PAM windings resistances
PAM windings resistances
(OP)
I operate in a 50 Hz country and have a PAM wound motor 30 and 60 kW for corresponding speeds 750 and 1000 RPM. The name plate lists the windings as 2 star / star. The motor is used as an induction generator.
The terminals are marked U1, V1, W1, U2, V2, W2. From the text books I understand that each of the three legs of the star consist of two windings in series and the star point is fixed and internal to the stator. The outer connections of the star are the U1, V1, and W1 and the U2, V2 and W2 are the intermediate connections between each winding.
For low speed we connect to U1, V1 and W1 only.
For high speed we bridge U1, V1 and W1 and connect to U2, V2 and W2
Generator runs well and all is ok.
The odd thing I would like assistance on is the understanding of the DC resistances I obtain from the terminals using a modern Ductor. All values in milli ohms.
From U1 to U2 258. From U2 to V2 180. From U1 to V2 90 I would expect U1 to V2 to equal 258 + 180 = 438. Same issue when the other leg combinations are measured.
Also using the series circuit U1 to V1 I would expect to obtain 258 + 180 + 258 = 696, but instead the Ductor reading is 344.
The two attachments show the resistances from two generators and what I believe to be the star connections except the subscript 3 should be read as subscript 1 for consistency with my description above.
The terminals are marked U1, V1, W1, U2, V2, W2. From the text books I understand that each of the three legs of the star consist of two windings in series and the star point is fixed and internal to the stator. The outer connections of the star are the U1, V1, and W1 and the U2, V2 and W2 are the intermediate connections between each winding.
For low speed we connect to U1, V1 and W1 only.
For high speed we bridge U1, V1 and W1 and connect to U2, V2 and W2
Generator runs well and all is ok.
The odd thing I would like assistance on is the understanding of the DC resistances I obtain from the terminals using a modern Ductor. All values in milli ohms.
From U1 to U2 258. From U2 to V2 180. From U1 to V2 90 I would expect U1 to V2 to equal 258 + 180 = 438. Same issue when the other leg combinations are measured.
Also using the series circuit U1 to V1 I would expect to obtain 258 + 180 + 258 = 696, but instead the Ductor reading is 344.
The two attachments show the resistances from two generators and what I believe to be the star connections except the subscript 3 should be read as subscript 1 for consistency with my description above.





RE: PAM windings resistances
With your measurements and your generator all is OK. Everything will be clear if you look at the attachment.
Zlatkodo
RE: PAM windings resistances
The attachment does not appear to solve my problem.
First thing I noticed is that U1 and U2 are not in the same phase legs similar for V and W.
Secondly if a measurement from U1 to V1 was made on the attachment you sent we would logically read 340 ohm. But my readings do not add like something in series.
Am I missing the point you have raised ?
RE: PAM windings resistances
Terminal markings are a matter of agreement, convention. You need to know how to connect the voltage at the winding, like you said:
If you know the terminal markings of an PAM motor, you can not know (in all cases) how looks like the internal connections in the winding. A similar case is with NEMA markings for PAM windings ( pairs: 1-4, 2-6, 3-5 not 1-4, 2-5, 3-6). PAM winding should be considered as a single winding (not three windings).
Your second question I do not understand:
In your table I've read the resistance U1 - V1 is 345 and 361.
By the way , see:
http://www.youtube.com/watch?v=FlCdyrSmLxc
Zlatkodo
RE: PAM windings resistances
Regarding the terminal markings yes there are standards and the standard method and published drawings etc align with U1 and U2 etc being part of the same phase etc. Although I follow IEC standards. I am not asking questions re the numbering only commenting on the drawing you supplied as not following convention.
Regarding the 340 milli ohm I replied to you with that came from your example where the measurement would be 4 X 85 = 340.
But with my data I would expect 258+180+258 = 696 but instead I measure 345. This suggests that I do not have a simple star winding but something else. This is the crux of my confusion.
Best wishes
David
RE: PAM windings resistances
If you change the marks : V2 with U2, W2 with V2, and U2 with the W2 and then recconect voltage in accordance with the new marks, you will have the same situation, physically nothing will change, only a resistance- measurement will be in accordance with your expectations.
Zlatkodo
RE: PAM windings resistances
Such a simple transposition solves the issue. This also indicates that each of the windings have the same resistance although I am not sure if that is just a coincidence in this case.
Once again, thank you for staying with me on this one.
Best wishes
David
RE: PAM windings resistances
for this valuable post!" to give him a coveted LPS (Little Purple Star) award.
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(2B)+(2B)' ?
RE: PAM windings resistances
I pressed it twice as you deserve it. Not sure if the system allows that. All the best.