75kva 208 wye load calculation "Balancing"
- Thread starter Scali
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Carultch
Senior Member
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- Massachusetts
In general, total KVA you can use in a balanced manner, is the total KVA of the transformer, no matter which voltage you apply to the connected loads. The sqrt(3) rule applies to keeping track of the relationship between phase-to-phase voltage, KVA, and amperes for a balanced 3-phase system. KVA = I*V*sqrt(3). I = per phase current assumed uniformly balanced, V=voltage phase to phase.
If you balance the lamp circuits uniformly on this system, you'll get 16 lamps per phase. Or per pair of phases, if they were individually single phase 208V circuits.
Assuming these are 120V circuits, you would connect phase-to-neutral. 16 qty 1kVA lamps = 133A at 120V. Which means you have 133 amps on each phase, which is 133 amps that you count for the full system.
If the circuits require 208V single phase circuits, then you would wire 16 across phases A&B, 16 across phases B&C, and 16 across phases C&A. The current of 16 qty 1kVA lamps at 208V = 76.9A. The equivalent "per phase" current doesn't add up directly when sourced from phase-to-phase loads. Instead, there is a square root formula to combine phase-to-phase currents into a per phase current for the particular phase:
Ia = sqrt(Iab^2 + Ica^2 + Iab*Ica)
Observe that each term in the equation for Ia contains "a" in the subscript. Square each contributing current to phase A, and then multiply a mixed term. Add up the three terms.
And continue this formula for the remaining three phases.
Ib = sqrt(Iab^2 + Ibc^2 + Iab*Ibc)
Ic = sqrt(Ica^2 + Ibc^2 + Ica*Ibc)
sqrt(76.9^2 + 76.9^2 + 76.9*76.9) = 133A, which is just as we had calculated before.
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- Location
- Massachusetts
75kva = 75,000 watts. You could light up seventy five, 1,000 watt lamps.
each light with ballast ~ 1.1 kva ~ 53 kva < 75 kva good to go
in essence you have 3 25 kva single ph xfmrs
48 lamps / 3 phases = 16 lamps per phase
put 16 lamps on each ph
each ph 16 x 1.1 ~ 18 kva < 25 kva good to go
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- Location
- Massachusetts
Is there a reason to add confusion to this thread?
More to the point, the fixture nameplate voltage times ampere values must be used for the calculation.
You find that confusing?
that was pretty much his question
he calc'ed the xfmr rated i on a 3 phase basis and the load on a single phase
yes, the 1.732 applies but I believe he thought the 230 A was multplied not divided
208V means your measuring the line-to-line voltage. Your line-to-neutral voltage will be 120V (208/1.732 = 120V)
On your problem, you are going to use 1000W lamps, that would mean a total actual wattage per fixture of 1,100 watts!
If you intend to wire your lamps line-to neutral, 4-wires (3-lines plus a neutral), then you will load 16 of your 48 lamps on each line:
Line amps = (16 x 1,100)/120 = 146.67 amps
That's the simplest way of answering your question, IMO.
Where did OP specify this value? I'm thinking it must be a "flavor of the month"...!!!
most modern energy efficient ballast have a pf of 0.9
using the 1000 W lamp rating is not accurate, but probably moot
90% https://www.1000bulbs.com/product/71061/PLUSRITE-7277.html
some are as low as 60%, so VA would be <1600, substantial vs 1000
https://www.1000bulbs.com/pdf/Advance-HID-brochure.pdf
page 8
some normal are as low as 50%
48 x 1000 / (208 x 1.732) = 133 A
48 x 1000/0.5 / (208 x 1.732) = 266 A
xfrm i rated is 208
the point is the lamp va must be considered, not lamp wattage
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All I'm trying to point out is that the calculation must use actual nameplate values... not speculated, uncertain values from this forum.
and all I was doing was using a typical representative value to illustrate his core question:
when he calculates xfmr I rated is it 208 (based on 3 phase)
and load is 230 (he did not consider the 3 phase component)
so yes, 1.732 is a factor
But you did not state you were using an arbitrary, estimated, or otherwise assumed value. Bob (iwire) brought it up, and you asked... it's "stuff" like this that adds confusion to the thread... and then after you use 1.1kVA, another poster uses it with no basis in fact, and that cascades the confusion.
the obvious does not need stated
anymore than the person (you) who said 1000 x 48 < 75000 is good to go
which is dead wrong since the ballast is not even considered
if he has a 50% ballast the power is 96000>75000 no go
you have muddied this thread far more than I by making it personal and focused on me
Whatr is obvious to you doesn't necessarily make it obvious to others. If the OP'er has to ask whether or not to use the 3Ø factor in his power formula, I definitely have to assume incorporating power factor into his equation is not obvious.
Not I... but I agree it is but an estimation method with the faults unstated.
Nothing personal on my part. I even thought we were getting to the bottom of the matter.
- Location
- Massachusetts
Really? I based my response on the factors given in the OP, I did not assume things.
https://www.grainger.com/product/GE-LIGHTING-Incandescent-Light-Bulb-5V849
But really, it always comes down to this, people come and ask a basic question and instead of taking them through it a step at a time many members can't seem to take it slowly.
kwired
Electron manager
- Location
- NE Nebraska
How about if we just say that if the 1000 watt lamp is an HID lamp the actual luminaire draw is going to be a little higher then 1000 VA, how much higher depends on power factor.
230 amps is what it would draw (if each luminaire is actually drawing 1000VA) on single phase 208 volt supply.
If you balance it across three phases though each line only sees 133 amps.
Scali, even if you don't fully understand the phase angles, just consider that instead of putting all 48000VA on two supply conuctors, you are balancing it across three conductors. This means all current on A is returning via both B and C, current on B is returning on A and C, current on C is returning on A and B. (this is if they are 208 volt loads) All you need to remember about the phase angle is it creates a 1.73 multiplier (which is the square root of 3). Now if they were all 120 volt loads and evenly divided across the lines you have 16000 VA each line which is still 133 amps per line.
230 amps is what it would draw (if each luminaire is actually drawing 1000VA) on single phase 208 volt supply.
If you balance it across three phases though each line only sees 133 amps.
Scali, even if you don't fully understand the phase angles, just consider that instead of putting all 48000VA on two supply conuctors, you are balancing it across three conductors. This means all current on A is returning via both B and C, current on B is returning on A and C, current on C is returning on A and B. (this is if they are 208 volt loads) All you need to remember about the phase angle is it creates a 1.73 multiplier (which is the square root of 3). Now if they were all 120 volt loads and evenly divided across the lines you have 16000 VA each line which is still 133 amps per line.
Whatr is obvious to you doesn't necessarily make it obvious to others. If the OP'er has to ask whether or not to use the 3Ø factor in his power formula, I definitely have to assume incorporating power factor into his equation is not obvious.
Not I... but I agree it is but an estimation method with the faults unstated.
Nothing personal on my part. I even thought we were getting to the bottom of the matter.
I can't eliminate the dunning-kruger effect
what is obvious to the majority will be lost on the minority, but is still obvious
you did not address his query: is the 1.732 factor involved in the load calc
it is
you seem to have a rep for nitpicking and redirection
matters not to me
I understand, you are never wrong
you assumed 1000 W incand fixtures
they ask a basic question that is never answered
yes, 1.732 is a factor in the load calc
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