filter harmonic

7/29/2019 filter harmonic

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PFC-Formulas

Real energy [kW]

Reactive energy [kvar]

Apparent energy [kVA]

Power factor

Required kvar-power [kvar]

Detuned power factor correction

Calculation of the capacitor nominalvoltage (minimum voltage withstand

capability) - Uc

Un = Grid voltageP = Detuning ratio

Calculation of the capacitor nominal

power - Qc

Nc = Filter-nominal powerUn = Grid voltageUcn = selected capacitor nominal voltage (Ucn

> Uc)

P = Detuning ratio

Filter frequency - fres f = Grid frequency

P = Detuning ratio

Filter star capacitance - Cy

Nc = Filter-nominal power

Un = Grid voltageP = Detuning ratio

f = Grid frequency

Filter inductance – L

P = Detuning ratio

f = Grid frequencyCy = Star capacitance

Calculation of resonance points - fr

ST =Transformator-nominal power [kVA]Qc = Nominal power of the capacitor bank

[kVAr]

uk = Transformator impedancy [%]

Resonance frequency - fres



Hello,

I just bought a harmonic filter to mitigate the distortion produced by a 400 HP VFD. The filter consists of some capacitors and 2 line reactors. Here's how it is wired.

One line reactor is connected immediately downstream of the main breaker. From there it is connectedto the VFD. Between the VFD and the line reactor, a second filter is connected phase to phase. The

second reactor is wired in series with some capacitors, and that series combination is connected line toline. So, between each two phases, there is an LC circuit (not including the resistance of the reactor).

So, I have a couple of questions. The first is very basic.

1. How does a line reactor absorb high frequency current? I understand that the impedance varies withfrequency. But in order to absorb the high frequency current, something has to absorb real (notreactive) power. There is a tiny DC resistance in the line reactor. Is this the load which absorbs the highfrequency current? Where does the high frequency energy go?

Question 2. Why would the circuit be wired such as it is? If I had designed this, I would have put an LCshunt off of each phase, tying one end to ground. The filter I bough makes the connection line toline. And why the two line reactors?

thanks in advance

ScottyUK (Electrical) 16 Apr 09

5:10

Sounds like the series L-C is a resonant trap and will be tuned close to a frequency of interest, likely the5th harmonic. The resistor will reduce the Q-factor of the tuned circuit and will limit current if the filteractually goes in to resonance under abnormal supply conditions. The series reactor is a low pass filterwhich presents a rising impedance as harmonic order increases and serves to decouple the drive and its

filter from the network so you're filtering harmonics from the drive and not trying to flter harmonics fromthe whole network.

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If we learn from our mistakes I'm getting a great education!

rockman7892 (Electrical) 16 Apr 098:37

ScottyUK

Are you saying that the 5th order harmonic is of interest assuming that the drive is a 6-pulse

drive? What about the 7th harmonic in this case?

Can you explain how the series reactor acts as a low pass filter. I understand that the reactor impedencerepresented by XL = jwL will increase with increased frequencies as you mentioned, however what willthis increased impedance do to this high frequency current?

eeprom (Electrical) 16 Apr 099:15

Thanks for your help. I spent all night studying that circuit and I had come to the same conclusion thatthe LC circuit is a bandpass with a very narrow band, assumed to be centered at 300Hz. I have theresults from a harmonic distortion reading, and the 5th is by far the most prevalent harmonic.

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ScottyUK (Electrical) 16 Apr 0910:12

Rockman,

Yes, low order odd harmonics are of interest if it's a 6-pulse front end. If it's tuned to the 6th and the Q-factor isn't too high then it will trap 5th and 7th without being resonant at either. It's reasonable to

expect that even order harmonics will be negligible in a properly behaving rectifier so the resonance at6th isn't a problem.

Low pass filter was perhaps not the best description. It will serve to slow down commutation edges andreduce the production of high order harmonics inherent in fast switching edges.

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VTer (Electrical) 16 Apr 0911:10

1. For a 6 pulse drive the filter is typically tuned for 5th order harmonic (typically the highest current

and % fundamental magnitude) which implies that the circuit resonates at approximately 300Hz withsuch a filter. Note, however, that if you do design the filter for that frequency theoretically you can getvery high resonant currents, therefore the filter is typically tuned at 4.7 rather than 5th order harmonic

which means that the circuit is tuned at approx 282hz to avoid extremely high currents. Now at relativelyhigh frequencies you can view inductance as "open circuit" and capacitance as a "short circuit".Either

way, Xl=2pifL indicates that effective reactance goes up with frequency therefore limiting current, but theopposite goes for a capacitor since Xc=1/(2pifC). With higher frequency, the Xc goes down reducingreactance. The basic theory behind tuning at a frequency is to match these two up per say (inductance

with capacitance) and have them cancel each other out which in turn provides path of least resistance forcurrents at those frequencies .2. The reason the circuit is being wired this way is to have these resonant frequency currents circulateinside the filter only, if you ground the filter you will provide a path back to the system canceling its verypurpose. .

