Powerfactor Improvement

Description is not crystal clear.SLD will give better picture to understand .Can you please send the SLD for study and suggestions.

Ramesh

cbr@cuppu.com

I do not know if this is actually done in real life anywhere, but I have considered the problems and I would say probably not as it would be, to my mind, far too complicated to carry out.

The Power factor is as a result of the type of loads being applied and I would have thought that any PF correction, would be better made at the load, not at the point of generation!!!

Otherwise you will be increasing the amps being carried by the whole link is what my head tells me.

Corrections made at the load itself will fix or allay part of the PF error, without causing any circulation load amps (if I am wrong on this point, please somebody correct my thinking, my knowledge is very, very old!!)

The user is the one paying the bill, so if he has a bad PF, then he is paying for more electricity than he is using!! His problem to my mind......!!!

I'm with you on that one Andy.

Although my knowledge of those little electron things buzzing around is fairly basic, I understood that ideally the power factor should be 1 and I believe that that is unachievable, so the aim is to have that power factor as close to 1 as possible.

I also understood that when the power factor falls off it was always due to load problems. The greater the power factor is away from 1 the greater the inefficiencies of the load and its system.

You've nailed it, Andy. Adding caps essentially creates a tank circuit (L&C in parallel). Since they are 180° out of phase, the inductance feeds the capacitance in one half-cycle, and the cap feeds the inductance in the other half. The closer the inductance & capacitance are, the shorter the section of line carrying the reactive current.

Losses are proportional to the square of the total current. By placing the capacitors at the source, loss through the line increases geometrically.

Thanks for the timely reminder, I knew the effects (roughly!) but I could not be certain of the exact cause after so many years....and I hate sticking my neck out!!!

So as I said before, PF correction can/should only be done at the load, not at the supply end!!!

It took several people to help me nail it down correctly, but that is what CR4 is all about!!!!

Many thanks Guys. Teamwork!!

Thankyou andy Germany 16.02.2008

This is the Wind generator as you told It ios not load Here we are using the INDUCTION GENERATOR So for Magnetising It required , Power from Grid , The Generation System vot is 690 , so we need a transformer to Step up this 690 to Grid voltage we are using 690/33000 volt transformer it is also giving lagging power factor .

I feel it is Enough

Regards

I. Dhanaraj Lazarus

Dear sir

Thank you

can you read my posting and give a solution please . give your E mail id sothat i can send SLD

General Write up of The APFC requiremant in 8X 1500 Kw Wind mills

The customer is having 8 X 1.5 Mw Wind mills in a 33 Kv feeder feeding from a rural feeding Power transformer of 10Mva 110 Kv / 33 Kv transformer

Wind mill is located at 13 Km away from the trans former .

Already 5to 8 mva load is there TNEB is feeding 8 MVA rural load And also receiving about 15 MVA generated power from Wind Mills connected to this 10 mva transformer . This is the detail of the loading of the transformer .

We are attaching the individual Wind mill HT ss 1.8 Mva 33/0.690 V transformer used for Steping up the 1500Kw power generated by the Variable speed Technology Wind mill . The stepped up power is connected to the ss to the 10 MVA Transformer thro a common feeder . all the 8 machies are having separate Individual Metering point with the 4 Quaderent Trivector meter to record the generation .

Our Requirement

The Wind Electric Generator is a variable speed asynchronous generator , It has winding in Both Rotor & in Stator From both winding Power is drawn and the total power is 1500 Kw

Normally from Stator 1000 Kw

Normally from Rotor 500 Kw power will be generated during the optimam wind speed , they are using the rotor feeding for injecting the field current & also receiving the power produced in it . Normally in this type of Wind mill the Power factor correction will be done by the controller it self .

Our Present site condition

Ø Wind mill supplier is telling that their machine is giving a 98 % leading pf .( they are having separate CT. in the panel incoming that is sensing only the power of the Generator )

Ø Wind mill supplier is not compensating the reactive drawl of the1800 Kva power trans former .

Ø Wind mill supplier is not allowing to use the same transformer to add the Capacitor Thro APFC rely compensate the VAR drawl .

Ø Wind mill supplier insists the Customer to Compensate in the VAR drawl in H T side alone

Ø Wind form owner's inconvenient is that they can't have the single point of supply , they have 8 nos point of supply .

Ø Having switching capacitor addition in 33 kv is costly, so we decided to put the compensation APFC Panel in LT side Say 690 V.

Ø For this we have to put one more 33/0.690 Kv transformer parallel to the Excisting main 1800 Kva transformer after the metering point of the Wind mill SS . To measure the combined Kvar requirement we have to put one no 33 kv CT in the 33 Kv Bus after the Main Metering Set before the 1800 Kva trans former .

Ø Your Requirement for design

Ø You have asked the no Load and full load current

Ø Since we are not maintaining the wind mill we could not give to you .

Ø But the wind mill will give the out put depending upon the wind flow , it will vary upto the maximum of 1500 Kw during the high wind season

Ø But the compensation transformer is required for the full out put only , we cannot change the transformer now& then , Since the Adding & deleting of the Capacitor will be in steps of 12.5 Kvar , 25, 37.5 insteps as per the requirement of the system .

