LEADING Power Factor Correction Problem

[oh_well1500]

QUESTION,
I can get into specifics later, but on a conceptual level, I have 277 V 60 Hz from a three phase source powering a low pass filter that has enough capacitance that dozens of them at the same facility are causing a LEADING power factor and the utility company is demanding a fix. Could I fix this simply by introducing a shunted inductor with the right amount of inductance to lower the apparent power to more acceptable levels?

 

[Easy peasy]

I fear FvM has missed the point - if you have L & C of a similar size to above but the L is sized for resonance at 60Hz then the currents in the L & C will be " Q " times the current to the parallel pair - even for a Q of 4 -5 this will cook the L & C very quickly.

What is confusing me FvM, is I built the circuit, with a 12.8 mH inductor, in parallel with a capacitor bank of 550 uF, and I am measuring 60 amps on the cap bank line, 60 amps on the inductor line, and 20 amps on the input current before the node.

Indeed - the confusion arises from a lack of engineering knowledge, resonant systems can have currents and voltages higher than the supplying circuit quite easily due to the physics involved.

 

[FvM]

I fear FvM has missed the point - if you have L & C of a similar size to above but the L is sized for resonance at 60Hz then the currents in the L & C will be " Q " times the current to the parallel pair - even for a Q of 4 -5 this will cook the L & C very quickly.

No. It's a voltage driven resonator. I = V/X.

--- Updated Aug 3, 2024 ---

In the specific case: You have 550 uF filter capacitors parallel to grid. Leading reactive current Ix = 277 * 6.28 * 60 * 550e-6 = 57.4 A
A matched parallel PFC inductor (12.8 mH) consumes a lagging reactive current of same magnitude, both add up to ideally zero.

 

[Easy peasy]

No. It's a voltage driven resonator. I = V/X.

I fear you are ignoring the Z of the lines back to the transformer . . . very few things are purely volt driven - and even then it doesn't matter - if there is any current to the parallel pair ( 60 Hz resonant ) then there will be Q times this inside the LC.

 

[FvM]

The original scenario is a system with 550 uF filter capacitors. It consumes 57 A reactive current and causes a certain increase of local grid voltage (e.g. a few percent) due to grid inductance.

Adding a PFC inductor will cancel reactive grid current and respective voltage increase. That's all.

--- Updated Aug 3, 2024 ---

if there is any current to the parallel pair ( 60 Hz resonant ) then there will be Q times this inside the LC.

Sure. But current "inside LC" is V/X and not dependant on Q as long as V is near to nominal grid voltage.

 

[KlausST]

Hi,

from my previous work at a grid company I´ve learned that it may cause resonance problems when compensating to 1.0.

But indeed my technical understanding (may be wrong) is like FvM.

On parallel compensation you get predictable current values: for C and for L. And at resonsonance the total current is minimum. You neither get overly high voltage across the C and L nor do you get overly high currents.

This is the opposite at series compensation. Where the voltage across C and L rise way over input voltage (according Q) ... and the currents rise with the voltages.

***
On parallel compensation.
One may explain a small rise in voltage due to the minimized drop of mains lines. But it should never go higher than without load.
If I remember right, then the series grid impedance is about 1 ohms at our 230V outlets (may vary a lot from place to place).

***
Now thinking about where it may really cause problems.
1) as already said: series compensation. It may cause a resonance catastrophy
But we all talk about compensation of load currents. While I surely remember the grid company compensating the grid impedance.
Decades a go (the time where I got the information) the grid had inductive character due to it´s generators, transformators and the overhead lines.
2) if you do the grid compensation, then all inductivity is in series .. and then there is the compensation capacitor. It (again) may form a critical series compensation.

Nowadays our (may not be true for every region) gird is more of capacitive character (due to underground lines with their protective earth layer around the conductor) and capacitive character electronic power supplies.

***
Conclusion:
I´ve got the information that compensation to 1.0 may cause resonance problems ... but indeed I still miss a prove in from of a technical experiment / simulation.
But I did not dive very deep into this field.