"Throughout space there is energy. Is this energy static or kinetic! If static our hopes are in vain; if kinetic — andthis we know it is, for certain — then it is a mere question of time when men will succeed in attaching theirmachinery to the very wheelwork of nature". – Nikola Tesla

ScottyUK (Electrical) 16 Apr 09

11:18

Agree 5th harmonic trap is most common. Is this installed in the 50Hz world or the 60Hz world? We'remaking assumptions up to now. 300Hz is 6th harmonic where I work, but 5th harmonic in 60Hz world.

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VTer (Electrical) 16 Apr 0911:28

I was referring to 60hz as fundamental, my apologies for not mentioning that.

"Throughout space there is energy. Is this energy static or kinetic! If static our hopes are in vain; if kinetic — andthis we know it is, for certain — then it is a mere question of time when men will succeed in attaching theirmachinery to the very wheelwork of nature". – Nikola Tesla


If you're right - which the OP's noting of drive rating in HP rather than kW seems to suggest - thendisregard my last post. I was trying to acount for reasons to tune a filter to 300Hz on a 50Hz system and

missed the most obvious reason. Ah well...

Came across this while looking for something else on CDA's website: CDA filter link . I haven't read it

properly but might be of interest.

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VTer

If I understand you correctly the LC circuit is wired inbetween phases so that the resonant harmoniccurrents will be trapped and fluculate between the reactor and the cap. It sounds like you are sayingthat this LC circuit is designed with values such that at the 5th harmonic there would be a resonantfrequency between the cap and the reactor and thus set up a resonant current trapped inbetween these

two devices? Do I understnad this correctly?

If you ground this LC circuit would the harmonic resonant current not still flow back and forth between

the cap and reactor or would it flow to ground and back to the source?


The harmonic trap presents a low impedance near resonance so current at 5th harmonic (60Hz)preferentially flows down that path rather than into the source.

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VTer (Electrical) 16 Apr 09





http://www.copperinfo.co.uk/power-quality/downloads/pqug/331-passive-filters.pdf

















16:09

rockman,

"It sounds like you are saying that this LC circuit is designed with values such that at the 5th harmonicthere would be a resonant frequency"

Not exactly to resonate at 5th for the reasons mentioned above, but close to it, and yes that is myunderstanding of it. And as far as grounding the filter, Yes the current will flow back and forth through your system, but the questions is do you really want thiscurrent in your grounding system? Note that you cannot completely eliminate these harmonic currents

rather just reduce the THD.

Refer the attachment for a diagram that shows a phase to phase connection.

"Throughout space there is energy. Is this energy static or kinetic! If static our hopes are in vain; if kinetic — andthis we know it is, for certain — then it is a mere question of time when men will succeed in attaching theirmachinery to the very wheelwork of nature".

–Nikola Tesla

http://files.engineering.com/getfile.aspx?folder=0518828f-e752-42b5-a98d-89

Skogsgurra (Electrical) 16 Apr 0916:29

It is unfortunate that this thread is centered on low order harmonics. That usually is not a problem with VFDs. True, they are there, but it is the HF that is the big problem in VFDs and that is mostly why filters

are used between grid and VFD.

If you look at the components, they are probably not tuned to a frequency close to 250 or 350 Hz (in 50

Hz grids) or 300 or 420 Hz (in 60 Hz grids), but a frequency that is a lot higher.

The filter simply shorts out frequencies starting at around 50 - 100 kHz (150 kHz is where the Europeanstandards atart).

Also, to answer the OPs first question; they do not absorb much of the energy. They just short thefrequency components so they cannot propagate out into he grid.

Another point to consider is that real good filters also have capacitors going to ground in order to keepcommon mode interference low - not just normal mode interference.

The resistors seen in some filters are mostly there to discharge the capacitors so the filters don't carry

lethal voltages after disconnection from grid.

It is only in some motor reactor applications (other side of the inverter) that you find damping resistorsparallel to the coils. They are there to avoid ringing between reactor and cable capacitance.

There is a lot of confusion about these filters. Manufacturers, salespersons, users are among theconfused. Not saying that all are confused, but a large percentatge.

