In an article ("Selecting Motors for Adjustable Frequency Drives") I read: "..., the drive will probably need to be oversized to supply the increased current usually required by motors of 8 or more poles". I cannot find the reason why 8 pole motors should require increased values of current. Can anyone help me?
As the number of poles increase, so does the percentage of reactive current vs total current.
If you measure both KVAR and KVA on motors of equal HP and loading you will see this effect. I got burned on this ounce when another department failed to inform me that they were using low speed motors on waste water pumps at a new station. I was able to work around it by adding power factor correction capacitors, but this is not an option with VFDs. You will need to get the name plate data from the motor, and work with the VFD people. If you can not get the motor nameplate data, as an aproximentation use the tables from a power factor correction manual.
Ralph McDonald, P.E.
More poles mean more inductance which means more current to maintain voltage. V=IR
Guy H. Looney
Guy . . . .
Last time I checked . . .inductance is "L" . . . . current is "I" . .
. . ;)
An 8 pole motor is typically less efficient than a 4 pole motor. For very small motors (less than 2HP) typical efficiency is around 60 - 70 percent. For larger motors it ranges from 80 - 90 percent. In contrast, 4pole motors run in the 85 percent range for the smaller motors and 95 percent in the larger motors. Data is from the Baldor CD.
For a typical 20HP 4-pole motor, current will be ~24..25A at 460V For a typical 20HP 8-pole motor, current will be ~29A at 460V, a 15 - 20 percent increase.
If more poles means more inductance then you need LESS current to maintain the same voltage V = IR, i.e. if R increases, V stays the same, then I decreases.
If the poles are wired in parallel then the inductance will actually be lower, so I increases.
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I posted the original question and I thank you all for your answers.
I agree with those of you who say that "more inductance" means "more currents" under the important condition that we want to keep the same active power output and not the same phase motor voltage.
Ken & Simon,
I agree w/ both of you, however you're assuming only the inductive properties of the windings. The windings do have an overall impendance. The more windings you have, the more impendance you will have. Motors do have an internal resistance spec as well as an inductance spec. Also, I THINK (not positive mind you) motors w/ higher poles use windings w/ lower inductances (more impendance). I guess I didn't state my case clearly enough.
The motor will generally get hotter due to increased current draw. Heat is energy & in this case the heat can be considered losses or inefficiency. This would help to explain the lower efficiency.
Then again, I could be wrong.........I'm sure I'll hear about it if I am.
Guy H. Looney
Regan Controls, Inc.
475 Metroplex Dr.
Nashville, TN 37211
phone: (615) 333-1940
fax: (615) 333-1941
"Guy H. Looney" wrote:
> Ken & Simon,
> I agree w/ both of you, however you're assuming only the inductive
> properties of the windings. The windings do have an overall
> impendance. The more windings you have, the more impendance you
> will have. Motors do have an internal resistance spec as well as an
> inductance spec. Also, I THINK (not positive mind you) motors w/
> higher poles use windings w/ lower inductances (more impendance). I
> guess I didn't state my case clearly enough.
Less inductance, less resistance, less impedance Z = the square root of the sum of Xl squared + R squared.
In response to Sr. P. Moschetti's query and Guy Looney's reply:
An AC induction (asynchronous) motor is a remarkable machine in its simplicity. But, that very simplicity is a result of a design
compromise. High efficiency under normal operating conditions requires low rotor resistance. But, low rotor resistance results in several deficits: low starting torque; high starting current for across-the-line starting; and low starting power-factor.
BTW, the wound-rotor motor eliminates the need for compromise but at a substantial first-cost increase. However, it does fill the need for
certain applications as was described by Ken Brown's lift-bridge application.
Returning to discussion of the original question. Neglecting physical constraints, the following explains why the a size increase in the VSD
drive rating may be required.
For the constant horsepower application where speed is reduced, ie, from the 2-pole to the 4, 6, 8-pole case, then the torque requirement
increases proportionately. But, the rotor's power is derived from the stator's electrical winding via energy transfer thru the air-gap separating the stator and rotor magnetic structures. Thus, the required increase in torque must comes from an increase in "slip", that is, the difference between synchronous and operating speed.
The power absorbed by the rotor divides between mechanical power (torque x speed) and electrical power (rotor winding I^2xR losses). Thus, there
is an additional increase in the stator current that reflects the change in rotor current. One more detriment is the lowering of efficiency due
to additional stator and rotor winding I^2xR losses.
Phil Corso, PE
Deerfield Beach, FL
The Physics of...
"AC Induction (asynchronous in Europe) Squirrel-Cage Motors."
Further to my 4-July reply on the subject, I apologize for the redundancy and superfluidity. The penultimate sentence should be replaced with:
"... Thus, there is an increase in stator current corresponding to the increase in rotor current. Simply stated, increased torque requirement
---> increases "slip" ---> increases rotor current ---> increases stator current. Furthermore, the additional I^2xR losses in both stator and rotor windings reduces efficiency.
These characteristics are illustrated in the mathematical model for AC induction motors and the resultant Speed vs Torque and Ampere curves.
In conclusion, suggestions advanced to explain the need for increasing the VSD drive rating... are in error. However, Ralph McDonald did
provide a pointer about nameplate data. In fact the NEC contains the following cautionary note:
"... Motors built for especially low speed or high torques may require more running current, in which case the nameplate rating should be used." At the very least, the motor manufacturer (if available) should be consulted.
Phil Corso, PE
Deerfield Beach, FL
I missed the beginning of this thread, but I thought HP was HP, and poles just determined the RPMs of the motor...
So if HP remains constant, so should the Current...assuming same efficiency.
beyond four poles it is the case that the power factor and efficiency of electric motors drops. you can see the data from handbooks on motors, or try to download the motormaster software from the us dept of energy website. the physical reason i am not sure but might be the need to have more torque to produce same horsepower at lower speed, torque is usually produced by more current flowing
thru the conductors. i think this leads to more losses and the worse power factor.
Looking through some motor catalogs, I was unable to see any definite relationship between pole count and the resistance/inductance of the stator windings. There are probably too many other variables.
However, one effect is undeniable. Take two motors with identical electrical properties, but one with twice the pole count of the other. To spin at the same mechanical speed, the higher pole-count motor requires twice the electrical frequency. At this higher frequency, the inductive lag due to the L/R time constant will be larger.
As Ralph MacDonald pointed out in the first reply, this results in greater reactive power requirements -- really current that is not generating useful work.
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