Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations waross on being selected by the Tek-Tips community for having the most helpful posts in the forums last week. Way to Go!

DC motor Field 2

Status
Not open for further replies.

John62JRJ

Electrical
Jan 10, 2023
4
0
0
US
We have a 180kw siemens motor. The tag on the motor has 90vdc at 27 amps for the field. The power supply for the field voltage is 140vdc. Getting 138vdc at the motor. We have been running this way for years, but we just had the motor at a shop and now the field is drawing high current, 51 amps. Thanks for any help John
 
Replies continue below

Recommended for you

The field was burnt. Had an ohm reading, But was not correct. They rewound the field and installed new brushes. They have the motor back at their shop. He said that the 140vdc field power was to high.
 
Most likely. Wrong connection by rewinder. Changed original series winding to parallel. (one ckt to two ckts or two ckts to four ckts etc)

Less likely. Half the turns in the field winding.

The field winding resistance should be about 2.8 ohms at room temperature. Do you get this value?

Muthu
 
That is what I was getting 2.8 ohms. He said that the 140vdc is what was causing the high current. Have 30 amp fuses on field supply. Field was pulling 51 amps. Motor is a siemens 1GG6206-0ZZ90-6jV1-Z.
 
So you've been supplying the original 90V-rated field at 138VDC for years and it was not blowing the 30-amp fuse, and now it's been re-wound and it's supplying 51 amps at the same(ish) 140VDC?

I'm thinking the original winding probably had a high-resistance bad spot in it "for years" and this led you to compensate "for years" by over-supplying it beyond rated voltage, until it blew, and now that the winding is correct, it now draws more current because it no longer has that bad spot in it.

Reduce your field voltage supply until the voltage is within ratings (90 VDC) and current is hopefully what it's supposed to be, and within the ratings of your fuses.

What happens?
 
Brian and Edison, let me backup. This motor has been on this press for 20 years. It originally had a 120vdc power supply feeding the ABB drive. We had a few problems with terminals on power supply getting hot. About 8 years ago they had me install an upgrade to the 140vdc power supply. So my question is could we be feeding the drive with the 140vdc, and the drive is controlling the output for the 90vdc field ? A factory tech had me hook up the field power supply direct to the motor. That is the reason I had the 140vdc and 51 amps on the field.

 
" So my question is could we be feeding the drive with the 140vdc, and the drive is controlling the output for the 90vdc field ?"

Sure, if the drive is capable of adjustment of the output field voltage. (and that would normally be the case)

"A factory tech had me hook up the field power supply direct to the motor. That is the reason I had the 140vdc and 51 amps on the field."

... which is beyond the nameplate rating on the motor. And it also explains why "He [motor rebuild shop] said that the 140vdc field power was to high." Probably ... he's not wrong. Might explain "We had a few problems with terminals on power supply getting hot", too.

It still sounds, to me, like you have been using this motor beyond its nameplate ratings for 20 years. Why? Undersized / overloaded for what it's doing? Wrong speed?

N.B. The only experience I have with DC drives and motors has involved ripping them out and replacing them with modern 3-phase AC motors driven by modern VFDs.
 
There are two ways to set a field control. One is to regulate the applied voltage: if you set it for 140 V, it will draw enough current to match the resistance of the field circuit (140/(2.8*1.3)=38.5 A). If you're wondering where the 1.3 comes from: the field should increase by roughly 80-100 C from room temperature under normal operation. The 1.3 is the ballpark multiplier for the difference between hot and cold resistance. The second is to regulate the applied current: set the current to full load nameplate (27 A) and let the voltage go where it needs to go based on the winding resistance. This means that as the winding heats up, the voltage reading at the supply will increase slightly (roughly that 1.3 multiplier again).

The volt and amp listed on the motor nameplate are - normally - for operating temperature. The supply can be a higher voltage for a couple of reasons: A) the required voltage falls between two "standard" supply voltages, or B) the nature of the application is such that there is occasionally a requirement for "field forcing", where the field is bumped to a higher magnitude to provide a short-term torque increase.

Converting energy to motion for more than half a century
 
Field current and armature voltage are the only two parameters that control DC motor operation (flux, speed and torque) and that's why all manual/automatic (drives) controls focus on them, not on the field voltage. The field voltage is merely adjusted to compensate for the field winding resistance rise from room temp to operating temp, which is around 20 to 30%. The field current remains constant (up to base speed) and is the one monitored & controlled by the drive along with armature voltage.

If OP's claim that the field was fed directly 140 V DC bypassing the drive (don't know why or how this was done) is correct, then the field current will be nearly 55% more than the rated thereby frying the field winding (already happened) and the speed will be proportionally reduced. The drive (presumably controlling only the armature voltage at this point) then will try to boost the armature voltage (and current) to bring the speed back to the rated thereby possibly damaging the armature too.

And we have not even addressed overfluxing issues at 140 V.

Muthu
 
With the field at 140 Volts, the drive will have to push the armature to about 140 Volts to get rated speed.
Well, if the field saturates, you won't need so much voltage on the armature.
The armature voltage will have to be close to the voltage at which the field saturates.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
Status
Not open for further replies.
Back
Top