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Vfds And Pf Correction


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#1 Amal

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Posted 03 September 2007 - 07:21 AM

I am designing a bulk p.f. correction system for a cotton spinning unit. The installed power is around 1150 kW. Almost 90% load is VFD driven. The biggest drive is rated 45kW.

What should be the kVAr of capacitors required if I want a p.f. of 0.996?

Also, whether the capacitors would require line chokes? If yes, of what value?

Would it be necessary to install harmonic filters if I am connecting all drives through line chokes?

Thanks.

Amal

#2 marke

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Posted 03 September 2007 - 10:55 AM

Hello Amal

Welcome to the forum.

QUOTE
I am designing a bulk p.f. correction system for a cotton spinning unit. The installed power is around 1150 kW. Almost 90% load is VFD driven. The biggest drive is rated 45kW.
Why are you considering bulk pf correction?
Bulk correction can only be used to reduce the inductive current in the supply. As the load is almost 90% VSDs, there is very little inductive current, so little or no correction is required.

VSDs do not draw inductive current which is why the Cos(phi) is quoted as better than 0.95
The true power factor, is the ratio of KW /KVA and on this score, the VSDs do not look as good.
The problem is that there is a lot of harmonic current flowing into the VSD, and this results in a poor distortion power factor. Distortion power factor can not be corrected by capacitors. It can only be corrected by filters and the filters are very expensive.

QUOTE
What should be the kVAr of capacitors required if I want a p.f. of 0.996?
I would say that this is an unrealistic target and would require a very large number of very small steps for no gain. The normal target for displacement power factor, is better than 0.95 and this is considered both practical and achievable.

QUOTE
Also, whether the capacitors would require line chokes? If yes, of what value?
Where capacitors are installed on supplies that have high harmonic currents, it is important to use detuning reactors. These reactors are specially rated for the capacitor values. NB if you use detuning reactors, you must also use special capacitors with higher internal operating voltage limits.

Have a look at http://www.LMPhotonics.com/pwrfact.htm

Best regards,

#3 mariomaggi

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Posted 03 September 2007 - 01:33 PM

Amal,
in your application you will have benefits using some big passive harmonic filters, non-tuned type.
You could install few of these filters, reactors are already embedded in these filters.

Regards
Mario

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#4 Dr Win

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Posted 03 September 2007 - 05:44 PM

QUOTE(Amal @ Sep 3 2007, 02:21 AM) View Post

I am designing a bulk p.f. correction system for a cotton spinning unit. The installed power is around 1150 kW. Almost 90% load is VFD driven. The biggest drive is rated 45kW.

What should be the kVAr of capacitors required if I want a p.f. of 0.996?

Also, whether the capacitors would require line chokes? If yes, of what value?

Would it be necessary to install harmonic filters if I am connecting all drives through line chokes?

Thanks.

Amal


Dear Amal

You only need to correct 10% load. For bulk pf correction, intelligent controllers difficult to correct to get your goal pf of 0.996. Actually bet 0.95 - 1 is satisfied. Instead this, you better choose for statistic pf correction. For harmonics, capacitors cannot correct them. You need to install filters.

Best regards

Dr Win



#5 Amal

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Posted 06 September 2007 - 02:06 PM

QUOTE(marke @ Sep 3 2007, 04:25 PM) View Post

Hello Amal

Welcome to the forum.

Why are you considering bulk pf correction?
Bulk correction can only be used to reduce the inductive current in the supply. As the load is almost 90% VSDs, there is very little inductive current, so little or no correction is required.

VSDs do not draw inductive current which is why the Cos(phi) is quoted as better than 0.95
The true power factor, is the ratio of KW /KVA and on this score, the VSDs do not look as good.
The problem is that there is a lot of harmonic current flowing into the VSD, and this results in a poor distortion power factor. Distortion power factor can not be corrected by capacitors. It can only be corrected by filters and the filters are very expensive.

I would say that this is an unrealistic target and would require a very large number of very small steps for no gain. The normal target for displacement power factor, is better than 0.95 and this is considered both practical and achievable.

Where capacitors are installed on supplies that have high harmonic currents, it is important to use detuning reactors. These reactors are specially rated for the capacitor values. NB if you use detuning reactors, you must also use special capacitors with higher internal operating voltage limits.

Have a look at http://www.LMPhotonics.com/pwrfact.htm

Best regards,


Thanks for the reply.

In India, where we are doing this project, achieving a p.f. of 0.996 gives a large economic incentive. So, it is good if we can get this value.

The metering unit is most probably incapable of measuring distortion p.f. It is measuring only the inductive currnent. In that case, should I assume only 10% inductive load and calculate the capacitors requird?

I am perfectly happy to get a distortion p.f. of around 0.95. Can you recommend sources about harmonic filter design?

Thanks once again.

