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Power factor and measurement

Heres a copy of a post i made to tesla-at-usa-dot-net for those of you considering
powerfactor at the moment.


>First of all the basic rules of power, the amount of power in a circuit doing
>useful work, i.e heat from an electric element, is calculated as watts.
>In a D.C. circuit this is found simply by multiplying supply voltage by the
>load current.
>
>In an A.C. circuit this is also true BUT only for the current flowing in phase
>with the supply voltage.
>
>If a perfect inductor is placed across the A.C. supply the current will lag
>behind the voltage by 90 deg. (This is going to be hard to explain without
>diags).
>If a complete cycle of A.C. is taken from 0 volts then as the voltage rises,
>a back
>E.M.F. is induced in the coil which opposes the flow of current untill the
>voltage reaches its peak. As the voltage starts to decrease again the
>magnetic flux in the
>coil begins to collapse, reducing the back E.M.F. and allowing the current
>to start
>to flow. As the voltage reaches 0 volts again the current is at its maximum,
>then the voltage starts to increase in its negative half cycle which starts
>the back E.M.F.
>opposing the flow of current again untill at the negative voltage peak the
>current
>is zero again. As the negative peak starts to return towards zero the
>collapsing
>flux allows the negative cycle of the current to start flowing, reaching
>it's peak
>as the voltage once more returns to zero.
>
>This current produces no useful work, it just sits, running 90 deg. behind the
>voltage.
>
>If this inductor is placed in parrallel with the heating element then the
>resultant current flowing in to the circuit will consist of two components.
>The current producing the useful work, in phase with the voltage and the
>current doing no work in the inductor, lagging the voltage by 90 Deg.
>
>If we measure the current in order to calculate the circuits load, then the
>meter
>will show the vectorialy added current of the in-phase and out-phase
>currents, not
>the actual power doing useful work i.e. current in phase with voltage.
>
>This measurement is expressed in Volt-Amps/1000 or kVA to distinguish it
>from the
>proportion of power doing useful work, Watts/1000 or kW.
>
>In practice, the inductor itself has resistive losses caused by the
>resistance of
>the copper windings, Copper loss and induced eddy currents in it's core,
>Iron loss.
>These losses manifest themselfs as heat, which means the current lost is in
>phase
>with the supply voltage, or "usefull work" and is measureable in watts i.e. the
>inductor gets warm.
>
>The ratio kW/kVA is called the power factor (P.F.) of the cicuit, ranging in
>value of 0 for a purely inductive load to 1 for purely resistive.
>
>As an example, if in a circuit having a P.F. of 0.5 the current is 10 Amps
>at 230 Volts, the power consumption will not be 2.3 kW, but 0.5 by 2.3, or
>1.15kW.
>Or in other words, only 5 Amps of the current is doing any work. The 10 amps
>is definitely present in the circuit and would be recorded by ammeter.
>
>This example shows that a load taking a supply at 0.5 P.F. Will take double the
>current of a supply using the same amount of power at unity P.F. although
>the indication on the electricity Companys meter will be the same. It
>follows that the supply with the low P.F. will require larger cables to
>supply it.
>
>All systems employing coils of wire , i.e. transformers, inductive ballasts
>etc.
>Will have power factors of less than 1.
>
>Capacity currents are 90 deg. In advance of the voltage and are diametrically
>opposite to purely inductive currents. When a circuit contains inductance and
>capacity currents of equal value they will netralise each other, so that the
>resultant current will be in phase with the voltage.
>
>My old college notes contain a table which will take a month of sundays try and
>tabulate here, but basicaly gives a value of capacitive kVA per kW of load for
>raising the power factor from a given value to a range of values between 0.8
>and 1. 0.8 usually being the minimum value that most power co's insist on for
>industrial users.(i could possibly type up this table if enough poeple were
>interested but may take some time).
>
>An example of the current savings possible is shown below:-
>
>The P.F. of a 10kW load is to be raised from 0.65 to 0.95
>
>Therefore, (using the table), Required reactive kVA=10x0.84=8.4
>
>(0.84 is the trigonometricaly derived value for changing the phase angle
>of the current)
>
>the resultant kVA would be 10/0.95=10.53 kVA
>the uncorrected kVA=10/0.65=15.38 kVA
>and the reduction in load=15.38-10.53=4.85 kVA
>
>or assuming a mains supply of 230v -at- 50hz 4.85/0.23=21.09 Amps
>
>Remember though that despite saving this current your house meter will still
>show the same amount of power (kW) consumed.
>
>If you are running multi-kilowatt power supplys for your tesla coils then by
>increasing the power factor of your transformers will save a lot of heavy
>cabling.
>
>The relationship between kVA and micro-farads is given as:-
>
>                   2
>         c x 2(pi)v^
>     kVA=-----------
>               9
>             10^
>
>
>
>and can be used to give the value of 632 micro-farads -at- 50hz to correct the
>above
>example.
>
>It is not always preferable to correct the P.F. to unity because if the load is
>reduced and the correction cap. stays fixed then the P.F. will become
>leading and
>the useless proportion of current increases again, although this time
>leading the
>voltage. It may be preferable to select a correction value that strikes a
>compromise
>between maximum leading and lagging P.F.s and for this reason the table is
>useful rather than using the above equation for determining a capacitance
>value to give
>unity correction.
>
>I havn't sat down and calculated the effect of varying the input power
>before the corrected load i.e. a variac in the supply so if anybody who may
>have tried this can let me know of their results i would be greatful, but
>varying
>primary circuit gap distances will effect this for a given input power.
>
>The power factor can easily be measured to a fair degree of accuracy, by
>using an ammeter and an domestic eletricity supply meter (providing its
>calibration is still fairly accurate)
>
>kVA is obtained simply by   Amps x Volts
>                            ------------
>                                1000
>
>
>kW is obtained as follows:-
>
>On the meter name plate is engraved a "meter constant" which is in either
>"revolutions per kW hour" or "full-load revs. per minute." If the latter:
>
>               Full load revs. per min x 60
>Revs. per kWh= ----------------------------
>               Normal full load of meter in kW
>
>The kW load can be measured by timing a number of revolutions made by the meter
>disk and applying the formula:-
>
>              3,600 X n
>          kW= ---------
>                t X k
>
>Where:-
>        n= number of revolutions timed
>        t= duration of test in seconds
>        k= meter constant in revs. per kWh
>    3,600= No. of seconds in one hour
>
>Incidentaly for those following the A.C./D.C. thread the A.C. transmission
>system
>can only maintain efficiancy if the P.F. can be kept close to unity. Any
current
>drawn by the consumer out of phase with the voltage, whilst doing no "useful
>work"
>will cause volts drop across any resistance in the network in phase with
itself,
>causing loss of power through heat. The power station alternators are usually
>rated at their maximum output in kW at a maximum P.F. of 0.8 After which their
>efficiancy falls off due to the windings having to provide current for the
>useless
>proportion of load.
>
>P.S. some conventions refer to the powers as "real" and "apparent"
>
>Real power is that doing useful work i.e. kW
>Apparent power is that calculated by measured current i.e. kVA
>
>*******************************************
>Ian Hopley ---->  i_hopley-at-wintermute.co.uk
>Aberdeen
>Scotland
>*******************************************
>
>
*******************************************
Ian Hopley ---->  i_hopley-at-wintermute.co.uk
Aberdeen
Scotland
*******************************************