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Power and Energy

  1. Jun 5, 2006 #1
    I have two questions
    a)Water flows over a waterfall of height 75m of volume flow rate 6.0ms-1 and then lands in a pool below.
    Outline the energy changes that take place


    b)It is suggested that installing a turbine of the waterfall could provide a source of electrical energy. Estimate the power that might be available, stating one assumption that you make.
    (The denstity of water= 1000 m-3)
     
  2. jcsd
  3. Jun 5, 2006 #2

    siddharth

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    Do you have any thoughts/ideas on this problem? You need to show your work to get help.
     
  4. Jun 5, 2006 #3
    for sure i thought about it but i dont know...
     
  5. Jun 5, 2006 #4

    Hootenanny

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    Would you mind posting what you have done thus far? The guidlines of these forums state clearly that one should show one's work before receiveing assistance.
     
  6. Jun 5, 2006 #5
    actually first i thought what energy transfer and i put 75/6 and but then i thought you just have to explain meaning that it goes from vaporisation to convection
     
  7. Jun 5, 2006 #6

    Hootenanny

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    What :confused: ?? I was thinking more along the lines of potential energy ---> kinetic energy?
     
  8. Jun 5, 2006 #7
    ho yeh so actually the energy changes are from potential energy ---> kinetic energy, so thats actually the answer...
    thank you
     
  9. Jun 5, 2006 #8

    Hootenanny

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    What are your thoughts for question (b)...?
     
  10. Jun 5, 2006 #9
    for question B i thought that because they have to install a turbine, looking that the density of water is 1000 than you have to do 1000/6 which is the flow rate so then to find the estimated power
     
  11. Jun 5, 2006 #10

    Hootenanny

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    You have just basically rephrased the question there :rolleyes: . Here's a hint: Think about mass and the relationship in question (a).
     
  12. Jun 5, 2006 #11
    yeh but what do you to find the power
     
  13. Jun 5, 2006 #12

    Hootenanny

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    HINT Power is the rate at which energy is transfered;

    [tex]P = \frac{dE}{dt}[/tex]
     
  14. Jun 6, 2006 #13
    ok got it thank you
     
  15. Jun 6, 2006 #14

    Hootenanny

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    Your welcome :smile:
     
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