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Problem In Metering Kw & Kwh


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#1 AB2005

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Posted 02 October 2007 - 09:29 AM

Dear Friends
We have a power distribution system in our new plant where I have recently erected the Corrugator and a Flexo printing machine. The system is fed the power through a distribution transformer 11kv/400v, 2500kva. We have divided the distribution panels in to two phases, phase 1 for power distribution to Corrugator machine and phase 2 for printing and other small machines. The main ACB is in the centre of panels where incoming cables are connected. There is an energy analyzer (EA1) Socomac brand DIRIS A20 with CTs 4000/5A connected at mains. There is also an 800Kkvar PFI plant with Nokian capacitors and PFC relay (PFCR1).

The Corrugator machine is about 1MW which has almost 60% DC motors and 20% VFD driven motors. The Flexo machine is about 100kw with one DC motor (49kw) and other AC motors.

I have observed that when only Flexo machine and other small machines work, the power factor reading of EA1 and PFCR1 almost equals; 95-98 (with capacitors on). But as Corrugator machine starts working, the current increases and the power factor reading of EA1 drops at about 0.70-0.8 while PFCR1 reading remains 95-98 (with more capacitors on). Dropping of power factor disturbs the KW and KWH readings of EA1.
Can anyone help me where is the problem? Is there harmonic in the supply line? If yes then why the readings of both EA1 and PFCR1 are not equal as the CTs of both are mounted at same place.
Please help me.

"Don't assume any thing, always check/ask and clear yourself".

#2 mariomaggi

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Posted 02 October 2007 - 04:02 PM

Dear AB2005,

QUOTE(AB2005 @ Oct 2 2007, 11:29 AM) <{POST_SNAPBACK}>
The Corrugator machine is about 1MW which has almost 60% DC motors and 20% VFD driven motors. The Flexo machine is about 100kw with one DC motor (49kw) and other AC motors.
...........
Is there harmonic in the supply line?


your machines generates harmonics. Probably you don't have any harmonic filter in your systems.
Readings could be wrong for this reason, if your meters are not TrueRMS type.

Regards
Mario

Mario Maggi - Italy - http://www.evlist.ithttps://www.axu.it


#3 kens

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Posted 04 October 2007 - 11:33 PM

I think the PFC controller will be reading displacement pf only and the energy meter will be reading total pf. Like Mario says you will have a harmonics issue of some sort.
An expert is one who knows more and more about less and less until he knows absolutely everything about nothing

#4 AB2005

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Posted 06 October 2007 - 02:12 AM

Thanks for your posting
It means I have now two options;
(1) Install a harmonic filter.
(2) Install an energy meter which must be “True RMS” type like Nokian PFC Relay.

For the first option, I haven’t budget in my account and management wouldn’t give me permission for spending a huge amount of money at that stage of project.

Can anyone guide me any kind of other solution so that I would be able to use the existing energy meter?

Thanks and Regards

"Don't assume any thing, always check/ask and clear yourself".

#5 marke

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Posted 06 October 2007 - 04:00 AM

Hello AB2005

The Nokian Power Factor meter will only measure displacement power factor as that is all it can control. The energy meter will almost definitely be measuring total power factor which is a combination of displacement and distortion power factor. It will measure it by using the ratio of KW / KVA and due to the distortion present (harmonics) it will read a lower power factor than the Nokian. This does not mean that either unit is faulty, just that they are making different measurements.

Best regards,

#6 AB2005

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Posted 06 October 2007 - 10:10 AM

QUOTE(marke @ Oct 6 2007, 09:00 AM) <{POST_SNAPBACK}>
Hello AB2005

The Nokian Power Factor meter will only measure displacement power factor as that is all it can control. The energy meter will almost definitely be measuring total power factor which is a combination of displacement and distortion power factor. It will measure it by using the ratio of KW / KVA and due to the distortion present (harmonics) it will read a lower power factor than the Nokian. This does not mean that either unit is faulty, just that they are making different measurements.

