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Homework Help: Wind turbine energy HW Question

  1. Nov 16, 2007 #1
    Hi everyone. I am having extreme trouble finding the answer to this problem. I thought I originally had it, but nope. The original Q is: "Assume a wind turbine with a hub 50 meters above the ground, a rotor diameter of 65 meters and a wind-conversion efficiency of 25 percent. The turbine operates in an area with an average wind-power density of 300 watts/sq meter at 50 meters altitude. How much electricity (kwH) can the turbine generate per year? <--- First Q.

    Wind densities greater than or equal to 500 watts/sq. meter at an altitude of 50 meters are present on 0.700 percent of the land area of the US. If, on average, wind farms contain 8 turbines/sq km, how much electricity (kwH/yr) could be generated from these wind turbines? (The area of the US is 7,827, 989 sq km)." <-- Second Q.

    Here's my reasoning & thought:

    The energy generated = efficiency * (power per unit area) * area * time
    (In one year) = 0.25 * 300W/m2* (65/2)2m2*86400*365 Joules = 0.25 * 300 * 3318.31 * 86400 * 365 J
    = 7.8485*10^12 J (1 J = 2.778*10^-7KWH)
    = 2180313.3 KWH/Yr <--- 1st ANSWER

    The total no. of turbines can then be installed by:

    N = (Total area available) * no of turbines per unit area
    = (7,827, 989 sq km * 0.007) * 8 per sq Km
    = 438367(.384)

    The energy generated by one turbine = efficiency*(power per unit area)*area* time
    = 0.25 * 500W/m2* (65/2)2*86400*365 J = 0.25 * 500 * 3318.31 * 86400 * 365 J
    = 1.308*10^13 J
    = 3633855 KWH/year

    therefore total power generated = 3633855*438367 = 1.592*10^12 KWH/year <--- 2nd ANSWER.

    But the teacher tells me "this is not correct," and won't tell me why or what to do. I'm helplessly lost & confused. I thought I had it down. Please tell me what you guys think.
     
  2. jcsd
  3. Nov 19, 2007 #2
    Any help please? I think I have gotten the numbers correct, but unsure of the formula or derivation. Thank you.

    And...I realized I made a silly typo. Forgot to add in pi... for some reason my paste does NOT work. The pi is already factored into my original post as the #'s should come out with pi:

    "(In one year) = 0.25 * 300W/m2*pi (65/2)2m2*86400*365 Joules = 0.25 * 300 * 3318.31 * 86400 * 365 J
    = 7.8485*10^12 J (1 J = 2.778*10^-7KWH)
    = 2180313.3 KWH/Yr <--- 1st ANSWER

    The total no. of turbines can then be installed by:

    N = (Total area available) * no of turbines per unit area
    = (7,827, 989 sq km * 0.007) * 8 per sq Km
    = 438367(.384)

    The energy generated by one turbine = efficiency*(power per unit area)*area* time = 0.25 * 500W/m2*pi (65/2)2*86400*365 J = 0.25 * 500 * 3318.31 * 86400 * 365 J"

    Sorry for the confusion.
     
    Last edited: Nov 20, 2007
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