So where does extra current go with bad distortion PF (Standard VFD input)?

[mike_kilroy]

A recent thread examined displacement PF and VFDs. It was shown this displacement PF stayed inside and on the output of the VFD so did not effect the AC input to the VFD.

So many say since the input displacement PF to the VFD is nearly 1, there is no extra input current, like out of phase, or 'imaginary.'

Others point out the input RMS current can be 2,3,4,5 times HIGHER that to produce power in the motor at output; yet since displacement PF is effectively 1.0, how can this be? So where does this extra rms current GO? It is caused by distortion PF, even though it is effectively in phase current.

Example: Our 460vac 3 ph input VFD outputs 460Vac 3ph into a fully loaded motor pulling 6 amps RMS in phase current (forget the out of phase aka magnetizing aka imaginary current flowing back and forth from VFD to motor). So assuming near 100% efficiency of the VFD, we expect to see same 6 amps RMS at VFD input terminals. But the input RMS current is measured at 12 amps!

Where does the extra 6 amps go?? Is it real? Imaginary? Power Factor?

Understanding THIS will go a long way to helping eliminate confusion and helping some folks grasp the whole picture.

Where does the extra 6 amps go??

 

[Ingenieur]

Sounds implausible

???
motor hp
pf
eff
total motor current
any line filters or reactors
line side pf caps
???

something does not compute
unless the vfd has a bad pf ~0.70

 

[mike_kilroy]

motor hp - irrelevant
pf- irrelevant
eff- irrelevant
total motor current- irrelevant
any line filters or reactors- irrelevant
line side pf caps- irrelevant

 

[topgone]

motor hp - irrelevant
pf- irrelevant
eff- irrelevant
total motor current- irrelevant
any line filters or reactors- irrelevant
line side pf caps- irrelevant

Things are "converted" in a vfd. You can't expect the same values at its input and the output.

 

[gar]

170706-2027 EDT

mike_kilroy:

Measure the real power at the motor input or VFD output.

Measure the real power input to the VFD at its input terminals.

Measure the RMS input current to the VFD at one input terminal.

Measure the RMS input voltage to the VFD at its input terminals.

Use a scope to simultaneously measure input current and voltage at the VFD input.

Show us this data.

.

 

[Ingenieur]

motor hp - irrelevant
pf- irrelevant
eff- irrelevant
total motor current- irrelevant
any line filters or reactors- irrelevant
line side pf caps- irrelevant

sure is relevant for the conditions stated
you seem to imply the 6 A to the motor is all real power 'forget imaginary current'
or that there is no reactive current

1.732 x 12 x 460 = 9.6 kva in
1.732 x 6 x 460 = 4.8 kw out
vfd pf = 0.50

if there are caps or reactors after the input measurement location it will have an effect

 

[Smart $]

...

Where does the extra 6 amps go??

I'm with Ingenieur as to 12A-rms in and 6A-rms out being plausible, at least with a highly-efficient, modern VFD. Perhaps a much older VFD? IDK. Not my forte.

Nevertheless, you cannot violate the physics of the matter.

Power_IN – internal losses = Power_OUT

Break it down from there...

FWIW, what appears as missing power is obscured by the waveforms, output voltage included... and that is where the distortion power factor enters the picture.

 

[Ingenieur]

Drive is 100% eff
P in = P out = 4.8 kw
9.6 kva = 4.8 kw + jQ kvar
Q = 8.3 var
S = 9.6 / 60 deg (can't determine lead or lag)
pf = cos 60 = 0.5

something beyond the measurement point is consuming/supplying reactive power
since the motor apparently needs none pf=1

 

[Coppersmith]

Heat?

 

[Smart $]

Heat?

Not possible if the drive is [near] 100% efficient.

 

[Ingenieur]

Heat?

not if 100% eff and the mystery current is reactive :happyno:

 

[topgone]

not if 100% eff and the mystery current is reactive :happyno:

Heat will always be there if there is current, reactive or not. Too much current will get you nearer a thermal runaway though.

 

[Smart $]

Heat will always be there if there is current, reactive or not. Too much current will get you nearer a thermal runaway though.

But heat shows up in the efficiency rating because it is an "internal" loss.

 

[Ingenieur]

Heat will always be there if there is current, reactive or not. Too much current will get you nearer a thermal runaway though.

Not if eff is 100%
no thermal losses, i^2 R
ie, no resistance, only L and C

 

[Sahib]

One thing we used to is power, whether it is active or reactive, always flows ie input power=output power. But this not the case with distortion power ie it is not conserved. May be that is the point OP is trying to make!

 

[Besoeker]

A recent thread examined displacement PF and VFDs. It was shown this displacement PF stayed inside and on the output of the VFD so did not effect the AC input to the VFD.

So many say since the input displacement PF to the VFD is nearly 1, there is no extra input current, like out of phase, or 'imaginary.'

Others point out the input RMS current can be 2,3,4,5 times HIGHER that to produce power in the motor at output; yet since displacement PF is effectively 1.0, how can this be? So where does this extra rms current GO? It is caused by distortion PF, even though it is effectively in phase current.

Example: Our 460vac 3 ph input VFD outputs 460Vac 3ph into a fully loaded motor pulling 6 amps RMS in phase current (forget the out of phase aka magnetizing aka imaginary current flowing back and forth from VFD to motor). So assuming near 100% efficiency of the VFD, we expect to see same 6 amps RMS at VFD input terminals. But the input RMS current is measured at 12 amps!

Where does the extra 6 amps go?? Is it real? Imaginary? Power Factor?

Understanding THIS will go a long way to helping eliminate confusion and helping some folks grasp the whole picture.

Where does the extra 6 amps go??

It doesn't go any where.

 

[Sahib]

It doesn't go any where.

It is present on line side and causes losses.

 

[Besoeker]

It is present on line side and causes losses.

Indeed it does.
Common sense kicks in.

 

[kwired]

But heat shows up in the efficiency rating because it is an "internal" loss.

100% efficiency means there is no losses doesn't it?

Not if eff is 100%
no thermal losses, i^2 R
ie, no resistance, only L and C

Have we developed conductors with no resistance? AFAIK there is always some loss, we just have found some ways to keep them pretty minimal.

 

[mike_kilroy]

It doesn't go any where.

Ah, but it does.