Single Phase current draw for a 3 phase output VFD: technical discussion

[Ingenieur]

170611-1705 EDT

If I have a circuit consisting of a DC battery of constant voltage with a resistive load of constant resistance, then what is the power factor of that load.

Next add some inductance in series with the resistor. What is the power factor of the series resistor and inductor under steady state conditions?

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Is this a trick question?
Xl = j 2Pi f L
f = 0

 

[Ingenieur]

ABB
Derate by 50% (or double the HP rating)
https://library.e.abb.com/public/610f2d332765615e85257d8d005c1475/LVD-EOTN71U-EN_REVA.pdf

 

[gar]

170611-1747 EDT

Ingenieur:

Not a trick question.

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[mike_kilroy]

There is little reactive input current for a diode bridge.


Components have to be rated for actual current, not just one component of it.


Not really. The current can be anything from a square wave to something very peaky as shown earlier. Diode dissipation from nearly level continuous current to peaky discontinuous is quite different.


There's the crunch and why you cannot specify a fixed multiplying factor.

Guess you didn't even bother to look at the cool link I listed. OK. And if you do not believe rms exactly describes heating value of the diode current input, ok again. You are wrong.

 

[mike_kilroy]

That's just not true.
With decent filtering, the Idc is approximately level. Distortion PF is about 0.958.
The last motor I dealt with (225kW 400V) had a PF of 0.7. The currents can't be nearly the same.

With minimal filtering the input PF could be as low as 0.3. The output power factor would still have been 0.7. Again, the currents can't be nearly the same.
Like I said, "there is no one size fits all" factor.

nope. Did you bother to look at the link I provided? You will see your error.

 

[mike_kilroy]

you need to take that up with Hitachi's vfd design engineers (that is an excerpt from the doc)
I agree with them, though

Hitachi says next size up from a factor of 2

pf does not factor into S = sqrt3 x v x i

As a Hitachi hi tech distributor, I have had this discussion with them too. They have models they sell as constant torque and also rerated as variable torque. They had to agree with my analysis that since they rate the currents HIGHER on the variable torque models of the same units, that upsizing only 1.5x was perfectly acceptable.

 

[Besoeker]

nope. Did you bother to look at the link I provided? You will see your error.

I looked at the link. It corroborates what I have been saying.

 

[Ingenieur]

As a Hitachi hi tech distributor, I have had this discussion with them too. They have models they sell as constant torque and also rerated as variable torque. They had to agree with my analysis that since they rate the currents HIGHER on the variable torque models of the same units, that upsizing only 1.5x was perfectly acceptable.

That makes sense
if the drive has a i rating > nominal hp by say 10-20%
1/1.15 x 1.732 = 1.5
plus motors if applied properly are not loaded 100%

 

[mivey]

That makes sense

Except for the following issue:

I put my money on Hitachi's 40 years experience and team of PhD's

I hardly qualify as a 'layman'
grad degree in EE with several courses on power electronics
working with drives since the mid 80's

I would never consider you a layman and was not implying such. My point was that it would be unusual to find write-ups like that to be fully vetted and certainly not by all their experts over the past 40 years including their team of PhD's.

I'm sure you know and have seen as well as I that all publications even from reputable companies are not always correct nor are they always technically accurate. Many are written as general guides and must be taken as such. If you want a detailed/precise answer to complex topics, you usually have to dig deeper than consumer guides as they are simplified on purpose.

 

[Ingenieur]

Except for the following issue:



I would never consider you a layman and was not implying such. My point was that it would be unusual to find write-ups like that to be fully vetted and certainly not by all their experts over the past 40 years including their team of PhD's.

I'm sure you know and have seen as well as I that all publications even from reputable companies are not always correct nor are they always technically accurate. Many are written as general guides and must be taken as such. If you want a detailed/precise answer to complex topics, you usually have to dig deeper than consumer guides as they are simplified on purpose.

EVERY company says basically the same thing
1.732, use 2
I doubt those companies put out application notes that are incorrect
I'm sure their document qc/qa is good

 

[gar]

170611-2017 EDT

mike_kilroy:

In post #225 you reference a link in some other post that is supposed to prove something.

What is that other post, and link? What is it supposed to prove?

I believe you are trying to say that the RMS value of current thru a diode can tell one the power dissipated in the diode. It can not. A diode is not a linear constant valued component.

Because a diode is not a linear resistor, to determine the power dissipated in a diode, you must multiply the instantaneous values of the current through the diode with the corresponding voltage across the diode, integrate (sum the small values of v*i*dt), this must be done over an adequate time, and divide that integral result by the time period of integration. Most convenient to use the time period of one cycle.

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[mivey]

EVERY company says basically the same thing
1.732, use 2
I doubt those companies put out application notes that are incorrect
I'm sure their document qc/qa is good

I don't doubt the notes are adequate for their purpose.

I doubt they were vetted by the staff of 40 years and by all of their team of PhD's.

I also doubt their vfd design engineer would disagree with Besoeker's or gar's points.

 

[Smart $]

...
I also doubt their vfd design engineer would disagree with Besoeker's or gar's points.

:rotflmao:

Are you sure you understand the nature of design engineers (and educators)? :slaphead:

 

[Phil Corso]

Gentlepeople... look at the problem from an energy viewpoint (assuming a lossless system) that is:

The energy conversion in the output’s 3-phase full-wave, AC circuit, is essentially, Vm*Im (ignoring ripple)... call it 1-unit! But, the energy conversion in the input’s single-phase, DC-circuit, is essentially, Vm*Im*[1-Cos(wt], resulting in an input energy deficiency of Vm*Im*[Cos(wt)] or 1-Cos(wt)!

Thus, the importance of the capacitor, and the reason for de-rating as indicated in one of the attachments!

Phil Corso

 

[gar]

170612-14124 EDT

Phil Corso:

I have no idea what you are trying to say in post #234.

Further, how would your #234 comments relate to where the weak spots exist in a VFD when supplied from single phase vs three phase power source?

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[Sahib]

The highlighted statements are a contradiction for a 100% efficient black box considering there are no reactive components in the input stage. If any of the reactiveness of the capacitor is leaked back to the input, then the black box is not 100% efficient. Keep in mind, when I say power in equals power out, I am talking real power (W), not apparent power (VA).

Are you saying if an equipment has no resistance, it cannot generate harmonics? If yes, incorrect!

 

[Sahib]

170612-14124 EDT

Phil Corso:

I have no idea what you are trying to say in post #234.

Further, how would your #234 comments relate to where the weak spots exist in a VFD when supplied from single phase vs three phase power source?

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IMO, Phil is trying to say three phase power is steady whereas single phase power is varying and so a derating is necessary in case of single phase power.

 

[Smart $]

Are you saying if an equipment has no resistance, it cannot generate harmonics? If yes, incorrect!

No... I'm saying you can't have a 100% efficient black box with a non-reactive power factor less than 1. It defeats the concept of "ideal" by placing limited and imposed conditions on the result of the exercise.

 

[gar]

170613-0838 EDT

A lossless nonlinear inductor.

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[Sahib]

No... I'm saying you can't have a 100% efficient black box with a non-reactive power factor less than 1. It defeats the concept of "ideal" by placing limited and imposed conditions on the result of the exercise.

'Ideal' here refers to active power input and output of black box proper and not that of its associated power lines. In that case input power and output power equal but power factors need not be as no power loss in the lossless black box as a result.