Wind turbine energy HW Question

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SUMMARY

The discussion centers on calculating the annual electricity generation of a wind turbine with a hub height of 50 meters, a rotor diameter of 65 meters, and a wind-conversion efficiency of 25%. The first calculation yields approximately 2,180,313.3 kWh/year based on an average wind-power density of 300 watts/sq meter. The second calculation estimates that if wind densities of 500 watts/sq meter are present on 0.7% of the U.S. land area, with 8 turbines per square kilometer, the total energy generated could reach approximately 1.592 trillion kWh/year. The user expresses confusion regarding the correctness of these calculations, particularly the formulas used.

PREREQUISITES
  • Understanding of wind turbine specifications (hub height, rotor diameter)
  • Knowledge of wind-power density and its implications
  • Familiarity with energy conversion calculations (Joules to kWh)
  • Basic grasp of area calculations for turbine installations
NEXT STEPS
  • Review the formula for energy generation from wind turbines, focusing on efficiency and area.
  • Study the impact of wind-power density on energy output in different geographical areas.
  • Explore the relationship between turbine spacing and total energy generation in wind farms.
  • Investigate common errors in energy calculation for renewable resources, specifically in wind energy.
USEFUL FOR

Engineers, renewable energy analysts, and students studying wind energy systems will benefit from this discussion, particularly those involved in energy generation calculations and wind farm design.

JuggerNauto
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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.
 
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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:

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