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Math Help - Application of Differential Equations

  1. #1
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    Application of Differential Equations

    The rate at which a solid dissolves varies directly as the amount of undissolved solid present in the solvent and as the difference between the saturation concentration and the instantaneous concentration of the substance.
    a saturated solution of salt in H2O will hold approximately 3 lb salt/gal. A block of salt weighing 60 lb is placed into a vessel containing 100 gal of H20. In 5 mins., 20 lb of salt were dissolved. (a) How much salt will be dissolved in one hour? (b) When will 45 lb of salt be dissolved?
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  2. #2
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    Your first sentence allows you to set up the DE. First thing to do is assign variable names to relevant quantities. Then you want to write out your DE. What do you get when you do that?
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  3. #3
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    DE = k (a-x)(Cs-Ci)dt is that correct?
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    Hmm. That might be right, but it's a bit confusing. Set up your variables first. Try constructing a list like this:

    t = time
    C_{s} = saturation concentration = const
    .
    .
    .

    What do you get?
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  5. #5
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    k = constant of proportionality
    a = total solid
    x = dissolved solid
    Cs = saturation concentration
    Ci = instantaneous concentration

    may i ask what is instantaneous concentration? is constant?
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    Ok, nice list. Some thoughts, then:

    1. The instantaneous concentration is going to be related to the amount of dissolved solid somehow, in, I think, a linear fashion. That is, I think you're going to have x = m C_{i} + b, where m and b are constants.
    2. The instantaneous concentration at time t is defined as the amount of dissolved solid divided by the total volume of the brine, in your case. The units would be lb./gal. It definitely changes with time. While you're going to try to find x(t) as the solution of your differential equation, I think C_{i} is going to play an important role.
    3. The DE you posted in post #3 is almost correct. Hint: look at the LHS.
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  7. #7
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    im still having trouble solving this problem, but i already have a solution but still can't figure out what i'm doing wrong

    dQ/dt = kQ(Cs -Ci)
    where dQ/dt = the rate of which the solid dissolves
    k = constant of proportionality
    Q= number of salt at any given time
    Cs = saturation concentration = 3 lb/gal
    Ci = Q/V= Q/100

    dQ/dt = kQ(3-Q/100)
    through bernoulli's equation i'll be able to get
    1/Q = 1/300 + ce^(-3kt)
    when t=0 Q=60
    therefore c=1/75
    so, the equation would be 1/Q =1/300 + (1/75)e^(-3kt)
    when t=5 Q=60-20=40
    k= -0.0323718772
    so, when t=60 Q=?
    1/Q = 1/300 +(1/75)e^[(-3)(-0.0323718772)(60)]
    1/Q = 4.469
    Q=.224

    b. 1/15 = 1/300 + (1/75)e^[(-3)(-0.0323718771)t]
    t = 16.07 mins

    i keep on rechecking my solution but still have the wrong answer
    the answer should be
    a. Q= .885
    b. t = 18.21
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  8. #8
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    The problem is your DE. You're confusing dissolved salt with undissolved salt. Let Q be the quantity of dissolved salt. Then you're going to have

    dQ/dt = k(60-Q)(Cs-Q/100), I think.

    Now, before you plunge ahead and solve this DE (which might be wrong!), I think you should justify this DE to me. Explain why it's correct or incorrect.
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  9. #9
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    well, i thought of that also, but im not sure which one is correct but i don't know how to solve that differential equation.
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  10. #10
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    Don't worry about solving the DE: we can deal with that one later. Right now, the important thing is to make sure we're solving the right problem. Take this sentence: "The rate at which a solid dissolves varies directly as the amount of undissolved solid present in the solvent and as the difference between the saturation concentration and the instantaneous concentration of the substance." and translate that directly into a differential equation, or compare to the one I gave you. Does it make sense from that perspective?
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  11. #11
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    The equation u gave me was the right one. I already solved it. thank u!
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  12. #12
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    You're very welcome. Have a good one!
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  13. #13
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    excuse me... i have the same problem as well... can you show me the solution on the D.E. dQ/dt = k(60-Q)(Cs-Q/100). thank you...
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  14. #14
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    \frac{dQ}{dt} = k(60 - Q)\left(C_S - \frac{Q}{100}\right)

    \frac{dt}{dQ} = \frac{1}{k(60 - Q)\left(C_S - \frac{Q}{100}\right)}

    t = \frac{1}{k}\int{\frac{1}{(60 - Q)\left(C_S -\frac{Q}{100}\right)}.

    Now solve the integral using Partial Fractions.
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