Calculating motor full load current with power factor correction

[DrDrreeeaaa]

Hello All

When inspecting a 3ph induction motor nameplate a value for FLA is given for rated power and speed.

Does this value take into account the motor's power factor?

In which case, when calculating a plant total amps where power factor is fitted, one needs to consider motor rated watts only.

Is this correct? Thanks

Michael

 

[itsmoked]

The FLA is what you will measure when you clamp an amp probe around a wire feeding the motor. The intent, of course, is so any technician can check out the setup. It does however include the magnetic component. It doesn't take in the pf.



Keith Cress
kcress -

http://www.flaminsystems.com

 

[DrDrreeeaaa]

If you have a 40kW 3ph induction motor on a power-factor corrected circuit (i.e. say for argument's sake a 1.0 PF) then it will draw approximately 70% of the nameplate FLA. Is this not correct?

 

[waross]

The motor full load amps is at full load and does not consider power factor correction.
There are a number of reasons for this.
Correction may be applied at the motor, at the starter, at the service, or at some other location.
There are many philosophies on power factor correction.
Power factor may be corrected to unity or it may be corrected to a value that avoids penalties.
It may be corrected continuously or it may be corrected on the basis of KVARHrs per billing period.
It may be bulk corrected, corrected by a real time controller, capacitors may be arranged to switch in with selected motors, or some other scheme may be used.
In the event of a failure of the correction capacitors, the system must be able to safely supply full load current as determined by the motor nameplates.
A typical power factor value for a loaded motor may be in the order of 90%. In many jurisdictions there is no penalty for a 90% power factor.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[DrDrreeeaaa]

So, in an industrial plant with many motors, you would size your power-factor corrected service to serve the nameplate FLAs, with diversity as determined.

In reality you would be drawing anywhere from 70-90% of this number depending on the motor pf, but the service needs to be able to supply the load when PF correction is not working...

So an asset owner would not get any benefits in service size (augmentation , TF size, etc.) for PF correction, only regulatory benefits in terms of the distirbutor's rules...?

 

[davidbeach]

You can never have a service size that is "just right". 10%, 15%, 25% over sized doesn't add much cost and provides a great deal of flexibility plus allows for inevitable expansion. Undersized can get very expensive.

 

[waross]

In North America, The code requires feeders, panels and transformers to be sized at 100% of the sum of the motor full load currents plus 25% of the current of the largest motor. (This is for customer owned equipment, utility owned transformers are sized at the discretion of the utility and may be less. At one time our local utility used a factor of 500 VA per horsepower to size transformers for industrial plants.) You can only use a diversity factor when adding new equipment to an existing installation. You must have several months or more of load records or monthly peak load records to justify to the AHJ that the proposed addition will be safe.
The penalties for poor power factor are severe enough that the pay back period for the correction equipment is most often less than a year.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[NAZ55]

Bill,

I think you meant 125%

Thanks
Zazmat

 

[NAZ55]

Never mind, I see what you were saying

 

[davidbeach]

No, it was stated correctly. 100% of all plus 25% of the largest is the same as 125% of the largest plus 100% of all others.

 

[tomad]

The so-called FLA rating indicates the total current drawn by the motor, at nominal (rated) voltage and frequency, to develop the rated mechanical power (100%). Since you determinate the imput (electrical)power as Input Power[kVA} = Voltage [V} x sqrt(3) x Current (A, FLA)
and the output power (mechanical) as
P [ kW/HP] = Input Power [kVA} x efficiency [%] x power factor, the FLA current takes care of the power factor.

 

[DrDrreeeaaa]

So we can agree that when sizing the service it is best practice to assume an uncorrected power factor.

Has anyone had experience justifying this to their client?

It is only the infrastructure costs though, not the ongoing $/kWh

 

[itsmoked]

Not only best practice it's the rules.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[DrDrreeeaaa]

And what about with a motor controlled by a VFD? do you use nameplate amps or power-factor corrected amps?

 

[waross]

Actually, in Canada you use the motor nameplate amps for motors. In the US you look up the HP in the NEC and use the value in the code. That way if a high efficiency motor is replaced with a standard motor at a later date, the installation is still adequate.
For VFDs, use the nameplate amps of the VFD.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[DrDrreeeaaa]

Thanks all

 

[jghrist]

The input to a VFD is close to unity pf, so pf correction would not be applied.

 

[waross]

I was called in to a plant to find the power bill included thousands of dollars of penalties.
Turns out the word had gone out from the head office to economize and turn off all unused equipment. Someone turned off the main breaker to the automatic power factor correction equipment.
The moral.
1> Forget power factor.
2> Design your system.
3> Remember power factor and add correction.
You don't want a situation where it is possible to overload your feeders and mains by turning something off, even though the actual possibility is nil.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[DrDrreeeaaa]

Hi All, I have come across this problem again and would like your opinion.

We have several (20+) large motors (>50kW anf 400V) fitted with VFD. When determining estimated maximum demand I have considered them operating at 0.9PF, while all other DOLs operate at 0.75. If I ignore the PF benefits of the VFD the load is too big for the TF.

I am not comfortable with my approach since:

1) While displacement PF quoted by the manufac. is 0.98, distortion PF will be poor. Does distortion power factor contribute to greater current draw?

2) What I have done is take motor kW and converted to Amps at 0.9PF to get the VFD line current. Will the VFD draw more than motor kW amps from the line (even if it is at a good disp. PF)

Thanks all

MB