Wind turbine torque

  • #1
neonwarrior
2
0
Hello everyone,

I am reading a book about wind power turbines and found a calculation. I tried it myself and the numbers doesn't match.

Here is an image of the data and result.
Capture.PNG


I don't get to the same torque.

And also what do you think about the moment of inertia value?

Before the data the author gave the formula:

1665195381364.png


Thanks in advance
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
15,268
4,244
Hi,

I don't get to the same torque.
I don't get the same torque either. Do we agree on some value ?
[edit] It might just be a simple power of 10 the author 'forgot')

And also what do you think about the moment of inertia value?
Again, what do you get for a reasonable size windmill ?

[edit] The author makes a mess of symbols and dimensions. Uses ##J## for moment of inertia (the 7500 kg/m2) AND for polar moment (m4) .
I find no way to balance dimensions in ##(5-4)##

##\ ##
 
Last edited:
  • #3
neonwarrior
2
0
Hi BvU,
Thank you for your reply.

I wasn't aware that there was another moment of inertia (the polar). It's time to read myselft a physics textbook first LOL.

What I get is :

In the first one: converting first the rpm to rad/s

Net torque = J . (DELTA w)/ 5 seconds = 785.4 Nm

And the other question:
I calculated from some papers of windmills a moment of inertia of 113,000 kgm2

from a paper.PNG


And from this paper there are some values: https://core.ac.uk/download/pdf/70596755.pdf
values.PNG


Yes, as you said, I think also there is a 10 factor missing or something.

Neonwarrior
 
  • #4
BvU
Science Advisor
Homework Helper
15,268
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Net torque = J . (DELTA w)/ 5 seconds = 785.4 Nm
Same here -- from the given data.

##I## is the symbol of choice for moment of inertia. With three blades times ##{1\over 3} ML^2##, 50 m long wings you would have left over 1 kg of mass for each wing :oldlaugh: if the total ##I## is 7500 kg m2

These guys mention 12.5 tonne per blade... so more like a factor of 10000 ...

##\ ##
 
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