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

[mike_kilroy]

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!

First time in my life I AGREE WITH YOU SAHIB!!!!!

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[mike_kilroy]

There may have been some confusion also with the peak input current rather than the RMS, and with the current from a single phase source when driving a three phase output. And finally, perhaps, the Code requirement to size the input circuit by the VFD rated current rather than by the actual load.
Regardless of any of those points of potential confusion the input power is greater than the output power only by the inefficiency of the conversion circuit.
Finally some VFDs allow a higher output voltage than input voltage using voltage multiplier circuits to produce a higher DC bus voltage. In that case equal power means a higher input current than output current by the voltage conversion ratio.

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steve, my post has nothing to do with OUTPUT side.

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[Ingenieur]

Irrelevant. If one says efficiency of the black box is 100% by definition for this discussion. I did.

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wrong
if 100% eff
input P = output P, goes for real i too
we know
i total =sqrt(re i^2 + img i^2) = 12
we can ignore v and sart 3, they cancel
and re i in = re i out since 100% eff and no img i out (no reactive component)
so 12 = sqrt(6^2 + img i^2)
img i = 10.4
total i in = 6 + j10.4
total i out = 6 + j0
not possible unless the box is dissapating Q (not exchanging)
but 100% eff means no losses

not possible w/o reactive devices in the ckt which you say there are not

 

[Ingenieur]

steve, my post has nothing to do with OUTPUT side.

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then why do you define the output in detail?
V
i
no reactive

 

[Ingenieur]

Matched vfd/motor at 100% load and eff
displacement pf ~0.98
tdh 45% max
total pf 0.91
if total i = 12
then re i = 10.9
but out re i is only 6?
and eff = 100%
how???

your example is pf = 0.5

 

[kwired]

steve, my post has nothing to do with OUTPUT side.

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:?

How does one assign any kind of efficiency if we don't compare input to output?

If the thing gives off heat there is inefficiency in there - unless it's primary function is to create heat.

I've never seen a VFD that doesn't give off heat when in use, there is some that are higher efficiency then others but none that are 100% efficient at transferring all input energy to the output terminals.

 

[mike_kilroy]

There may have been some confusion also with the peak input current rather than the RMS, and with the current from a single phase source when driving a three phase output. And finally, perhaps, the Code requirement to size the input circuit by the VFD rated current rather than by the actual load.
Regardless of any of those points of potential confusion the input power is greater than the output power only by the inefficiency of the conversion circuit.
Finally some VFDs allow a higher output voltage than input voltage using voltage multiplier circuits to produce a higher DC bus voltage. In that case equal power means a higher input current than output current by the voltage conversion ratio.

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No confusion. None of your other ideas either. Just simple question as asked.

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[mike_kilroy]

We are talking about the vfd losses
if we assume vfd to be 100% eff it has no losses
in reality on the order of 2%
for the purpose of this discussion it has none

Thank you. Not significant.

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[mike_kilroy]

My first guess would be that you have mislabeled input and output. My second, if that is not true, is that you have single-phase input; have you checked input current in all 3 legs?

Not 1 ph. Same exact current all 3 legs of input. Will post scope pix proof soon.

This is why I posted this thread. We all (me included) have a lot to learn/understand

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[mike_kilroy]

If we accept it as reactive it oscillates between L and C elements gradually dissapating and needing replacement if R elements exist in the path

I do not believe the op's scenario plausible as described

It is NOT reactive current.

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[mike_kilroy]

I asked
he said 'irrelevant'

what happens if the 'motor' is over-driven, ie, generating?

Irrelevant to discussion. And not happening.

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[mike_kilroy]

We do not know what is being discussed
that is the problem
all we know
460/3 in/out
12 A in apparently at 0.5 pf (lead or lag undetermined)
6 A out pf 1 100% loaded (no reactive components)
vfd eff close to 100%
no info on drive type; regen, braking resistor, etc
no motor data:hp, pf, etc
calculated tdh 150%
no external caps or reactors (since that question is 'irrelevant')

so regeneration IS on the table
but this would imply motor is negatively loaded
He did not spec power angle/flow direction, lead/lag, etc

can this scenario exist?
does the current exist?
if so what is the source and snk, supply/load, etc

Don't keep changing it....

I said output in phase current was 6a rms. I said IGNORE the reactive output - it is irrelevant. Reactive output stays in vfd - motor so has no bearing on my OP.

I did not say output was pf 1. I said INPUT displacement PF is 1.

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[mike_kilroy]

We do not know what is being discussed
that is the problem
all we know
460/3 in/out
12 A in apparently at 0.5 pf (lead or lag undetermined)
6 A out pf 1 100% loaded (no reactive components)
vfd eff close to 100%
no info on drive type; regen, braking resistor, etc
no motor data:hp, pf, etc
calculated tdh 150%
no external caps or reactors (since that question is 'irrelevant')

so regeneration IS on the table
but this would imply motor is negatively loaded
He did not spec power angle/flow direction, lead/lag, etc

can this scenario exist?
does the current exist?
if so what is the source and snk, supply/load, etc

No regen, no braking - all irrelevant anyway.

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[mike_kilroy]

I have a black box
v in/out 460/3
12 A in
6 A out 100% load with negligible reactive component and with mechanical conversion, mech torque delivered
the box is 100% efficient

where does the 6 A go?
how and why?

Not what I said. I said IGNORE output reactive component - it is irrelevant to this question!

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[mike_kilroy]

Blown fuse on the 480V input, making the input EFFECTIVELY single phase? I've seen it happen, albeit on a VFD that was sufficiently over sized to avoid tripping on excess DC bus ripple. Some VFD mfrs put Phase Loss detection on drives designed for 3 phase input, some rely on DC bus ripple monitoring instead. Allen Bradley drives for example will not trip off line on a blown input fuse if the load on the drive is low enough (50% or less of the VFD rating) to avoid excess ripple. On several occasions where people have complained about drives tripping on DC bus ripple, I have found blown fuses. The users ASSUMED the drive would trip on Phase Loss, so they ASSUMED the fuses were good, but there is no Phase Loss Detection on an AB drive. If the drive was over sized, or the load was low, they will continue on as if nothing is wrong.

But IF, at that time, you read the input current and the output current, I now believe the input WILL be roughly 2X* the output current

*(Prior to last month I would have said 1.732x, but I have learned my lesson... :slaphead

No blown fuse. Someone needs to READ OP more carefully.

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[kwired]

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??

Are you certain you are measuring true RMS?

 

[mike_kilroy]

wrong
if 100% eff
input P = output P, goes for real i too
we know
i total =sqrt(re i^2 + img i^2) = 12
we can ignore v and sart 3, they cancel
and re i in = re i out since 100% eff and no img i out (no reactive component)
so 12 = sqrt(6^2 + img i^2)
img i = 10.4
total i in = 6 + j10.4
total i out = 6 + j0
not possible unless the box is dissapating Q (not exchanging)
but 100% eff means no losses

not possible w/o reactive devices in the ckt which you say there are not

All not so. You are so missing the obvious!

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[mike_kilroy]

then why do you define the output in detail?
V
i
no reactive

Just to show volts out =volts in, so question of 200v out, 400vdc in so input current expected to be different and that is reason.

Ditto current. Ignore reactive current as it does not come from vfd input.

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[mike_kilroy]

Are you certain you are measuring true RMS?

Yes

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[mike_kilroy]

Enough. Can distortion PF increase Irms? If so, where does that increase GO?

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