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

[Smart $]

170614-0858 EDT

Smart $:

Displacement power factor that is created by a linear inductive or capacitive load is not normally considered to be corrected by adding a lot of power wasting resistive load. However, added resistive loading does somewhat improve power factor. But that is only because of the increased real power load in the numerator of the PF eqauation.

To correct reactive power factor one has to use the opposite type of reactive component.

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I have no idea where you are getting this "adding a lot of power wasting resistive load."


We have a powered VFD and a motor. The power factor exhibited on the input is typically less than the power factor exhibited between VFD and motor. Added load (VFD losses) is minimal if not insignificant. I, as most, call this power factor correction. :happyyes:

 

[Besoeker]

We have a powered VFD and a motor. The power factor exhibited on the input is typically less than the power factor exhibited between VFD and motor.

Typically it isn't.

 

[Ingenieur]

http://ecmweb.com/power-quality/improving-power-factor-variable-speed-ac-drives
The benefits of VSDs To produce the required rotating torque and speed, an induction motor takes both active current and reactive current from the power supply. The rotating torque of the motor is created as an interaction between the active current component and the magnetic field. The field is produced by the reactive current component. Light load takes less active current, but the magnetic field, as well as the reactive current, stays constant. This means that the power factor decreases with decreasing load. The current is mainly active at the full load but mainly reactive at the light load.
By installing variable speed AC drives, you can improve process controls, increase energy savings, and reduce wear on the machinery. Such drives also have the added benefit of improving PF.


http://www04.abb.com/GLOBAL/seitp/s...r+Factor+Improved+Via+AC+Drives+-+Release.pdf

 

[Besoeker]

Such drives also have the added benefit of improving PF.

Which is contrary to post#261........

 

[gar]

170614-2019 EDT

This is directed at Sahib's posts 250 and 252 and my 256.

A large filter capacitor on the output of a rectifier in no way is reflected back thru the diode to look like any capacitance to compensate, to any extent at all, any inductive current on the input side of the rectifier.

A simple experiment will verify said statement.

Using a simple DC power supply that I have previously mentioned for this experiment the results are:

With the filter capacitor disconnected.
Vin = 120.6 V, Iin = 0.09 A, Pin = 6.4 W, VAin = 11.8, and PF = 0.54 .

Adding a 1 ufd PF corrrection capacitor in parallel at the power supply input the measurements changed to:
Vin = 120.7 V, Iin = 0.07 A, Pin = 6.3 W, VAin = 8.8, and PF = 0.72 .
I lacked another capacitor box and no readily available fixed capacitor with adequate voltage rating to more completely compensate the inductive current.


With the filter capacitor connected, and no compensation capacitance.
Vin = 120.1 V, Iin = 0.09 A, Pin = 6.5 W, VAin = 11.8, and PF = 0.55 .

Adding the 1 ufd PF corrrection capacitor.
Vin = 120.1 V, Iin = 0.07 A, Pin = 6.4 W, VAin = 8.6, and PF = 0.73 .

The filter capacitor made no difference, but PFC capacitor made a big difference.

The measurements were made with a Kill-A-Watt EZ meter.

The power supply consists of a Stancor RT transformer with a center tapped secondary, two diode rectifiers, and a 50,000 ufd filter.

No DC load was applied in the above experiment.

.

 

[Sahib]

Which is contrary to post#261........

Smart$ is right (I, wrong). The ABB article of ingeniur has a flaw that power factor used in it is actually displacement power factor and not real power factor.

 

[Besoeker]

Smart$ is right (I, wrong). The ABB article of ingeniur has a flaw that power factor used in it is actually displacement power factor and not real power factor.

A well-designed 3 phase VFD has an input power factor of around 0.958. The last motor we installed (225kW) had a power factor of 0.7.

 

[Smart $]

A well-designed 3 phase VFD has an input power factor of around 0.958. The last motor we installed (225kW) had a power factor of 0.7.

While to a non-electrical scientist, that would be a greater power factor on the input... but to a person in the electrical trade, the input power factor is less than the output power factor. Makes me wonder which way you are looking at my earlier statement. When I say a lower or lesser power factor, I mean closer to a value of 1.
:blink:

...The power factor exhibited on the input is typically less than the power factor exhibited between VFD and motor. ...