***
The information above is all about compensation and problems regarding fundamental mains frequency.
I know (especially nowadays) there are problems with overtones, especially 3rd and 9th in three phase systems .. but this was not subject of my writing.

Klaus

 

[Easy peasy]

I like the '' that's all " in fact it is rarely performed ( parallel L ) without suitable resistors to limit Q precisely because of the effects I have mentioned.

--- Updated Aug 3, 2024 ---

any simple simulation confirms this.

 

[FvM]

Simple simulation. Low pass filter, optional load, optional parallel PFC inductor.
First comparison currents with 10 kW load and unloaded, PFC inductor connected.

Second comparison. No load, with and without Inductor (Q=50). Vgrid and filter current I(L1) don't change.

 

[mtwieg]

I fear FvM has missed the point - if you have L & C of a similar size to above but the L is sized for resonance at 60Hz then the currents in the L & C will be " Q " times the current to the parallel pair - even for a Q of 4 -5 this will cook the L & C very quickly.

I think you're confusing parallel and series resonant circuits. What FvM (and I think everyone else) is proposing is a parallel inductor. For parallel LC circuits, Q=R*sqrt(C/L), where R is the resistance of the drive source. For typical AC mains, R will be so low as to completely dampen the resonance. Even with a relatively high R of 0.5ohm, you'd get Q=0.5*sqrt(550e-6/12.8e-3)=0.10, and that's assuming the L and C themselves are perfectly lossless.

I don't see any specific reason why operating at resonance would be problematic in this case.

 

[Easy peasy]

again - you don't see because you haven't done the work in practice - there is no confusion - except by others - your idea of the Z of the mains is misguided, as is your statement of Q for this case - every power engineer who has done this work knows the dangers.

 

[KlausST]

any simple simulation confirms this.

So - to get a constructive discussion - what needs to be corrected in the simulation of post#27 to show "the problem"?

Klaus

 

[Easy peasy]

it does appear there is insufficient series Z to the L//C combo to let the parallel currents increase - for the simple case here:

however - this is often far from the case on the real mains supply - so - designer beware.

 

[FvM]

Your simulation shows that there's no "resonance" problem with the setup, even for an inductor Q of 250 and higher grid impedance.

There are still possible problems with PFC inductor, DC inrush current on startup and kickback voltage on disconnection.
I presumed that problems of huge filter capacitors, particularly inrush current have been solved before.

 

[KlausST]

Hi,

It´s not obvious to me...
So, do you see any overly high "resonance" current?

Klaus

Oh, FvM already posted .. while I did my calculations.

 

[Easy peasy]

Ah yes - as said above - it's what you don't see that trips the unsuspecting soul.

 

[FvM]

any simple simulation confirms this.

Ah yes - as said above - it's what you don't see that trips the unsuspecting soul.

Contradicting statements. I'd appreciated at least a hint which parameters or conditions are missing in the considerations.

 

[Easy peasy]

The problem is we do not know exactly the mains supply Z - it can be a lot more complex than all the assumptions - this regularly catches people out . . .

 

[KlausST]

The problem is we do not know exactly the mains supply Z - it can be a lot more complex than all the assumptions - this regularly catches people out . . .

Yes, true, we don´t know source side.
Again: we try to compensite the load side. And even for PF 1.0 (load side) we don´t get maximum current, but we get minimum current. In case of "no resistance" the resulting current would be zero. (not infinite!)
The capacitor current would be the same: V / Xc .. with or without compensation, with or without "resonance".
Also the inductor current would be the same: V / XL .. with or without compensation, with or without "resonance".

It´s totally the opposite to series compensation.

***
For sure our capacitor ... or our inductor, or our complex, compensated load ... may interact with source side impedance.
I guess it can resonante.
But this has nothing to do with PF 1.0 load side.
The same source side resonance may happen if the load side has any other PF.

Again: A resonance catastrophy (overly high voltage and current) can happen with series resonance ... but at parallel compensation the single currents would stay about the same, only the resulting current becomes minimal. Nothing catastrophy like.