QUOTE(mariomaggi @ Sep 3 2007, 07:03 PM) View Post

Amal,
in your application you will have benefits using some big passive harmonic filters, non-tuned type.
You could install few of these filters, reactors are already embedded in these filters.

Regards
Mario


Thanks for the reply.

Can you recommend sources about harmonic filter design?



QUOTE(Dr Win @ Sep 3 2007, 11:14 PM) View Post

Dear Amal

You only need to correct 10% load. For bulk pf correction, intelligent controllers difficult to correct to get your goal pf of 0.996. Actually bet 0.95 - 1 is satisfied. Instead this, you better choose for statistic pf correction. For harmonics, capacitors cannot correct them. You need to install filters.

Best regards

Dr Win


Thanks for the reply.

In India, where we are doing this project, achieving a p.f. of 0.996 gives a large economic incentive. So, it is good if we can get this value.

The metering unit is most probably incapable of measuring distortion p.f. It is measuring only the inductive currnent. In that case, should I assume only 10% inductive load and calculate the capacitors requird?

I am perfectly happy to get a distortion p.f. of around 0.95. Can you recommend sources about harmonic filter design?

Thanks once again.



#6 Amal

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Posted 10 September 2007 - 07:32 AM

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According to my calculations, the kVAR required for getting a p.f. of 0.9 is around 450. So, I now have following problems:

1. Should I use tuned or detuned filters?
2. If detuned filters are used, what frequency should be used? (Siemens handbook says that these filters take out around 20% of 5th harmonic if frequency is 189 Hz and 50% id frequency is 210 Hz)
3. Where should I locate the filters?

Thanks.


#7 marke

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Posted 10 September 2007 - 08:47 AM

Hello Amal

How did you calculate that figure of 470KVAR?

If 10% of the load is not VSD load, that would suggest that the non VSD load is in the order of 115KW.
You should only be correcting this load, not the VSD load. I would not expect to apply 450KVAR to 115KW of load unless the power factor is very low. Do you know the actual power factor or the uncorrected KVAR of the load?

The detuning reactors are connected in series with the capacitors only and must match the capacitors. You purchase detuning reactors rated in KVAR where the KVAR equals the capacitor that you are detuning. For example, if you install a 50KVAR capacitor, you order a 50KVAR detuning reactor. The reactors are either 5%, or 7% reactors. What this means is that they reduce the effective KVAR of the capacitor by 5% or 7% of it's KVAR. The detuning reactor adds some inductive reactance in series with the capacitive reactance and thereby reduces the net reactance.

The capacitor and reactor series circuit is connected across the supply as a means of reducing the reactive current drawn from the supply. If the supply voltage is a very clean sine wave, there is no need for detuning reactors. If there is any waveform distortion, there will be harmonic currents through the capacitors. The magnitude of this harmonic current is dependent on the amount of distortion on the supply. If the level of harmonic voltage on the supply is quite low, the 5% detuning reactors will be adequate. If the harmonic voltage is higher, then you need to use the 7% reactors.

I would recommend that you i) measure the existing displacement power factor before determining the amount of bulk correction to be added, and ii) measure the harmonic voltages on the supply to determine the potential harmonic current through the capacitors.

Best regards,

#8 Amal

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Posted 17 September 2007 - 03:00 AM

Thanks a lot. Now, I have calculated as under:

a) For the 120 kW load which is not on VFDs:

Uncorrected Power Factor: 0.8 Expected Power Factor: 0.997

Required Capacitors: 120 * 0.75 = 90 kVAR

cool.gif For the 1000 kW load on VFDs:

Uncorrected displacement power factor = 0.9 ( Because of line chokes)
Expected P.F. = 0.997

Required Capacitors: 1000 * 0.46 = 460 kVAR

Total Capacitors required = 460+ 90 = 550 kVAR

I will be adding reactors of 25 KVAR each for each capacitor of 25 kVAR. This would be a 7% reactor. So, Actual capacitance in the circuit would be (550 * 0.93) = 511.5 kVAR. So, I will put 600 KVAR to get (600*0.93) = 558 kVAR.

Is this accurate?

Thanks.

#9 marke

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Posted 17 September 2007 - 03:55 AM

Hello Amal

a) I calculate to correct 120KVA from 0.8 to 0.997 would require 77KVAR of capacitance.

cool.gif You can not correct the inputs to drives. There is a discontinuous current flow. You can only correct sinusoidal currents!!

I would not apply more than 77KVAR to this load, and I would personaly be reluctant to apply any correction at all. I am concerned that if the supply impedance is not very low, you are going to have resonance issues. I would only correct to maximum of 0.95 This would require 47KVAR of correction.

I assume that the 0.80 is a measured figure. What is the accuracy of this figure? your target implies an accuracy of better than 3%. I do not believe that this is in any way realistic.

Best regards,




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