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Power Consumption

  1. Aug 21, 2011 #1
    1. The problem statement, all variables and given/known data
    Okay, all I want to know is, what are the equations/information I need to solve this question. I'm not too sure.


    2. Relevant equations

    Not sure

    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Aug 21, 2011 #2

    dynamicsolo

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    Average power is given by the energy or work per unit time. The sort of energy we could make a calculation for, using the available information about the air flow, is the kinetic energy of the air. We know the area of the cross-section of the air duct and the average speed at which the air passes through the duct. What volume of air passes by the fan every second?

    We are given the density of air, so what is the mass of air passing the fan in one second?

    We now have a mass of air and its speed. What is the kinetic energy of that mass of air? This energy flows by in one second, so what power does that represent?
     
  4. Aug 22, 2011 #3
    density = mass / volume

    Work done by fan = KE of air = 1/2 mv2

    power = workdone / time, where your time is 1s if you use the values given in the question
     
  5. Aug 22, 2011 #4
    so the volume flow rate is simply the avg velocity times the cross sectional area. so volume flow rate is 1 x 7 = 7m^3 / second right?? which then allows us to calculate mass flow rate as you said which is density times the volume flow rate which is 1.2 x 7 = 8.4 kg/second. So the kinetic energy in one second becomes 0.5 x 8.4 x 7^2 = 205.8 Watts. ? that right?
     
  6. Aug 22, 2011 #5

    dynamicsolo

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    A more precise way to say it is that the kinetic energy in one second is 206 Joules, so the estimate for the average power is 206 Watts. I agree with your value.

    It is a minimum because we are assuming for the problem that 100% of the electrical power consumption goes into the mechanical work of moving the air at that flow rate. The actual value would be larger because the efficiency of energy transfer will not be perfect.
     
  7. Aug 22, 2011 #6
    okay thankyou very much for your help
     
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