# power factor measurement for single phase supply at any power point ?

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Hi, is it possible to measure power factor on the single phase supply at any single power point ? I need to make a simple PF electronic circuit with capacitor to save energy on my house by 24 hours without worried about over compensated which may cost my electricity bill more. I use less energy at home during daytime and more on the night time but using single capacitor would sometime over compensated during day time since it is not regulated automatically. My house is 240V, single phase with 30A suppy, during day time using 6 A and nigth time using 15A. How many microfarad needed ???
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Hello kongkc

Why do you want to power factor correct your house?? In all house situations that I have come across, the power is charged for KWHrs only. Power factor correction will NOT reduce the KWHrs used and therefore will not produce a rduction in your electricity bill.

Power factor correction will only be of value to you if you pay a KVA or power factor penalty. These are usually only applied to industrial consumers.

You must measure the power factor at the same place as the metering. You can not measure the power factor at a single phase outlet. Once you know the power factor, and the the connected load, and the basis of the power factor penalty, you can then determine the best strategy for power factor correction.

Best regards,

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I am supprised since I have bought an energy saving device and install at our house and it is just consist of capacitor and surge protector only. Current draw from the networks is actually lower by 10-20%. Is it consider saving ? The charges based on kwhrs.

I am confuse whether this device is cheating ?

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Hello kongkc

If you have a power factor of less than unity, then adding capacitance to the supply will reduce the current drawn from the supply, but will not reduce the KW. The current you measure consists of two parts, a resistive current that creates KW and a reactive part that does nothing. If you power factor is close to unity, then the resitive current is the highest. The total current drawn, is the square root of the sum of the squares of the two currents. A poowr power factor is due to inductive current from motors, transformers and discharge lighting (flourescent lighting). Adding capacitance effectively neutralises the inductive current, so the sum of the squares reduces, but the resistive current (making KW) stays the same.

Power factor correction can reduce current draw, reduce losses in the electrical supply, but does not slow your meter down.

Best regards,

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• 4 months later...

Hi Mark, I have actually save quite a lot of energy bill from using the capacitor correction device for my house. I been told that our KWH meter is actually a disk type and measure based on current. Therefore, it saves me quite a lot of energy.

Now, My factory is having a bulk capacitor correction bank. Could I save more energy by installing individual capacitor at the induction motor by increase the PF and reduce the current drawn ? I have the central mail PF automatic bank already, will it works ?

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Hello kongkc

Well what can I say? in a regular domestic environment, you pay for KWHrs measured by a rotating disk KWHr meter that measures power, not current. In this situation, you will not save any appreciable power by the addition of power factor correction. If the metering is faulty, or measures current rather than KWHrs, then it is possible that you will get a saving, but in that situation, I would suggest that you are being over charged in the first place, and I personaly would be getting some check metering done to make sure.

In regards to your factory, the addition of static power factor correction will reduce the current drawn, and therefore will reduce the i2R losses in the cables between the switchboard and the motor. This loss should be very small. If it is significant, then your cables must be well undersized. - I would not expect to see any significant savings.

Best regards,

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• 1 year later...
Surely, a domestic power meter measures the product of the current and the voltage. Therefore, a reduction in the current drawn from the supply (by whatever means) would result in a lower measured power.
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The domestic power meter measures the KW consumed. If you have a power factor of 1.0, then yes this is the product of V x I, but if the power factor is less than 1, then it will be less.

The actual formula for KW is V x I x Cos(thi) where cos(thi) is the power factor.

If you use capacitors to improve the power factor, the current will drop, but the power factor will improve and the KW will stay the same.

Best regards,

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In a residential application, if power factor correction is installed on mass, can it reduce the amount of generation required to supply these homes?
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Yes,

If the collective power factor is low, there will be higher losses in the transmission system than needed. Corection of the power factor in this case will reduce the electrical supply losses in the cables, lines and transformers. This will not affect the KWhr consumed by the homes, but will reduce the losses in getting the KW to the homes.

Generally, dommestic installations have a faily high power factor because most of the load is heat and light, although today there is an increase in airconditioning units that run for an extended period of time causing a lagging power factor, and electronic devices suchs as TV and computer that have switchmode power supplies causing a poor harmonic power factor.

Best regards,

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• 4 weeks later...

Hi, i also install a so called energy saving device, or perhaps it is function like a voltage stabilizer...

well, i dun see my electricity bill is being reduced. but in fact, my bill increase quite a lot...

could someone explain to me ???...

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• 5 weeks later...
;a;
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• 2 weeks later...

In many cases, these "energy saving" devices are either not applied correctly, or have no scientific justification. Sometimes referred to as snake oil!!

