/(10t+100)
    by solving this DE :
    e^∫(1/ t+10) dx = t+10
    Multiplying through by both sides gives:
    (t+10)c`(t)+c(t)=1
    ∂/∂t{c(t) (t+10)} =1

    ==> by integral
    c(t) ( t+ 10) = t
    ==> c(t) = t/(t+10) +k
    c(0) = 0.1
    ==>
    0.1= 0 + k
    so c(t) = t/(t+10) +0.1
     
  2. jcsd
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