# Induction motor problem

• Engineering
• Connorm1
In summary, the conversation discusses windage/friction losses and how they are related to power. The speaker has reworked their calculations and determined that the windage/friction loss is approximately 220 watts. They have also calculated the magnetization loss per phase and found the phase angle to be 84.4 degrees. Using this information, they have calculated the values for R0 and X0 to be 84 ohms and 18.77 ohms, respectively. They are unsure about how to approach the question about the locked rotor.

#### Connorm1

Homework Statement
4. A four-pole, star-connected, squirrel-cage induction motor operates from
a variable voltage 50 Hz three-phase supply. The following results were
obtained as the supply voltage was gradually reduced with the motor
Stator line voltage 220 164 112 88 42
Stator line current (amperes) 6.8 5.4 3.9 3.8 3.7
Stator power (watts) 470 360 278 244 232
(a) By plotting a suitable graph from these results, determine the total of
windage and friction losses, the no load magnetising current I0
(assume no load current I0 is magnetising losses) and the equivalent
circuit magnetising circuit parameters R0 and X0.
Relevant Equations
P = √3 x VL x IL x CosФ
So I've attached my graph and found that the windage/friction losses occur roughly around 230Watts. I just don't know how i further explain the magnetising current I0. I know I0 = input current on no load. So do i assumed from P = √3 x VL x IL x CosФ that CosФ=230/√3*220*6.8= 0.0849 or 85.1degrees (220 and 6.8 is what i assume voltage and current are on load). I0=IL*CosФ = 6.8/0.08 = 0.577Amps. Just wish to see if I am going right so far otherwise my R0 and X0 will be wrong.

#### Attachments

• Graph.pdf
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Connorm1 said:
found that the windage/friction losses occur roughly around 230Watts.

Where do windage/friction losses come from? How should they depend on power?

Hi @anorlunda ,

So i know from the graph windage/friction losses constitute the power lost when the stator voltage=0V. I have actually reworked this question and got something I am more happy with. I've actually said the windage/friction loss is 220W from extending the graph to 0V. I know that VL is 220/√ 3 and the magnetising current is 6.8A, from what I've read i could have used any line voltage and current from the graph. I then found the magnetization loss per phase which wsa 470-220/3 giving 84W per phase (roughly). Then found the phase angle which was Φ= cos-1(Pm/VL*Io) so Φ=cos-1(84/((220/√ 3)*6.8)= 84.4 degrees. Then using R0= VL/(I0*CosΦ) and X0=VL/(I0*CosΦ) giving R0=84ohms and X0=18.77ohm. Does this look right?