Power Factor

Dont look the same!

Maybe the input is different.

Back to you!

Because your formulas are wrong. Time (hours) is not used in calculating power factor, it's based on instantaneous values.

I agree with RAM. PF = kw/kva . Where did you find the formula you used?

If you get rid of all the H's, how are the results sometimes different?

Although the dimensions are the same (time/time), the periods (15 min. vs 30 min.) or the units (24 hrs. vs. 1 day) could be different. Anyway, when was the last time you saw a kVARH meter?

Actually, they do exist, or are a separate element of an electronic kWH meter. They are usually used in power generation plants where one of the things they can generate is kVAR for system voltage support. The generating plant needs to have a way to quantify and record the kVAR they generate for the purchasers (usually the transmission utilities and energy service providers).

kW = kilo Watt (Watt by definition is "real" power only)
kVA = kilo Volt*Amp (both a "real" power and an imaginary or "reactive" power component may be present)
kVAR = kilo Volt*Amp Reactive (only the reactive power component)

The addition of time only changes the internal units from power to energy. Since power factor is dimensionless and all the added time units will cancel out, the end result will be the same using either equation.

The equation you posted is basically correct. For power factor calculation, a more common usage would be the power equation (no "hours"):

PF = kW/kVA = kW/[sqrt{kW² + kVAR²}]

If you are getting variable results, maybe you are not using the equation correctly?

ref.
http://en.wikipedia.org/wiki/Power_factor#Definition_and_calculation
http://en.wikipedia.org/wiki/Phasor
http://en.wikipedia.org/wiki/Volt-ampere_reactive

When you use an energy meter which has time factor also built in you do get KWh and KVAh or KVARh. Your formulas are right. I too learnt that what they teach in college is incomplete. Pl take a true rms voltmeter and a true rms ammeter, multiply and see whether you get KVA and KW and now you use these get KVA. Power factor consists of two components (not taught in college) = Displacement component * distortion component.

Displacement component = cos (Θ) where Θ = angular shift between voltage and current of the FUNDAMENTAL COMPONENTS. Assuming voltage - from supply is not distorted, the curernt due to load can be distorted. So distortion component = rms of fundamental current / (√ (sum of squares of all harmonic currents). Hence your second formula assumes KVA = √(KW^2+KVAr^2) - which is not true when the waveforms are distorted (as it takes care of fundamental components only), which is true in real life (may not be in a college lab, in controlled conditions)

xyz, you get a GA because u opened up new thinking on pf and harmonics.

Are you saying kva (n) = kw (n) + kvAR (n)

here n= the harmonic number

and then ∑kva = ∑kw + Σkvar

or

kva (total) = √ [(∑kw)^2 + (Σkvar)^2]

or something similar?

The numerators are the same so let's look at the denominator. To make things less messy, let z=KVA, x=kw and y=KVAR, in other words, z = total power, x = real power, and y = reactive power.

z=sqrt(x^2+y^2) or z^2=x^2+y^2

KVAH=sum of z over time

kwh=sum of x over time

kVARH=sum of y over time

KVAH^2 = kwh^2 + kVARH^2 will be true if the power factor does not change during the time interval.

Here is an example using x,y, and z with unchanging power factor:

x=3, y=4, z=5

x=6, y=8, z=10

sum(x)=9, sum(y)=12, sum(z)=15

15^2 = 9^2 + 12^2

225=81+144

Here is an example with changing power factor:

x=3, y=4, z=5

x=5, y=12, z=13

sum(x)=8, sum(y)=16, sum(z)=18

324 not equal to 64+256

Very well demonstrated. GA

For the OP: You get an average P.F. over the time span when using the H terms.

In the examples: 1) P.F. = 0.6 instantaneous

2) P.F. = 0.38 Instantaneous

When Adding them (as accumulation over a period), then P.F. = 0.44

Must use the instantaneous values.