Typically it isn't.

 

[Smart $]

170614-2019 EDT

This is directed at Sahib's posts 250 and 252 and my 256.

A large filter capacitor on the output of a rectifier in no way is reflected back thru the diode to look like any capacitance to compensate, to any extent at all, any inductive current on the input side of the rectifier.

A simple experiment will verify said statement.

Using a simple DC power supply that I have previously mentioned for this experiment the results are:

With the filter capacitor disconnected.
Vin = 120.6 V, Iin = 0.09 A, Pin = 6.4 W, VAin = 11.8, and PF = 0.54 .

Adding a 1 ufd PF corrrection capacitor in parallel at the power supply input the measurements changed to:
Vin = 120.7 V, Iin = 0.07 A, Pin = 6.3 W, VAin = 8.8, and PF = 0.72 .
I lacked another capacitor box and no readily available fixed capacitor with adequate voltage rating to more completely compensate the inductive current.


With the filter capacitor connected, and no compensation capacitance.
Vin = 120.1 V, Iin = 0.09 A, Pin = 6.5 W, VAin = 11.8, and PF = 0.55 .

Adding the 1 ufd PF corrrection capacitor.
Vin = 120.1 V, Iin = 0.07 A, Pin = 6.4 W, VAin = 8.6, and PF = 0.73 .

The filter capacitor made no difference, but PFC capacitor made a big difference.

The measurements were made with a Kill-A-Watt EZ meter.

The power supply consists of a Stancor RT transformer with a center tapped secondary, two diode rectifiers, and a 50,000 ufd filter.

No DC load was applied in the above experiment.

.

Here you are again, looking at the details (trees) and not seeing the bigger picture (forest). Forget about the capacitor. The motor requires reactive current to function. If you do not see this reactive current on the input of the VFD, then the VFD has effectively reduced the reactive power as realized by the power source.

ETA: Did I read your post [more] correctly the second time... no load connected? That's the whole point of argument. We don't care about what the VFD contributes to power factor by itself. All we care about is whether the input power factor is closer to 1 than the output power factor... with a motor connected!

 

[gar]

170515-0909 EDT

Smart $:

My post #265 was directed to Sahib and his post where he said (my interpretation of what I thought he said) that the filter capacitor on the output side of the rectifier contributed to improving power factor preceding the rectifier. Apparently something like a PFC capacitor on the input side would do.

If we exclude some sort of active power factor modifier at the input to a DC supply where there is a very large filter capacitor after the rectifier (near zero ripple), then increasing power consuming load on the DC supply will improve power factor on the input side of the power supply. But, this is a change in power factor as a result of load change on the DC side. It is not a power factor corrector.

Power Factor = Real Power/(Vrms*Irms) by absolute definition. Historically this appears to date from earlier than 1900, possibly 1892. This is not an equation based on experimental data, but simply a defined mathematical equation.

A high power factor is good, low is bad. To do something that increases power factor is good. Good because it reduces power losses, and the requirement for larger equipment on the supply side for a given amount of load power.

.

 

[Sahib]

A well-designed 3 phase VFD has an input power factor of around 0.958. The last motor we installed (225kW) had a power factor of 0.7.

Is the VFD input power factor of about 0.958, you are talking about, displacement power factor?

 

[Besoeker]

While to a non-electrical scientist, that would be a greater power factor on the input... but to a person in the electrical trade, the input power factor is less than the output power factor. Makes me wonder which way you are looking at my earlier statement.

The way most electrical engineers do.

When I say a lower or lesser power factor, I mean closer to a value of 1.

A high power factor would be close to unity. You use PFC capacitors to make the power factor higher, not lower.

From the IET:

Most Network Operating companies now penalize for power factors below* 0.95 or 0.9.

*i.e. lower than 0.95 or 0.9.

 

[Smart $]

170515-0909 EDT

Smart $:

My post #265 was directed to Sahib and his post where he said (my interpretation of what I thought he said) that the filter capacitor on the output side of the rectifier contributed to improving power factor preceding the rectifier. Apparently something like a PFC capacitor on the input side would do.

