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Can you help me ?

  1. Jan 13, 2012 #1
    can you help me ??

    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?
     
  2. jcsd
  3. Jan 13, 2012 #2
    Re: can you help me ??

    you can first sole the top one and then solve the lower one.

    for what v(t) is the derivative 10. This will lead to an answer up to an additive constant (since the derivative of a constant is 0) but you know v(0)=100 so you can solve for that constant.

    then write c'(t) out in terms of c(t) and solve that differential equation. and again you know the concentration for t=0 (this time you will actually find a multiplicative constant.

    If it is the actual differential equations you're having trouble with I suggest you find a good introductory physics book or of course an introductory differential equations book.
     
  4. Jan 14, 2012 #3
    Re: can you help me ??

    i tray to solve it and i find this resolute, can any one tell me are my way right ??

    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
     
  5. Jan 14, 2012 #4
    Re: can you help me ??

    any one know how to solve b
     
  6. Jan 15, 2012 #5
    Re: can you help me ??

    c(t) = 0.5 does not make any sense since the concentration of salt does not remain the same for all time. To solve b simply calculate when the tank overflows (you know the volume of the tank). Then plug that time into c(t)
     
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