XYLENE POWER LTD.

ELECTRICITY POWER FACTOR

By C. Rhodes

INTRODUCTION:
Power Factor is a measure of reduction in electricity system power delivery capability caused by a grid customer having a finite behind the meter reactance at the line frequency. However, if the grid customer has a behind the meter power inverter that generates voltage and/or current harmonics, the situation becomes more complex. It is necessary to measure received and transmitted power to properly determine transmission/distribution usage. This web page reviews basic electrical engineering concepts relating to power factor and its effect on power transmission capability and then shows how transmission/distribution usage can be determined from the measured value of:
[(Etb - Eta) / (Erb - Era)],
as defined in the section titled Electricity Metering. This methodology allows for arbitrary current and voltage waveforms that may be introduced by power inverters and similar devices.

MAXIMUM TRANSMITTED POWER:
The maximum instantaneous power Pmax that can be transmitted via a two wire circuit is given by:
Pmax = Vo Io
where:
Vo = maximum voltage
Io = maximum current

Apply the trigonometric identity:
(sin WT)^2 + (cos WT)^2 = 1
to get:
Pmax = Vo Io [(sin WT)^2 + (cos WT)^2]
Note that for all values of WT the instantaneous power never exceeds Pmax.

At the metering point in an AC system with no harmonics:
V = Vo sin(WT)
where:
V = instantaneous voltage
W = angular frequency
T = time

The corresponding instantaneous current I is given by:
I = Io sin(WT - B)
where B is the phase angle by which the current lags the voltage due to inductance (reactance)
Expanding this relationship using the trigonometric identity:
sin(WT - B) = sin(WT)cos(-B) + cos(WT)sin(-B)
= sin(WT)cos(B) - cos(WT)sin(B)
gives:
I = Io [sin(WT)cos(B) - cos(WT)sin(B)]

The corresponding instantaneous power P is given by:
P = V I
or
P = Vo Io [(sin(WT))^2 cos(B) - sin(WT)cos(WT)sin(B)]

From time T = 0 to time T = 2 Pi / W the net energy E that is transmitted is:
E = Integral from T = 0 to T = 2 Pi / W of P dT
or
E = Integral from T = 0 to T = 2 Pi / W of Vo Io [(sin(WT))^2 cos(B) - sin(WT)cos(WT)sin(B)]dT
= Integral from A = 0 to A = 2 Pi of (Vo Io / W) [(sin(A))^2 cos(B) - sin(A)cos(A)sin(B)]dA
or
E = (Vo Io / W) Pi cos(B)

The average power Pave transmitted between T = 0 and T = 2 Pi / W is:
Pave = E / (2 Pi / W)
= (Vo Io / W) Pi cos(B) / (2 Pi / W)
or
Pave = (Vo Io / 2) cos(B)

When cos(B) = 1, the power transmission capability of a single phase AC circuit takes its maximum value of (Vo Io / 2). The actual power transmission capability is determined by the phase angle B which determines the size of the factor cos(B). Hence:
cos(B) = power factor
This result is well known and can be found in most introductory electrical engineering textbooks.

For an individual load cos(B) can take any value in the range 0 to 1, so that the transmission capacity that must be assigned to that load is:
[(Vo Io)/ 2] = Pave / cos(B)
This power transmission capacity has units of kVA.

Recall that:
Pave = [(Erb - Era) - (Etb - Eta)] / (Tb - Ta)
where:
Ta = time at beginning of measurement interval
Tb = time at end of measurement interval
Erb - Era = energy received during the measurement interval
Etb - Eta = energy transmitted during the measurement interval

Hence the transmission capacity used to deliver the net received power is:
Pave / cos(B) = [(Erb - Era) - (Etb - Eta)] / [(Tb - Ta) cos(B)]

This expression can be rearranged to give:
Pave / cos(B) = [(Erb-Era)/(Tb-Ta)]{1 - [(Etb-Eta)/(Erb-Era)]}/ cos(B)
or
Pave / cos(B) = [-(Etb-Eta)/(Tb-Ta)]{1 - [(Erb-Era)/(Etb-Eta)]}/ cos(B)

In order to properly allocate transmission/distribution costs it is necessary to find an expression for:
{1 - [(Erb-Era)/(Etb-Eta)]}/ cos(B) in terms of only (Etb-Eta)/(Erb - Era).