Ø Also the capacitors are added through the capacitor Switching contactor s if the system or the grid fails Immediately all the contactors connected thro this system will also open also the capacitors added will be disconnected immediately .

Ø So the system may not disturb due to this .

Ø We request you to consider these we also attach the single line diagram of the 8 X 1500 Kw machines arrangement s & one Single wind mill 33 Kv yard , Excisting & new proposed in separate drawing .

Can you help me to get a good AUTO MATIC POWER FACTOR CONTROLLER / ^() V Gel Filled Capacitors of Good Quality

Thank You

IDHANARAJ LAZARUS

It seems to me that there is some confusion in the use of terms and the understanding of power factor. Here is something from an expert source that may help.

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Frequently Asked Questions FAQ0002-0297 OSRAM SYLVANIA National Customer Support Center Visit our website: www.sylvania.com

18725 N. Union Street 1-800-LIGHTBULB Westfield, IN 46074 USA © 2000 OSRAM SYLVANIA Power Factor

What is power factor?

This is a very involved subject that will be dealt with in terms of field application and typical questions from end-users.
Power factor is characteristic of alternating current (AC) circuits. Always a value between (0.0) and (1.0), the higher the number the greater/better the power factor.
Circuits containing only heating elements (filament lamps, strip heaters, cooking stoves, etc.) have a power factor of
1.0. Other circuits containing inductive or capacitive elements (ballasts, motors, personal computer, etc.) usually have a power factor below 1.0. Normal power factor ballasts (NPF) typically have a value of (0.4) - (0.6). Ballasts with a power factor greater than (0.9) are considered high power factor ballasts (HPF).
The significance of power factor lies in the fact that utility companies supply customers with volt-amperes, but bill them for watts. The relationship is (watts = volts x amperes x power factor). It is clear that power factors below 1.0 require a utility to generate more than the minimum volt-amperes necessary to supply the power (watts). This increases generation and transmission costs. Good power factor is considered to be greater than 0.85 or 85%.
Utilities may impose penalties on customers who do not have good power factors on their overall buildings.
Watts, or real power, is what a customer pays for. VARS is the extra "power" transmitted to compensate for a power factor less than 1.0. The combination of the two is called "apparent" power (VA or volt-amperes).
Consider this popular analogy to clarify the relationship between real and apparent power.
We all know a glass of draft beer generally has a "head" on it. Let's say your favorite pub institutes a new policy - you only pay for the beer, not the foam. While the foam is just aerated beer, it is not really usable in that form. If the glass of beer is half foam, you pay half the price.
This is the same principle as electricity generation - the consumer only pays for the beer (real power), not the foam (the "VARS" mentioned above).
FAQ0002-0297 1-800-LIGHTBULB Visit our website: www.sylvania.com - 2 - © 2000 OSRAM SYLVANIA Illustration of Effect of Power Factor The effect of power factor on both the utility and the customer. Note that the utility bills the customer for watts but generates volt-amperes (VA), and that (watts = volts x amperes x power factor).
Four applications will be discussed:
1. 60 watt incandescent lamp (PF=1.0)
2. 15 watt medium-based compact fluorescent lamp, electronic ballast, normal power factor (PF = 0.6)
3. 15 watt medium-based compact fluorescent lamp, electronic ballast, high power factor (PF = 0.95)
4. 13 watt medium-based compact fluorescent lamp, magnetic adapter unit (PF =0.25)
For the 60 watt incandescent lamps:
· Customer pays for 60 watts of power · Utility generates 60 watts ÷ 1.0 PF = 60 VA For the 15 watt medium-based compact fluorescent lamp, electronic ballast, normal power factor:
· Customer pays for 15 watts of power.
· Utility generates 15 watts ÷ 0.6 PF = 25 VA For the 15 watt medium-based compact fluorescent lamp, electronic ballast, high power factor:
· Customer pays for 15 watts of power · Utility generates 15 watts ÷ 0.95 PF = 15.8 VA For the 13 watt medium-based compact fluorescent lamp, magnetic adapter unit:
· Customer pays for 16 watts of power · Utility generates 16 watts ÷ 0.25 PF = 64 VA While the consumer saves watts with any of the compact fluorescent alternatives, the utility must generate nearly the same amount of VA for the magnetic adapter unit as for the incandescent lamp. While this is an extreme case, it does illustrate the effect of power factor. Some "better" retrofit units have power factors ranging from 30% - 50% (0.3 to 0.5). High power factor products do offer the customer an additional benefit as well. On a one for one replacement/retrofit onto existing circuits for incandescent lamps (15 or 20 amperes), multiple normal power factor units may trip the breakers. High power factor ballasts allow more lamps per circuit than do normal power factor ballasts. Note that these high power factor units will be slightly higher in cost.
Typical Questions and Answers From End-Users Q: What is meant by a "High Power Factor" ballast?
A: Ballast designated High Power Factor (HPF) must have a power of 90% (0.90) or greater. Ballast factors lower than a 0.90 are called Normal Power Factor (NPF).
FAQ0002-0297 1-800-LIGHTBULB Visit our website: www.sylvania.com - 3 - © 2000 OSRAM SYLVANIA Q: Will large scale use of NPF compact fluorescent lamps in a building have any detrimental effects?
A: The harmonic generation and resulting power factor of these lamps may cause the neutral wire in a four wire system (WYE) to be overloaded but only if the "maximum number per circuit" specification is exceeded. HPF compact fluorescent lamps will have no detrimental effect on the electrical system.
Q: Is an HPF ballast more efficient than an NPF Ballast?
A: Yes, but only in terms of how well it uses the apparent power (volt-amperes), not the metered power (watts).
Remember that the user pays for watts, not volt-amperes. Therefore, if the wattage ratings are the same for both the HPF and NPF units, then the user will not see a difference in the electric bill. However, if the user's electric rate includes a low power factor penalty and if enough NPF units are in use, there may be an increase in the electric bill.
The other issue concerns circuiting. Use of HPF ballast will allow more fixtures per circuit.