Best regards,

Thank you Marke
Any solution?
"Don't assume any thing, always check/ask and clear yourself".

#7 marke

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Posted 10 October 2007 - 04:34 AM

Hello AB2005

QUOTE
Any solution?

What are you looking for? you have an energy meter that should read true pf as the ratio of KW/KVA and a power factor relay that shows displacement power factor only.
If you wish to have the energy meter read the same as the power factor relay, I do not belieive that it should!!
Cos(phi) and True Power factor are not the same. Cos(phi) will usually read higher than True Power Factor and it should if there are any harmonics on the supply.
You can only correct for displacement power factor, not distortion power factor. This is why a power factor controller will measure displacement power factor only.

Best regards,

#8 AB2005

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Posted 12 October 2007 - 06:38 AM

QUOTE(marke @ Oct 10 2007, 09:34 AM) <{POST_SNAPBACK}>
Hello AB2005
What are you looking for?
Best regards,

Hello Marke

This is the time to clarification.
Actually, I am assuming that the KW and KWH reading in Energy Analyzer is not correct because it shows a low power factor as compare with PFC relay (during working of dc motors and VFDs). If the power factor is low then its calculated KW would be less than actual KW and same with KWH. Am I right or mistaking?

"Don't assume any thing, always check/ask and clear yourself".

#9 marke

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Posted 12 October 2007 - 06:53 AM

If the meter measured the RMS current and the RMS voltage and the power factor and multiplied them together, then you could be correct, however the meter will measure the instantaneous voltage and instantaneous current, multiply them and sum the results so the KW will be very accurate and take into account any distortion in the voltage and/or current waveforms.
The power factor will then be calculated from the ratio of KW/KVA
I would believe the energy meter unless it is a real back yard unit manufactured by a company with no understanding of metering.
KW is the integral of the product of current and voltage over the period of one cycle with a higher number of samples leading to a higher accuracy.

RMS current multiplied by RMS voltage, multiplied by displacement power factor will give correct results only for pure sinusoidal voltages and currents.

Best regards

#10 AB2005

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Posted 18 October 2007 - 07:00 AM

Dear Mark

Check in the following link in which you would find some information about energy analyzer DIRIS A20. The manufacturer states that this meter is TRMS type.
Please help.
http://www.socomec.f...DirisA20-gb.pdf
"Don't assume any thing, always check/ask and clear yourself".

#11 marke

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Posted 18 October 2007 - 08:41 AM

Hello AB2005

The energy meter is rated to the IEC specs class 0.5. I believe that this is giving you the correct answers.
The KW is calculated by making many measurements of voltage and current within a cycle and multiplying the instantaneous current and voltage and averaging them.
The KVA is calculated by multiplying the RMS current by the RMS voltage.
Power factor is calculated by dividing the KW by the KVA.

If the load current includes current supplied to DC drives and AC drives, there will be a high level of harmonic current. This will reduce the true power factor but will not reduce the displacement power factor.
Your meters are indicating exactly what is happening. You are correcting the displacement power factor, but not the distortion power factor. If you wish to improve the distortion power factor you will need to install large harmonic filters between the meter and the load to reduce the harmonic content, or install an active filter system. Both options are very expensive, but are the only ways to improve the harmonics. - it is not a metering problem, it is a load problem.

Best regards,

#12 AB2005

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Posted 20 October 2007 - 05:33 AM

Thank you Marke

All is now cleared. It means that there is no problem in metering.

Can you please explain that what happen if we don't install the harmonic filter?
"Don't assume any thing, always check/ask and clear yourself".

#13 marke

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Posted 20 October 2007 - 05:54 AM

Hello AB2005

QUOTE
Can you please explain that what happen if we don't install the harmonic filter?

Basically, nothing will happen unless you are paying extra for the poor power factor.
From my experience, many of the Power Factor Penalties are based on displacement power factor only so there are no issues under those conditions.

