Phasor diagram help needed

  • Thread starter Cassy85
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  • #1
Cassy85
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Hi, I'm struggling to find out how to create a phasor diagram. I have attached a photo of the question any help would be greatly appreciated.
Screenshot_20211003-194735.png

This is what I have attempted so far.
It's question 2b I am really struggling with.
I'm not looking for the answers just looking for some direction.

  1. Calculate phase current
Phase current = 250003 x 415 x 0.8
Phase current = 25.1 Amps
cos-10.8=36.87
I = 25.1 ∠36.87° Amps
  1. Calculate the stator reactive voltage drop IXs

Voltage drop = IaXs
25.1 x 4 ∠36.87° + 90°
Voltage drop (IXs) = 100.4 ∠126.87° volts
 

Answers and Replies

  • #2
Babadag
537
147
First no such motor "synchronous induction motor" does exist .Since the power factor is leading it is about synchronous motor when the excitation current is more than rated.
The motor has an a.c. winding in stator part and a d.c. winding [poles] on the rotor. Since it is a separately E noted, the generated voltage is equal to supply voltage.
The generated voltage has to be in the load case.
If the power mentioned is "input power" the apparent power it is the ratio 25/0.8.
overexcited synchronous motor phasor diagram.jpg

The loaded current it is 25000/0.8/√3/415=43.475 A
 
  • #3
Babadag
537
147
Actually this is the current from the network. The current in the motor winding in triangle it is only 25.1A, indeed.
 

Attachments

  • Triangle currents.jpg
    Triangle currents.jpg
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  • #4
Babadag
537
147
In my opinion, the power factor and Xs are calculated from the network point of view so for an equivalent star motor. In this case the current IA= 43.475 A and is leading the phase-to-neutral voltage VAN with φ= 36.87 degrees. Then the voltage drop [total] is IA*[cosφ+jsinφ]*jXs and the reactive part is jIA*cosφ*Xs.
 
  • #5
Babadag
537
147
The vectorial diagram it could be like this one:
 

Attachments

  • synchronous motor vectorial diagram.jpg
    synchronous motor vectorial diagram.jpg
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  • #6
Babadag
537
147
My mistake. The voltage considered for equivalent motor is not
“the leading the phase-to-neutral voltage” as I said but the phase-to-phase voltage. So, the vectorial diagram on the trigonometric circle it has to be like this:
 

Attachments

  • synchronous motor vectorial diagram.jpg
    synchronous motor vectorial diagram.jpg
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