Kindly see the explanation given by me for the powerfactor http://cr4.globalspec.com/comment/176148

This explanation will be understood by an Electrical Engineer.

I personally have no problem with understanding PF., it was explaining the reason NOT to correct at the point of generation that I could not remember clearly enough!!!

hi, i hope i can make worthy a contribution.

normally for electrical loads, andy is correct in saying that pf correction is done at the load side and not at the generation side. benefits being the reduction of losses.

however, the inquiry relates to an induction generator, wherein the equipment generates kw but at the same time consumes kvar. the kvar is supplied via the grid, making the induction generator operable (ie generate kw power) only when connected to another power source and not on island mode.

thus, their concern on pf improvement, which is possible via capacitors. however, since the pf of the system would continually be changing due to fluctuations of the wind, this has to be controlled.

Dear sir

Thank you

can you read my posting and give a solution please . give your E mail id sothat i can send SLD

General Write up of The APFC requiremant in 8X 1500 Kw Wind mills

The customer is having 8 X 1.5 Mw Wind mills in a 33 Kv feeder feeding from a rural feeding Power transformer of 10Mva 110 Kv / 33 Kv transformer

Wind mill is located at 13 Km away from the trans former .

Already 5to 8 mva load is there TNEB is feeding 8 MVA rural load And also receiving about 15 MVA generated power from Wind Mills connected to this 10 mva transformer . This is the detail of the loading of the transformer .

We are attaching the individual Wind mill HT ss 1.8 Mva 33/0.690 V transformer used for Steping up the 1500Kw power generated by the Variable speed Technology Wind mill . The stepped up power is connected to the ss to the 10 MVA Transformer thro a common feeder . all the 8 machies are having separate Individual Metering point with the 4 Quaderent Trivector meter to record the generation .

Our Requirement

The Wind Electric Generator is a variable speed asynchronous generator , It has winding in Both Rotor & in Stator From both winding Power is drawn and the total power is 1500 Kw

Normally from Stator 1000 Kw

Normally from Rotor 500 Kw power will be generated during the optimam wind speed , they are using the rotor feeding for injecting the field current & also receiving the power produced in it . Normally in this type of Wind mill the Power factor correction will be done by the controller it self .

Our Present site condition

Ø Wind mill supplier is telling that their machine is giving a 98 % leading pf .( they are having separate CT. in the panel incoming that is sensing only the power of the Generator )

Ø Wind mill supplier is not compensating the reactive drawl of the1800 Kva power trans former .

Ø Wind mill supplier is not allowing to use the same transformer to add the Capacitor Thro APFC rely compensate the VAR drawl .

Ø Wind mill supplier insists the Customer to Compensate in the VAR drawl in H T side alone

Ø Wind form owner's inconvenient is that they can't have the single point of supply , they have 8 nos point of supply .

Ø Having switching capacitor addition in 33 kv is costly, so we decided to put the compensation APFC Panel in LT side Say 690 V.

Ø For this we have to put one more 33/0.690 Kv transformer parallel to the Excisting main 1800 Kva transformer after the metering point of the Wind mill SS . To measure the combined Kvar requirement we have to put one no 33 kv CT in the 33 Kv Bus after the Main Metering Set before the 1800 Kva trans former .

Ø Your Requirement for design

Ø You have asked the no Load and full load current

Ø Since we are not maintaining the wind mill we could not give to you .

Ø But the wind mill will give the out put depending upon the wind flow , it will vary upto the maximum of 1500 Kw during the high wind season

Ø But the compensation transformer is required for the full out put only , we cannot change the transformer now& then , Since the Adding & deleting of the Capacitor will be in steps of 12.5 Kvar , 25, 37.5 insteps as per the requirement of the system .

Ø Also the capacitors are added through the capacitor Switching contactor s if the system or the grid fails Immediately all the contactors connected thro this system will also open also the capacitors added will be disconnected immediately .

Ø So the system may not disturb due to this .

Ø We request you to consider these we also attach the single line diagram of the 8 X 1500 Kw machines arrangement s & one Single wind mill 33 Kv yard , Excisting & new proposed in separate drawing .

Thank You

IDHANARAJ LAZARUS