What if it is not usual?
The drive/motor manufacturer, as weressl noted, have necessary software to recommend the VFD output cable in all cases. Let the OP refer to them for his problem.
VFD Long Lead Lengths - Correctly Calculating Voltage Drop
- Thread starter wonderdave
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The drive/motor manufacturer, as weressl noted, have necessary software to recommend the VFD output cable in all cases. Let the OP refer to them for his problem.
mike_kilroy
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Perhaps his Xl voltage drop addition is not due to the cable, but due to this series inductor? What is the model/specs on the TCI "filter?" The are often speced as "X% impedance;" which is really X% voltage drop at rated current flow.
I have seen no indication of including this voltage drop in your OP, and it certainly may be a significant portion - depending on the filter design.
Then it would be unusual.
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Assume 6 pulse drive
pu current FFT (typical and for calculation reference only) ABB VFD
freq/harm...pu
f 0.6
5 0.20
7 0.14
11 0.09
13 0.07
17 0.06
19 0.05
X = L/2 x 2 x D/1000 x 2 x Pi x 60 x (0.6x1 + 0.2x5 + 0.14x7 + 0.09x11 + 0.07x13 x 0.06x17 + 0.05x19)
= 0.377 L D (6.45) = 2.43 L D
this tells us harmonics increase X by 645% over a fund only
L = 3 ph inductance for 1 set, we have 2) 4/0 per 1000' calculated from NEC table steel cnd
= 135 uH = 135 x 10^-6
D = distance 2200'
X = 2.43 x 2200 x 135 x 10^-6 = 0.72 Ohm
R is easily calculated 2 x 2200/1000 x 0.063/2 = 0.14 (26 vdrop at 190 A)
Z = sqrt(R^2 + X^2) = 1.1 ohm
will be much lower for a 12 (or 18) pulse since fund 0.8+ (0.9+) and harmonic sum
muh lower but of higher freq
I figure the op's engineer did something like this for his calc
pu current FFT (typical and for calculation reference only) ABB VFD
freq/harm...pu
f 0.6
5 0.20
7 0.14
11 0.09
13 0.07
17 0.06
19 0.05
X = L/2 x 2 x D/1000 x 2 x Pi x 60 x (0.6x1 + 0.2x5 + 0.14x7 + 0.09x11 + 0.07x13 x 0.06x17 + 0.05x19)
= 0.377 L D (6.45) = 2.43 L D
this tells us harmonics increase X by 645% over a fund only
L = 3 ph inductance for 1 set, we have 2) 4/0 per 1000' calculated from NEC table steel cnd
= 135 uH = 135 x 10^-6
D = distance 2200'
X = 2.43 x 2200 x 135 x 10^-6 = 0.72 Ohm
R is easily calculated 2 x 2200/1000 x 0.063/2 = 0.14 (26 vdrop at 190 A)
Z = sqrt(R^2 + X^2) = 1.1 ohm
will be much lower for a 12 (or 18) pulse since fund 0.8+ (0.9+) and harmonic sum
muh lower but of higher freq
I figure the op's engineer did something like this for his calc
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Not going to happen.
That is lame. The VFD output has harmonics with 5th and 7th predominant. If you do not agree, please show with your own meaurement from field.
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Done a lot of work on drives have you?
Thank you once again for your kind, informative, and erudite response.
With your keen powers of observation you will have noted that the magnitudes given are the exact reciprocal of the harmonic number. That would require the input current to be a rectangular 120deg wave. Exactly level DC out of the bridge. This would be approximately true on a DC dive. Not really very likely on a variable frequency drive unless it has an approximately infinite DC choke and many of the cheap and nasty ones don't have any DC choke at all. So you get discontinuous current pulses.
Fifth is often quoted as 40% or more.
winnie
Senior Member
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- Springfield, MA, USA
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- Electric motor research
Those numbers are suggestive of a 'current source inverter', where the DC link between rectifier and output switches is a large DC inductor rather than a capacitor bank.
In a CSI the inductor acts to maintain constant current through the switching elements and the load. A six step CSI will have output _current_ pulses that are approximately 120 degree square pulses with huge harmonic content.
Beo can comment about how likely it would be to find a six step CSI on a low voltage 200 hp drive installed in the past couple of years. I was under the impression that they were used at higher voltages, and are an older technology.
IMHO it is very unlikely that the drive that the OP is dealing with is a CSI, and unlikely that the OP is dealing with such high harmonic current. However the calculation showing the high voltage drop associated with such an inverter is possibly very useful to the OP because the 'other engineer' might be using the same sort of assumptions to come up with a voltage drop calculation.
-Jon
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- Massachusetts
How about this, I will ask of you what we ask of all folks here.
You put your own figures up, not a link, your own figures to prove the point you want to make. Before asking others to show you how it is done.
I have only ever dealt with one and that was probably somewhere around two or possibly even three decades ago.
For drives in the MW range, we had our own design of the Static Kramer.
mike_kilroy
Senior Member
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Didn't the OP current feedback picture prove it was 6 step? Maybe ya'll need to go look at that cool 6 step sine wave he posted again?
Don't forget the voltage drop on the output LCI inductor!
Incorrect.
Don't recall seeing that.
Are you saying there is no harmonics in output wave form of VFD PWM drive?
Of course not! But 5th and 7th refer to the input. Not the output.
lol
big difference between selling and applying a converter and designing one
they are used for more than xxx HP motor drives
such as 1 MV DC links and xxx MW pf correction and fault transient control
in these applications the OUTPUT is injected onto the grid
this requires harmonic distortion mitigation
iso transforrmers
tuned LC filters based on the output characteristics
variable switching timing/angles
etc
you never explained what method you think the op's engineer used to arrive at a 15-20% drop
we know the R drop is 3-6% depending of whose calcs you use
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winnie
Senior Member
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- Springfield, MA, USA
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- Electric motor research
First you must differentiate between voltage harmonics and current harmonics.
A voltage source inverter (the kind with an input rectifier, capacitor filter, and output switches) will have a very ugly waveform at the switching frequency, and will have 'harmonics' associated with that. (Harmonics in quotes because the switching components of the output waveform are generally not harmonic multiples of the desired output frequency.) However the inductance of the motor (and circuit conductors) will act as a much higher impedance to these high frequency components. So the output _current_ of the inverter has much lower 'harmonic content' than the output _voltage_.
There will be _some_ harmonic current in the output of a modern PWM VFD, however below the switching frequency, such harmonics (both voltage and current) will be minimal, certainly not 20% of the fundamental. Below the switching frequency, any harmonic current has to do with imperfections in the PWM generation, and proper design should minimize this. At the switching frequency there is pretty significant current flow, however inductive reactance in the conductors to the motor will act to reduce this switching frequency ripple.
The waveforms in this article match what I've seen on my own research inverters; you can see the huge ugly waveform for a voltage source six step drive.
http://powerelectronics.com/motion-systems/motors-efficiency-and-adjustable-speed-drives
The waveforms in this article match what I've seen for output voltage on inverters:
http://www.inverter.co/inverter-troubleshooting-guide-547067.html
-Jon
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