Power factor

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mivey said:
Well I ain't from 'round there so...

Are you implying that the Scots can't all agree on the pronunciation?
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Not really although regional accents abound. An Aberdonian and a Glaswegian don't sound at all alike.
The difference between Chic Murray and say Billy Connoly is very marked.

But lest we stray off topic too far and invoke moderaters' intervention........
Distortion power factor is a real issue these days.
The waveform measured on my incoming supply:

HomeVrev00_zpsf7551ef6.jpg


The flattened top tells the story.
 
mivey said:
It is not pronouced "lock ness monster"? How is it pronounced?
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mivey said:
So proper is eck-lee-fetch-in? or eck-lee-feck-in? How does she say it?
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Besoeker said:
Loch Ness Monster of course.............:)
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mivey said:
Here we go:

How to pronounce the 15 most difficult Scottish place names
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Besoeker said:
Except that they don't have it right.
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mivey said:
The internet of truth.:D
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mivey said:
Well I ain't from 'round there so...

Are you implying that the Scots can't all agree on the pronunciation?
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Punch drunk. humourous, and somewhat silly arguments abound...:)

Y'all train drivers amuse me and provide a somewhat pleasant diversion....

Hint: Neither Scots nor Americans speak English actually.:D
 
mbrooke said:
Awesome incite, thank :):cool:

But how does on go about net power factor? Consider an HID lamp with a reactive ballast. Its essentially a resistor in series with a reactor, yet the power factor can vary all over the place.
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I'm getting in on this late and didn't read all 19 pages, but an HID arc is not purely resistive, the current waveform of the arc is highly non-linear.
 
Besoeker said:
I think you wil find that Robert Louis Balfour Stevenson made a pretty good fist of it.
Samuel Clemens too.
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Bes, it was a riddle. To continue it, such as:

What is common or not between British and Americans?

Hint 2 :For the above statement see Oscar Wilde, Bernard Shaw, and Winston Churchill.

P.S. Both authors you cite are outstanding IMHO.:thumbsup:
 
I want to keep this thread on track, politely saying :) And yes the British and Americans are different in different ways.


Anyhow, why is it that ballasts are parallel X and C circuits and not series circuits?
 
mbrooke said:
I want to keep this thread on track, politely saying :) And yes the British and Americans are different in different ways.


Anyhow, why is it that ballasts are parallel X and C circuits and not series circuits?
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As you vary X and C to get close to unity power factor the series impedance approaches zero, which is not much help in limiting the current through the lamp. :)
 
GoldDigger said:
As you vary X and C to get close to unity power factor the series impedance approaches zero, which is not much help in limiting the current through the lamp. :)
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Perhaps... If you take say a 5 milli henry inconductor and a 4uF capacitor, would these small components really behave like a solid piece of #12 when driver a 400 watt Metal Halide?
 
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GoldDigger said:
As you vary X and C to get close to unity power factor the series impedance approaches zero, which is not much help in limiting the current through the lamp. :)
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Here is what I have in mind (second picture in)


http://www.candlepowerforums.com/vb/showthread.php?260956-Fluorescent-to-LED-night-light-conversion


The ballast in this case us an capacitor and inductor in between the line. Perhaps not unit PF, but close. Yet even though its said this circuit would not provide impedance it still does somehow. It is this series circuit which puzzles me. The parallel aspect I understand from the equations posts (awesome help btw) :cool: :)
 
mbrooke said:
I found this but still can't grasp total series impedance:


http://www.allaboutcircuits.com/textbook/alternating-current/chpt-5/series-r-l-and-c/
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You can calculate the impedances as shown but you can't add them up arithmetically - as simple numbers.

If you look down the page a bit you will see calculations for ZR, ZL, and ZC.
You will note that impedances are expressed as two components.
If you take ZL, and ZC you will not that they have opposite signs, one positive, one negative. They subtract.

Do that and you end up with two components, R and X. But note the 90deg difference. You can't simply add them.
Imagine two sides of a triangle with a 90 deg angle between them. A shape like a corner if you like. Join the outer sides to get the diagonal. That diagonal represents the total impedance.

Obviously, it is longer than the two component parts. But not simply the arithmenic sum. You need to use pythagorous.
 
mbrooke said:
Here is what I have in mind (second picture in)


http://www.candlepowerforums.com/vb/showthread.php?260956-Fluorescent-to-LED-night-light-conversion


The ballast in this case us an capacitor and inductor in between the line. Perhaps not unit PF, but close. Yet even though its said this circuit would not provide impedance it still does somehow. It is this series circuit which puzzles me. The parallel aspect I understand from the equations posts (awesome help btw) :cool: :)
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I do not see an inductor anywhere in the schematic. Nor in the picture.
Using a series capacitor to limit current is common for small loads and it does not provide anywhere near unity PF.
 
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