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Average useful power

  1. Aug 8, 2015 #1
    1. The problem statement, all variables and given/known data
    A windmill is used to raise water from a well. The depth of the well is 5.0 m. The windmill raises 204 kg of water every day. What is the average useful power extracted from the wind?
    A. 12 mW
    B. 120 mW
    C. 690 mW
    D. 1700 mW

    2. Relevant equations
    P = W/t

    3. The attempt at a solution
    Work done by windmill daily = 204 kg x 9.81 x 5.0 = 10006.2 kJ
    Question doesn't provide info about time interval though
     
  2. jcsd
  3. Aug 8, 2015 #2

    russ_watters

    User Avatar

    Staff: Mentor

    Welcome to PF!

    The time interval is in the 3rd sentence in the problem and you correctly stated it in your answer...use it!
     
  4. Aug 8, 2015 #3
    Hint :
     
  5. Aug 9, 2015 #4
    I got 204 x 9.81 x 5 = 10006.2 kJ
    Power = 10006.2 kJ / 1 = 10006.2 kW = 10.0062 mW which doesn't fall under any of the options. Did I misinterpret something here
     
  6. Aug 9, 2015 #5
    you need to convert everything to SI units

    Kilograms
    metres
    seconds
    watts


    You need to convert a day to seconds.








    A very useful thing to do here, when you have multi-choice and the choices are quite different in size, is to use a mental arithmetic trick called zequals, which we were taught at school 30 years ago but is becoming quite trendy again.

    Zequals is the trick of rounding everything up or down to make the maths as easy as possible to do in your head. It does not give you an exact answer, but in engineering, it is very important to have an idea of the size of an answer before you go near the calculator.


    So you could say the depth of the well is 10m and the mass of water is 100kg and acceleration due to gravity is 10m/s^2. You could say the approximate number of a seconds in a day is about 100,000

    so,

    10 x 10 x 100 / 100,000

    cancel all the zeroes

    1 / 10

    so I would estimate, using zequals, that the answer is about 0.1 Watts


    And I would say my answer is 'more' correct than available answers, because you can (usually) only give the answer to the same number of significant figures as the least accurate piece of data supplied - which in this case is one sig figure.




    It looks like SI engineering unit prefixes might be confusing too (you have equated 10 MW with 10 mW which is an easy mistake to make, so make it all Watts, and if necessary use powers of 10 until you can remember the prefixes).

    So convert the answers to base units (and round up and down if it makes it easier)
    A ) 0.01 W
    B ) 0.1 W
    C ) 1 W
    D ) 2W
     
    Last edited: Aug 9, 2015
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