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GSXR600

Senior Member
If i had 20, 1 P, 480 volt, 250 watt matal halide fixtures. That would be

20 * 25 = 5000 watts

5000 * 125% = 6250 watts

6250 / 1000 = 6.25 KVA

6250 / 480 = 13.03 amps

Is this correct way of doing this?
 

GSXR600

Senior Member
What i have is a 208 volt service and i am installing 20-480 volt, 250 watt metal halide fixture approximately 1300' away from the panel so i am boosting the voltage to save on wire am i looking at this correctly?
 

skeshesh

Senior Member
Location
Los Angeles, Ca
kVA*pf = kW

Look around. There's tons of decent stuff on power factor, power triangle, power quality, etc. on the internet.

Charlie has a good post on that here:

http://forums.mikeholt.com/showthread.php?t=120696

I always liked the GSXs. Don't kill yourself.

Oh wow Charlie, good job indeed... good job making it so I won't even be able to sip on some beer without thinking about electricity and this forum! jk:D
 

charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Retired Electrical Engineer
6250 / 1000 = 6.25 KVA
No. If you divide 6250 watts by 1000, you get 6.25 kilowatts (KW), not 6.25 KVA. In order to convert KW to KVA, you need to divide by the power factor. Presuming (and I am not certain about this) that metal halide fixtures are purely resistive in nature, the power factor will be 1.0. In that case, 6.25 KW will be equivalent to 6.25 KVA.
6250 / 480 = 13.03 amps
Definitely not! You are dealing with three phase system. There will always be a factor of the square root of three (about 1.732) somewhere in the equation. In this case, 6250 divided by 480 and then divided again by 1.732 gives you about 7.5 amps. That will be the current on your 480 volt portion of the run.
 

charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Retired Electrical Engineer
Oh wow Charlie, good job indeed... good job making it so I won't even be able to sip on some beer without thinking about electricity and this forum! jk:D
I'll take credit for coming up with "Charlie's Rule," but not for the "beer and foam analogy" for power factor. Fortunately, my drink is bourbon, not beer. ;)

 

MD88

Member
Hi, I was actually wondering about this power factor stuff too. I know how the calcs work and all that, but I still can't find a good explanation of what is physically happening!

Is this what happens? The windings on the motor set up some sort of resonant circuit with the windings on the generator, so you have this current flying between the two and just burning up power due to line losses? So on a current/voltage vs. time chart, when the current and voltages are opposite polarities, it's the resonant circuit b/w the windings pushing more current "backwards" than is moving "forwards" resulting in a net difference in polarity?

Also, as far as a wattmeter measures.. it measures volts and amps over a cycle and gives you the peak I(t)xV(t) over the cycle? So if you measured volts and amps separately with a voltmeter and an ammeter, and multiplied then you would get the full VA?

Thanks!
 

Besoeker

Senior Member
Location
UK
Definitely not! You are dealing with three phase system. There will always be a factor of the square root of three (about 1.732) somewhere in the equation. In this case, 6250 divided by 480 and then divided again by 1.732 gives you about 7.5 amps. That will be the current on your 480 volt portion of the run.
Well, isn't that true only if the loads are equally balanced on the three phases? Bit difficult to arrange that with 20 units.....;)
And maybe they are all between say, L1 and L2 in which case the sqrt(3) would not be applicable.
 

charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Retired Electrical Engineer
Well, isn't that true only if the loads are equally balanced on the three phases? Bit difficult to arrange that with 20 units.....;)
We must do our best to balance the loads. Put 7 on A-B, 7 on A-C, and 6 on B-C, and I would call that good.

And maybe they are all between say, L1 and L2 in which case the sqrt(3) would not be applicable.
And in which case we will have done our worst, in our duty to balance the loads.:roll:


 

skeshesh

Senior Member
Location
Los Angeles, Ca

Besoeker

Senior Member
Location
UK
Hi, I was actually wondering about this power factor stuff too. I know how the calcs work and all that, but I still can't find a good explanation of what is physically happening!
Maybe you need to think about what happens in real time.
When you start your car, the battery first provides voltage and current in the same direction. It provides output power to crank the engine. When the engine fires up, the alternator kicks in and the current flow (but not the battery voltage) is reversed to recharge the battery. The power flow also reverses. +Vx-I = -P Negative power with the power going back to the source if you see what I mean.

Alternating current, ac, as you know, is a sine wave. It alternates between positive and negative.
For a power factor of 1.0, both the current and voltage go positive and negative at exactly the same time. Power always flows from the supply to the load. Like the battery cranking the engine.

When the PF isn't 1.00 the current and voltage don't go positive and negative at the same time. There are periods where the voltage is positive and the current negative. The power flow is reversed and goes from load to supply. Like the alternator charging the battery.
Heres a picture of around 0.7 pf.

PF07.jpg


Mostly the power is positive but sometimes it is negative.
The current lags behind the voltage.
This is what happens in an inductive circuit - typically motors but nearly anything that has a wound coil.
When you apply voltage to that, it takes time for the current to rise. It also takes time for the current to fall. That's why the currents lags the voltage.
It's a bit like putting effort into getting a flywheel moving. It takes time to accelerate it. And time to stop it.
 

Besoeker

Senior Member
Location
UK
We must do our best to balance the loads. Put 7 on A-B, 7 on A-C, and 6 on B-C, and I would call that good.

Quite so. But it does mean that you can't simply apply the sqrt(3) which would require balanced loads which we can't have in this case.
You really need to calculate the line currents to arrive at the actual currents.
FWIW they are 7.89A, 7.34A, and 7.34A allowing for the 125%.
Not a big difference but there is an important point here.
Suppose we had five 1000W units to disperse around the three phases. That would be the same total power.
The highest current would then be 20% more than your 6250/(480*sqrt(3)) calculation yields.
As a professional engineer, I'm sure you understand my point and I don't want to labour it.
 
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charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Retired Electrical Engineer
I understand, and I agree that the math is not as easy when the loads are not perfectly balanced. But for "practical purposes," so long as a reasonable effort is made to balance the loads, calculating the current by taking VA/(480*1.732), declaring that value to be the same on all phases, and using that value as the basis for selecting conductor sizes, will not result in a design that risks failure due to an overload on one of the phases.
 

Besoeker

Senior Member
Location
UK
I understand, and I agree that the math is not as easy when the loads are not perfectly balanced.
Yes, not as easy but not but, as you know, not altogether complicated. Trignometry rather than advanced mathematics.

But for "practical purposes," so long as a reasonable effort is made to balance the loads, calculating the current by taking VA/(480*1.732), declaring that value to be the same on all phases, and using that value as the basis for selecting conductor sizes, will not result in a design that risks failure due to an overload on one of the phases.
That's not an assumption I'd want to risk.
 
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