Gunnar Englund

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VTer

Thanks for the circuit diagram. I'm trying to understand how the harmonic current propogates in thiscircuit. I'm assuming Ih in this circuit is the total harmonic current and If is the current that is passedthrough the filter, while It is the current that goes back to the transformer.

Is the Ih current the current consisting of the 5th and 7th harmonic?

Why does only the harmonic currents flow as If and not back through It. In other words how does theLC combination divert and trap the harmonic current. When we talked above about the resonant current

going back and forth between the cap and reactor, does it travel back and forth on the short path on thebranch that they are on, or does it travel back and forth back through the transformer.

VTer (Electrical) 17 Apr 09

10:16

rockman,

Ih Current is the total current drawn by the vfd which includes the fundamental and harmonic currents.Now by getting the different values for Xt, Xl and Xc at different frequencies (5th, 7th, 11th, 13th...),you can calculate approximately how much current will be "diverted" by the filter using the equationprovided in previous attachment. For example, lets assume 1MVA xfmr 5%Z, 60Hz, 4160V pri/480V sec.feeding a VFD driving a 350hp motor. 6 Pulse drive.

Ifla=1200A Im=414A

Calculate Zeff = %Z(Im/Ifla) = 1.75%

From Table (Prediction of Harmonic Spectrum based upon Effective Source Impedance) I get(at Zeff=1.75, Individual Harmonic magnitudes (% fundamental) gives 1st at 100%, 5th at 48%, 7th at27%, 11th at 10%, 13th at 6%, and THD at 59%.)

Now these are your expected harmonics before the filter is applied.

Next step then would be to design the filter.First thing to do is to convert the hp to approx. Kvars.Kvar=hp*0.30 = 105

Then calculate CCf=Var/(2pi60V^2) = 1209 uF

Xc then would be approx 2.195ohms in our setup.

Now having calculated the C you can calculate L and design it to resonate at 282Hz (4.7*fund.)

Lh=1/[Cf(2pif)^2] = 263uHXl=.099

Those are your values for the filter.Now to check the harmonic distortion after the filter has been applied. Refer to diagram previously

attached.

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Now next thing to do is to find Xt.(from previous diagram)

Xt=(kV^2/MVA)*ZpuXt=0.01152ohms at fundamental

Now all of the values are in respect to the fundamental.Calculate Xt, Xl and Xc for 5th harmonic by multiplying fundamental values by 5.

You get Xt= 0.0576, Xl=0.495, and Xc=.438Do the same for 7th, 11th, 13th....

Now again refer to the diagram and equation attached. Using the equation you can calculate It current,and from this example I get5th at 23%, 7th at 20%, 11th at 8%, 13th at 5% and total with 33%.

So as you can see you reduced but not completely eliminated the current distortion.

FYI:I also get that the total voltage distortion with at approx. 3.98% which meets IEEE-519 voltagedistortion standard, I believe.



How to Calculate Harmonics Filters

In power supply systems based on alternating current (AC) -- such as the main power distributionnetwork from electric utilities -- non-linear loads can feed some amount of power back into the wiring.This feedback typically occurs in the form of harmonics: multiples of the frequency of the original AC

wave. Harmonics need to be eliminated from a power circuit by a harmonic filter to prevent them fromcausing voltage distortions and excessive currents in grounding connections. A harmonic filter consistsof a power capacitor connected in series with a tuning reactor, with both of them placed between the

power line and ground. The parameters for a harmonic filter depend on the electrical circuit in whichharmonic elimination needs to happen.

Instructions1.

o 1

Measure, using the harmonic analyzer on the circuit at 30% load, the load LD in kilowatts and the powerfactor PF.

o 2

Calculate the phase angles for both the actual and desired power factors (a typically desirable power factoris 0.97) by evaluating:

PAActual = arccos(PF)

PADesired = arccos(0.97)

o 3

Calculate KVAR, the kilo-volt-amperes required to raise the power factor from PF to, for example, 0.97 by evaluating:

KVAR = LD x (tan(PAActual) - tan(PADesired))

o 4

Calculate the capacitance required for the capacitor in the harmonic filter by evaluating:

C = KVAR / ((KV)^2 x 2 x Pi x F x 0.001)

Replace KV by the power line's voltage in kilovolts, and F by the power line's frequency in Hertz.

o 5

Calculate the reactance required for the tuning reactor in the harmonic filter by evaluating:

X = 1 / (2 x Pi x F x C)

Read more: How to Calculate Harmonics Filters | eHow.com http://www.ehow.com/how_7920770_calculate-harmonics-

filters.html#ixzz2I7sq5LsL

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