Resultant magnetic flux in a 3-phase induction motor

  • #1
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Homework Statement


I have some attachements below from my textbook. According to it the value of resultant flux is constant in magnitude and 1.5 times the maximum value of flux.
But i have a doubt regarding how it is proved.
The magnetic flux waveforms are 120° apart from each other and at time 0, the instantaneous value of flux waveform of 1st phase is zero. Whereas 2nd phase has magnitude -√3/2.Φm and the third phase √3/2.Φm.The textbook says that:
Φr = 2.(√3/2)Φm.cos60/2
Here is my doubt

If the fluxes are 120° apart, shouldn't the equation be: Φr = 2.(√3/2)Φm.cos120/2 ?
Please explain me in the simple way how this works.i appreciate the help:) [/B]

Homework Equations


I know that when forces have same magnitudes then for resultant force we use the formula:
R=2.F.cosΘ/2[/B]


The Attempt at a Solution


I have been trying to think about it and draw a waveform to figure it out but couldn't understand it.
 

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Answers and Replies

  • #2
cnh1995
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Looks like you're studying from B.L.Theraja.

You can do it yourself. Just draw two phasors of magnitudes √3/2 and -√3/2 , and 120° apart. Use the parallelogram rule of vector addition and you'll get Φ=1.5Φm.
 
  • #3
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3
Looks like you're studying from B.L.Theraja.

You can do it yourself. Just draw two phasors of magnitudes √3/2 and -√3/2 , and 120° apart. Use the parallelogram rule of vector addition and you'll get Φ=1.5Φm.
Hi! Can you please solve this and send a picture right here? I will appreciate that. And also explain this little bit more.
 
  • #6
cnh1995
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Please explain me how to derive it to get resultant flux. I really need help.
Do you know the parallelogram law of vector addition?
Just draw two phasors of magnitudes √3/2 and -√3/2 , and 120° apart. Use the parallelogram rule
That's how you can derive it. I can't make it any simpler.
 
  • #7
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Do you know the parallelogram law of vector addition?

That's how you can derive it. I can't make it any simpler.
This is what im getting as answer
 

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  • #8
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This is what im getting as answer
But the same one when i solve using cos60/2 :
 

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  • #9
cnh1995
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The formula you used in #7 works for two 'equal' vectors. The fluxes here are √3/2 and
-√3/2, which are not equal. You can rotate the second vector by 180° and its magnitude will be √3/2. Now the angle between the two phasors is 60° and the formula works as it did in #8.
 
  • #10
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The formula you used in #7 works for two 'equal' vectors. The fluxes here are √3/2 and
-√3/2, which are not equal. You can rotate the second vector by 180° and its magnitude will be √3/2. Now the angle between the two phasors is 60° and the formula works as it did in #8.
Oh i got it! Here is one more attachment. This is how i need to think about it?
 

Attachments

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  • #11
cnh1995
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Oh i got it! Here is one more attachment. This is how i need to think about it?
That's right.
 
  • #12
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The formula you used in #7 works for two 'equal' vectors. The fluxes here are √3/2 and
-√3/2, which are not equal. You can rotate the second vector by 180° and its magnitude will be √3/2. Now the angle between the two phasors is 60° and the formula works as it did in #8.
Thanks so much i really got this concept clear today. I realized that what i didn't know is that vectors cant have negative magnitude now if i keep this in mind my concept about resultant flux is clearly understood. :)
Im so thankful to you :)
 
  • #13
cnh1995
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Thanks so much i really got this concept clear today. I realized that what i didn't know is that vectors cant have negative magnitude now if i keep this in mind my concept about resultant flux is clearly understood. :)
Im so thankful to you :)
You're welcome!:smile:
 
  • #14
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You're welcome!:smile:
Hi !
Sorry, but one more question is overwhelming me right now.
Does the magnitude of resultant flux not get affected when we move -√3/2 by 180°.
Is it okay to do so ?
 
  • #15
cnh1995
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Hi !
Sorry, but one more question is overwhelming me right now.
Does the magnitude of resultant flux not get affected when we move -√3/2 by 180°.
Is it okay to do so ?
Yes.
Rotating a vector by 180° and changing its sign gives the same vector. It is like multplying a number by -1 twice.
 
  • #16
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3
Yes.
Rotating a vector by 180° and changing its sign gives the same vector. It is like multplying a number by -1 twice.
Okay. So what does negative sign mean ? Does it mean opposite direction? Like if right side is positive then left is negative?
 
  • #17
cnh1995
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Okay. So what does negative sign mean ? Does it mean opposite direction? Like if right side is positive then left is negative?
Kind of like that, yes.
 

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