Son Agro Ballast Q for Electrical Engineers

[Aquattro]

I'm thinking he means your fingers are able to type again?? Good to see the coma wore off!! How ya been?? :P

[Aquattro]

Originally Posted by DJ88

Finishing the plumbing right now.

And we know you have a camera

[DJ88]

And we know you have a camera

Kamera? Huh? Don't own anything of the sort. No modern fandangled stuff like that around here.. I don't like technology... Don't you know it will kill ya..

Maybe someday soon I'll have pics.. Maybe..

[BC_Grl]

Nice to see you back, Darren.

[stephane]

Hi Darren! glad to see you

[reefburnaby]

Hi,

Apparant power (VA) = voltage * current.

True power consumption (W) = power factor * apparant power.

True power consumption is the power that hydro charges us.

With that... Volt = 120V.

Magnatek/Universal 4.3 amps...we will say PF=0.9
True power = 464W

Venture 4.55 amps 0.9 power factor
True power = 491W

Advance 5.3 amps 0.8 power factor.
True power = 508W

Motors sit around the pF of 0.6 to 0.9. Electronic ballasts have a pF of near 1.0

And Darren is correct about the industrial surcharge of 25% for a PF of 0.7 and the PF is cause by the load...not hydro. Hydro provides power factor correction for households on the telephone poles (near the large hanging garbage cans on the telephone pole) and transformer cabinets.

- Victor.

[titus]

Hello,

Hey last time I went to the IEEE dinner gathering. Some of the old timer (huh huh) EE from BC Hydro said they even have to monitor harmonics coming from large industrial and shopping mall type of customers and the customers have to provide proper correction.

Titus

[DJ88]

We could get into calculating the entire power triangle for these ballasts. Why not??? You Forgot VAR's Victor.

Why not make it easy and use cos^-1of the pf gives the phase angle. Then you can figure out all three for each ballast.

And I'd be hesitant to guess that the pf of the Magnatek/universal is .9. We can make the assumption that the VA for the ballast is 516VA(simple math as shown earlier). But beyond that. Not a clue with the data provided. For all we know it is .75. And you know what happens when we make assumptions. You make an A$$ out of U and ME

But once we take a close look on the PDF file for the you can figure it out quite readily. It is given to you. But hidden. kinda.

Lo and behold the wattage consumed is given. And for that aren't familiar with power in the real world there are three components. Watts(which we all know and love to hate), VA(Volt Amps) and VAR(Volt Amp Reactive). These three work together in a interesting way. All of you should remember what a right triangle is from high school right? Getting chills yet? NO? You will.

Ok so we put Watts along the x axis VA turns out to be the Hypotenuse and VAR's are the y axis(positive or negative depending on if the load is inductive or capacative). With that said we can find out the pf of that ballast since we know that the VA is 516 due to the fact that we are making the assumption that out voltage is 120VAC and the rated amperage of the ballast is 4.3amps. Multiply the two together and we get... 516VA. From the data sheet we can find out that the Watts consumed by the ballast is 490. Now's where some simple math comes in. Take your VA and divide it into your wattage. 490W/516VA=0.949612403.
So we have a pf of 0.9496 or say 0.95. A little better than the .9 Victor assumed.. Luckily this time it went to our favor. Or we could start doing some work with sines and cosines and figure out the numbers that way.

Easy huh?

Titus,

Have you taken a look at tha harmonics that are on a regular power grid. Too cool. What a mess.

[StirCrazy]

:shock: Hmmm... how can you tell Darren is back in school

good to see you have some time free now Darren.. whare are the pics??

Steve

[titus]

Hello,

Titus,

Have you taken a look at tha harmonics that are on a regular power grid. Too cool. What a mess.

Nope. Didn't even know they were an issue until the issue was brought up at the dinner.

Titus