Delta Transformer Calculation

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dabatman

Member
I have an installation where the customer wants to use a 480 D / 240V D transformer which i haven't dealt with that much. I have a bunch of loads that are 7000VA and will be fed with 2wire+gnd circuits (hot,hot,gnd) to be run off 230V.

I am looking at the calculation to balance the loads across all three phases and seem to be running into problems. It seems like the current should be

I= VAline/(sqr(3)*Eline) which gives me 17.6 amps. I think this is what you would see at the load possibly but when balancing the system, you would end up with your VA higher. I think then it works out like below because it is a 3phase system.

If you think about it say in a panelboard (in VA with three loads):

A B C
3500 A
3500 B
3500 C
3500 A
3500 B
3500 C
----------------------
7000 7000 7000

to amps

A B C
15.2 A
15.2 B
15.2 C
15.2 A
15.2 B
15.2 C
----------------------
30.4 30.4 30.4

It seems like I may be missing something but all of my numbers seem to work out except I am a bit leary on the first number of 17.4. Just looking for someone either to confirm this is correct, or show me the error of my ways.
 

Smart $

Esteemed Member
Location
Ohio
Hmmm...

Three balanced 7000VA Line-to-Line loads is 21000VA.

21000VA ? (240V ? 1.732) = 50.5A on each Line

Seems to me you are trying to work this same as you would a wye supply. You can, but you will have to use a "virtual" Line-to-Neutral voltage of 240 ? 1.732 = 138.564V, which may or likely will confuse others... ;)

7000VA ? 138.564V = 50.5A
 
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dabatman

Member
Still Confused I guess

Still Confused I guess

That somewhat makes sense but still a little bit confused on sizing the circuit breaker then. Say for one load:

3500 on A
3500 on B

then looking at this from your reply,

3500/138.564 = ~25.2 on each phase

so then I need a 35A 2pole breaker to feed each load.

Then I end up with the 50.5 on each phase for 21000VA which works out.

You are correct in that I was trying to figure this out like a wye configuration. I guess my final question would be then, how do you figure out the load on each line for a 2pole feed. The comment was not to use the "virtual" voltage of 138.564 but then what is the actual calculation?? It ends up being

I = (3500*sqr(3))/240 but this doesn't seem to make sense as VA=E*I*sqr(3) ?? Seems like the sqr(3) is in the wrong spot on the first calculation.

Again, I am probably approaching this wrong, but just trying to learn the correct way to look at this. I kind of searched around on the internet looking for the calculation but came up empty.
 

beanland

Senior Member
Location
Vancouver, WA
Single-phase Load

Single-phase Load

The 7000VA is 230V load, 30.4A. So a 40A 2-pole breaker will work. For three 7000VA 230V loads, you will have three 40A 2-pole breakers on the 230V transformer. You want to connect each breaker so there is balance meaning AB, BC, CA.
 

jumper

Senior Member
The 7000VA is 230V load, 30.4A. So a 40A 2-pole breaker will work. For three 7000VA 230V loads, you will have three 40A 2-pole breakers on the 230V transformer. You want to connect each breaker so there is balance meaning AB, BC, CA.

Isn't a 35 amp breaker a standard size?

240.6 Standard Ampere Ratings.
(A) Fuses and Fixed-Trip Circuit Breakers. The standard
ampere ratings for fuses and inverse time circuit
breakers shall be considered 15, 20, 25, 30, 35, 40, 45, 50,
60, 70, 80, 90, 100, 110, 125, 150, 175, 200, 225, 250, 300,
350, 400, 450, 500, 600, 700, 800, 1000, 1200, 1600, 2000,
2500, 3000, 4000, 5000, and 6000 amperes. Additional
standard ampere ratings for fuses shall be 1, 3, 6, 10, and
601. The use of fuses and inverse time circuit breakers with
nonstandard ampere ratings shall be permitted.
 

kwired

Electron manager
Location
NE Nebraska
Isn't a 35 amp breaker a standard size?

240.6 Standard Ampere Ratings.
(A) Fuses and Fixed-Trip Circuit Breakers. The standard
ampere ratings for fuses and inverse time circuit
breakers shall be considered 15, 20, 25, 30, 35, 40, 45, 50,
60, 70, 80, 90, 100, 110, 125, 150, 175, 200, 225, 250, 300,
350, 400, 450, 500, 600, 700, 800, 1000, 1200, 1600, 2000,
2500, 3000, 4000, 5000, and 6000 amperes. Additional
standard ampere ratings for fuses shall be 1, 3, 6, 10, and
601. The use of fuses and inverse time circuit breakers with
nonstandard ampere ratings shall be permitted.

If it is continuous load the 35 amp is not large enough.

If there is no line to neutral load why are we even considering the neutral. This calculation without any neutral load is same whether supplied from a wye or a delta.

You don't have half load of one 7000 single phase on A and other half on B you have 1/1.73 of the load on each phase as well as same amount of other loads connected to the same phase.

Easiset way to figure this (no matter if wye or delta supply) is figure out how many (single phase) amps are balanced and multiply that by 1.73 to get the net balanced three phase load. Then add the unbalanced load to the applicable lines.
 

Smart $

Esteemed Member
Location
Ohio
That somewhat makes sense but still a little bit confused on sizing the circuit breaker then. Say for one load:

3500 on A
3500 on B

then looking at this from your reply,

3500/138.564 = ~25.2 on each phase

so then I need a 35A 2pole breaker to feed each load.
The branch circuit load is 7000VA. You divide that by L-L voltage of 240 for a current of 29.2A. Sidenote: You determine current using line voltage, not the voltage specfied on equipment (such as 230V). If necessary, adust the VA to match 240V supply.

When you enter 3500VA on a panel schedule, it is simply a math shortcut. It assumes there will be a counterpart of 3500VA across another line pair. Say you have three 7000VA loads cannected AB, BC, and CA. With each column having 3500VA twice, the columns total 7000VA each and then all three columns total 21000VA. This is for determining feeder load, not branch circuit load (which you already know).

Note this method is used for wye supplies. I am uncertain how engineers actually write up delta supply schedules. Perhaps one will chime in and make us privy to such information.

Back to the 29.2A for each 7000VA load. If you have one connected AB and another connected AC. The current seen by feeder conductor "A" would be 29.2 ? 1.732 = 50.5, yet the current seen by feeder conductor "B" and "C" would be 29.2A each, provided there are no other loads.
 
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