Hi every one,
I have a case :
-an Y-delta open transistion
- 6 wires Induction motor
- 240V- 600V
- 300-375 rla
when the 3-phase motor is disconnected completely (for some time interval) from the power lines, to go from Y into Delta config. several things happens:
1-there are some discontnuinity in time for that motor.
2-The motor is changing its speed when disconnected
3-the motor terminal holding a tremendous high voltage while disconect from line voltage.

The question is:
what dictates or affects the operation of the motor? and how i can reduce the high "voltage jam" there (someone call it residual voltage on the motor leads)

also, if contacts are needed, please give me some advice

Thanks, you all

Hello koi6456

Welcome to the forum.
This is a good question!
When you use a star / delta (or wye / delta) starter, you begin by connecting the motor in star connection. The motor draws 1/3 locked rotor current and produces 1/3 of Locked Rotor Torque. The motor accelerates until it either reaches full speed, or reaches a point where the load torque equals the shaft torque developed by the motor.
You then open the star contactor, interrupting the current flow through the motor windings and then close the delta contactor reconnecting the motor in delta connection.
During the open transition, you yave a small period of time when a) the motor is producing no torque, so it slows down, and b) the motor behaves as a generator due to the rotating magnetic field from the rotor spinning within the stator windings. The rotor will retain it's magnetic field for quite a period, with the time constant being a function of the inductance and resistance of the rotor. The frequency of the voltage generated by the motor is directly related to the speed that the rotor is spinning and this is not the same as the supply frequency.
When you close the delta contactor, you are closing onto a generator that is out of synch with the supply. If you are very lucky, you may close when the generated voltage and the supply voltage are in phase, in which case there is not problem. If however, when you cloase the delta contactor, the generated voltage is exactly out of phase with the supply voltage, and the two are equal in magnitude, you are effectiviely closing at twice supply voltage, resulting in a very high current and torque transient.
This transient causes far more damage to the supply and machine than Direct On Line starting.
You can reduce this transient by converting the starter to a closed transition star / delta starter. This is achieved by connecting a second contactor across the delta contactor with large resistors in series. The operation is to start the machine in star, then close the auxiliary delta contactor. This results in current flowing through the motor windings to the star point, and also through the resistors to the star point. Next you can open the star contactor. Current continues to flow through the motor windings via the resistors. Next the resistors are shorted by the delta contactor.
The resistors need to similar to those that you would use for a primary resistance starter, chosen to withstand the high current into the short circuit (while the star contactor is closed), and selected to ensure that significant current flows through the windings when the resistors are in series. If the resistance is too high, the generated voltage will increase and the problem occurs.

If I get a chance, I may develop a full description and diagram on our web page.

You can also reduce the open transition transient (reclose effect) by lengthening the open time to 0.5 - 1 second, but the motor could have slowed significantly in this time.)

The problems with a closed transition star / delta are that a) you are effectively using a primary resistance starter within a star /delta starter (might as well consider a standardy primary resistance starter) and b) if there is insufficient torque available in star to get to full speed, you are still changing over at part speed and causing the same current and torque as full voltage starting. (might as well use a full voltage starter, it's cheaper!!)

Best regards, ;e;

[Edited on 7/11/02 by marke]

Hi, Marke
You are the only one can see my problem sofar. I put the same question in other forum, and they have no idea what i talk about.

I knew there is several methods of starter outhere to iliminate the transient as you described, but this customer is prefering only the open transistion. below 240V, the motor seems to be ok, but above that voltage , the motor is acting up. That why i ask if there is any minor method to reduce the transient like playing with the time as you susgested. Or the open circuit time costant as in the old text book of an inductance circuit?
How can you or will you figure out that .5 to 1 sec? Is there any relationship between the speed of the motor and the time to reclose the contact?

Hello koi6456

Yes, the time constant of the rotor is the usual time constant as in all text books, the lower the resistance, the longer the time. The difficulty in trying to work it out is that we usually can't get the actual vaules for a standard motor. I rely on experience here. From what I have found, generally you will need to have a significant time for the generated voltage to decay. I nominate 0.5 - 1 second, but have seen situations, particularly on high inertia loads where there is still appreciable voltage generated after 1 second. From my experience, the decay is also related to the rate of deceleration of the motor.
Unfortunately, most applications will not give a good result with a 1 second delay between star and delta. After a one second open time, the motor will have slowed to the point where you are going to draw close to locked rotor current when you reconnect to delta, and in this case, you may as well just use a Direct On Line starter.
If you have a purely inertial load, with a modertely high inertia, then you can start in star, open the star contactor, wait for a second or two and then close in delta. With an inertial load, them motor would not have time to slow down in the one second change over, so you will get a useful result.
Remeber, that you must only switch to delta at 90% full speed or better if you are to avoid the full voltage start current of 6 times Full load current or higher!!

If you make the rule that the motor must be at 90% speed or better, and there must be a 1 second transition time, you can not use a start /delta starter in many applications. If you do not apply that rule, then you would be better to use a direct on line starter.

Best regards,

Thank you Mark, I have found the solution and I am very happy