Increased Current from motor wired in wye configuration

[philly]

If I have a 480V 6 lead motor that is intended to be wired in Delta, but is wired in wye would I expect to see more current for any given load (assuming motor is able to be started)

I am thinking that in a wye configuration the voltage at the motor is .58% of rated and therefore the torque is reduced to somewhere around 33% of rated. So since the torque is reduced shouldn't the motor be drawing higher current for any given loading on the motor under the new maximum avaliable torque value? Since the flux is reduced wont the motor try to make up for toque by pulling more current?

If the load now requires more than the new 33% rated torque value, wont the motor pull a higher than rated amount of current and trip?

So althgouh torque capability is reduced the motor will try to make up for it by pulling more current?

 

[Jraef]

It will try. But at the same time that you are reducing the effective voltage and torque,you are reducing the peak currents as well. So if, let's say, your LRC was 600% of FLC, then in Wye, it would be 33% of 600% of FLC, or roughly 200% of the original Delta FLC. If the motor completely stalls, that would be the scenario and the OL will trip, but not as quickly as it would at LRC if in Delta. If the motor does not completely stall, i.e. it continues to run at Break Down Torque and current, then it may take even longer. Lets say BTD is 250% FLT, so now you are at 33% of that 250% in Wye, that is 82.5% of Delta FLT, which may be enough to spin the load. Current will still be a lot higher, because in a Design B motor, BDT occurs at roughly 75-80% speed (20-25% Slip), so current will still be as high as 450%. So now you are at 33% of 450%, or 150% of the original FLC and if the OL is set for the Delta FLC, it will take a while to trip.

So looking at the above curve plotted for both torque and current, you basically just shift everything lower on this curve when in Wye, the actions and reactions are the same. But if the load remains the same as it was in Delta, you hit the thermal limits of the motor more quickly.

 

[GeorgeB]

If I have a 480V 6 lead motor that is intended to be wired in Delta, but is wired in wye would I expect to see more current for any given load (assuming motor is able to be started)

Yes. Specific case, customer had a motor capable of being wye-delta operated who OEM installers wired wye ... effectively a 796V motor operating on 460V.

Load was hydraulic fixed pump, unloaded for starting and much of running. Frequently breakdown torque would be reached, motor would slow WAY down and overloads trip. Theoretical load (hydraulic pumps make good dummy loads) was about 85% of motor rated HP. The 2nd question I asked was if they were sure it was wired correctly ... answer was that OEM had checked that 3 times.

OEM had made 4 service calls and not found problem ... my employer was told to charge double, pay me at least double. They backcharged the OEM for all 4 of their service calls and for ours for their ineptitude. OUCH!

 

[Besoeker]

It will try. But at the same time that you are reducing the effective voltage and torque,you are reducing the peak currents as well. So if, let's say, your LRC was 600% of FLC, then in Wye, it would be 33% of 600% of FLC,

I have seen such statements elsewhere.
In star (Y) the LRC is 0.577 of what it would be in delta. and the torque about one third (being proportional to the square of the voltage).

 

[philly]

If the motor does not completely stall, i.e. it continues to run at Break Down Torque and current, then it may take even longer. Lets say BTD is 250% FLT, so now you are at 33% of that 250% in Wye, that is 82.5% of Delta FLT, which may be enough to spin the load. Current will still be a lot higher, because in a Design B motor, BDT occurs at roughly 75-80% speed (20-25% Slip), so current will still be as high as 450%. So now you are at 33% of 450%, or 150% of the original FLC and if the OL is set for the Delta FLC, it will take a while to trip.

So with this you are saying if we are operating at the new reduced breakdown torque? At this new breakdown torque we are at 82.5% of the origonal Full load torque. If that is the case then why do we not operate at the current associated with the origonal 82.5% of full load torque? I'm not sure how you take it from here? You mention 33% of 450 gives us 150% FLC? I dont follow how you arrive at this?

What if the load was operating below the breakdown torque at some value within the normal torque operating range? Could the current be under full load, but just be greater than the delta current for that particular load?

We have an instance where a drive on a VFD was stalling. The motor was stalling and the encoder was tripping out the drive. The strange part is the motor has a full load of 17A ans when stalling we only noticed the motor drawing about 4-5A. The motor was operating at very low speeds when this happened. We found the motor to be wired in a wye configuration. We changed the motor to a delta configuration and it seems to now work fine drawing about 7A or so. This doesnt make sense to me. I would have expected to see FLA or above when motor was stalling. I also would have expected to have seen current get reduced when we changed to the delta config. Any thoughts?