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Alex Pel
February 24th, 2006, 11:48 PM
Hello,

Does anyone have any idea what is going to happen with a 3phase AC motor if it connected in Wye when on the motor junction box was specified Delta connection?

Actually, this situation just happened. I did that mistake yesterday, today I found that motor burnet out. In my theoretical knowledge it is not gonna happen:

WWYE = V2 /R

WDELTA = 3(V2)/R

Wye connection with the same Voltage supply will drawn 3 times less power then Delta, so current will be 3 times less. In this case the motor has 3500 RPM and loaded by a fan.

For me it should produce low airflow but never burn out. What I am missing?

Thanks for any help, company about to charge me $1500 for that mistake.

CroCop
February 24th, 2006, 11:53 PM
Looks like you lost $1500.
Good link, about machine tools, but still relevant.
http://www.cncmagazine.com/archive01/v2i07/v2i07i.htm

Lancie1
February 25th, 2006, 12:26 AM
If it is a 9-lead or 12-lead, 6-coil motor, as shown below, then if you connect it in the parallel "star" or parallel wye, but apply high voltage, then it will burn up naturally, because each coil has twice the voltage, and twice the current than if connected in the series star or series wye.

If it is only a 6-lead, 3-coil motor, then in the Delta configuration, with same line volts connected as for Wye, each COIL gets twice the voltage and twice the current and will probably melt down.

http://www.plctalk.net/qanda/uploads/Motor_Wiring.JPG

Alex Pel
February 25th, 2006, 01:04 AM
Well, if Delta increases speed by 1.75 and draws more current by 75% therefore Wye will reduce speed and current by the same values. So still not clear why the motor should overheat?

Alex Pel
February 25th, 2006, 01:19 AM
Lancie1

I connected exactly like your Fig. 3

9-6 7-4 8-5 10-11-12

1 – L1

2 – L2

3 – L3



Motor has 12 leads and I known difference between HIGH and LOW voltage.

But on the motor cover was specified Delta connection:

HV

9-6 7-4 8-5

1-12 – L1

2-10 – L2

3-11 – L3



So then I connected in Wye where suppose to be Delta it should be less voltage for winding and less current. Am I right?

keithkyll
February 25th, 2006, 01:46 AM
In this case, forget about Delta or Wye. It doesn't matter. There are 2 connection methods shown on the motor plate. One for low voltage (230) and one for high voltage (460). Both connections will run at the same speed with the proper voltage.

Vic
February 25th, 2006, 01:57 AM
Alex

You are right when you say that the Wye connection will give a lower voltage and a lower current than a Delta connection at a given motor speed. However, the torque is also reduced to about .33 of the Delta torque. If the load needs the full motor torque at running speed, the motor will never get to normal running speed and the current will be much higher than normal. Remember that the motor current is determined by the motor speed. The lower the speed, the higher the current.

With the motor drawing starting current continuously, the overload should have tripped fairly quckly and saved the motor. Did the overload trip? Is the overload correctly set for the motor FLA. Without the overload, the motor will burn out quickly.

Lancie1
February 25th, 2006, 02:01 AM
Alex,
I think your confusion is that the motor is not a "2-speed" motor as you are thinking, hence your reference to speed.. Instead it is a "dual voltage" motor, designed to run a a single speed, but using either LOW or HIGH voltage. No, you can NEVER swap the voltages on this type of motor. It has been tried many times, and the result is always the same, not good!

keithkyll
February 25th, 2006, 02:20 AM
I agree with everything here. The main question is how the motor burned up. If the motor wasn't designed for lower speed, the deviation from nameplate wiring would be the cause. The load is a fan. The load would be very low at half speed.
I suspect an internal short in the motor. It may have gone undetected in the original config, but caused a serious problem when rewired.
If one winding is burned up more than the others, this would prove it.

At $1500.00, I expect a motor that would shut down from overtemp or trip the overload long before total meltdown.

Alex Pel
February 25th, 2006, 02:30 AM
Actually there were 4 connections shown:

CONNECTION

LV

12 10 11

| | |

6 4 5

| | |

7 8 9

| | |

1 2 3

| | |

L1 L2 L3

HV

9 7 8

| | |

6 4 5



12 10 11

| | |

1 2 3

| | |

L1 L2 L3





FOR ACROSS-THE-LINE SATRT



12—10—11



6 — 4 — 5



7 8 9

| | |

1 2 3

| | |

L1 L2 L3



12—10—11



6 4 5

| | |

9 7 8



1 2 3

| | |

L1 L2 L3





I chose the last one. What does it mean “FOR ACROSS-THE-LINE SATRT”? I understood that it means start commutation. Start Wye and switch to Delta to reduce a starting current. So I thought to connect in Wye and see how it will work. I checked without load (the fan should blow scrim in normal operation) and didn’t see any problem. At day shift somebody play more and burnt the motor.

