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

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topgone

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It would be useless using a vfd if there is no advantage using it. That said, nobody uses vfd if the voltage out will be the same as the input voltage. Voltage into the vfd will have to be "different" from the output voltage.
I am of the opinion that the conservation of energy holds, only the voltages and currents varies.
 

GoldDigger

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Very simple. Irms only corresponds to real power when it is going into a linear load. As well as only when it is phase if the load is linear but reactive.
The power going out is exactly equal to the power going out, by assumption.
But the measure of power in is precisely the integral of instantaneous voltage times instantaneous current. And the instantaneous product is never negative, unlike the case where the input also has a displacement power factor.
The bottom line purely and simply, is that neither Irms nor Iave is a valid measure of power in this situation. So the very question of "where it went" is based on an incorrect assumption.

Thanks for making this point, Mike, even if it took awhile to get there.

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Sahib

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I am of the opinion that the conservation of energy holds,

Yes but the conservation of distortion power does not hold because it does not flow (it does not contradict energy conservation law anyway). Reference: Electrical power system quality by Dugan, McGraw Hill publishers.
 

gar

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mike_kilroy:

If we have a Q charge flow in time T, then we can calculate an Iave for that time period.

Iave = the integral of i(t) dt from 0 to T divided by T. This is directly related to total charge flow during the time T, and is independent of waveform. Assuming a rectified source.

Irms is a mathematical value obtained from the equation:
Irms = (1/T) * the sq-root of the integral of i(t)^2 dt from 0 to T. This does not have a direct relationship with charge flow. Its value is a function of both charge flow and wave shape.

When charge flow into a capacitor in combination with the power load on the capacitor produces very little ripple on the capacitor, then the average power input to the capacitor is close to equal to Vcap*Iave independent of the input current wave shape. Note, Vcap is approximately constant. Here we are dealing with a rectified source to the capacitor.

The input power to the capacitor is equal to whatever power is consumed by the load on the capacitor no matter what the load current wave shape is.

It is not a question of where the current went. Irms is a mathematically determined value from actual current flow and is waveform dependent. Differences simply result from the math processing, and the waveform.

Irms is a synthesized value, but it does have a real world use.

.
 

Ingenieur

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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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wrong
reactive current in the output can't be ignored and is not irrelevant
the scenario you describe is not possible
 

Ingenieur

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You should have read this before posting


Current THD increases partially because the harmonic component increases, but, also because thefundamental component decreases. THD increases to what appears to be a significant level, but, theactual overall harmonic current component itself is quite small as compared to the full load current.Overall, VFD input current will be nearly zero as the largely reactive current flowing into the lightlyloaded motor does not get reflected back to the supply. The reactive current is supplied by the VFD DClink circuit (bus capacitors) and circulates within the VFD’s output section. It does not flow from thesupply side. Therefore, even though drive input PF is low with the lightly loaded condition, the VFD inputcurrent is also much lower because only the “real” current (torque producing) component is reflectedback to the supply. The resulting harmonic current is quite small even with the high THD level and doesnot represent any significant issue for the supply and associated components. Therefore, the low PFencountered with a lightly loaded VFD is not a detriment to the system and effective use of power ismaintained.


For example, a lightly loaded motor operating across-the-line will continue to draw 25-35% of ratedcurrent as magnetizing current (reactive current). This current is at nearly 90 degrees displacement fromthe applied voltage making for a very low power factor system. The supply and associated componentswill be exposed to this substantial current at low power factor. This same current will flow to the motorwhen operating from a VFD, however, the VFD input current will be near zero because this reactivemotor current is supplied solely by the VFD DC link circuit. Only the real (torque producing) componentis measured on the supply side. Even though drive input PF is lower due to increased harmonics at lightload, the input current is also much lower. Subsequently, the low input power factor does not negativelyimpact the overall supply components (transformer, wiring, etc.) that are already sized to accommodatefull load operation in the first place. Over sizing of supply components that is typical for low powerfactor systems is not required because the VFD input current is also low.


Expanding on this, motors with low power factor will exhibit a large magnetizing current (reactivecurrent). An additional benefit of using a VFD is elimination of the magnetizing component from beingreflected back on the supply. VFD input current can be less than the motor current even at 60Hzoperation. This will hold true for motors with relatively low full load power factor (less thanapproximately 0.85) and VFDs that maintain high overall input power factor (0.89 or greater.


if you had not said the motor/vfd were loaded to 100% it may make sense
if motor loaded <40%
vfd is way oversized for motor
and motor i mostly reactive
 

Sahib

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Ingeniur: Total KVA input to VFD is square root of sum of squares of active power, reactive power and distortion power. For each component power, there is a corresponding current. So I think I see why mike does not agree with you.:)
 

Ingenieur

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Ingeniur: Total KVA input to VFD is square root of sum of squares of active power, reactive power and distortion power. For each component power, there is a corresponding current. So I think I see why mike does not agree with you.:)


It's not me he doesn't agree with
it is the law of energy conservation

the i input is not sinusoidal due to rectification
but its sum and the sum of harmonics (distortion component) can't be larger than the rms value of the undistorted fundemental
 

Sahib

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the i input is not sinusoidal due to rectificationbut its sum and the sum of harmonics (distortion component) can't be larger than the rms value of the undistorted fundemental
Individual harmonics are normally smaller in magnitude than fundamental, unless there is resonance and resultant total harmonic current is always greater than fundamental.
 

Ingenieur

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Individual harmonics are normally smaller in magnitude than fundamental, unless there is resonance and resultant total harmonic current is always greater than fundamental.

thd = sum harmonics/fund and is seldom greater than 1
for a loaded drive/motor it is <50% with diodes for rectification, even lower with igbts
as harmonics sum increases fund decreases, offsetting and reducing impact

it only approaches 1 if motor is lightly loaded

if thd is 50% pf adjustment is 90%
so if pf is 0.95 tdh lowers it to 0.85
nowhere near 0.5

his model implied current is being created
12 in/6 out, 100% eff
it is not
pf is affected so more reactive current is required (but this also shows up on the load side since the motor is ightly loaded)
it goes where it came from...the grid
distortion and all
 
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Ingenieur

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V in = v out = 460/3
vfd hp rating = motor hp = 10
assume vfd eff 100%
fla = 12 pf 0.9
operating at 4 A pf 0.3 (33% fla)
input rated pf 0.92 at 35% thd
thd = 85% at 33% fla

what is input i/ang at 100% fla, what is net pf?
what is input i/ang at 33% fla, net pf?
same for output i at 100% and 33% fla

imho that is how you illustrate a point
make the 'student' see it in numbers, and why

time to get some physical vs mental excercise
wash the AMG :)
 
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