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

1. Nov 28, 2012

### rico22

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:
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?

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. Nov 28, 2012

### Staff: Mentor

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.

3. Nov 28, 2012

### 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.

4. Nov 28, 2012

### 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.

5. Nov 28, 2012

### Staff: Mentor

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.

6. Nov 28, 2012

### rico22

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

7. Nov 28, 2012

### Staff: Mentor

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.

8. Nov 29, 2012

### rico22

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

9. Nov 29, 2012

### Staff: Mentor

Yes. Nominally, 15 minutes or so.