Presented By N.E.M Business Solutions  
Power
factor


The Power factor is a complicated concept to grasp and is often over complicated by people trying to explain a subject they don't fully understand. So in this section of the site I will try to help the situation ! 



Explanation for the Business manager (The big picture) 
If you use electricity in your factory, dairy, mill, shop, then you have a POWER FACTOR and it effects the cost of the electricity you use. The lower the power factor the higher the cost. Power factor range from 0.0 to 1.0 (normally). You should do something about a power factor less than 0.94.......................Its costing your business money !  


The Explanation for the Technical manager (working knowledge) 
What we call "Electricity" is actually made up of three parts.
Real Power (Kw, Mw),
These 3 parts form the "Power Triangle"  


For the Engineering manager and electricians and those interested in this sort of thing  
This section is reproduced with the permission of the 

Power factor is one of those subjects about which there seems to be a great deal of confusion and many misconceptions. The culprit is the the assertion that power factor = cos(Phi), Phi being the phase shift between voltage and current. While this is true under certain ideal conditions, there are many realworld instances where it is quite incorrect. This article will give you a basic introduction into power factor, and explain why cos(Phi) is not the whole answer.  
The BasicsIf you connect a sinusoidal voltage source to a resistor, current will flow, power will be dissipated in the resistor and the resistor will heat up. The current is given by I=V/R and the power is given by P=I*V or P=V²/R. The voltage and current are the rms values.  
Figure 1 shows the waveforms for this experiment. The top blue
waveform is sinusoidal voltage. The voltage is 1V rms giving a peak
voltage of 1.414V, The red waveform is the current. It is 1A rms, 1.414A
peak. (If you are awake you will deduce that the resistor is 1Ohm). The
green waveform is the instantaneous power, i.e. the product of voltage and
current from moment to moment. For example, at the left hand vertical line
the voltage and current are both at their peaks, so the power is
1.414V * 1.414A = 2W 
Fig. 1  
At the right hand vertical line we are at the negative peaks
of voltage and current. Here the instantaneous power is
1.414V * 1.414A = 2Wi.e. the product of two negatives giving a positive. It should not take too much imagination to see that the average of the power waveform is 1W.  
Suppose we now replace the resistor with an inductor. The
current in an inductor lags exactly 90° behind the applied voltage
(at least for a perfect inductor, which we assume we have). This is
illustrated by the red current waveform in figure 2. Now look at the
instantaneous power waveform. Between the vertical lines we have a
negative voltage being multiplied by a positive current, giving a negative
power. Negative Power!? What does that mean?
What the negative power means is that during that part of the cycle, energy is actually transferred from the inductor (load) back into the voltage source. An inductor is an energy storage device. The energy stored in an inductor is ½I²L. 
Fig. 2  
If the load is a perfect inductor the negative power will exactly cancel the positive power, and we get a net power dissipation of zero. However, the voltage is still 1V and the current is still 1A, giving a product which is definitely not zero. Our supposed 1W of input is not producing 1W of heat. This is commonly called "wattless watts". The correct term is VoltAmps or VA. We say the circuit is drawing 1VA but consuming no power.  
The ratio between power and VA is the power factor.  
The pure inductor dissipates no heat, so it has a power factor of zero.  
Imagine now that we use an inductor and a resistor simultaneously, connected in parallel with each other. (this whole argument can be developed with them in series, but parallel connection is easier to visualise).  
The same voltage as before has been applied to both inductor and resistor. Hence, each one must be drawing the same current as before, with the resistor drawing 1A in phase with the supply voltage and the inductor drawing 1A at a phase lagging the voltage by 90°. The overall current (Fig 3) is the moment to moment summation of these two currents. It can be shown graphically or mathematically that the total current has a phase exactly half way between the two individual currents (i.e. lagging the voltage by 45°) and a magnitude of 1.414A (for unequal component currents the composite is the hypotenuse of a right angle triangle drawn with the lengths of the two other sides equal to the two components).  Fig. 3  
So, where does this leave us with our power factor
calculation? The input voltage is 1V and the current 1.414A, so the input
VA is 1.414VA. The power consumed is 1W, the same as in the resistoronly
case. Hence, the power factor is 1W/1.414VA = 0.707. It just so happens
that the cosine of the phase angle (Phi) between voltage and current is
also 0.707 (cos45°=0.707). This is informal proof of the generally
accepted assertion that PF=Cos(Phi). We will now show that that is not
always the case.  
It ain't necessarily so!  
Imagine now that we make a load which always draws exactly ±1A, with the same polarity as the voltage but with a magnitude independent of the voltage. (How make such a load is merely an engineering problem.) With this load the voltage is 1V RMS, and the current is 1A RMS, giving an input of 1VA. The average power in the green waveform is 0.9W (1.414*2/pi). And yet, there is no negative power (the load cannot store power and return it to the supply like an inductor can) and the load current is seemingly in phase with the voltage. The power factor of this circuit is 0.9, yet there is no phase shift. What gives?!  Fig. 4  
The answer lies in the shape of the current waveform, which in this instance is a square wave. A square wave can be shown to be the summation of a whole number of waveforms at different frequencies, each frequency being a multiple of the fundamental. (If this is unfamiliar to you, you can download a spreadsheet which illustrates the generation of a squarewave as the summation of many sinewaves. This ZIP file unzips to a Lotus WK1 file, which can be imported into virtually every known DOS and Windows spreadsheet. HARMONIC.ZIP, 14K.) It can be shown that in a squarewave with amplitude 1V (2V_{pp}) the amplitude of the fundamental frequency is 0.9V. Hence where harmonics are present in the load current (but not in the voltage) the harmonic currents contribute nothing to the power in the load but do reduce the power factor. Is such a situation likely to occur in the real world? Yes it is. In fact, the only loads on the mains which are not likely to introduce harmonic currents are purely resistive loads such as heaters and incandescent lamps.  
 
 
Power Factor InstrumentsAn important consequence of the realization that cos(Phi) and PF are not synonymous is in instrumentation, i.e. how we measure power factor. In the power industry power factor meters are quite common. The misconception about cos(Phi) goes so far that many meters are actually labelled "cos(Phi)", perhaps because it looks more technical than Power Factor. Many very expensive instruments have absolutely nothing in their specifications to define exactly what it is they measure. Imagine: An instrument could be designed which merely measures the phase angle using zero crossing detectors. This could be labelled cos(Phi) and sold for a lot of money. We don't know if this has ever been done, but would not be surprised to discover that it had. Do not trust an instrument if you don't know exactly what it measures One excellent instrument is the Fluke Model 40 (we are not in any way associated with Fluke, but have used the Model 40). This instrument is for use on mains circuits at 110V and 240V 50/60Hz. It has a voltage input and a current clamp. Internally it uses a DSP chip and carefully synchronized dual A to D converters to analyse the voltage and current waveforms. It displays harmonics of current and voltage up to 31st harmonic. Fluke define two separate power factor measurements: Displacement Power Factor (DPF) and plain Power Factor (PF). DPF is the cosine of the phase angle between the fundamental frequency components of voltage and current. The Model 40 actually mathematically extracts the fundamental waveforms from the inputs and then measures the phase angle between them. In the Model 40, PF is the true ratio between VA and power with all harmonics factored in.  
We are aware that other instruments exist which are equally correct in their operation, and will list them if brought to our attention. One such instrument is a setup we made ourselves with a PC plugin A to D card and some custom designed input circuitry. Our results correlate very well to the Fluke Model 40, giving a good feeling of confidence in both.  
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