# Help with question regarding surface area

1. Nov 24, 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
First I converted my flow of rate into kg/s by multiplying it by its density. But the rate of dissolution and rate of evaporation give are per square meter so I guess my question would be if I multiply it by the Area of the slick once it becomes balanced would I be able to then get the rate of dissolution and evaporation in kg/s???

So I could set up my equation like this:
rate of oil to slick from the source = (rate of dissolution + rate of evaporation) A

all in kg/s except the area of course. And if I divide then the rate of oil slick from the source by the rate of dissolution and the rate of evaporation... would this give me, then the balanced area?

Or am I seeing this all wrong?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 24, 2012

### haruspex

Yes, that will give you the steady-state area. But the question asks how long it will take to get to that, so I think you'll need a differential equation for the growth period.

3. Nov 24, 2012

### frogjg2003

You're doing everything right.

4. Nov 24, 2012

### rico22

Thank you guys for the replies... for the amount of time couldn't I take the area that I found from the first question multiply it by the thickness to get the Volume and then divide this by the flow rate to get the time? Or am I missing something?

5. Nov 24, 2012

### frogjg2003

At t=0, A=0 and therefore dissolution and evaporation don't have an effect.
Over time, the rate of change in surface area decreases as the area increases due to the second and third terms in dM/dt.

You need to set up a differential equation and use the initial conditions to find the solution.