3 phase motor current calculation - from winding resistance

[Dave0098]

Hi,

hopefully someone can help or put me straight on this.

i have a 3 phase/415v motor which is 11kw - 15kw, 15A - 28A depending what voltage is applied in either star or delta.

measured the winding resistance which is 0.55ohms between the 3 phases.

now i do think this winding resistance is a good result, but i am a little confused as volts/resistance=amps (1.73 x 415/0.55 = 1305amps!!!

how can this be right?

 

[telectrix]

because the impedance of the motor winding are mostly inductive . you can't measure this with an ohm meter or multimeter.

 

[Dave0098]

yes, i see so because inductive resistance is between the magnetic forces of the rotor and stator.

 

[yellowvanman]

You've got impedance (inductance + resistance) in the coils which is a bit more difficult to measure.

Bit slow with Tel around!

 

[Lucien Nunes]

Exactly the same is true of a transformer - the DC resistance of a winding will be much lower than V/I. The resistance does not tell you anything about the characteristics of the device as a whole, only about the wire with which it is wound.

The lower the DC resistance the better because winding resistance wastes energy, but there is a practical minimum resistance that can be achieved with copper wire for a given size of motor or transformer. The aim is always to make it a very small fraction of the inductive reactance, so that most of the supply voltage is used to do work and very little is wasted producing heat.

Taking your motor as an example, if it has a resistance of 0.55Ω between lines, each coil is 3/2*0.55=0.825Ω. At 15A line current, the current per coil will be 15/sqrt(3)=8.7A. The total resistive heat loss will be 3 x 8.7 x 8.7 x .825 or 185W of heat wasted in overcoming the resistance of the copper.

 

[Engineer54]

A precise and clearly explained post Lucien.... I'd of taken at least twice as much space to say the same!! lol!!