I believe you misinterpreted his post. I believe he is saying that the filter capacitor improves the power factor as measured on the input compared to the power factor measured on the output (with a motor connected ). Remove the filter capacitor and there is no reactive component in the "converter". The reactive component of the motor current would have to be passed upstream to the input.

If we exclude some sort of active power factor modifier at the input to a DC supply where there is a very large filter capacitor after the rectifier (near zero ripple), then increasing power consuming load on the DC supply will improve power factor on the input side of the power supply. But, this is a change in power factor as a result of load change on the DC side. It is not a power factor corrector.

Power Factor = Real Power/(Vrms*Irms) by absolute definition. Historically this appears to date from earlier than 1900, possibly 1892. This is not an equation based on experimental data, but simply a defined mathematical equation.

A high power factor is good, low is bad. To do something that increases power factor is good. Good because it reduces power losses, and the requirement for larger equipment on the supply side for a given amount of load power.

.

We're not saying our purpose is solely to improve power factor of a motor load, but if the input power factor is better (closer to 1) than without the device, what we call it is moot. We accept the inadvertent improvement and move on.

 

[Besoeker]

We're not saying our purpose is solely to improve power factor of a motor load, but if the input power factor is better (closer to 1) than without the device, what we call it is moot. We accept the inadvertent improvement and move on.

I was objecting to your post:

When I say a lower or lesser power factor, I mean closer to a value of 1.

It's just wrong.

 

[Besoeker]

Is the VFD input power factor of about 0.958, you are talking about, displacement power factor?

Yes. It's the value you get when the DC current is approximately level.

 

[Sahib]

Yes.

But VFD real input PF=Displacement PF*Distortion Factor
where distortion factor is lower than unity. So the real input PF would be lower. A suitably designed active/passive filter in VFD can make the input PF almost equal to displacement PF.

 

[Besoeker]

But VFD real input PF=Displacement PF*Distortion Factor
where distortion factor is lower than unity. So the real input PF would be lower. A suitably designed active/passive filter in VFD can make the input PF almost equal to displacement PF.

The majority of VFDs have a plain diode rectifier as the input AC to DC conversion stage. Displacement doesn't come into it.

 

[gar]

170615-1704 EDT

Smart $:

From your post #273:

---- I believe he is saying that the filter capacitor improves the power factor as measured on the input compared to the power factor measured on the output (with a motor connected ).

Why would you believe that the power factor looking at the input to a diode rectifier with a large output DC filter capacitor is better than the power factor at the input to a sine wave fed induction motor fully loaded?

Remove the filter capacitor and there is no reactive component in the "converter". The reactive component of the motor current would have to be passed upstream to the input.

Why do you believe that the power factor of a reactive or resistive component on the output of an unfiltered rectifier is going to propagate to the rectifier input unchanged.

Study the defining equation for power factor I gave in my post #270, and if you analyze input currents and output currents you will usually find they have different shapes. Thus, different RMS values for the same average current. One important point I did not state in #270 is that the equation only applies to a two terminal load or source. But if you assume a balanced system, then you can extend the use of the equation. Further you need to look at whether input voltage is the same as output voltage.

.

 

[Smart $]

170615-1704 EDT

Smart $:

From your post #273:
Why would you believe that the power factor looking at the input to a diode rectifier with a large output DC filter capacitor is better than the power factor at the input to a sine wave fed induction motor fully loaded?


Why do you believe that the power factor of a reactive or resistive component on the output of an unfiltered rectifier is going to propagate to the rectifier input unchanged.

Why would you not I think is the better question?

Did you read the articles which Ingenieur linked in post #263?

Study the defining equation for power factor I gave in my post #270, and if you analyze input currents and output currents you will usually find they have different shapes. Thus, different RMS values for the same average current. One important point I did not state in #270 is that the equation only applies to a two terminal load or source. But if you assume a balanced system, then you can extend the use of the equation. Further you need to look at whether input voltage is the same as output voltage.

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Still seeing only trees? Is there such a thing as a generalization in your world?

 

[Smart $]

I was objecting to your post:

It's just wrong.

I agree. I let myself get suckered into using the upside down terminology in a discussion and through my own fault, it stuck. My bad. I should have just corrected the other person as you are me now. Thank you!