DETERMINATION OF (Erb - Era) AND (Etb - Eta) AS FUNCTIONS OF B:
Recall that for the period from time Ta = 0 to time Tb = 2 Pi / W:
[(Erb - Era) - (Etb - Eta)] = Integral from A = 0 to A = 2 Pi of (Vo Io / W)[(sin(A))^2 cos(B) - sin(A)cos(A)sin(B)]dA
= Integral from A = 0 to A = 2 Pi of (Vo Io / W)[(sin(A) (sin(A) cos(B) - cos(A)sin(B))]dA
= Integral from A = 0 to A = A1 of (Vo Io / W)[(sin(A) (sin(A) cos(B) - cos(A)sin(B))]dA
+ Integral from A = A1 to A = Pi of (Vo Io / W)[(sin(A) (sin(A) cos(B) - cos(A)sin(B))]dA
+ Integral from A = Pi to A = Pi+A1 of (Vo Io / W)[(sin(A) (sin(A) cos(B) - cos(A)sin(B))]dA
+ Integral from A = Pi+A1 to A = 2 Pi of (Vo Io / W)[(sin(A) (sin(A) cos(B) - cos(A)sin(B))]dA
Where angle A1 is given by:
sin(A1) cos(B) - cos(A1) sin(B) = 0
or
tan(A1) = tan(B)
or
A1 = B

Consider the case of: (Erb - Era) > (Etb - Eta). Then in the above 4 line integration the 1st and 3rd lines are negative and contribute to transmitted power (Etb - Eta). The 2nd and 4th lines are positive and contribute to received power (Erb - Era).

A table of integrals gives:
Integral (sin(A))^2 dA = [(A / 2) - (sin(2A)) / 4]
and
Integral -sin(A)cos(A)dA = -(sin(A))^2 / 2

FIND (Erb - Era):
Integral from A = B to A = Pi of (Vo Io / W)[(sin(A) (sin(A) cos(B) - cos(A)sin(B))]dA
= (Vo Io / W)cos(B)[(Pi / 2) - (B / 2) + (sin(2B)) / 4]+(Vo Io / W)sin(B)[(sin(B))^2 / 2]
and
Integral from A = Pi+B to A = 2 Pi of (Vo Io / W)[(sin(A) (sin(A) cos(B) - cos(A)sin(B))]dA
=(Vo Io / W)cos(B)[(2 Pi/2) - ((B+Pi)/2) + (sin(2B))/4] + (Vo Io / W)sin(B)[(sin(Pi + B))^2 /2]
Thus:
(Erb - Era) = (Vo Io / W)cos(B)[(Pi) - (B) + (sin(2B)) / 2]+(Vo Io / W)[(sin(B))^3]

FIND (Etb - Eta):
Integral from A = 0 to A = B of (-Vo Io / W)[(sin(A) (sin(A) cos(B) - cos(A)sin(B))]dA
= (-Vo Io / W) cos(B)[(B / 2) - (sin(2B)) / 4] + (Vo Io / W)sin(B)[(sin(B))^2 / 2]
and
Integral from A = Pi to A = Pi+B of (-Vo Io / W)[(sin(A) (sin(A) cos(B) - cos(A)sin(B))]dA
= (-Vo Io / W) cos(B)[(B / 2) - (sin(2B)) / 4]+(Vo Io / W)sin(B)[((sin(B))^2 / 2]
Thus:
(Etb - Eta) = (-Vo Io / W)cos(B)[ (B) - (sin(2B)) / 2]+(Vo Io / W)[(sin(B))^3]

FIND [(Erb - Era) - (Etb - Eta)]:
[(Erb - Era) - (Etb - Eta)] = (Vo Io / W)cos(B)(Pi)
Hence:
Pave = [(Erb - Era) - (Etb - Eta)]/ (2 Pi / W)
= (Vo Io / W)cos(B)(Pi) / (2 Pi / W)
= (Vo Io / 2)cos(B)
as expected.