You need to be careful with this type of equipment.

The NASA type energy saver does work in the right envirmonment, very small lightly loaded motors running for a very long period of time.

Other "energy savers" are actually power factor correction and this will not "save" energy in a domestic environment.

Best regards,

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• 2 weeks later...

This site is very informative, thanks Marke...i had a second thought of buying "energy saving device"...a very simple computation for domestic environment...current drawn of the system is inversely proportional to its power factor!

rgds

Moises

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;p; hi u may also use active power factor correction for pf.
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• 2 weeks later...
Saving kWhr is the secondary benefit by using static capacitor next to the individual load, as the current reduced from the load the energy meter also sense less current to run the disk slower. It is my opinion. Please correct.
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If you found the bill got increasing after installing a capacitor because you installed at a wrong location. You should installed at the load side but not the line side otherwise the capacitor still consumed power even u turned off the appliance at home. I am still believe installing capacitor can reduce the kWhr because the distribution loss will be reduced because of reduction in current. Of course for a domestic used, you will not see the significant result but in a factory, it is a different story, you will see reduction in both of kVA and kWhr.
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Hello toprange

A KWHr meter or the "energy meter" is sensitive to KW, not current. If you connect a capacitor only to the meter, you will measure amps, but not KW and the disk will not rotate.

The addition of a power factor correction capacitor neutralises the laging powerfactor caused by inductive loads. This reduces the current flow, but does not reduce the KW in the load. The reduced current flow will reduce the power loss in the cables upstream of the correction bank, but this should be very low anyway. If this loss is significant in a short run of cable, then the cable must be well undersized.

Adding the capaciotr will not make the disk run slower!!

Best regards,

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If the "energy meter" is snesitive to kW, Please tell me the principle of disk turning to calculate the kWhr.

Also, for example, after installing a capacitor bank , the current dropped from 1000A to 900A in 380V, 3 phase system, What will you expect the difference in kVA and kWhr?Please giving the detail explanation and calculation is much appreciated . Thank you.

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Hello toprange

Do a search on the internet for a detailed explanation of how a rotating disk KWHr meter works, my memory on the theory is a little scratchy.

If you have an industrial load or 100A 380 volts, and adding power factor correction drops the current, I would expect the KWHr to remain essentially the same, but the KVAR would drop from 658KVA to 592KVA.

You can calculate the KVA by multiplying the amps by the voltage by Root 3.

You can calculate the KW by multiplying the KVA by the power factor.

Adding power factor correction will improve the power factor and drop the current, but will not alter the KW of the load. If there is a significant distance between the metering and the correction, there will be a drop in the line losses in the cable and this will reduce the KW, but in most practical installations, this should be almost insignificant.

Best regards,

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Hi, mark

In your post in regarding of the PFC, you say PFC will not reduce kWh unless the copper loss is significant like the cable undersize or the lenght of cable is very long. Can you please tell me what's you estimation of the loss in real life like in a factory? 5%? 10% or 20%?

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Any one will help me to answer my previous post? Thank you.
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Hello toprange

Cables are usually selected for a worst case voltage drop of less than 5%. This would suggest that the losses will be less than 5%, depending on the powerfactor of the load.

Short run cable voltage drop would be considerably less than this.

Best regards,

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• 1 month later...

Being an HVAC tech I didn't understand the difference in what you were saying right at the start... ie: the difference between kVA and kW, and why power factor correction won't have any bearing on residential electric bills.

However, it's very clear now. (I think).

In industrial installations, when a motor wastes power generating a magnetic field, that power is not returned to the power source. (The power company). So if they supply a certain VA to the industial location, they receive less VA back. This is not hard to understand. The wasted power (especially in motors and transformers where there is a lot of heat and other waste, such as magnetic resistance) consumes power in large quantaties but doesn't do any useful work.

But in residential, that is not taken into account. (Tell me if I haven't gotten this yet). In residential -- if I understand correctly, in layman's terms -- the difference in lost VA is not measured. The power company doesn't care if your loads consume VA (possibly because there are too many residential locations and each one wastes too little, so the resources required to keep track of them all individually would be enormous, so what they do instead, probably, is just divide the overall lump residential cost of increased capacity required by the amount of residential customers and put it on their bill as a "transmission charge" or some such thing?)

However in industrial locations the difference is very noticeable, and costly to the power company, so they actually calculate the VA loss and bill for it.

It seems that they clearly know how many VA they are supplying, and how many are returning, from any given installation... they COULD do this with any given residential installation, however they just don't bother. They just bill pure usage.

Am I getting this straight?

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• 2 weeks later...
How many microfarads in 1 KVAR???, is it 729??.

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