Calculating Electric Power from Wind Turbines

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SUMMARY

The discussion focuses on calculating the electric power generated by wind turbines with an 80 m blade span, operating in wind speeds ranging from 4 to 32 m/s. The overall efficiency of each turbine is established at 25%, and the air density is given as 1.112 kg/m³. The key equation used for power calculation is W = ρπ(d²/8)V³η, which demonstrates that power is a function of wind speed (V) rather than a single average value. The conversation emphasizes the importance of understanding the range of wind speeds and the implications for power generation.

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  • Understanding of wind turbine physics and efficiency calculations
  • Familiarity with the equation for kinetic energy in fluid dynamics
  • Knowledge of integration techniques for averaging functions
  • Basic principles of air density and its impact on power calculations
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  • Explore the derivation of the power equation W = ρπ(d²/8)V³η in detail
  • Learn about the impact of varying wind speeds on wind turbine performance
  • Research methods for calculating average power output over a range of velocities
  • Investigate advancements in semiconductor technology for optimizing power generation
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Engineers, physicists, and students in renewable energy fields who are involved in wind energy calculations and turbine design optimization.

Bizkit
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Homework Statement


Consider a wind turbine farm that uses turbines with a blade span diameter of 80 m at a site subjected to wind speeds over the range of 4 to 32 m/s. Taking the overall efficiency of each turbine to be 25% and the air density to be 1.112 kg/m3, determine the electric power generated by a single turbine.

Homework Equations


I know that, for an average wind speed, the equation I need is: W = m(V2/2)*η (the bold variables are per unit time), but I'm not sure exactly how to modify this equation for a range of velocities. I at first thought that I could take the integral of the equation with respect to velocity, but then I end up with the wrong units.

W = m(V2/2)*η = ρVA(V2/2)*η = ρπr2(V3/2)*η = ρπ(d/2)2(V3/2)*η = ρπ(d2/4)(V3/2)*η = ρπ(d2/8)V3

The Attempt at a Solution


Since I'm not sure what equation to use, I can't attempt to find the solution.If someone could help me, I would really appreciate it. Thanks.
 
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I think they just want the power as a function of v. You already gave this in your last equation.
You might think that you could find the average power by integrating, but this only works if
- the wind is always between 4 and 32 m/s
- all wind speeds between 4 and 32 m/s are equally likely.
Only if these two are valid you could find the avergage power by integrating from v_min to v_max and then dividing by the range of v_max - v_min.
 


Thanks for the help. I still have a question, though. How does leaving the power as the function of V I gave above (W = ρπ(d2/8)V3*η) show that V has a range of values, rather than just being an average? I mean, shouldn't there be like a Vlow and Vhigh or something like that to show that V has a range values, or is that not how it works? Hopefully this question doesn't confuse you. Sorry if it does.
 


Bizkit said:
Thanks for the help. I still have a question, though. How does leaving the power as the function of V I gave above (W = ρπ(d2/8)V3*η) show that V has a range of values, rather than just being an average? I mean, shouldn't there be like a Vlow and Vhigh or something like that to show that V has a range values, or is that not how it works? Hopefully this question doesn't confuse you. Sorry if it does.

You kinda answered your own question there. By defining the power as a function of V it follows that V can take a range of values and not just one set value.
 


Ok, I understand what you are saying, but I just thought of something else. If the people who wrote the problem wanted me to just define the power as a function of V, then why didn't they just ask me to find the power over a range of wind speeds, without giving any specific values. Since they gave specific values, that makes me think that I need to integrate and divide. What do you think?
 


Mark: Semiconductor manufacturers are now producing chips that will generate a constant voltage from a source that has varying voltage or current capacity. Some even match their input impedance to maximize total power, rather than just current.

Although you may still wish to explore the mathematical theory regarding such things, these new chips will help eliminate some of those calculation problems.
 

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