FIND [(Etb - Eta) / (Erb - Era)]:
[(Etb - Eta) / (Erb - Era)] = {-cos(B)[B - (sin(2B))/2]+[(sin(B))^3]}/{cos(B)[Pi - B + (sin(2B))/2]+(sin(B))^3}
= {-B cos(B) + sin(B)(cos(B))^2+(sin(B))^3}/{(Pi - B)cos(B) + sin(B)(cos(B))^2+(sin(B))^3}
= {-B cos(B) + sin(B)}/{(Pi - B)cos(B) + sin(B)}
or
[(Etb - Eta) / (Erb - Era)] = {-B + tan(B)}/{(Pi - B) + tan(B)}

FIND F(B):
F(B) = {1 - [(Etb-Eta)/(Erb-Era)]}/ cos(B)
= {1 - {-B + tan(B)}/{(Pi - B) + tan(B)}}/cos(B)
= Pi /{(Pi - B) + tan(B)}cos(B)
= Pi / {(Pi - B)cos(B) + sin(B)}

Thus for B in the range 0 < B < (Pi / 2) the functions:
[(Etb - Eta) / (Erb - Era)] = {-B + tan(B)}/{(Pi - B) + tan(B)}
and
F(B) = Pi / {(Pi - B)cos(B) + sin(B)}
are easily tabulated.

Thus we can make a lookup table which for 0 < B < (Pi/2) tabulates:
B,
[(Etb - Eta) / (Erb - Era)]= {-B + tan(B)}/{(Pi - B) + tan(B)}
and
{1 - [(Etb-Eta)/(Erb-Era)]}/ F = Pi / {(Pi - B)cos(B) + sin(B)}.

The calculation procedure is to first measure (Etb - Eta) and (Erb - Era).
Then calculate [(Etb - Eta) / (Erb - Era)] and use the lookup table to find the corresponding value of B and hence F(B) = [Pi / {(Pi - B)cos(B) + sin(B)}].

Since the input data is just [(Etb - Eta) / (Erb - Era)], this procedure will work for determining power factor for 1, 2 or 3 phases.

If (Etb - Eta) > (Erb - Era), (Erb - Era) and (Etb - Eta) interchange roles, so for:
(Etb - Eta) > (Erb - Era):
[(Erb - Era) / (Etb - Eta)] = {-B + tan(B)}/{(Pi - B) + tan(B)}

SUMMARY:
The transmission capacity in kVA used by a load customer is given by:
[(Erb - Era) /(Tb-Ta)][F(B)]
where the value of the parameter:
F(B) = [ Pi / {(Pi - B)cos(B) + sin(B)}]
is obtained from a table of:
B, [(Etb-Eta) / (Erb-Era)] = {-B + tan(B)}/{(Pi-B) + tan(B)}, F(B)=[Pi/{(Pi-B)cos(B)+sin(B)}]
using as input data the measured value of:
[(Etb - Eta) / (Erb - Era)].

Similarly,the transmission capacity in kVA used by an operating generator is given by:
[(Etb - Eta) /(Tb-Ta)][F(B)]
where the value of the parameter:
F(B)=[ Pi / {(Pi - B)cos(B) + sin(B)}]
is obtained from a table of:
B, [(Erb-Era) / (Etb-Eta)] = {-B + tan(B)}/{(Pi-B) + tan(B)}, F(B)=[Pi/{(Pi-B)cos(B)+sin(B)}]
using as input data the measured value of:
[(Erb - Era) / (Etb - Eta)].

