Power factor question

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puckman

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ridgewood, n.j.
At our switchgear we have a pf meter and I believe the pf is low .86 , amps = 1200 @480v. What would we do to increase the pf ? Not sure if there is a penalty for low pf . I like to get a better understanding of pf conditions.
 
Power factor is nothing excep the angle between Real and apparant power kw/kVA. When the power factor is high the real power increases and by low power factor reactive power kVAR increases. In industrials and motors load kVAR always is high and power factor is low.


Do you want to increase the power factor in switchgear. But the reactive power is needed according to load. In my opinion, it is better to increase load power factor.


Please follow the below examples, I hope it will help you




I have attached the file with all its calculation as below.


a) When the capacitor is absent, what is the current through the source?

b) What is the voltage across the motor?

c) What is the complex power of the motor?

d) What capacitance should be added to correct the power factor at the motor?

e) What is the current through the source after that capacitance is added?



--------------------------------------------------------------------------------
a)
Zeq = (10.0 + j) ohm, so is = vs / Zeq
= 450 V / (10.0 ohm \/ 5.7 deg)
= 44.8 A \/ -5.7 deg

b) Zm = (7 + j) ohm
= 7.07 ohm \/ 8.1 deg
vm = imZm
= (44.8 A \/ -5.7)(7.07 ohm \/ 8.1)
= 317 V \/ 2.4 deg


c) Sm = vm . im*
= (317 V \/ 2.4)(44.8 A \/ 5.7)
= 14.2 kVA \/ 8.1 deg
= 14.0 kW + j2.0 kVAR


d)
XC should be added such that 317 V across it will contribute a reactive power of -2.0 kVAR, thereby eliminating that from the inductive reactance. Since Q = V2 / XC, Xc = V2 / Q
= (317 V)2 / 2000 VAR
= 52 ohm


A capacitance of 51 microfarad yields a reactive capacitance of 52 ohm at 60 Hz.

e)
In this part, we check that we accomplished what we wanted in part d). There are numerous ways to proceed; perhaps the easiest is to evaluated the equivalent impedance of the network after the capacitance is added. The capacitor in parallel with the motor yields an equivalent impedance of Z|| = ZmZc / (Zm + Zc)
= (7+j)(-j52) / (7-j51)
= (7.07 \/ 8.1)(52 \/ -90) / (51.5 \/ -82.2)
= 7.1 ohm


Note that the equivalent impedance of the parallel combination has become real, with no net reactance. In general, this is the goal in power factor correction.

The total impedance of the simple circuit is 10.1 ohm; the current through the source is now (450 V) / 10.1 ohm) = 44.4 A.
 
puckman said:
At our switchgear we have a pf meter and I believe the pf is low .86 , amps = 1200 @480v. What would we do to increase the pf ? Not sure if there is a penalty for low pf . I like to get a better understanding of pf conditions.
Click to expand...

I fyou do not have a penalty I would not worry about 0.86, thats not bad for an industrial plant.
 
puckman said:
At our switchgear we have a pf meter and I believe the pf is low .86 , amps = 1200 @480v. What would we do to increase the pf ? Not sure if there is a penalty for low pf . I like to get a better understanding of pf conditions.
Click to expand...

.86 is not really that bad considering if its industrial most of the load is inductive, but I would bet the farm that you're being panalized. Ours is .84 and yes we are being penalized as to dollar signs I'm not sure. Add some caps or you may have a synchronous motor driven generator that can be run unloaded.
 
I'm with the others; even if you're paying a penalty, 0.86 is not bad at all. The last plant I worked at was in the 0.40-0.50 range before correction (if that dang meter was right. I never totally trusted that thing.) Either way, IMO, anything above 0.80 is well enough. Leave it alone.
 
100212-1138 EST

Hameedulla-Ekhlas:
Power factor is nothing excep the angle between Real and apparant power kw/kVA.
Click to expand...
I want to clarify your definition.

The definition of power factor is PF = POWER/VOLT-AMPERES. It is a ratio and not an angle. Also volts and amperes are the RMS values. For pure sinusoidal waveforms and with both the current and voltage of the same frequency you can relate power factor to an angle between the two sine waves by arc cos (PF).

For non-sinusoidal current waveforms there is no useful angular relationship between an angle and power factor. A bridge rectifier with an RC load produces a sharp current pulse located at about the peak of the applied voltage. I might choose to call this a 0 deg phase shift because the current peak corresponds to the voltage peak. My scope has a PF = 0.62 . The calculated angle from arc cos 0.62 is 51.7 deg. Doesn't mean much relative to my load.


puckman:

Study the discussion at
http://en.wikipedia.org/wiki/Power_factor
It is quite well written.

.
 
Yes, you are right it is a ratio but if you express it as a triangle among apparant power, real power and reactive power then there is an angle.
 
