Uncoupled Motor Load
Uncoupled Motor Load
(OP)
We are testing out a pulverizer motor (800 HP @4.16kV) and it is running uncoupled. The amperage reading is about 80% of the rated FLA. This seems high to me and was wondering what the rule of thumb (if any) regarding comparing the running current coupled versus uncoupled?
Thanks
Thanks





RE: Uncoupled Motor Load
When you operate a motor uncoupled, you effectively get the magnetising current of the motor.
On very small motors, this can be as high as 60% of the rated current of the motor. On large motors it is typically around 20% of the motor rating. Liquid cooled motors will often have a higher magnetising current due to increased operating flux in the iron.
I would say at 80% of the rated current, you have a problem and I would definitely not operate that motor without confirmation from the manufacturer.
An excessive magnetising current is usually caused by the V/Hz ratio being higher than the design ratio. This increases the iron flux and results in core saturation and great heat loss in the core.
Typically, this is caused by operating a 60Hz motor at it's desing voltage but at 50 Hz, or by operating at the correct frequency and a higher voltage. On low voltage machines, this can be caused by operating a star designed motor in delta on the star designed voltage.
Let us know the outcome,
Best regards,
Mark Empson
http://www.lmphotonics.com
RE: Uncoupled Motor Load
If it is new, as Mark says, check whether your supply matches the name-plate values (Voltage, frequency, stator connection etc.)
* Basically, I would like a full-time job on part-time basis *
RE: Uncoupled Motor Load
no-load current ~ 20% of FLA for 2-pole
no-load current ~ 50% of FLA for slow speed motor (ex 10-pole)
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RE: Uncoupled Motor Load
the reactive current as a percentage of the full load current may be calculated from the power factor rating.
(1-(P.F.^2))^.5
For a full load Power Factor of 75%
(1-(.75^2))^.5
(1- .5625)^.5
Reactive current equals 66% of full load current.
If we assume 30% losses at no load we get
(.3^2 + .66^2)^.5 = 72% of full load.
These assumptions are getting unrealistic for normal motors and we are still short of 89% of full load current.
I vote with the others, check the connections. It may be connected delta instead of wye.
respectfully
RE: Uncoupled Motor Load
I think you have a Synchronous Motor. It could be operating without exitation at all or being set to opearte around 80% leading PF at full load. Under both conditions the motor reactive current is very high, taking or returning reactive power to the line. Check the exitation setup while running under no load condition. If you confirm the description above, the motor is OK and the high no load current is normal. As a test change the excitation current up and down, the motor line current must decrease to a minimum and then increase again, that is known as the "V" curve of the motor current.
RE: Uncoupled Motor Load
I guess if you have an efficiency rating on your nameplate, you can use it to calculate the power factor
pf = sqrt(3)*FLA * Vnameplate / (eff * nameplate Horsepower)
with suitable conversion from horsepower to watts to make the units work.
Then plug p.f. into waross' equation to estimate no-load current.
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RE: Uncoupled Motor Load
pf = nameplate horespower/(eff * sqrt(3)*FLA * Vnameplate)
again with unit conversions needed.
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RE: Uncoupled Motor Load
You need to confirm motor type and speed. We have a 16 pole, 800HP, 4.16kV induction machine with no load current exactly 61.5% of full load current.
RE: Uncoupled Motor Load
Could your reading be an instrumentation or measurement error ? (Improper CT ratio or meter multiplication factor ?)
Many questions but no replies !!!
* Basically, I would like a full-time job on part-time basis *
RE: Uncoupled Motor Load
RE: Uncoupled Motor Load
respectfully.
RE: Uncoupled Motor Load
Eng-Tips: Help for your job, not for your homework Read FAQ731-376
RE: Uncoupled Motor Load
I am wondering if this is a "High slip" motor for pulverizer duty.
I would feel comfortable with these readings if the motor is a special motor designed for shock loading and high momentary overloads.
The motor on a "Hammer hog" type pulverizer may need to withstand shock loading and momentary overloads.
respectfully
RE: Uncoupled Motor Load
Eng-Tips: Help for your job, not for your homework Read FAQ731-376
RE: Uncoupled Motor Load
I'm not disagreeing but I don't see why it follows.
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