electronic ballasts, THD, and non-linear loads

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GoldDigger said:
20% 5th and still a PF of .95+? How does it do that?
When I said "ideal" I was including your BIG choke.

Tapatalk!
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Pasting a table may not work....

540 Vdc
100 Adc
54 kW
81.6 Aac
400 Vac
56.6 kVA
0.955 pf

OK. Narrative version.
Take 400Vac and rectify it with a three phase bridge.
You get 540Vdc. (400*1.35)

6pConversionderivation_zpse8714fc1.jpg

Apologies - not best resolution. I used equation editor in Word to do it and then stuck it in photobucket. It is what it is.

Load it to 100Adc. So you have 54kW. (100*540/1000)

The input current is a 120deg rectangular pulse in each half cycle thus is 100*sqrt(2/3) or 81.6A

Input kVA is sqrt(3)*VL*IL/1000 = 56.6kVA
Defining pf as kW/kVA you get 54/56.6
which is >0.95

This is for level DC. That big choke again.

Clear as mud?
 
Besoeker said:
Pasting a table may not work....540 Vdc100 Adc54 kW81.6 Aac400 Vac56.6 kVA0.955 pfOK. Narrative version.Take 400Vac and rectify it with a three phase bridge.You get 540Vdc. (400*1.35)View attachment 10266Apologies - not best resolution. I used equation editor in Word to do it and then stuck it in photobucket. It is what it is. Load it to 100Adc. So you have 54kW. (100*540/1000)The input current is a 120deg rectangular pulse in each half cycle thus is 100*sqrt(2/3) or 81.6AInput kVA is sqrt(3)*VL*IL/1000 = 56.6kVADefining pf as kW/kVA you get 54/56.6which is >0.95This is for level DC. That big choke again.Clear as mud?
Click to expand...
The supply voltage VL is distorted by nonsinusoidal current draw of the rectier. So your formula for KVA is invalid.
 
Sahib said:
Then what is the value of VL given ?t is distorted by rectifier current?
Click to expand...
If it the RMS value............it's the RMS value.

Now, if you have nothing constructive to add might I suggest that you add nothing?
Otherwise this thread will go the same way as many of the others where you have "contributed".
 
Besoeker said:
If it the RMS value............it's the RMS value.Now, if you have nothing constructive to add might I suggest that you add nothing?Otherwise this thread will go the same way as many of the others where you have "contributed".
Click to expand...
RMS value of a sinusoidal quantity may be different from its nonsinusoidal value.Just acknowledge that.
 
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What, pray tell, is the nonsinusoidal value of a sinusoidal quantity? You refer to that above....
Also, the inclusion of voltage drop into the calculation may provide better (although not necessarily justified) real world accuracy but does not have any effect on the theoretical analysis.
What you appear to be saying is that the representation of the current as a superposition of harmonic terms is not valid because the applied voltage waveform is not sinusoidal.
That objection fails because VD appears as a *linear* term!

Tapatalk!
 
GoldDigger said:
What, pray tell, is the nonsinusoidal value of a sinusoidal quantity? You refer to that above....Also, the inclusion of voltage drop into the calculation may provide better (although not necessarily justified) real world accuracy but does not have any effect on the theoretical analysis.What you appear to be saying is that the representation of the current as a superposition of harmonic terms is not valid because the applied voltage waveform is not sinusoidal.That objection fails because VD appears as a *linear* term!Tapatalk!
Click to expand...
Suppose 100v is RMS value of purely sinusoidal voltage, then if some harmonics are added tn it, itbecomes nons?nusoidal and its RMS value is different say 100.6v.Same way harmonic voltage drops superimposed on VL make it nons?nusoidal.
 
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140512-1735 EDT

electrofelon:

With reference to your post identified by #19.

Two 8' Slimlines I measured, both magnetic ballast, gave the following results for power factor:
1958 unit 0.88
1962 unit 0.94 .

