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Homework Help: Problem with solving differential equation to solve for time

  1. Nov 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Oil is released from a submerged source at a constant flow of rate of Q=(0.1 m3)/s. Density of oil (p=870 kg/m3) is less than that of water and since oil is only sparingly soluble in water a slick will form on the water surface. Once its formed it will have a tendency to spread as more oil is added but also have a tendency to shrink as the oil both evaporates to the air and dissolves in the water. Eventually it would reach a constant size when these two tendencies became balanced.

    A mass balance on the slick provides the following relationship:
    dM/dt = rate of oil slick from the source - rate of dissolution - rate of evaporation

    where each term is in kg/s and where M is the mass of the slick. For this particular oil the rate of dissolution per square meter is 0.0000011 kg/m2s and the rate of evaporation per square meter is 0.004 kg/m2s.

    The mass M of the slick is related to its size by:

    where p is the density of the oil (given above), A is the surface area of the slick in m2, and W is the thickness of the slick in m. The thickness of a shrinking spill will decrease with time. In this case however, you will be looking at the time period between when it starts to grow and when it reaches a constant size. Assuming a constant slick thickness is probably reasonable for this situation and W= 0.001 m is a realistic thickness to use.

    What surface area ( in m2) will the slick ultimately have?

    Assume that the constant release of 0.1 m3/s of oil begins at t=0 and that A=0 at t=0. About how long will it take for the slick to reach a constant size?

    2. Relevant equations

    3. The attempt at a solution
    Ok so the first question is pretty straight forward... the set up for the area was as follows:
    A= (rate of oil to slick from source)/(rate of dissolution + rate of evaporation)... which gives A = 21744.02 m2; which I can then use to find the Mass = 18917.3 kg.

    The second question is the one that im having problems with... I've set up my differential equation and solved for time,

    t= 1/0.0046 ln(87) - 1/0.0046 ln (87 -0.0046M)
    the problem is that when I plug in M this gives ln(0). What am I missing? did I completely screw up the differential equation set up? Any type of ideas or tips would be greatly appreciated.
  2. jcsd
  3. Nov 28, 2012 #2
    Lets see the differential equation you came up with. If you feel like it, you might also walk through how you came up with that equation. This is a pretty straightforward problem, so we should readily be able to help.
  4. Nov 28, 2012 #3
    the equation that I based everything from was:

    dM/dt = 87 kg/s - .0040011 M/ρw ⇔ dM/dt = 87 kg/s - .0046M... then I used this to set up my integral:

    ∫dM/(87 - 0.0046M) = ∫dt ⇔ -1/0.0046 ln (87 - 0.0046M) = t + C... and since A = 0 at t=0, C= -1/0.0046 ln(87)... and then I solved for t from the equation on my first post...thanks in advance.
  5. Nov 28, 2012 #4
    I forgot to mention that I based that equation from the first equation on the first post and then I took A = M/ρW.
  6. Nov 28, 2012 #5
    The idealized model is such that it takes an infinite amount of time to accumulate the final steady state mass. What the problem description is really asking is that, on a practical basis, how much time is required to closely approach the final steady state mass. It might help to re-express your solution in a little different way:

    M = 18917 (1 - exp(-t /2174))

    From this, you can see that the characteristic time constant for the mass growth is 2174 sec. This means that, after a time equal to one time constant, the mass is 63% of its final steady state value; after 2 time constants, the mass is 86% of its final steady state value; after 4 time constants, the mass is 98% of its final steady state value. Four time constants is 8700 sec., or about 2.4 hours.
  7. Nov 28, 2012 #6
    this might be a silly question but where did you get 2174 from?
  8. Nov 28, 2012 #7
    Oops. My mistake. 1/0.0046 = 217.4 sec, so

    M = 18917 (1 - exp(-t /217.4))

    and the time constant is 217.4 sec.
  9. Nov 29, 2012 #8
    so this means that to get 99% of the mass it would take about 1000 seconds correct?
  10. Nov 29, 2012 #9
    Yes. Nominally, 15 minutes or so.
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