# Evaporation differential problem

1. Oct 15, 2012

### ElliottG

1. The problem statement, all variables and given/known data

2. Relevant equations
Given in problem

3. The attempt at a solution

Hey guys

So I have this problem I'm working on and I just can't seem to get too far with it. I've gotten half of the first part...I'm just having trouble integrating the equation I got...and I don't even know if the equation I got is right in the first place!

I set:

Rin = pi
Rout = pi(20h-h^2)

dh/dt = Rin - Rout

I'm trying to solve the DE right now but don't know how to proceed...

Any help would be great,
Thanks.

2. Oct 16, 2012

### ElliottG

Anyone? Sorry for bumping this so early!

3. Oct 16, 2012

### jackmell

I looked at it and it looks very messy to include the evaporation term. But maybe I'm missing something. So instead, I worked an analogous problem by filling an inverted right triangle (like the cross-section of a trough made with two pieces of wood) with area in the same way: We can fill the area at a rate of $k$ and let the evaporation rate be proportional to the length of the base (on top of it) with proportionality constant $c$. So, I would write the change in area as:

$$\frac{dA}{dt}=f(A,t)=k-cL$$

but I can get the length of the base as a function of A. For a right triangle, that's $L(A)=\sqrt{4 A}$ so then we have including evaporation:

$$\frac{dA}{dt}=k-c\sqrt{4A}$$

Integrating:

$$\int_0^A \frac{dy}{k-c\sqrt{4y}}=\int_0^t dt$$

And even that simple case is still very messy and the function A(t) is in terms of a Lambert function.

Now, if you used the same principle, in your case, expressing area of the pond surface as a function of volume beneath it, the DE I think would be:

$$\frac{dV}{dt}=\pi-0.01 A(V)$$

but the function A(V) looks way too messy to do it symbolically.

Maybe someone else can help.

Last edited: Oct 16, 2012
4. Oct 16, 2012

### ElliottG

Thanks I greatly appreciated your time and effort.

Hopefully someone else will chime in soon.

5. Oct 17, 2012

### jackmell

We know from part (a) that $A(h)=\pi(20 h-h^2)$ and:

We can write:

$$\frac{dh}{dt}=\frac{dh}{dV} \frac{dV}{dt}$$

and first without evaporation we have:

$$\frac{dh}{dt}=\frac{dh}{dV} \frac{dV}{dt}=\frac{1}{2\pi Rh-\pi h^2} \pi$$

we can integrate that:

$$\int_0^h (2Rh-h^2)dh=\int_0^t dt$$
and get:

$$Rh^2-1/3h^3=t$$
and use that to figure out how long it takes to fill without evaporation.

Now with evaporation, we have:

$$\frac{dV}{dt}=\pi-kA(h)$$

so substituting, we obtain:

$$\frac{dh}{dt}=\frac{1}{2\pi Rh-\pi h^2}\left[\pi -k(20\pi h-\pi h^2)\right]$$

Now the question is, can we integrate that and is it correct? Gotta' try things in math you know, and learn to tolerate making wrong turns. :)