Note that in the presence of harmonics B does not mean the phase difference between the voltage and current waveforms. Instead B is a function of:
[(Etb-Eta)/(Erb-Era)].
However, if there are no harmonics present then B equals the phase angle between the voltage and the current waveforms.

For a load customer (Etb-Eta) < (Erb-Era) and numerical tabulation gives:
TABLE OF B (RADIANS), [(Etb-Eta)/(Erb-Era)], F(B):
B = 0.00000000 [(Etb-Eta)/(Erb-Era)] = 0.00000000 F(B) = 1.00000000
B = 0.01745327 [(Etb-Eta)/(Erb-Era)] = 0.00000056 F(B) = 1.00015176
B = 0.03490655 [(Etb-Eta)/(Erb-Era)] = 0.00000451 F(B) = 1.00060503
B = 0.05235983 [(Etb-Eta)/(Erb-Era)] = 0.00001524 F(B) = 1.00135708
B = 0.06981311 [(Etb-Eta)/(Erb-Era)] = 0.00003617 F(B) = 1.00240563
B = 0.08726638 [(Etb-Eta)/(Erb-Era)] = 0.00007072 F(B) = 1.00374884
B = 0.10471967 [(Etb-Eta)/(Erb-Era)] = 0.00012236 F(B) = 1.00538523
B = 0.12217294 [(Etb-Eta)/(Erb-Era)] = 0.00019461 F(B) = 1.00731374
B = 0.13962622 [(Etb-Eta)/(Erb-Era)] = 0.00029100 F(B) = 1.00953369
B = 0.15707950 [(Etb-Eta)/(Erb-Era)] = 0.00041516 F(B) = 1.01204477
B = 0.17453278 [(Etb-Eta)/(Erb-Era)] = 0.00057073 F(B) = 1.01484704
B = 0.19198606 [(Etb-Eta)/(Erb-Era)] = 0.00076148 F(B) = 1.01794093
B = 0.20943933 [(Etb-Eta)/(Erb-Era)] = 0.00099120 F(B) = 1.02132721
B = 0.22689261 [(Etb-Eta)/(Erb-Era)] = 0.00126380 F(B) = 1.02500702
B = 0.24434589 [(Etb-Eta)/(Erb-Era)] = 0.00158327 F(B) = 1.02898183
B = 0.26179917 [(Etb-Eta)/(Erb-Era)] = 0.00195371 F(B) = 1.03325348
B = 0.27925244 [(Etb-Eta)/(Erb-Era)] = 0.00237932 F(B) = 1.03782416
B = 0.29670572 [(Etb-Eta)/(Erb-Era)] = 0.00286442 F(B) = 1.04269638
B = 0.31415900 [(Etb-Eta)/(Erb-Era)] = 0.00341345 F(B) = 1.04787302
B = 0.33161228 [(Etb-Eta)/(Erb-Era)] = 0.00403100 F(B) = 1.05335730
B = 0.34906556 [(Etb-Eta)/(Erb-Era)] = 0.00472180 F(B) = 1.05915282
B = 0.36651883 [(Etb-Eta)/(Erb-Era)] = 0.00549072 F(B) = 1.06526350
B = 0.38397211 [(Etb-Eta)/(Erb-Era)] = 0.00634282 F(B) = 1.07169365
B = 0.40142539 [(Etb-Eta)/(Erb-Era)] = 0.00728330 F(B) = 1.07844793
B = 0.41887867 [(Etb-Eta)/(Erb-Era)] = 0.00831758 F(B) = 1.08553138
B = 0.43633194 [(Etb-Eta)/(Erb-Era)] = 0.00945125 F(B) = 1.09294943
B = 0.45378522 [(Etb-Eta)/(Erb-Era)] = 0.01069012 F(B) = 1.10070788