Maj. Toxido
February 25th, 2006, 02:54 AM
I think even if you make your connection in wye your motor will not burn-out provided that it has an thermal-overload relay on it & ckt. breaker. In my experience i had a problem just like what you have said,my fellow comrades make a wrong connection it should be in delta but he connect it on wye.The thermal-overload keeps on tripping i found out that he make a wrong connection.

By the way across the line or direct on line connection has a high ampere starting & high torque,once you start the motor it will draw a high reading of ampere also big transient in contactor.

I suggest that safety factor & components should have if you are running a 3-phase motor.

Note: Always set the thermal overload with the motor FLA

Hope it will hep!

Alex Pel
February 25th, 2006, 03:09 AM
Thanks Maj. Toxido

Looks like the troth, but I don’t understand why? Wye connection should drain less current. Why in your experience the overload keeps on tripping? In my case O/L was set 20 Amps, when FLA is 17A.

BTW “FOR ACROSS-THE-LINE SATRT” which one connection is fist Wye or Delta?

Maj. Toxido
February 25th, 2006, 03:31 AM
The motor that I`ve example has an service factor of 1( SF 1 )
that`s why I made the setting of my TOR to it`s full load current.
If you check the name plate of your motor there is a service factor of that motor.

Now in your problem you said that your TOR is not tripping did you consider the service factor of your motor?

Note: If the service factor of your motor is 1 adjust it to FLA
of your motor, meaning your TOR

If your sercive factor is greater that 1 multiply
your FLA with 1.25 this will be your setting of overload

If there is no service factor in the name plate of the
motor it means that consider it 1.


About your question in across the line which comes first wye or delta? it depends in your connection.

If you are using wye, 80% of current is drawn
And if you use delta, 54% of current is drawn

Alex Pel
February 25th, 2006, 04:09 AM
About your question in across the line which comes first wye or delta? it depends in your connection.

If you are using wye, 80% of current is drawn
And if you use delta, 54% of current is drawn
Can you please explain that does it mean? 80% of what?
You are stating that Wye drawing more current than Delta under the same line voltage. Can you explain why?

Alex Pel
February 25th, 2006, 05:53 AM
Just found that confirm my fault but still doesn’t explain why.
http://www.usmotors.com/Service/Bulletins/Issue7-May03.pdf

Alex Pel
February 25th, 2006, 06:14 AM
Finally it’s clear. The motor never reached a full speed and burnt to death.
************************************************** ******************
When a motor is started at reduced voltage, the current at the motor terminals is reduced in direct proportion to the voltage reduction while the torque is reduced by the square of the voltage reduction. If the "typical" NEMA B motor is started at 70% of line voltage, the starting current would be 70% of the full voltage value (that is, 0.70 x 600% = 420% FLA). The torque would then be (0.70)2 or 49% of the normal starting torque (that is, 0.49 x 150% = 74% Full Load Torque). Therefore, reduced voltage starting provides an effective means of reducing both inrush current and starting torque.
If the motor has a high inertia or if the motor rating is marginal for the applied load, reducing the starting torque may prevent the motor from reaching full speed before the thermal overloads trip. Applications that require high starting torque should be reviewed to determine if reduced voltage starting is suitable.
************************************************** ******************

Lancie1
February 25th, 2006, 10:25 AM
Alex,

Apologize a lot, grovel some, talk about your starving kids that need braces on their teeth, then offer to have the motor rebuilt at your local Motor Rewinding Shop. It usually costs about half of a new one, or less.

Alex Pel
February 25th, 2006, 04:53 PM
Thanks a lot Lancie1 and everybody.

Apologize a lot, grovel some, talk about your starving kids that need braces on their teeth, then offer to have the motor rebuilt at your local Motor Rewinding Shop. It usually costs about half of a new one, or less.

That is what I can only do.

elevmike
February 25th, 2006, 06:40 PM
.... then offer to have the motor rebuilt at your local Motor Rewinding Shop. It usually costs about half of a new one, or less.

We used to have motors rewound, but for the most part in the last 10 years or so the cost seems to be about the same. As of late the only benifit to rewinding is because it's a special motor and turnaround time on rewinding might be far less then new.

Before jumping to rewind, get a quote on the rewinding and a quote for new. If its a standard motor (Frame, HP & RPM) getting a new motor might be both less expensive and more timely.

Tark
February 25th, 2006, 11:57 PM
I would question who ordered the motor and who ordered the starter. Why would you pay for a Wye/Delta motor but not install a starter to run the motor as it was designed?