Problem with solving differential equation to solve for time

[rico22]

Homework Statement

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:
M=pAW

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?

Homework Equations

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 I am 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.

 

[Chestermiller]

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.

 

[rico22]

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.

 

[rico22]

I forgot to mention that I based that equation from the first equation on the first post and then I took A = M/ρW.

 

[Chestermiller]

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.

 

[rico22]

this might be a silly question but where did you get 2174 from?

 

[Chestermiller]

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.

 

[rico22]

so this means that to get 99% of the mass it would take about 1000 seconds correct?

 

[Chestermiller]

Yes. Nominally, 15 minutes or so.