B = 0.47123850 [(Etb-Eta)/(Erb-Era)] = 0.01204023 F(B) = 1.10881295
B = 0.48869178 [(Etb-Eta)/(Erb-Era)] = 0.01350782 F(B) = 1.11727124
B = 0.50614506 [(Etb-Eta)/(Erb-Era)] = 0.01509942 F(B) = 1.12608981
B = 0.52359833 [(Etb-Eta)/(Erb-Era)] = 0.01682178 F(B) = 1.13527612
B = 0.54105161 [(Etb-Eta)/(Erb-Era)] = 0.01868195 F(B) = 1.14483809
B = 0.55850489 [(Etb-Eta)/(Erb-Era)] = 0.02068727 F(B) = 1.15478408
B = 0.57595817 [(Etb-Eta)/(Erb-Era)] = 0.02284536 F(B) = 1.16512295
B = 0.59341144 [(Etb-Eta)/(Erb-Era)] = 0.02516420 F(B) = 1.17586403
B = 0.61086472 [(Etb-Eta)/(Erb-Era)] = 0.02765209 F(B) = 1.18701718
B = 0.62831800 [(Etb-Eta)/(Erb-Era)] = 0.03031770 F(B) = 1.19859277
B = 0.64577128 [(Etb-Eta)/(Erb-Era)] = 0.03317007 F(B) = 1.21060172
B = 0.66322456 [(Etb-Eta)/(Erb-Era)] = 0.03621866 F(B) = 1.22305554
B = 0.68067783 [(Etb-Eta)/(Erb-Era)] = 0.03947333 F(B) = 1.23596630
B = 0.69813111 [(Etb-Eta)/(Erb-Era)] = 0.04294441 F(B) = 1.24934672
B = 0.71558439 [(Etb-Eta)/(Erb-Era)] = 0.04664270 F(B) = 1.26321014
B = 0.73303767 [(Etb-Eta)/(Erb-Era)] = 0.05057948 F(B) = 1.27757061
B = 0.75049094 [(Etb-Eta)/(Erb-Era)] = 0.05476659 F(B) = 1.29244283
B = 0.76794422 [(Etb-Eta)/(Erb-Era)] = 0.05921640 F(B) = 1.30784229
B = 0.78539750 [(Etb-Eta)/(Erb-Era)] = 0.06394188 F(B) = 1.32378521
B = 0.80285078 [(Etb-Eta)/(Erb-Era)] = 0.06895662 F(B) = 1.34028864
B = 0.82030406 [(Etb-Eta)/(Erb-Era)] = 0.07427487 F(B) = 1.35737047
B = 0.83775733 [(Etb-Eta)/(Erb-Era)] = 0.07991158 F(B) = 1.37504948
B = 0.85521061 [(Etb-Eta)/(Erb-Era)] = 0.08588243 F(B) = 1.39334537
B = 0.87266389 [(Etb-Eta)/(Erb-Era)] = 0.09220387 F(B) = 1.41227882
B = 0.89011717 [(Etb-Eta)/(Erb-Era)] = 0.09889319 F(B) = 1.43187155
B = 0.90757044 [(Etb-Eta)/(Erb-Era)] = 0.10596855 F(B) = 1.45214636
B = 0.92502372 [(Etb-Eta)/(Erb-Era)] = 0.11344901 F(B) = 1.47312718
B = 0.94247700 [(Etb-Eta)/(Erb-Era)] = 0.12135464 F(B) = 1.49483913
B = 0.95993028 [(Etb-Eta)/(Erb-Era)] = 0.12970652 F(B) = 1.51730863
B = 0.97738356 [(Etb-Eta)/(Erb-Era)] = 0.13852683 F(B) = 1.54056339
B = 0.99483683 [(Etb-Eta)/(Erb-Era)] = 0.14783894 F(B) = 1.56463254
B = 1.01229011 [(Etb-Eta)/(Erb-Era)] = 0.15766742 F(B) = 1.58954671
B = 1.02974339 [(Etb-Eta)/(Erb-Era)] = 0.16803818 F(B) = 1.61533808
B = 1.04719667 [(Etb-Eta)/(Erb-Era)] = 0.17897850 F(B) = 1.64204048