Hameedulla-Ekhlas said:
Yes, you are right it is a ratio but if you express it as a triangle among apparant power, real power and reactive power then there is an angle.
Click to expand...

Only if you are considering displacement power factor, caused by having sinusoidal current flowing out of phase with the voltage.

Distortion power factor is not represented by a phase angle difference.

-Jon
 
winnie said:
Only if you are considering displacement power factor, caused by having sinusoidal current flowing out of phase with the voltage.

Distortion power factor is not represented by a phase angle difference.

-Jon
Click to expand...
Yes, that's true. But it just doesn't generally seem to be a well understood concept.
 
Meaning of power Factor

Meaning of power Factor

What Power Factor is
Power Factor is an indication for the efficiency of your facility. Utility charges penalties in order to force you help them maintain an efficiency network. Low Power Factor means your own facility can run more efficiently.

Can Power Factor Save me money?
A PF of 0.86 means you can be at least 16% more efficient. As typical sites has 5-10% losses, this means you can reduce your bill by 1.5-3%, in addition to penalties, if applicable. In addition, it will let you use more power with your existing infrastructure and reduce the heat in the cables and transformers.

How to correct
The easiest way is to place central power factor correction system. It will reduce your overall PF value. It will improve the transformer's efficiency but will not improve the cables efficiency. If you have large sub MCCB - consider putting the correction unit over there to save on the cable to the the main cabinet. It is also possible to improve each load, but it is more complicated. A nice solution and easy to implement for motors is manufactured by the company in which I work is called SinuMEC which improves the motor kW and kVAr consumption. This will improve your PF, reduce your kW and save on the cables as well.

Note on Capacitors
If you choose to use capacitors for improvement - put inductors to protect against harmonic resonance. It is a small extra investment that will prevent future problems.

Meters' accuracy and measurements
Note that there are two different values - DPF and TPF. DPF is for first harmony only. TPF is for all the harmonics. TPF is what utility measure, as this is the value for efficiency. Note that some loads, such as frequency inverters, improve the DPF but worsen the TPF. This means that they can NOT help you with PF issues.
 
There are losses on each plant. The losses varies with the investment on infrastructure and quality of maintenance. Common figures are from 5% to 15%, but I measured figures of more than 30%.

The losses on cables are linear with the square of I. The losses on transformers are more complicated to calculate, but some of it are also linear with square of I. When you improve the PF, the current is reduced. Improving from 0.86 to 0.95 will reduce the current by 10% and will reduce the losses by 20%.

How much it will save on your bill depends on two parameters:
* What are your actual losses
* Where will you implement the correction

The minimum is implementing on the secondary of the main power transformer. As transformers has 1.5 to 5% losses, the saving will be 0.3% to 1%.

When I measured the over 30% losses, the power factor was 0.50. I have installed the SinuMEC energy saver which improved the power factor to 0.80. It also of the saving on reduced kW and many other benefits that the product provides, the improvement of PF by 60% reduces the current (almost) respectively and reduces the losses by more than half, which reduces the overall bill be 15%.

To this end, improving PF reduces the electricity bill. The best is to improve the load operation, but it requires more efforts.
 
Well yes power factor is important as we see it but a factor of 1# is not what it is who benefits the power company does as a low power factor is wasted power which is not heat loss but reactive in the lines returning to there plant so yes your helping there return and they can resell the positive voltage by not correcting the negative return power which is a bad or low power factor they dont make money on your bill . So actually a 95 % pf is better for you than a 1# .


The loss is in the lines back to the plant /


Question when you check a service for power factor correction what do you do meaning do you just come out and install you magic box or do you monitor it for a week or days and check odd times and load swings of the system .

Adding to much capacitance in the wrong way meaning too much XC reactance is not good its better to have a little of both worlds thats just my input the non comercial power factor correction boxes i see are worthless you need to add XC in steps under load changes only .

Each service is a calculation by itself no two services are the same when you design PF correction and if your selling these devices do you tell the home owner we need to monitor your system and the cost for this will not save them dollars just pennys each year but it will benefit the power company more so on there resell of power to others .


This is nothing new and i dont understand how they give out US patents on these devices its been around since mr ohms mr watts what inventive today making free power .

Did you know that harmonics are bad but they can save you watts on that watt meter as it can not measure harmonics and the chopping up of the wave thur your watt meter it can not give a correct wattage just a thought ya saving money by chopping up the line voltage .

Its bad and good and at a lower PF of say 91% your saving dollars and the money you save can by more stuff to add to your toys each year .


.
 
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Well yes power factor is important as we see it but a factor of 1# is not what it is who benefits the power company does as a low power factor is wasted power which is not heat loss but reactive in the lines returning to there plant so yes your helping there return and they can resell the positive voltage by not correcting the negative return power which is a bad or low power factor they dont make money on your bill . So actually a 95 % pf is better for you than a 1# .


