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Measuring waterwheel efficiency

  1. Jan 29, 2005 #1
    How did British engineer John Smeaton measure the efficiency of an overshot waterwheel?

    I am doing an experiment with overshot waterwheels to find their efficiency, and found that John Smeaton measured their efficiency at about 65%. However I have no idea how he got this number, and I cannot think of an accurate way of calculating its efficiency.

    My latest idea is to use a dynamometer to find the difference in forces, then multiply by the distance travelled (2prn (where n is number of revolutions in t seconds, and r is the radius of the axle)) Then divide it all by t; to get Power. This would be compared with a theoretical power output calculated from gravitational and kinetic energy going in.

    Is this the correct way to go about this? Or not?
  2. jcsd
  3. Jan 29, 2005 #2


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    That's a good question. I couldn't find anything in my research into it. However, I did find a link to a professor at Univ. of Minnesota. She has specifically called out Smeaton and his waterwheel work and she is researching exactly what you are asking about. A link to her page is here:

    http://www.me.umn.edu/research/faculty/alexander.shtml [Broken]

    Perhaps you could contact her and ask her directly. She has an e-mail link on her page.

    From an engineering standpoint, it seems that you are following the basic premise of power out/power in to get your efficiency. I think the hard part would be determining the actual power entering the wheel accurately, especially if you are using a natural source like a river.

    Hope this helps a bit.
    Last edited by a moderator: May 1, 2017
  4. Jan 29, 2005 #3
    Thanks alot.

    Initially i am using a model and the water is flowing down a 'slide' coming from the tap. As the model is not very good the water sprays all over the place, so energy entering and leaving the system is hard to find
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