6 Lead Motor Connection

[Pitt123]

I have a 200hp 4.16/2300V 6-Lead motor I need termination information for. This motor is going on a 4.16kV system and I'm not sure weather I should connect it in a wye or Delta fashion?

I always thought wye was for wye-delta starting and most motors were always connected delta but I'm not sure in this case. Any advice?

 

[don_resqcapt19]

With a 6 lead, dual voltage motor, you connect it in wye for the higher voltage and in delta for the lower voltage. In this case your motor will have windings rated at 2300 volts and when connected in delta to a 2300 volt source each winding will see its rated voltage. When you connect the motor in a wye for the higher voltage each winding will see the supply voltage divided by the square root of 3. 4160/square root of 3 = 2400 volts.

 

[triplstep]

With a 6 lead, dual voltage motor, you connect it in wye for the higher voltage and in delta for the lower voltage. In this case your motor will have windings rated at 2300 volts and when connected in delta to a 2300 volt source each winding will see its rated voltage. When you connect the motor in a wye for the higher voltage each winding will see the supply voltage divided by the square root of 3. 4160/square root of 3 = 2400 volts.

Don,

I would like to live inside your brain for 30 days

 

[Pitt123]

With a 6 lead, dual voltage motor, you connect it in wye for the higher voltage and in delta for the lower voltage. In this case your motor will have windings rated at 2300 volts and when connected in delta to a 2300 volt source each winding will see its rated voltage. When you connect the motor in a wye for the higher voltage each winding will see the supply voltage divided by the square root of 3. 4160/square root of 3 = 2400 volts.

O.K. so it looks like the answer is to wire it in the "wye" configuration so that the windings will see 2400V which is close to the rating of 2300V.

 

[GeorgeB]

O.K. so it looks like the answer is to wire it in the "wye" configuration so that the windings will see 2400V which is close to the rating of 2300V.

And, regarding your "I always thought wye was for wye-delta starting", that is also true, but with THIS motor, that would be done only on a 2400 supply. The wye connection starts a "4160" motor on "2400" reducing torque and inrush current.

 

[don_resqcapt19]

O.K. so it looks like the answer is to wire it in the "wye" configuration so that the windings will see 2400V which is close to the rating of 2300V.

Yes...the motor nameplate really should read 4.16kv/2.4kv. I don't know why it says 2300 volts.

 

[Pitt123]

Yes...the motor nameplate really should read 4.16kv/2.4kv. I don't know why it says 2300 volts.

Should this be a cause for concern?

 

[Rick Christopherson]

The way I read the original question is similar to a comment I just made in another thread. Some people confuse the configuration of the internal windings of the motor with how it should be connected to supply power. Even if the motor is configured as wye connected, it does not mean it should be connected as wye to the power supply. The wye configuration is internal to the motor only, and it should not be connected to the grounded conductor of the supply power. Doing otherwise can result in unbalanced currents in the windings.

 

[don_resqcapt19]

Should this be a cause for concern?

No...it is not an issue. I just find it strange that they used those numbers as that combination of voltages is not possible.

 

[LarryFine]

Yes...the motor nameplate really should read 4.16kv/2.4kv. I don't know why it says 2300 volts.

For the same reason 120v motors say 115v, 240's say 230v, etc.?

 

[don_resqcapt19]

For the same reason 120v motors say 115v, 240's say 230v, etc.?

But if they are dual voltage they don't say 240/115 or 480/230...the nameplate on this motor is the equivalent of that.

 

[Pitt123]

I also noticed on this motor that the LRC was listed as 430/246A. I always thought that during starting that the starting current was proportional to the voltage. In this case with this motor being our our 4.16kV system then the starting current would be greater than the 2300V value, but indeed the value of 246A is less.

Is this because the motor is connected in star to our 4.16kV system and only using 2300V on each phase? When connected in Delta at 2300V then each winding has full system voltage and thus current however in star each winding still has 2300V but this is only 1.73% of system voltage therefore maybe current would be less? Can anybody explain?

 

[dkarst]

The locked rotor current on your nameplate is the same ratio as the FLC, i.e. on your system you may expect ~25A FLC for that motor when connected to your 4160 system. If connected for a 2400V system, the supply current would have to be 25 X Sqrt(3) = 43A. If you look at the diagram below and sketch where the sqrt(3) factors are, you can see that the LRC is just a simple multiple of the Full load current.

A connection diagram shown here may help.
http://www.firstelectricmotor.com/images/motcon10.jpg

 

[Pitt123]

The locked rotor current on your nameplate is the same ratio as the FLC, i.e. on your system you may expect ~25A FLC for that motor when connected to your 4160 system. If connected for a 2400V system, the supply current would have to be 25 X Sqrt(3) = 43A. If you look at the diagram below and sketch where the sqrt(3) factors are, you can see that the LRC is just a simple multiple of the Full load current.

A connection diagram shown here may help.
http://www.firstelectricmotor.com/images/motcon10.jpg

Yes but when the motor is operating at full load, it can be seen as a constant kVA device trying to mainitain constant power, where as at LRC the motor is seen as a constant inpedance such that increased voltage would lead to a higher current. But I guess it is because of the 1.73 factor of the system voltage.

Are you suggesting to draw the sqrt factors between the ends of the wye points?

 

[dkarst]

Apologies if I'm confusing the issue.... As for the sqrt(3), I was just pointing out in the left diagram if you think Vline = 1.73 Vphase and in the right diagram you think Iline = 1.73Iphase, it may help to see what is happening.

Using your impedance rationale, even though the winding currents are the same, the impedance presented to the source varies depending on lead connection. You could calculate the V/I ratio for each connection and you will get a factor of three. I think this is the same 3X factor you would get if you performed a Y-delta transformation using 3X Zwye = Zdelta.

 

[LarryFine]

But if they are dual voltage they don't say 240/115 or 480/230...the nameplate on this motor is the equivalent of that.

Then they would say 230/115 and 460/230. The high voltage is twice the low voltage.

The winding pairs are either in series or in parallel, just like a dual-voltage transformer.

 

[don_resqcapt19]

Then they would say 230/115 and 460/230. The high voltage is twice the low voltage.

The winding pairs are either in series or in parallel, just like a dual-voltage transformer.

That is my point, there is a mathematical relationship between the two voltages of a wye/delta 6 lead dual voltage motor....the 4.16kV/2.3kV is not possible...the lower one should equal the higher one divided by the square root of 3. The 4.16/2.3 nameplate is the equivalent of one that would say 240/115.

 

[dkarst]

Yes but when the motor is operating at full load, it can be seen as a constant kVA device trying to mainitain constant power, where as at LRC the motor is seen as a constant inpedance such that increased voltage would lead to a higher current. But I guess it is because of the 1.73 factor of the system voltage.

Maybe a clearer statement would be: If you have the motor in your plant connected as left diagram above on your 4160 V system, and I have the same motor in my plant connected in the right diagram on my 2400 V system, at t=0 -> Each motor sees the identical 2400V across the windings and a rotor at standstill. Assuming they are attached to equal loads, the winding currents would then be equal. From the diagram above, in the right diagram, the line current would be equal to sqrt(3) times the winding current whereas in the wye connection, the winding current is equal to the line current, hence the sqrt(3) in your LRC you listed.