The loss is in the lines back to the plant /


Question when you check a service for power factor correction what do you do meaning do you just come out and install you magic box or do you monitor it for a week or days and check odd times and load swings of the system .

Adding to much capacitance in the wrong way meaning too much XC reactance is not good its better to have a little of both worlds thats just my input the non comercial power factor correction boxes i see are worthless you need to add XC in steps under load changes only .

Each service is a calculation by itself no two services are the same when you design PF correction and if your selling these devices do you tell the home owner we need to monitor your system and the cost for this will not save them dollars just pennys each year but it will benefit the power company more so on there resell of power to others .


.
 
I am a little bit confused. Why 95% PF is better than unity?

The main discussion was about PF - central and distributed compensation. I have mentioned the device that I am selling just as one alternative (and I must say a good one :)) for distributed reactive energy improvement. It is fitted to motors, according to the motor load. It changes the way the motor behaves and in any way does not turns into capacitive range. Other solutions include adding local capacitors, selecting different loads etc.

The problem with reactive energy is that it causes losses, both for the utility and the site itself. This is correct for large facilities - not residential site, which have minimal losses.

The affect of harmonics to energy measurement is very complicated. It is correct that mechanical meters do not measure all the harmonic power, but the extra losses harmonics create are much higher. Note that the active power in harmonics is very small. For example, if you have the maximum allowed harmonic pollution (depends on standard being used) of 5% in the voltage and 20% in the current, the power will be not more than 1%. If the pollution is in different orders - not even this. The extra losses for 20% current harmonics are much higher. Moreover, utilities around the world move to electronic meters which measure all the harmonics.
 
PowerQualityDoctor said:
I am a little bit confused. Why 95% PF is better than unity?

The main discussion was about PF - central and distributed compensation. I have mentioned the device that I am selling just as one alternative (and I must say a good one :)) for distributed reactive energy improvement. It is fitted to motors, according to the motor load. It changes the way the motor behaves and in any way does not turns into capacitive range. Other solutions include adding local capacitors, selecting different loads etc.

The problem with reactive energy is that it causes losses, both for the utility and the site itself. This is correct for large facilities - not residential site, which have minimal losses.

The affect of harmonics to energy measurement is very complicated. It is correct that mechanical meters do not measure all the harmonic power, but the extra losses harmonics create are much higher. Note that the active power in harmonics is very small. For example, if you have the maximum allowed harmonic pollution (depends on standard being used) of 5% in the voltage and 20% in the current, the power will be not more than 1%. If the pollution is in different orders - not even this. The extra losses for 20% current harmonics are much higher. Moreover, utilities around the world move to electronic meters which measure all the harmonics.[/QUOTE

Well harmonics are not measured with a watt meter and i think a little harmonics can save on the bill more as it chopps it up and the meter can not measure it .
Click to expand...
 
PowerQualityDoctor said:
To this end, improving PF reduces the electricity bill. The best is to improve the load operation, but it requires more efforts.
Click to expand...

I read your explanation and I still don't buy it. I may improve the efficiency of the wiring system but I will not reduce my bill beyond any penalties that may be eleimnated.
 
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ohmhead said:
Well harmonics are not measured with a watt meter and i think a little harmonics can save on the bill more as it chopps it up and the meter can not measure it .
Click to expand...

There are two ways to measure energy consumption - electro-mechanical and electronic meters. Electro-mechanical meters measure some of the harmonics in some going down amplitude. Common figures are around 20% of the 5th harmony. Usually, electronic meters measure all of them (up to the maximum frequency they measure - some of them are for 50/60Hz only and do not measure harmonics).

The energy that you consume at higher frequencies can be utilized by some of the loads, such as heaters or switched power supplies, but hearts the operation of rotating machines, such as air conditions.

The bottom line - harmonics increase your costs by increasing your losses. Minor "tweaking" of the electricity meter may be possible, but it is way less than the meter in-accuracy.
 
PowerQualityDoctor said:
There are two ways to measure energy consumption - electro-mechanical and electronic meters. Electro-mechanical meters measure some of the harmonics in some going down amplitude.
Click to expand...

No one is charged based on harmonics, so even if the meter can record it nothing changes with the KWH consumed.

You might as well no right off the bat that in my opinion you are a snake oil salesman.
 
iwire said:
I read your explanation and I still don't buy it. I may improve the efficiency of the wiring system but I will not reduce my bill beyond any penalties that may be eleimnated.
Click to expand...

Improving the power factor reduces losses. The saving depends on the losses and the available improvement. If you have single cable to single load with 2% of voltage drop over it, the losses are 2% divided by the PF. You can see the explanation of this formula on page 3 in my paper presented during EEMODS conference, discussing the efficiency of system.

The question is how long your cables are and how low your power factor is. For losses in transformers - the question is what is their quality (copper, aluminum) and how much they are loaded (the lower the load, the higher the losses).
 
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