Power factor

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And to further muddy the waters- the relations of excitation, shaft speed, and power output of a generator differ whether it's operated on- or off-grid. On-grid the shaft speed can't vary, and IIRC, you control the real and reactive power by varying the throttle and excitation. When off-grid, the throttle controls both the frequency and real power but the excitation controls the voltage. (Unless I have that backwards, it's been a while.)

I recall some discussion here about this a couple of years ago, but I'm too lazy at the moment to find that.

There are some also nice slides at http://www.drhenrylouie.com/uploads/34-Synchronous_Generator_Operation.pdf and a description of governors for hydro generators at https://www.usbr.gov/power/data/fist/fist2_3/vol2-3.pdf.

But I digress D:
 
mbrooke said:
What determines low power factor or more precisely why some loads have varying power factor compared to others? Say I had an inductor. The current lags by 90*, and no other load. My power factor would be zero, correct? But if I add a resistor in parellel drawing equal current my power factor would be 50%?
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I think it would be 0.707.
 
mbrooke said:
Can you go further in a generator producing leading power factor?
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Besoeker said:
The 0.8 PF tells you the maximum power in kW. (kW/kVA is PF).
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The PF at which a generator may operate safely with maximum power depends on its point of operation on its capability curve and its prime mover output limit. It is not possible to operate a conventional generator in leading PF mode.
 
mbrooke,

note that harmonics (distortion) also moves the power factor away from unity. If you consider the C & L components being on the y-axis and the real on the x-axis, distortion is on the z-axis and also contributes to the overall length of the kVA vector. In any case, power factor is always kW/kVA whether you consider the 2-dimensional case or the complete 3-dimensional case.
 
mbrooke said:
The current lags by 90*, and no other load. My power factor would be zero, correct?
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Correct.
mbrooke said:
if I add a resistor in parellel drawing equal current my power factor would be 50%? And if I add a capacitor of equal current (in parallel) my power factor is 100%? Does this accurately represent what goes on in most loads being a mixture of capacitance, inductance and resistance hence VA?
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Use phasor diagram in each case to find out PF clearly for yourself.
 
mivey said:
mbrooke,
In any case, power factor is always kW/kVA whether you consider the 2-dimensional case or the complete 3-dimensional case.
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It is one thing to say power factor is always kW/kVA. It is quite another matter to compute it, when distortion is present. Any universally acceptable method to compute distortion PF?
 
Sahib said:
It is one thing to say power factor is always kW/kVA. It is quite another matter to compute it, when distortion is present. Any universally acceptable method to compute distortion PF?
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kVA = sqrt(kW^2 + kvar_displacement^2 + kvar_distortion^2)

so

pf = kW / kVA = kW / sqrt(kW^2 + kvar_displacement^2 + kvar_distortion^2)
 
mivey said:
kVA = sqrt(kW^2 + kvar_displacement^2 + kvar_distortion^2)

so

pf = kW / kVA = kW / sqrt(kW^2 + kvar_displacement^2 + kvar_distortion^2)
Click to expand...

Thanks. How to calculate kvar_displacement^2 and kvar_distortion^2' from a given KVA^2 and KW^2,? A numerical example please.
 
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zbang said:
And to further muddy the waters- the relations of excitation, shaft speed, and power output of a generator differ whether it's operated on- or off-grid. On-grid the shaft speed can't vary, and IIRC, you control the real and reactive power by varying the throttle and excitation. When off-grid, the throttle controls both the frequency and real power but the excitation controls the voltage. (Unless I have that backwards, it's been a while.)

I recall some discussion here about this a couple of years ago, but I'm too lazy at the moment to find that.

There are some also nice slides at http://www.drhenrylouie.com/uploads/34-Synchronous_Generator_Operation.pdf and a description of governors for hydro generators at https://www.usbr.gov/power/data/fist/fist2_3/vol2-3.pdf.

But I digress D:
Click to expand...

On Grid shaft speed controls KW and excitation controls KVAR ( I think), but a stand alone unit has me really confused.
 
Sahib said:
Thanks. How to calculate kvar_displacement^2 and kvar_distortion^2' from a given KVA^2 and KW^2,? A numerical example please.
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You will need total harmonic distortion (THD) info as well.

If you have THDv and THDi then you have this relationship between the fundamental frequency values and true values:

V1rms = Vrms / sqrt(1 = (THDv /100)^2)
I1rms = Irms / sqrt(1 = (THDi /100)^2)

With true rms and distortion info you can get the power factor relationships in detail. However, we can usually make some simplifying assumptions.

We know that pf_disp = P1avg / (V1rms * I1rms). In most run-of-the-mill cases we know Pavg ~ P1avg. Also in most run-of-the-mill cases the THDv is relatively small and we can just assume most distortion is captured by the THDi value (THD ~ THDi) and Vrms ~ V1rms. Then we have:

pf = pf_disp * pf_dist = pf_disp / sqrt(1 + (THDi /100)^2)

I'll leave the numerical example up to you.
 
mivey said:
Plot one unit on the x-axis and one unit on the y-axis. The resulting vector is sqrt(2) long at a 45 degree angle. Cos(45d)= 0.7071
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Alternately:

1 W and 1 var.

VA = sqrt(1^2 + 1^2) = sqrt(2)

pf = W / VA = 1 / sqrt(2) = 0.7071
 
Just for fun, take a look at a 'synchronous condenser'. This is essentially a synchronous motor without an output shaft, operated with excitation to look like a capacitor.

-Jon
 
mivey said:
Plot one unit on the x-axis and one unit on the y-axis. The resulting vector is sqrt(2) long at a 45 degree angle. Cos(45d)= 0.7071
Click to expand...

mivey said:
Alternately:

1 W and 1 var.

VA = sqrt(1^2 + 1^2) = sqrt(2)

pf = W / VA = 1 / sqrt(2) = 0.7071
Click to expand...

Thank you :)

So if an inductor by itself has a zero power factor, but a resistor of equal current gives 0.7071? Just want to make sure I understand this.
 
mbrooke said:
Thank you :)

So if an inductor by itself has a zero power factor, but a resistor of equal current gives 0.7071? Just want to make sure I understand this.
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Possibly easier if you draw out the vectors.
You have two components, resistive (R) and reactive (L)

Draw a triangle, resistance on the horizonltal, reactance on the vertical.
That gives you the two components. The resistance gives you the kW, the hypotenuse gives the kVA. That's where the sqr(t2) comes from.
 
Besoeker said:
Possibly easier if you draw out the vectors.
You have two components, resistive (R) and reactive (L)

Draw a triangle, resistance on the horizonltal, reactance on the vertical.
That gives you the two components. The resistance gives you the kW, the hypotenuse gives the kVA. That's where the sqr(t2) comes from.
Click to expand...

a squared + b squared= c squared?
 
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