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

[Ingenieur]

To sustatin that you'd need the input bridge to be regererative. I don't think that is being discussed here.

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

 

[Ingenieur]

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?

 

[Smart $]

To sustatin that you'd need the input bridge to be regererative. I don't think that is being discussed here.

It's all irrelevant. :slaphead:

 

[Ingenieur]

It's all irrelevant. :slaphead:

:lol:

FEM
Freakin'
Electrical
Magic

 

[Jraef]

1 ph to 3 ph was my thought too (also lower v in than out)
but he clearly stipulates 3 ph/460 in/out

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

 

[Ingenieur]

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

that would explain it also

I don't know if the op posed this as an actual measured operating condition or as a hypothetical and we are to assume the given parameters of in=out =460/3 to be valid

if it is 'single phasing' the current is gong to the motor lol

 

[Ingenieur]

What would happen in this case:
10 kw motor at 100% losd
0.9 pf
S = 11.1 kva/26 deg = 10 kw + j4.8 kvar

vfd caps
sized at 4 kva
oversized at 10 kvar
???

 

[mike_kilroy]

This makes sense

Wrong.

Of course, but irrelevant to this discussion. I specifically said ignore output of vfd for good reason


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[mike_kilroy]

Depends on whether the power factor is displacement only or displacement and distortion.

https://mobile.yaskawa.com/delegate...D.10&cmd=documents&documentName=WP.AFD.10.pdf

Read OP post again please. Distortion vs displacement PF clearly discussed.

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[mike_kilroy]

Where?

Where indeed? That is the question to answer.

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[mike_kilroy]

Yes... but the VFD has a dc link choke which is added to the VFD design intentionally to improve power factor. Not all VFD's are designed with power factor improvement in mind.



Mostly reactive current? 0.2 pf is practically as low you can go.

VFD input won't see that, which is why it is 1-3%.

All irrelevant to OP question. I cannot be any clearer than saying output side of VFD is irrelevant.

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[Ingenieur]

Wrong.

Of course, but irrelevant to this discussion. I specifically said ignore output of vfd for good reason

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Wrong
It is one of only things that make your scenario plausible
lost phase
mistake : output i is >!than input
your scenario is impossible if everything is 'irrelevant'
output
reactive components
etc

there is no way you measured these as actual values unless you are giving incomplete or misleading data

your question is theoretical and implausible

 

[mike_kilroy]

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

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

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[mike_kilroy]

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.

.

Irrelevant. Reread OP please.

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[mike_kilroy]

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

Please do not change the question. No added hidden stuff involved.

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[Ingenieur]

All irrelevant to OP question. I cannot be any clearer than saying output side of VFD is irrelevant.

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wrong
otherwise you would not have included it
it is related to input, especially at 100% eff
absurd
clear as mud
it's made up and can't exist in reality

 

[mike_kilroy]

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.

Getting warm! (And not about older stuff)

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[mike_kilroy]

Heat?

Hmmmm.... maybe?

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[mike_kilroy]

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

You may assume 100% drive efficiency for this.

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[mike_kilroy]

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

This OP is totally NOT related to either C or L.

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