B = 1.06464994 [(Etb-Eta)/(Erb-Era)] = 0.19051714 F(B) = 1.66968952
B = 1.08210322 [(Etb-Eta)/(Erb-Era)] = 0.20268445 F(B) = 1.69832264
B = 1.09955650 [(Etb-Eta)/(Erb-Era)] = 0.21551241 F(B) = 1.72797925
B = 1.11700978 [(Etb-Eta)/(Erb-Era)] = 0.22903481 F(B) = 1.75870084
B = 1.13446306 [(Etb-Eta)/(Erb-Era)] = 0.24328730 F(B) = 1.79053110
B = 1.15191633 [(Etb-Eta)/(Erb-Era)] = 0.25830757 F(B) = 1.82351608
B = 1.16936961 [(Etb-Eta)/(Erb-Era)] = 0.27413542 F(B) = 1.85770427
B = 1.18682289 [(Etb-Eta)/(Erb-Era)] = 0.29081295 F(B) = 1.89314685
B = 1.20427617 [(Etb-Eta)/(Erb-Era)] = 0.30838467 F(B) = 1.92989774
B = 1.22172944 [(Etb-Eta)/(Erb-Era)] = 0.32689770 F(B) = 1.96801389
B = 1.23918272 [(Etb-Eta)/(Erb-Era)] = 0.34640191 F(B) = 2.00755540
B = 1.25663600 [(Etb-Eta)/(Erb-Era)] = 0.36695013 F(B) = 2.04858571
B = 1.27408928 [(Etb-Eta)/(Erb-Era)] = 0.38859835 F(B) = 2.09117191
B = 1.29154256 [(Etb-Eta)/(Erb-Era)] = 0.41140592 F(B) = 2.13538486
B = 1.30899583 [(Etb-Eta)/(Erb-Era)] = 0.43543580 F(B) = 2.18129955
B = 1.32644911 [(Etb-Eta)/(Erb-Era)] = 0.46075481 F(B) = 2.22899530
B = 1.34390239 [(Etb-Eta)/(Erb-Era)] = 0.48743389 F(B) = 2.27855608
B = 1.36135567 [(Etb-Eta)/(Erb-Era)] = 0.51554841 F(B) = 2.33007084
B = 1.37880894 [(Etb-Eta)/(Erb-Era)] = 0.54517849 F(B) = 2.38363386
B = 1.39626222 [(Etb-Eta)/(Erb-Era)] = 0.57640934 F(B) = 2.43934508
B = 1.41371550 [(Etb-Eta)/(Erb-Era)] = 0.60933161 F(B) = 2.49731056
B = 1.43116878 [(Etb-Eta)/(Erb-Era)] = 0.64404185 F(B) = 2.55764286
B = 1.44862206 [(Etb-Eta)/(Erb-Era)] = 0.68064289 F(B) = 2.62046156
B = 1.46607533 [(Etb-Eta)/(Erb-Era)] = 0.71924435 F(B) = 2.68589374
B = 1.48352861 [(Etb-Eta)/(Erb-Era)] = 0.75996315 F(B) = 2.75407455
B = 1.50098189 [(Etb-Eta)/(Erb-Era)] = 0.80292408 F(B) = 2.82514781
B = 1.51843517 [(Etb-Eta)/(Erb-Era)] = 0.84826039 F(B) = 2.89926666
B = 1.53588844 [(Etb-Eta)/(Erb-Era)] = 0.89611450 F(B) = 2.97659428
B = 1.55334172 [(Etb-Eta)/(Erb-Era)] = 0.94663867 F(B) = 3.05730466
B = 1.57079633 [(Etb-Eta)/(Erb-Era)] = 1.00000000 F(B) = 3.14159265

FOR A NET GENERATOR:
(Etb-Eta) > (Erb-Era)
and the table is identical except that the numbers in the second column represent:
[(Erb-Era)/(Etb-Eta)]

This web page last updated March 21, 2009.