pontussweden
Electrical
- Jun 12, 2014
- 8
Hello hello.
I have read the 50 - 60 Hz FAQ in this forum (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 (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