Best regards,

#14 AB2005

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Posted 20 October 2007 - 07:50 AM

We have our own 41MW power generation plant. It means no problem.

"Don't assume any thing, always check/ask and clear yourself".

#15 AB2005

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Posted 24 October 2007 - 05:16 AM

Dear Marke
I understand about the relation between Active Power, Reactive Power and Apparent Power. But one thing is confusing me is that this energy meter displays Active Power, Reactive Power and Apparent Power as you can see in the link posted above. If the reactive power is supplied by capacitors then why this energy meter displays the reactive power?
For example, when Corrugator machine was running, I noted the flowing readings from energy analyzer;

Active Power………..256KW
Reactive Power ……..136Kvar
Apparent Power……...286Kva

The Nokian PFC relay was displaying 0.99 Cos (ph).
It means approximate all reactive power was supplying by capacitors, then why energy meter (at mains) was showing 136kvar?

"Don't assume any thing, always check/ask and clear yourself".

#16 marke

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Posted 24 October 2007 - 06:25 AM

Hello AB2005

That is a very good question!!
If we consider a sinusoidal system only, then we can draw a vector diagram which has KW on the x axis, KVAR on the y axis and the resultant is the KVA, so we have real power, reactive power and apparent power.
If we now introduce harmonics, they do not fit on a vector diagram, so this simplification is no longer valid, except that the definition of pf = KW/KVA is definitely true where KVA is Vrms x Irms and KW is the integral of I x V for a large number of samples.
Where there is a high distortion current, the power factor will be low. Therefore there is an "imaginary" power which is the square root of the difference between KVA squared and KW squared. In a sinusoidal environment, this is the KVAR and is the reactive current either leading by 90 degrees or lagging by 90 degrees.
The problem with harmonics, is that they contribute to both the real power and to the imaginary power.
I would suggest that the term reactive power is correct for a sinusoidal system, but perhaps should be referred to as imaginary power in a distorted system.

The KVAR associated with capacitors is not the same as the "KVAR" associated with harmonics.
The Nokian is only sensitive to the capacitive KVAR and typically measures the angle between voltage and current. - hence only one CT. see Power Factor Measurement for examples on how the displacement power factor is measured.

Best regards,

#17 marke

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Posted 25 October 2007 - 01:06 AM

QUOTE
We have our own 41MW power generation plant. It means no problem.
If you own the generation and distribution equipment, then you are paying for the additional losses in your generation and distribution system due to the harmonics.
If you have plant that is well laded, then the effect o the harmonics is much worse than if the plant is under utilized.

Best regards,

#18 AB2005

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Posted 26 October 2007 - 10:28 AM

QUOTE(marke @ Oct 24 2007, 11:25 AM) <{POST_SNAPBACK}>
In a sinusoidal environment, this is the KVAR and is the reactive current either leading by 90 degrees or lagging by 90 degrees.

Yes, some time I have also seen a “-“sign with reactive power reading.

Now we can say that this reactive power is not true, only an “imaginary” power which has no effect on our system.

"Don't assume any thing, always check/ask and clear yourself".

#19 marke

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Posted 26 October 2007 - 07:43 PM

QUOTE
Now we can say that this reactive power is not true, only an “imaginary” power which has no effect on our system

NO I disagree!
This "imaginary power" or "reactive power" is still increasing the losses in your generation system, it is just that you can not correct it with power factor correction capacitors. It does not register as KW used in the load, but dissipates power in the supply.
Harmonics increase the losses in transformers, cables, capacitors, filters and generators.

Best regards,

#20 AB2005

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Posted 27 October 2007 - 04:37 AM

QUOTE(marke @ Oct 27 2007, 12:43 AM) <{POST_SNAPBACK}>
Harmonics increase the losses in transformers, cables, capacitors, filters and generators.

Ok, now I would collect the data and then calculate the losses. This would help me to convey the management to invest some amount in purchasing the Harmonic Filter.

"Don't assume any thing, always check/ask and clear yourself".




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