1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculating the ideal radius of a waterwheel given relevant constraints

  1. Oct 29, 2008 #1
    1. The problem statement, all variables and given/known data
    A solution to efficiently convert the potential energy of falling water into mechanical energy that will lift a weight is required. At the moment, I'm working to optimize my waterwheel design. A water hose with given dimensions will be placed above a platform 1 metre in height (therefore the max height of the waterwheel will be 1 metre).


    2. Relevant equations

    mgh = (mv^2)/2

    v=(2*Pi*r)/T

    3. The attempt at a solution

    Therefore I've taken two equations into consideration. Well, three, really. Assuming that 100% gravitational energy is converted to kinetic (I can dream) I've equated the two as:

    mgh = (mv^2)/2 (if only I could use maple notation) where m = mass in kg, v = speed in m/s, g = 9.8 m/s^2, h = height from the hose to top of wheel in metres

    Now looking at that I've determined that the actual mass is irrelevant to the kinetic velocity. Therefore,

    v=sqrt(2gh)

    But this doesn't entirely apply to uniform circular motion! So I dug up another equation for the tangential velocity of a point on a circle:

    v=(2*Pi*r)/T where T = the period of one revolution in seconds, r = radius

    Now for the constraints. The maximum height, that is from the base of the platform the waterwheel will be housed on to the tip of the hose, is precisely one metre. Therefore h=1-x, where x is the diameter of the wheel (it'd be more realistic to place some extra leeway since the wheel does have to turn and the water has to go somewhere...). Therefore, the radius of the wheel is x/2.

    Initially, I equated the two expressions.

    sqrt(2g(1-x))=(2*Pi*x/2)/T

    And solved for T.

    T=(2*Pi*r)/(sqrt(2gh))

    And then I took the derivative of the expression, dT/dx hoping to find a maximum or minimum value. Of course, this wasn't the correct variable to be looking for and I had found a maximum of 0.999~ metres (huge wheel!). The result T value for this radius was over a minute, and so of course this was absolutely incorrect for my purposes. (But if you ever wanted a wheel that hardly moved, there you go.)

    my dT/dx was (-19.6*Pi*sqrt(2g-2gx)-Pi*x)/(2*(sqrt(2g-2gx))^3) if you wanted to check my mechanics

    And so I realized that the variable I wanted to maximum was my velocity. But therein lies the problem and perhaps its just my algebra being miffed. I need an expression for velocity that will represent both the energy conversion and circular motion, but logically I'm finding it hard to come to a conclusion that incorporates both. I've taken the derivative of v=sqrt(2gh) where h is substituted with 1-x to fit the constraints, but this results in an answer of negative infinity. (I'd understand if this meant a very small number but this implies a very large negative number!)
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?



Similar Discussions: Calculating the ideal radius of a waterwheel given relevant constraints
  1. Adiabatic ideal gas (Replies: 0)

Loading...