mathematical modelling!

Jul 2011
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A conical tank of height 3m is full of water and the radius of surface is 1m. After 8 hrs the depth of water is 1.5m. If we assume that the water
evaporate at a rate proportional to the surface area expose to the air, obtain a mathematical model for predicting the volume of water in the tank
at any time t.

trying to solve this model since two days, and this is what i have been able to come up with.

factors/variables/parameters
1 radius of conical tank (r)
2 height (h)
3 cross-sectional area \(\displaystyle \pi r \sqrt{r^2 + h^2} \)
4 volume in t time V(t)
5 rate of evaporation \(\displaystyle \frac{1}{2} mv^2 = mgh , v = \sqrt{2gh} \)

then i came up with this model formulation

\(\displaystyle V(t) = \pi r \sqrt{r^2 + h^2} - \sqrt{2gh} t \) , v in volume, t in hrs


pls i will greatly appreciate any idea on this, thanks
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
A conical tank of height 3m is full of water and the radius of surface is 1m. After 8 hrs the depth of water is 1.5m. If we assume that the water
evaporate at a rate proportional to the surface area expose to the air, obtain a mathematical model for predicting the volume of water in the tank
at any time t.

trying to solve this model since two days, and this is what i have been able to come up with.

factors/variables/parameters
1 radius of conical tank (r)
2 height (h)
3 cross-sectional area \(\displaystyle \pi r \sqrt{r^2 + h^2} \)
4 volume in t time V(t)
5 rate of evaporation \(\displaystyle \frac{1}{2} mv^2 = mgh , v = \sqrt{2gh} \)

then i came up with this model formulation

\(\displaystyle V(t) = \pi r \sqrt{r^2 + h^2} - \sqrt{2gh} t \) , v in volume, t in hrs
This is obviously wrong because it has the wrong units. r and h have units of "length" (meters, feet, etc.) so \(\displaystyle r^2+ h^2\) has units of "length squared" and then \(\displaystyle \sqrt{r^2+ h^2}\) has units of length again. Multiplying that by r gives units of "length squared" which is an area unit, not volume. The second part is even worse. The only length in \(\displaystyle \sqrt{2gh}t\) is "h" and since it is under a square root sign that has units of "square root of length" which cannot even be added to "length squared", much less give "length cubed" which is a volume unit. Frankly, I cannot imagine how you got that!

You are told that this is a cone- if you do not know the formula for a cone and your textbook does not have it, you could "google" "volume of cone" and get \(\displaystyle V= \frac{1}{3}\pi r^2 h\). The surface of the water is a circle- it has area \(\displaystyle \pi r^2\).

Also you suddenly introduce the formula \(\displaystyle \frac{1}{2}mv^2= mgh\) as "rate of evaporation". Where did you get that? The left side is kinetic energy, the right side is potential energy under gravity. Evaporation, and this problem in general, has nothing to do with energy or gravity.

You are told that the water evaporates at a rate "proportional to the surface area". The surface area is the area of that top circle, \(\displaystyle \pi r^2\). The water evaporating reduces the volume so that says the \(\displaystyle \frac{dV}{dt}= k\pi r^2\) where "k" is the proportion.
Because \(\displaystyle V= \frac{1}{3}\pi r^2h\) that equation becomes \(\displaystyle \frac{1}{3}\pi \frac{d(r^2h)}{dt}= k\pi r^2\).

Now, that involves two dependent variables, r and h. We need a formula connecting the two so we can replace one. That comes from "A conical tank of height 3m is full of water and the radius of surface is 1m." The height is 3 times the radius and as the water level decreases that ratio stays the same: h= 3r so we have \(\displaystyle \frac{1}{3}\frac{d(3r^3)}{dt}= 3r^2\frac{dr}{dt}= \k\pi r^2\).

You should be able to solve that for r, as a function of time, then find h= 3r, and so find V as a function.


pls i will greatly appreciate any idea on this, thanks
 
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Jun 2019
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Please i need a full solution to this question. I mean need. Can someone help please