A cheap hardware store T8 4' two bulb unit with electronic ballast about 2012:
PF = 0.58 .

This sample of a cheap unit does not mean a new high quality ballast could not have a PF in the high .90s. Sylvania at http://www.lithonia.com/es8/pdfs/osram high eff qhe brochure.pdf specifies >0.98 PF and < 10% THD, lower right sidebar.

I believe you will need to instrument and run tests on units that you plan to use.

.
 
gar said:
140512-1735 EDT

electrofelon:

With reference to your post identified by #19.

Two 8' Slimlines I measured, both magnetic ballast, gave the following results for power factor:
1958 unit 0.88
1962 unit 0.94 .

A cheap hardware store T8 4' two bulb unit with electronic ballast about 2012:
PF = 0.58 .

This sample of a cheap unit does not mean a new high quality ballast could not have a PF in the high .90s. Sylvania at http://www.lithonia.com/es8/pdfs/osram high eff qhe brochure.pdf specifies >0.98 PF and < 10% THD, lower right sidebar.

I believe you will need to instrument and run tests on units that you plan to use.

.
Click to expand...

Thanks good info. It would be nice to get some more comments on my question/survey. This is a very common situation and I am surprised no one wants to throw down what they would do. Ok, how about this: Feel free to reply with "SWIM" (someone who isn't me), "My brother in law would......", or "I usually see.....". This will clear you of any embarassment from having to admit you dont anylize the wave form when you put ballasts on MWBC's ;)
 
electrofelon said:
Thanks good info. It would be nice to get some more comments on my question/survey. This is a very common situation and I am surprised no one wants to throw down what they would do. Ok, how about this: Feel free to reply with "SWIM" (someone who isn't me), "My brother in law would......", or "I usually see.....". This will clear you of any embarassment from having to admit you dont anylize the wave form when you put ballasts on MWBC's ;)
Click to expand...

Here's one I did earlier - mostly a combination of various electronic loads.

Curentforonecycle01_zps666d11e0.jpg
 
gar said:
140515-1058 EDT

Besoeker:

Could you show the details of how you calculated a power factor of > 0.95 . I get more like 0.78 .

.
Click to expand...

It's in post #21.
It's plain rectified three phase voltage and I based it on 400Vac, that being the standard 3-phase LV supply here in UK.
The designation of the rectifier is B6U - Bridge configuratiion with 6 diodes and uncontrolled. If SCRs were used it would be B6C as below (I didn't have a B6U but you get the idea, I'm sure)

B6C01_zps10f5d1ad.jpg

The resulting rectified Vdc is 540V.
Assume a 100Adc load and that equates to 54kW.
Assume also the the 100Adc is level. A DC choke ensures that but even without it the current would be pretty level anyway on a resistive load.

6Prectifier04_zps5e3a326b.jpg

This makes the current in each half cycle of the AC input a rectangular 120 deg current pulse with an amplitude of 100A. The RMS value of this is 100*sqrt(2/3) or 81.6Aac
For the kVA we then have sqrt(3)*400*81.6 which is 56.6kVA
pf is kW/kVA so 0.955.

I hope that this explains how I got the figures.
 
Wouldn't each current pulse be 60 degrees wide, not 120?
There are 3 differently phased current pulses in each 180 degree half cycle, yes?

Tapatalk!
 
GoldDigger said:
Wouldn't each current pulse be 60 degrees wide, not 120?
There are 3 differently phased current pulses in each 180 degree half cycle, yes?

Tapatalk!
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They are 120 deg wide.

Here's a pic of three phases:

Justthreephase01_zpsbf2acfae.jpg

If you look at the red phase for example, it is more positive than the other two from 30 deg to 150 deg. So you get current in the positive half cycle for that 120 deg period to each of the other two phases in sequence. Note that there are always two diodes conducting at any one time.

It's something I've measured dozens of times so I'll see if I can find an example that I've recorded.....
 
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