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Optimization of an evaporative cooler

  1. Nov 21, 2008 #1
    Hey all,
    I'm struggling to even start this problem:
    An evaporative cooler design uses a rotating wheel that is placed upright (vertical) and is partially submerged in water. The center of the circle is above the waterline. Let R be the radius of the wheel and x be the distance from the center to the waterline. As the wheel turns, the portion that was submerged remains we after it rotates above the waterline. The efficiency of the cooler is maximized when the wet area of the wheel above the waterline is the largest possible.

    a. Find the exact value of x which does this, in terms of R.
    b. What percent of the circle's diameter should be submerged?
     
  2. jcsd
  3. Nov 22, 2008 #2
    Re: Optimization

    Well, you want the greatest wet area above the waterline, so you want to find the area of the wet area in terms of R and x. If I understand the problem correctly, the region that will get wet is that band on the outside of the circle (R - x) thick, but then you'd have to subtract the area underneath the surface of the water to get the area that is wet above the surface of the water.
     
  4. Nov 22, 2008 #3
    Re: Optimization

    Yeah I've got that far, But would you use both the formula's A=pi(R^2-r^2) and A=piR^2? I'm having a difficult time trying to figure out what formulas to use.. and I'm almost sure I'll be taking the deriv of which ever it is.
     
  5. Nov 22, 2008 #4
    Re: Optimization

    Well, to find the area of the ring around the outside of the circle that will get wet, you can take the area of the whole circle, [tex]A = \pi R^{2}[/tex], and subtract the area of the inner circle that won't get wet, [tex]A = \pi x^{2}[/tex]. The area of that ring that will get wet will be [tex]A = \pi (R^{2} - x^{2})[/tex]. After that, you need to find the area of the submerged section and subtract that from the area of the ring.
     
  6. Nov 22, 2008 #5
    Re: Optimization

    maybe my mind is tired but I'm confused, should I be thinking about using derivatives to optimize the wet area above the waterline?
     
  7. Nov 22, 2008 #6
    Re: Optimization

    I believe so, you want to end up making a function that represents the wet area above the waterline. Then you want to find the maximum of the function, which you can find algebraically using derivatives.
     
  8. Nov 22, 2008 #7
    Re: Optimization

    Firstly, Thanks for your help and quick response...very much appreciated!
    So, if I'm understanding correctly, I'll find the function of the wet area above the waterline by combining the previous to functions of A..then, take its first derivative?
     
  9. Nov 22, 2008 #8
    Re: Optimization

    From the answers and the problem, R can be anything without affecting your answers. I think you should find a function that expresses the wet area above the waterline in terms of R and x, then substitute a 1 for the R so that it is now in terms of only x. Once you have the function, you can find its maximum because the maximum occurs when the first derivative of the function is 0.
     
  10. Nov 22, 2008 #9
    Re: Optimization

    Awesome! its starting to click =) now finding that function should be fun. Thank you again!
     
  11. Nov 22, 2008 #10
    Re: Optimization

    now if anybody knows how to tie in the area underneath the water with the formula for the area thats wet..
     
  12. Nov 22, 2008 #11
    Re: Optimization

    The area that will get wet is a ring:
    [tex]A = \pi (R^{2} - x^{2})[/tex]

    The area that is underneath the waterline is a segment and the area of a segment is something like:
    [tex]A = \frac{1}{2} r^{2} ( \theta - sin( \theta ))[/tex] where [tex]r[/tex] is the radius of the circle, and [tex]\theta[/tex] is the central angle of the segment.
     

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  13. Nov 24, 2008 #12
    Re: Optimization

    Hey Chaos,
    made a little progress with the problem. So I figured what I have to do it take both the above equations and combine pi(R^2-X^2)-1/2R^2(theta-sin(theta)) (sorry don't know how to do the eqations like you =) and First, Solve for angle theta. The angle theta will be the inverse cosine of X/R and since it's a double angle I'll end up with a Sin2(theta). Then, plug that value for theta into the original problem.
    Finally, I believe I need to take the derivative of pi(R^2-X^2)-1/2R^2(theta-sin(theta)) with the value of theta in it, set it equal to zero and solve for x. R is a constant.
    Any other suggestions? words of wisdom? Thanks again!
     
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