Power factor

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Besoeker said:
The current and voltage both go through a cycle. When both are positive at the same time you get positive power. When both are negative at the same time you get positive power. That happens twice per cycle. Hence double the fundamental frequency.
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Makes sense then. So 120 volts passing through a resistor left to right produces energy and then at the zero crossing electrons reverse going right to left producing another wave of power until the next zero crossing. So wave + wave = one complete cycle of power.
 
mbrooke said:
The twice part is clicking in, but what do you mean by positive offset?
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the v and i are symmetrical about 0 or the deg/time/x axis
power is not, it's pushed up, more above than below

if v and i were in phase, 0 deg ph ang or pf 1
it would all be above
the bottom swing would rest on 0 or the x axis
as power factor goes from 1 to 0 (ang 0-90 deg, cos ang 1-0) the P wave would drop until centered on the axis like the others
balanced above and below and cancelling, no net power
for reference the phase ang in the graph is 45 deg giving a pf of 0.71

I may try to model in matlab and video it
 
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mbrooke said:
The twice part is clicking in, but what do you mean by positive offset?
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Calling it an offset is a bit misleading in my opinion.
It's simply what you get by multiplying the instantaneous Volts and Amps.
When they are both positive or both negative at the same time you get a positive value for power.
+1 * +1 is positive.
-1 * -1 is positive.

but +1 * -1 is negative.

That's how some parts of the power waveform end up being negative.
 
Ingenieur said:
the v and i are symmetrical about 0 or the deg/time/x axis
power is not, it's pushed up, more above than below

if v and i were in phase, 0 deg ph ang or pf 1
it would all be above
the bottom swing would rest on 0 or the x axis
as power factor goes from 1 to 0 (ang 0-90 deg, cos ang 1-0) the P wave would drop until centered on the axis like the others
balanced above and below and cancelling, no net power
for reference the phase ang in the graph is 45 deg giving a pf of 0.71

I may try to model in matlab and video it
Click to expand...

This I do not understand to any degree. I mean to me negative power is energy being sent back to the generator... clearly not the case here but there whole thing is throwing me off. The numbers help btw, but in this case I am even more confused.
 
Ok I think I might get it...I think. So watt is always above, but any reactive power goes partly above and partly below where they cancel out at some point... I think :?
 
y
es
there still is voltage drop V drop because there is an Z
basic Z relationships
Xl or L as f increases so does Z
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So a reactor's impedance increases as frequency goes up?



Xc or C as f increases Z goes down
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And a capacitor's impedance in the circuit goes down (more power let through) as frequency goes up?

Going to the real world, a 50Hz under ground cable thus has a lower sheath current than the same cable on 60Hz?


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In so far I have reached this conclusion. PF, or reactive and capcitive power is simply "phantom" power. Energy is taken from the source at one point in the wave and then sent back at another point. Other than the resistance of the wire itself, said "phantom" power consumes no watts. Its like pumped hyrdo storage. Watts are consumer to pump water up hill, and then latter on those same watts are sent back into the grid, only difference being this happens every quarter cycle?
 
mbrooke said:
This I do not understand to any degree. I mean to me negative power is energy being sent back to the generator... clearly not the case here
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Actually, it is the case.
The waveforms I drew were for a lagging power factor which infers that the circuit would have inductance. An inductor stores - and releases energy in an releases during a cycle. That energy, into and from the inductor is exchanged back and forth with the supply. For pure inductance* there is no net energy consumed/expended.

*For the pedants among us, in practice, an inductor is usually a wound component, usually copper or aluminium, so there will be some losses.
 
mbrooke said:
And a capacitor's impedance in the circuit goes down (more power let through) as frequency goes up?
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More current, not more power. Ignoring increased losses that is.

And that has become increasingly important in recent years but that's another topic for another day.[/QUOTE]
 
PF = 1
Power signal 100% offset, never dips below zero and at freq x 2
as PF 1-0 offset decreases until centered on zero no real power
 
In the case of balanced three-phase circuits, application of the dqo transform reduces the three AC quantities to two DC quantities.

https://en.m.wikipedia.org/wiki/Direct–quadrature–zero_transformation




 
Ingenieur said:
yep
it represents the positive power flow
it is why the Parks dq transformation is useful
negates the cancelling ac component and leaves the dc
makes the math much simpler

Total signal = DC + sin(wt + ph ang)
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It's a power waveform. If the is no direct current in the power transfer, there can be no DC offset in the power waveform. It is simply a shifted or offset waveform... no DC offset.
 
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