My Math Forum Evaporation from changing surface

 Physics Physics Forum

 November 26th, 2014, 12:41 AM #11 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,130 Thanks: 716 Math Focus: Physics, mathematical modelling, numerical and computational solutions 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.
 November 26th, 2014, 06:28 AM #12 Newbie   Joined: Nov 2014 From: Slovakia Posts: 10 Thanks: 0 Thank you for your time.
 December 1st, 2014, 11:39 AM #13 Newbie   Joined: Nov 2014 From: Slovakia Posts: 10 Thanks: 0 Did you forgot at me?
 December 2nd, 2014, 01:17 AM #14 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,130 Thanks: 716 Math Focus: Physics, mathematical modelling, numerical and computational solutions 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!
 December 2nd, 2014, 01:43 AM #15 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,130 Thanks: 716 Math Focus: Physics, mathematical modelling, numerical and computational solutions 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 01:47 AM.
 December 3rd, 2014, 08:57 AM #16 Newbie   Joined: Nov 2014 From: Slovakia Posts: 10 Thanks: 0 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 09:03 AM.
 December 4th, 2014, 01:22 AM #17 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,130 Thanks: 716 Math Focus: Physics, mathematical modelling, numerical and computational solutions 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.

 Tags changing, evaporation, surface

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post skubo Applied Math 0 November 17th, 2014 04:44 AM Robertoo Calculus 1 June 19th, 2014 04:09 PM Ter Algebra 1 April 19th, 2012 11:09 AM uterfor86 Applied Math 2 January 28th, 2012 11:01 PM chuckienz Linear Algebra 1 May 6th, 2009 06:23 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top