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Torque on shaft of wind turbine

  1. May 7, 2013 #1
    Today this guy at work told me that a 2MW wind turbine required 20K ft lbs on the main shaft at 20 rpm. I thought that I had recently read that a 750KW wind turbine might need 20K ft lb torque. Im not looking for exact numbers but which seems correct?
     
  2. jcsd
  3. May 7, 2013 #2

    russ_watters

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    Welcome to PF!

    Do you know the equation for rotational power, relating rpm and torque?
     
  4. May 7, 2013 #3
    No I don't. I'm sorry. I was hoping someone here would understand.
     
  5. May 7, 2013 #4

    rock.freak667

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    You can calculate the torque using T=power/angular velocity.

    20 rpm = 20*π/30 radians

    and you can convert the result you will get (Nm) to ft-lbs and see which is closer to the torque you were told.
     
  6. May 7, 2013 #5
    It must be difficult to understand how non-math people find their eyes glaze over so quickly when given this information. It's not that I don't want to make the calculations, I just don't get it...really. 20* and radians...huh? I was just hoping that someone just knew enough to quickly know the right answer. Again I am sorry.

    "You can calculate the torque using T=power/angular velocity.
    20 rpm = 20*π/30 radians
    and you can convert the result you will get (Nm) to ft-lbs and see which is closer to the torque you were told. "
     
  7. May 7, 2013 #6

    rock.freak667

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    I assumed you had some basic understanding of rotation in relation to power. But having never worked on wind turbines or read up about their basic requirements, I would still need to do the calculation to tell you :redface:

    ω= 20 rpm = 2.0944 rad/s

    T = 20,000 ft-lbs = 27116.36 Nm

    So power, given this torque and angular velocity will give a power of P=T*ω= 56,792.5 W which is way less than 750 kW.
     
  8. May 7, 2013 #7
    Is this how the Physics Forum works? There is no interest in actually helping answer a question, but coaxing the questioner into finding his own answer? I am not looking for a homework answer. I understand that most of you are proficient in math. I am not. However, I am here with a question, and I do seek an answer.
     
  9. May 7, 2013 #8

    russ_watters

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    That is indeed how Physics Forums works. Whether it is homework or not, it is better for you to learn how to find the answer yourself than to be handed the answer.
     
  10. May 7, 2013 #9
    Well, I came here with no understanding and with a question. I still have no understanding and still have a question. What help can I expect from this community? I am more than willing to continue to try to understand, but I have very limited skills.
     
    Last edited: May 7, 2013
  11. May 7, 2013 #10

    russ_watters

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    [shrug] I don't know. The problem is you didn't "try to understand", you just complained about not being handed the answer. There is only so much we can do, so basic we can make the help. If you aren't willing to put in even the slightest effort, we can't really do anything about that. We don't have a way of uploading the knowledge directly to your brain.

    My suggestion would be to take a step back and a deep breath, then start over in this thread, reading the responses and addressing them with the attitude that you are trying to learn from them instead of from the attitude of 'why didn't you just hand me the answer'. You may find that if you try, you will learn. Unless your math skills don't go beyond middle school level, you should not have any problem with this.
     
  12. May 7, 2013 #11
    I just want to mention that I have diligently been here on this site for 2 hours hoping for an answer to my question. I had searched for an answer online prior to posting my question. I will remain on the site and wait for some help. I hope this community is able to help with this question.
     
  13. May 7, 2013 #12
    Don't assume. Truth is, I did go online to other sites with the replied information and did try to understand the calculations. There is no need to insult me, suggesting my lack of math skills (even though I know them). From your last reply, it seems we both have strengh in a subject and are weak in another.

    I will seek continue to seek information and advice online.
     
  14. May 7, 2013 #13

    russ_watters

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    Sheesh. I didn't assume. Your responses contained zero effort, just complaints about not getting the answer.
    It isn't an insult, it is a factual statement about how much math is required. And if yours are above that, it should give you confidence that you can handle this problem. If yours is below that, we may even be able to teach you that too!
    Good luck. If you decide you are willing to try to help us help you, please feel free to give it another shot here.
     
  15. May 7, 2013 #14

    lisab

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  16. May 7, 2013 #15

    SteamKing

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    HP = Torque * RPM / 5252

    Torque is in ft-lbs.
     
  17. May 7, 2013 #16

    Dale

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    I am not sure of the meaning of "diligently" in this context. You were given the answer very explicitly after 1 hour.

    Besides the learning material posted and the explicitly worked out answer, one other recommendation is Wolfram Alpha, which will perform unit conversions for you. E.g. Go to Wolfram Alpha and enter "20000 ft lbs * 20 rpm to watts"
     
  18. May 8, 2013 #17
    Apostrophe added by reviewer.

    I think it would depend on whether the requirement was slaved to producing the 2MW; or that it just took 20K ft pounds to move the shaft at 20 RPM's. There is a slight difference. The turning moment (torque) of the blades would of course require a discussion of the aerodynamic properties of same, and wind loading. After which, the RPM needs, and torque required to spin the rotor of the generator up to the minimal, or lowest, range value RPM, needed to supply the net< 2MW; up through until, net=2MW is reached could be discussed.

    [net] is used as; turning moment required to produce (< or = ; see in-sitio). Once again turning moment relates to torque; as distance from the desired value, times the applied force. As, the torque value applied by a one foot lever, where force is 5 pounds, is five foot pounds of torque. Of course a discussion of a rotating airfoil is complex. I hope, and do pray, an engineer of same would be so kind as to be able to provide a resultant force graph, at varied wind speeds.

    Thank you for the opportunity to practice and sharpen my technical writing process. I believe if you have to explain too much, you need to either sharpen your skills, or sharpen yo' skills, or stay off Usenet forums, at least. Hey!
     
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