Again: I know this rule from decades ago. But I never understood it in detail.
Switch_off ringing may be the reason.
Maybe one just wanted to electricians not to worry about the compensation method. (they don´t need to differentiate)
Maybe one recognized that compensation above 0.95 (or so) does not give much additional benefit (it does not pay for the extra effort)
Maybe it´s the "fast" jump (move) in phase shfit around the resonance frequency. (so that the phase continously jumps between lagging and leading and thus maybe upsets some regulation loops ... idk)
Maybe there are other reasons I don´t understand

From my current understanding: the load side catastrophy does not happen on parallel compensation.

Klaus

 

[Easy peasy]

If there are switchable PFC caps nearby on the 400V 3 phase line - i.e. in a nearby building - this can cause very interesting things to happen.

 

[KlausST]

Hi,

a very blury information...

If there are switchable PFC caps nearby on the 400V 3 phase line

Where else should they be? ... I mean they need a connection to the phase lines somehow.

i.e. in a nearby building

I don´t understand. Is it better outside a building? In the rain?
All compensation devices I´ve seeen are in a building .. and are connected to the phase lines.

it can be a lot more complex than all the assumptions

This is your assumption. Without proof so far.
We did the opposite of assumption: We used physics, math, simulation ... and nothing showed the "catastrophy"

this can cause very interesting things to happen.

What exactly? Change the color of snow?
What are these interesting things you talk about?

*****
You give big words aboout "knowledge of physics", "simple simulation" and so on.
But some of your statements simply are not true.

For example:

if you have L & C of a similar size to above but the L is sized for resonance at 60Hz then the currents in the L & C will be " Q " times the current to the parallel pair - even for a Q of 4 -5 this will cook the L & C very quickly.

There is a good reason why voltage driven LC parallel resonance also is called "anti resonance".
The name is, because the currents do the opposite to a series resosnance.
While in series resonance the total (and the single) currents go up, in parallel resonace the total current goes down (while the single currents do nothing special).
While in series resoanance the total impedance goes down, ... it goes up with parallel resonance.

For sure when you drive a parallel resonance from a current source, or any other very high impedance source (Megaohms and higher) then you will see: The higher the impedance, the higher the voltage. (then the Q factor comes into play ... since a "low loss" parallel anti resonance impedance will go in direction "infinite")
But the grid in our region is somewhere in the region of 1 Ohms (at the outlet) .. so, far away from critical Megaohms.

For sure the impedance of the grid depends on frequency. There may be high frequencies where the impedance becomes overly high. Unwanted, uncontrolled.
But the topic was "compensate the load to 1.0" which only is valid for mains frequency.
We don´t "compensate for a PF of 1.0 at 3MHz". (or whatever frequency)

***
Don´t get me wrong. Nothing against you. I´m still curious to learn why I´ve been told that compensation to 1.0 is critical.
I always got empty words without proof according physics and mathematics.
Show me math, show me a simulation, show me a vector diagram ... that proves the critical situation.

****
On a perfectly matched "parallel resonance for 50Hz" on a voltage source...
When looking at phase and amplitude of current at a frequency range of 45Hz to 55Hz
* then the current of the capacitor almost linearely rises from 90% to 110%, while the phase fix is at +90°
* then the current of the inductor almost linearely falls from 110% to 90%, while the phase fix is at -90°
And the total current starts at about 20% at 45Hz ... is zero at 50Hz and goes up to about 20% at 55Hz
Nothing crazy happens ... the only thing is: the total current goes down to zero ... way less than the single currents. (The total current is 1/Q for a lossy circuit)
The total current goes to infinite on 0Hz due to the inductor current
The total current goes to infinite on infinte Hz due to the capacitor current.
At resonancee the total current goes to zero.

Sadly I´m not experienced enough with circuit simulation to show the behaviour of phase and current of both devices (and maybe an additional resistor) in one combined simulation. Maybe someone else can do, please.

Klaus

 

[FvM]

Assuming a very "weak" grid, e.g. 128 uH impedance as in the simulation above. Parallel connecting 12.7 mH inductor reduces grid inductance by 1 percent.