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Wind speed

  • Thread starter Windseaker
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  • #1
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1. Wind turbine produces 8MWatts
(If wind speed is proportional to the 7th root of altitude)
(Wind power is proportional to the cube of wind speed
First the turbine is at 450ft then reconstructed at 650ft, how much faster is wind speed? and how much more power?


2.
Sw= 7√Altitude, Pw= (Sw)3


3.

a. 7√450 =2.39 and 7√650 =2.52 ,then output/input , so 2.39/2.52 =.948 or
95% more?

b. (2.39)3=13.6 and (2.52)3=16 , so 13.6/16 =.85 or
85% more?
 
Last edited:

Answers and Replies

  • #2
rl.bhat
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Sw= 7√Altitude, Pw= (Sw)3
Pw = [(A)^1/7]^3 = (A)^3/7
 
  • #3
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Pw = [(A)^1/7]^3 = (A)^3/7

your saying that power is:
(450')^3/7= Pw1
(650')^3/7= Pw2

and Pw2-Pw1= how much more power?

what happened to the differents in wind speed first?
 
  • #4
Redbelly98
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3.

a. 7√450 =2.39 and 7√650 =2.52 ,then output/input , so 2.39/2.52 =.948 or
95% more?
Not quite, you have things backwards. You can think of it in terms of final/initial (I guess that's what you mean by output/input). So it would be 2.52/2.39=___?, since the turbine was at 450 ft first and later moved to 650 ft.

b. (2.39)3=13.6 and (2.52)3=16 , so 13.6/16 =.85 or
85% more?
Again, take the value for 650 ft and divide it by the value for 450 ft.
 
  • #5
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Thank you for expanding the thought.




Not quite, you have things backwards. You can think of it in terms of final/initial (I guess that's what you mean by output/input). So it would be 2.52/2.39=___?, since the turbine was at 450 ft first and later moved to 650 ft.


Again, take the value for 650 ft and divide it by the value for 450 ft.
 

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