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Water Wheel

  1. Jul 1, 2008 #1
    The inventor of a water wheel claims that his design is better than a conventional overshot or breast wheel at extracting energy from water flows at low heads. He states that the design shown can generate up to 260 W of power from a head of 0.8 m. average household energy use over a 24-hour period is about 28 kW h.

    (a) The claimed efficiency of the device is 70%. With a head of 0.8 m, calculate the flow rate of water that would be required to deliver 260 W of power.

    Efficiency = 70%

    Gravity = 9.81 m/s

    Head size = 0.8 m

    Efficiency = Flow Rate x Gravity x Head Size

    E = Q x g x h

    So…

    Q = E
    gh

    Q = 70
    9.81 x 0.8

    Flow rate = 8.9 kg/s


    (b) When the wheel is operating, there are twelve full troughs of water on the downward-travelling side of the belt. The diameter of the toothed sprocket is given as 0.25 m.

    If any shaft rotates at 41 rpm when delivering 260 W of power, calculate how many litres of water each trough must be able to contain.

    Flow rate = 8.9 kg/s


    Power = 260 W

    Sprocket = 0.25 m

    41 rpm into angular speed = 41 x 2π
    60

    ω = 4.3 rad/s


    v = rω

    v = 0125 x 4.3

    v = 0.5375 rad/s



    Power = Torque x Angular speed

    Torque = Power
    Angular speed

    Torque = 260
    4.3

    Torque = 60.47 Nm

    Power = Gravity x Flow rate x Head size (P = G x Q x H)

    H = P
    GQ

    H = 260
    9.81 x 8.9

    H = 2.98 m

    unable to get the answer to part b)......any suggestions?
     
  2. jcsd
  3. Nov 20, 2011 #2
    You looking for work? Might have a job in Tidal Hydro Kinetic Turbine development.

    - BaNe
     
  4. Feb 22, 2013 #3
    Wouldn't you need to include the denisity of water in this equation?
     
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