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Tank Draining/Filling problem

  1. Mar 7, 2008 #1
    1. The problem statement, all variables and given/known data

    A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 12 L/min. The solution is mixed and drains from the tank at the rate 6 L/min.

    Find the amount of salt in the tank after 1.5 hours.

    2. Relevant equations

    Assorted Calculus stuff.

    3. The attempt at a solution

    rate in = (.045kg per L)(12L per min) = .54kg/min

    rate out = (y(t)/2000 kg per L)(6L per min) = (y(t))/(333.33)

    dy/dt = .54 - ((y(t))/333.33) = (180 - y(t))/333.33

    int(dy/(180 - y)) = int(dt/333.33)

    -ln(180 - y) = t/333.33 + C

    y(0) = 90

    So -ln(90) = C

    -ln(180 - y) = t/333.33 - ln(90)

    180 - y = 90e^(-t/333.33)

    y(t) = 180 - 90e^(t/333.33)

    y(90) = 62.103

    Actual Answer:

  2. jcsd
  3. Mar 7, 2008 #2
    rate in Ri=12 L/min

    rate out Ro=6L/min
    concentration in Ci=0
    Vo(initial volume)=2000
    amount of salt in at t=0 x(o)=90 kg
    concentraton out Co= x(t)*Ro/(Vo+(Ri-Ro)t) so

    dx/dt=RiCi-x(t)Ro/(Vo+(Ri-Ro)t now solve this one
  4. Mar 7, 2008 #3


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    Science Advisor

    Where did you get the ".045 kg per L"? That is the initial concentration of salt in the tank but has nothing to do with salt entering the tank. You are told that Pure water enters the tank at the rate of 12 L/min. There is 0 kg/min of salt entering the tank- none at all.

    Almost correct. If the volume of water in the tank were always 2000 Liters, then the concentration would be y/2000. But you have 12 L of water entering the tank every minute and only 6 L of water leaving. The amount of water in the tank is increasing by 12- 6= 6 L every minute. After t minutes, there will be 2000+ 6t Liters of water in the tank and that should be your denominator.

  5. Mar 7, 2008 #4
    This time I tried...

    rate in = 0 kg/min

    rate out = y(t)/(2000 + 3t)(3 L/min) = y(t)/(666.66 + t)

    dy/dt = -y/(666.66+t)

    int(dy/-y) = int(dt/666.66+t)

    -ln(-y) = t/(666.66+t) + C

    y(0) = 80

    So -ln(80) = C

    -ln(-y) = (t/(666.66+t)) - ln(80)

    y(t) = -80e^(t/333.33)

    No luck this way either. (this is a new problem of the same type.)
  6. Mar 7, 2008 #5
    have a closer look at my post #2, i set up all the necessary things you need for this problem,all you need to do is solve the diff. eq i set up!!
  7. Mar 7, 2008 #6
    Oh, sorry! It seems like you have another problem now, right? so is this a new problem? What does it originally say?
  8. Mar 7, 2008 #7
    this defenitely cannot be right??? You keep doing the same mistake as in your first problem!!!!!
    Last edited: Mar 7, 2008
  9. Mar 7, 2008 #8
    Okay, I regened the problem again.

    A tank contains `70` kg of salt and `1000` L of water. Pure water enters a tank at the rate `8` L/min. The solution is mixed and drains from the tank at the rate `4` L/min.

    Find the amount of salt in the tank after 2.5 hours.

    It's the same concept as the two above, so what was I doing wrong there?
  10. Mar 7, 2008 #9
    initial volume Vo=1000 L
    Rate in Ri=8 L/min
    Concentration in Ci=0
    Rate out Ro=4 L/min
    Concentration out = x(t)*Ro/(Vo+(Ri-Ro)t)


    dx/dt= Ri*Ci- x(t)*Ro/(Vo+(Ri-Ro)t)

    remember Co depends on the amount of the salt on the tank, and also on the volume of the tank, so

    Co= x(t)/V, but the volume on the tank V, depends on the initial volume, and also on how the volume is changing over time as we pipe in and out water so, V=Vo+(Ri-Ro)t

    Do you see your flaws now???
  11. Mar 7, 2008 #10
    Okay...before I try to punch in the problem, does this look right to everyone?

    dx/dt = -(x(t)*4)/(1000 + (4)t)

    Co = x(t)/(1000+(4)t)\

    int(dx/-x) = int(dt/(1000+4t))

    -ln(-x) = t/(1000+4t) + C

    y(0) = 70

    so -ln(70) = C

    -ln(-x) = (t/(1000+4t) - ln(70)

    x(t) = -70e^(t/(1000+4t))
  12. Mar 7, 2008 #11


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    The integration is wrong. You can't just ignore the "t" in the denominator!
    Do a substitution: Let u= 2000/3+ t (I hate using approximations to fractions!) so that du= dt and the integral on the right becomes [itex]\int du/u[/itex].

    Also the integral of 1/(-y) is -ln|y|. Since y is the amount of salt in the tank, which can't be negative, you should have -ln(y). I think it would be simpler to leave the negative sign outside the integral.

    Last edited: Mar 7, 2008
  13. Mar 7, 2008 #12
    you did not integrate correctly!!!
  14. Mar 7, 2008 #13
    I'm a little confused as to the integration...do I just take out the negative to get...

    x(t) = 70e^(t/(1000+4t))

    Here are my integration steps, what exactly did I do wrong?

    int(dx/x) = int(dt/(1000+4t))

    -ln(x) = t/(1000+4t) + C
  15. Mar 7, 2008 #14
    your right hand side is not correctly integrated, you are doing the same mistake there!!!!

    [tex]\int \frac{dt}{1000+4t}[/tex] let [tex]u=1000+4t, \ \ du=4dt=>dt=\frac{1}{4}du[/tex] so

    [tex]\frac{1}{4}\int\frac{du}{u}=\frac{1}{4}ln(u)+C[/tex] now go back and substitute for u, what do you get?
    Last edited: Mar 7, 2008
  16. Mar 7, 2008 #15
    So is this...

    x(t) = 70e^(.25ln(1000 + 4t)))

    The formula I want?
  17. Mar 7, 2008 #16
    x(0)=70, right?

    [tex]x(t)=e^{C-.25ln(1000+4t)}=Ae^{-ln(1000+4t)^{\frac{1}{4}}}=A(1000+4t)^{-(\frac{1}{4})}[/tex] now [tex]x(0)=70=A(1000+4*0)^\frac{-1}{4}[/tex], what do u get for A?

    EDIT: i forgot a minus sign on the power, i edited it!!!
    Last edited: Mar 7, 2008
  18. Mar 7, 2008 #17
  19. Mar 7, 2008 #18
    So are you saying the x(t) function is

    12.45(1000 + 4x)^(1/4)

    Because that function increases over time, I need one that decreases.
  20. Mar 7, 2008 #19

    I edited my post #16, i had forgotten a minus sign on the power, have another look at it!!
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