Questions about the 50 an 60Hz FAQ
Questions about the 50 an 60Hz FAQ
(OP)
Hello hello.
I have read the 50 - 60 Hz FAQ in this forum (http://www.eng-tips.com/faqs.cfm?fid=1224).
This got me thinking, especially on the lines below.
I would also like to say that i'm a happy amateur that find the induction motor very interesting, please have that in mind when you read this text.
"If the frequency drops the V/Hz goes up. This means that the motor needs a larger magnetic circuit. Without it the magnetic circuit can saturate. This leads to a rapid increase in current draw and a corresponding large increase in temperature.(A motor's chief enemy)
If the frequency increases the V/Hz drops. This is not a first order consideration. [The motor may have a worse power factor.]"
Ok, what does it say? If the frequency drops the V/Hz goes up.
400/50 = 8
400/40 = 10
No questions about that.
But now i come to a part that i cant figure out how its works.
"This means that the motor needs a larger magnetic circuit. Without it the magnetic circuit can saturate. This leads to a rapid increase in current draw and a corresponding large increase in temperature."
Ok lets see, If the V/Hz goes up the motor need a larger magnetic circuit.
-My thought is now this. If Hz drops the reactance(Q) decrease that leads to a decreasing Impedance (Z). Whit a constant voltage it will lead to a higher current in the stator package which leads to a greater magnetic field leaving the stator relativ the higher Hz which create a higher reactance ? - Is this correct ?
As i write this, it come to my mind perhaps it is not good whit a to high magnetic field in the coil due to that it will saturate (and what i understand saturation is no good ?) the iron core in the stator which will lead to the the current in the coils is forced to increase ?
If this is correct ? then the Power factor will decrease due to the higher magnetising current in the coil's.
Is my thought's any right ?
Best regards Pontus
I have read the 50 - 60 Hz FAQ in this forum (http://www.eng-tips.com/faqs.cfm?fid=1224).
This got me thinking, especially on the lines below.
I would also like to say that i'm a happy amateur that find the induction motor very interesting, please have that in mind when you read this text.
"If the frequency drops the V/Hz goes up. This means that the motor needs a larger magnetic circuit. Without it the magnetic circuit can saturate. This leads to a rapid increase in current draw and a corresponding large increase in temperature.(A motor's chief enemy)
If the frequency increases the V/Hz drops. This is not a first order consideration. [The motor may have a worse power factor.]"
Ok, what does it say? If the frequency drops the V/Hz goes up.
400/50 = 8
400/40 = 10
No questions about that.
But now i come to a part that i cant figure out how its works.
"This means that the motor needs a larger magnetic circuit. Without it the magnetic circuit can saturate. This leads to a rapid increase in current draw and a corresponding large increase in temperature."
Ok lets see, If the V/Hz goes up the motor need a larger magnetic circuit.
-My thought is now this. If Hz drops the reactance(Q) decrease that leads to a decreasing Impedance (Z). Whit a constant voltage it will lead to a higher current in the stator package which leads to a greater magnetic field leaving the stator relativ the higher Hz which create a higher reactance ? - Is this correct ?
As i write this, it come to my mind perhaps it is not good whit a to high magnetic field in the coil due to that it will saturate (and what i understand saturation is no good ?) the iron core in the stator which will lead to the the current in the coils is forced to increase ?
If this is correct ? then the Power factor will decrease due to the higher magnetising current in the coil's.
Is my thought's any right ?
Best regards Pontus





RE: Questions about the 50 an 60Hz FAQ
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Questions about the 50 an 60Hz FAQ
RE: Questions about the 50 an 60Hz FAQ
RE: Questions about the 50 an 60Hz FAQ
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Questions about the 50 an 60Hz FAQ
Keith Cress
kcress - http://www.flaminsystems.com
RE: Questions about the 50 an 60Hz FAQ
Thank you very mutch for the explanation. So if the iron reach saturation the reactance (Q) disappear, which will lead to a higher current.
But still will not the power factor decrease -because it will be a higher current which will create a larger magnetic field ??
RE: Questions about the 50 an 60Hz FAQ
What happens then depends on a lot of factors. The most common are:
1. Motor protection trips.
2. If it doesn't trip - motor will probably let out smoke. And blow a fuse or two.
3. If you stay at a moderate overvoltage (not tripping or burning the motor) then the saturation causes increased current with phase angle close to zero. So, it may look like you get a better P.F. - but that is nothing to strive for, because all that extra current translates to heat and doesn't get to the motor shaft. Wasted energy.
The fact that increased voltage leads to a worse power factor has been noticed all over Europe the last decades. The voltage harmonization (going from 380 V to 400 V) has actually decreased the P.F. and augmented the sale of compensation equipment like capacitors.
In the UK, the grid voltage was reduced from 415 - 420 V to 400 V (with some exceptions, UK is UK) and that has increased the P.F. Which is a good thing.
Gunnar Englund
www.gke.org
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Half full - Half empty? I don't mind. It's what in it that counts.