Induction Motor Auto-Transformer Starter - Power and Torque

AI Thread Summary
An auto-transformer starter is discussed in relation to its power input per phase and its proportionality to torque. The conversation revolves around understanding the expression for power input, which is given as proportional to torque, and how to manipulate it using the turns ratio of √3:1. Participants clarify the relationships between voltage, current, resistance, and impedance, emphasizing the importance of basic electrical principles. The discussion highlights the need to break down complex expressions to simplify the problem-solving process. Overall, the thread aims to guide the user toward a clearer understanding of the auto-transformer starter's operation and its mathematical relationships.
Stu80
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Homework Statement


Describe an auto-transformer starter. From the expression for power
input per phase (∝ torque) show that for a turns ratio (n) of √3:1 , for
the auto-transformer:

upload_2018-11-15_8-54-52.png

Homework Equations



Pin (phase) = 3 V I cos∅

I = V/Z

cos∅ = R/Z

T∝I∧2
T∝V∧2

For auto transformer with ration n:1, V=V/n, I=nI, T∝1/n∧2

The Attempt at a Solution



I don't understand the question fully, what would be my starting expression? Pin (phase) = 3 V I cos∅?

How do I relate this to being proportional to torque?

I presume in the expression for the power input to the motor/phase that the resistance and impedance is that of the phase with them being denoted with a p?

I've atempted a number of solutions by substituting I and cos∅ for V/Z and R/Z respectively into Pin (phase) = 3 V I cos∅ but I don't come out with anything relevant.

Could you help me get started please?
 

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Stu80 said:
From the expression for power
input per phase
Has your text given you a formula for "power input per phase " that includes terms Rp and Zp ?
 
No unfortunately it hasn’t, what I posted in the problem statement is everything I have.
 
Stu80 said:
I don't understand the question fully, what would be my starting expression? Pin (phase) = 3 V I cos∅?

i agree it isn't clearly stated .

How do I relate this to being proportional to torque?
i don't think you have to prove that because they already told you "the expression for power input per phase (whatever that is-jh) is (∝ torque) "

I presume in the expression for the power input to the motor/phase that the resistance and impedance is that of the phase with them being denoted with a p?
i think that's safe to assume.
what else do we know?
We know that a motor has resistance and inductance
so we might express the impedance Z of a phase as R +jXL.

And do we know that apparent power = V2/Z ? If not, check that out with Ohm's Law.

And do we know that R/Z = cosine Φ ? If not, check that out with Pythagoras.

Hmmm. Might that awful looking fraction break down into something less fearsome?
upload_2018-11-20_0-48-52.png

Break it into a product of individual terms and see if anything logical falls out...

##\frac{1} {n}## X ## \frac {1}{n}## X V X ## \frac {V} {Z} ## X ## \frac {R} {Z}##now ignore the 1/n 's for a moment and concentrate on what's left
does it resemble VIcosΦ ?

Now what to do with those n's ?

I've atempted a number of solutions by substituting I and cos∅ for V/Z and R/Z respectively into Pin (phase) = 3 V I cos∅ but I don't come out with anything relevant.
why the 3 ?

Grad students who write the homework problems for textbook authors often leave clues for you to decipher just like in a mystery novel.
Always take a close look at what they give you and ask "Why did he say that " ?
And, take just one thought step at a time.
.
Here in homework forums we're not supposed to give you the answer, instead to lead your thoughts toward it.
To that end i just went back to very basics .
Basics rock ... Trust them !

old jim
"Never trust a computer with anything important."
 

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Last edited:
Thanks for replying Jim, here's what I have snipped from MS Word:

upload_2018-11-26_13-29-54.png


Please let me know if it looks good.
 

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Stu80 said:
here's what I have snipped from MS Word:

Aha ! Same exact thing we figured out up above. from the basics.

But what MS Word document is that from? Your text ?

So what happens to that expression for power when you insert the turns ratio √3 ?

Might it make sense to divide the terms at the two applied voltages?
That's often a handy little algebra trick that reduces the number of unknowns.
 
It's my MS Word document, I'm just writing the answer to the question up, I find it easier to type it up in the equation editor in word as opposed to the text box provided.

I'm not sure I understand your comment, I thought I was simply proving the final equation or do I have to go a step further and substitute the (√3)^2 which would obviously replace the n^2 with 3?
 
For the next two parts of the question I have the following:

upload_2018-11-26_14-57-40.png
 

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  • Like
Likes jim hardy
I think you have now answered all three of the questions in your original post.

Bravo !

you might annotate your snip with comments
to help you remember should teacher ask you to explain it to class.
Be ready to clarify why cosine of current angle is the same as .cosine of impedance angle Rp/Zp.

I would rehearse a blackboard demonstration "just in case".
That might sound vain, but if he doesn't ask you don't have to admit to it.
Boy Scout's Motto "Be Prepared"

old jim
 
  • #10
Sorry for the late reply - Thank you for all your help Jim.
 
  • #11
You're welcome.

Did we get it right ? Did teacher have you work it on the blackboard ?

old jim
 
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