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November 26th, 2014, 01:41 AM   #11
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I came up with a solution but it was a bit rubbish. When I have time today I'll see if my second attempt fairs better than my first! Some simple calculus should be enough to get the surface area of the top face of the cone as a function of time.
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November 26th, 2014, 07:28 AM   #12
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Thank you for your time.
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December 1st, 2014, 12:39 PM   #13
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Did you forgot at me?
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December 2nd, 2014, 02:17 AM   #14
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Sorry dude, I'm just really busy at work at the moment. I'll see if I have a spare moment today to address your issue. I was kinda hoping someone else from the forum would help you out, but no one has!
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December 2nd, 2014, 02:43 AM   #15
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Okay, I think I got it now.

We know $\displaystyle \frac{dV}{dt} = 0.0883 mm^3/hour$ and we want $\displaystyle \frac{dA}{dt}$ so we can get $\displaystyle A = A(t)$, where A is the surface area of the base of a cone (the cone is upside down so that the surface is facing upwards).

The volume of a cone is $\displaystyle V = \frac{\pi r^2 h}{3}$. Using basic trigonometry, we know that $\displaystyle \tan\theta = \frac{r}{h}$, so $\displaystyle h = r \cot\theta$. Therefore

$\displaystyle V = \frac{\pi r^2 h}{3} = \frac{\pi r^3 \cot\theta}{3}$

Since the surface area is a circle, we have $\displaystyle A = \pi r^2$ and consequently $\displaystyle V = \frac{\pi r^3 \cot\theta}{3} = \frac{A^{3/2}\cot\theta}{3\pi^{1/2}}$. The volume of the cone is now purely a function of $\displaystyle A$ and $\displaystyle \theta$, which is good because $\displaystyle A$ is the independent variable we want to vary over and $\displaystyle \theta$ is a constant depending on the slope of the filling vessel.

We can calculate $\displaystyle \frac{dV}{dA}$, which is $\displaystyle \frac{dV}{dA} = \left(frac{A}{\pi}\right)^{1/2} \frac{\cot\theta}{2}$

Now, using the chain rule, $\displaystyle \frac{dA}{dt} = \frac{dA}{dV}\cdot \frac{dV}{dt} = \frac{dV}{dt}\left(\frac{dV}{dA}\right)^{-1} = 0.0833 \pi^{1/2} \tan\theta \cdot \frac{1}{A^{1/2}}$

We can now integrate to get $\displaystyle A=A(t)$:

$\displaystyle \int A^{1/2} dA = \int 0.0833\pi^{1/2}\tan\theta dt$
$\displaystyle \frac{2A^{3/2}}{3} = 0.0833t \pi^{1/2}\tan\theta + c$
$\displaystyle A(t) = \left(0.125t \pi^{1/2}\tan\theta + \frac{3c}{2}\right)^{2/3}$

We know that when $\displaystyle t=0$, the surface area is $\displaystyle A = \pi r_0^2 = \pi \times 15^2 = 225\pi$. Therefore, $\displaystyle c = (\frac{3}{2} \times 225\pi)^{2/3} = 103.98$. So our final equation is

$\displaystyle A(t) = \left(0.125t \pi^{1/2}\tan\theta + 103.98 \right)^{2/3}$

Try it in your evaporation function I am expecting the rate of evaporation in your system to either:

i) be constant (because only the surface area for air entering the top matters); or
ii) vary with $\displaystyle t^{2/3}$

Also note that this only works for the sloped part of the vassal. As soon as the water reaches the cylindrical part of the vassal, the surface area remains constant as a function of time and the above function doesn't work.
Thanks from skubo

Last edited by Benit13; December 2nd, 2014 at 02:47 AM.
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December 3rd, 2014, 09:57 AM   #16
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Thank you vary much.
Yes I understand it, just one thing confuse me. I do not know where we lost number 2 from equation dV/dA, when you used chain rule. Can you explain me it?

Last edited by skubo; December 3rd, 2014 at 10:03 AM.
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December 4th, 2014, 02:22 AM   #17
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Yep, that's a mistake. Well spotted! It will change the constant in front of your $\displaystyle t$ term from 0.125 to 0.25.
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