208Y 3-Phase Balanced L-L Amp Calculation for 1-mVa Transformer

Status
Not open for further replies.
C

cottora

Senior Member
Location
Atlanta, GA
The more I read on this the more I tie my brain in knots...Without considering derating of equipment for continuous load, if I have balanced L-L (208v) loads on a 1000kVa transformer, I get 1000 x 1000/(1.73 x 208) = ~2775-Amp rating for such transformer.

Does this mean that the maximum number of 208v L-L 6-Amp loads the transformer can handle is 462 (2775/6 = 462)?

This doesn't seem to add up because a 6-Amp 208v single-phase load is 1.248kVa (so we get 1000/1.248 = 801 allowable 6-Amp 208v loads)

I think my mistake is either 1) not using the neutral "derates" the transformer or 2) I must multiply the first calculation by 1.73 to account for phasing between the lines

Thanks in advance!
 
cottora said:
The more I read on this the more I tie my brain in knots...Without considering derating of equipment for continuous load, if I have balanced L-L (208v) loads on a 1000kVa transformer, I get 1000 x 1000/(1.73 x 208) = ~2775-Amp rating for such transformer.

Does this mean that the maximum number of 208v L-L 6-Amp loads the transformer can handle is 462 (2775/6 = 462)?

This doesn't seem to add up because a 6-Amp 208v single-phase load is 1.248kVa (so we get 1000/1.248 = 801 allowable 6-Amp 208v loads)

I think my mistake is either 1) not using the neutral "derates" the transformer or 2) I must multiply the first calculation by 1.72 to account for phasing between the lines

Thanks in advance!
Click to expand...

and 801 divided by 1.732 equals?
 
Each phase will get 2775 amps in balanced three phase.
If you want lots of single phase 208V line to line loads, then yes, 801 loads at 6 amps single phase.
 
462. So the L-L rating will see max 2775 amps, but on a balanced 3-phase we get to multiply that number by 1.73 = ~801 6-Amp loads?
 
cottora said:
462. So the L-L rating will see max 2775 amps, but on a balanced 3-phase we get to multiply that number by 1.73 = ~801 6-Amp loads?
Click to expand...
No.
801 single phase loads at 6 amps or 462 three phase loads at 6 amps.
Always start with KVA for single phase and divide thru from there.
Always start with KVA for three phase and divide thru from there.
 
I think I understand...if 6-Amp loads; 801 L-L 208v or 462 L-L-L 208v

So the maximum amperage on any given leg in this scenario is 2775 amps? If I use all of the legs at once (3-phase load), then 2775 is max. However, if I use two legs then we multiply by 1.73 due to phasing?
 
cottora said:
The more I read on this the more I tie my brain in knots...Without considering derating of equipment for continuous load, if I have balanced L-L (208v) loads on a 1000kVa transformer, I get 1000 x 1000/(1.73 x 208) = ~2775-Amp rating for such transformer.

Does this mean that the maximum number of 208v L-L 6-Amp loads the transformer can handle is 462 (2775/6 = 462)?

This doesn't seem to add up because a 6-Amp 208v single-phase load is 1.248kVa (so we get 1000/1.248 = 801 allowable 6-Amp 208v loads)

I think my mistake is either 1) not using the neutral "derates" the transformer or 2) I must multiply the first calculation by 1.73 to account for phasing between the lines

Thanks in advance!
Click to expand...
That is 801 such loads balanced across all three phases, and the phase angle makes an identical balanced set of three come out to 10.39 amps per phase. 267 of those sets (gives you 801 individual units) times 10.39 = 2774.13 amps per line, just barely under transformer rating.

You need that 1.73 factor in there for phase angle.
 
cottora said:
I think I understand...if 6-Amp loads; 801 L-L 208v or 462 L-L-L 208v

So the maximum amperage on any given leg in this scenario is 2775 amps? If I use all of the legs at once (3-phase load), then 2775 is max. However, if I use two legs then we multiply by 1.73 due to phasing?
Click to expand...
NO.
Almost.
Single phase power formula is P=IV
Three phase power formula is P=IV*1.73
Google three phase power formula, click images and then look at all the pretty pictures. Might help.
 
Let's say you have a 480V 3 phase panel and go thru a 75KVA transformer to make it 208/120V 3 phase panel.
You will drive yourself nuts trying to figure out how many 208/120V single pole, double pole and three phase breakers you might have if you focus on "480V"
However, if you forget about the 480V panel and just focus on the 75KVA transformer.
You now start with 75KVA and then get your total 3 phase amps from their using P=IV*1.73
If you want single phase double pole 208V loads you use P=IV. If you want single phase single pole 120V loads you also use P=IV.
 
cottora said:
I think I understand...if 6-Amp loads; 801 L-L 208v or 462 L-L-L 208v

So the maximum amperage on any given leg in this scenario is 2775 amps? If I use all of the legs at once (3-phase load), then 2775 is max. However, if I use two legs then we multiply by 1.73 due to phasing?
Click to expand...
Not exactly, if you want to connect to two legs of the supply only you can have 2775/6=462.5 units connected, but as soon as you add any phase to phase load to the third leg, you begin to overload portion of the source.

If you had just two 6 amp loads but connected A-B and B-C, total draw on A and C are 6 but B is 1.732 times 6 = 10.39. You have fixed load and fixed volts so current in the individual loads remains 6 amps, but that common B line is drawing some current from A some current from C and some from B, the 120 degree phase angle works out so that it is 1.732 factor on the balanced portion of the load, and since you have equal from each direction is simpler calculation than if you had two different sized loads involved.

Add one more 6 amp load between A and C and all three lines now see 10.39 amps, you still have three 1248 VA loads times 3 = 3744 VA total load.
 
Look at it this way too. If this stuff were easy to figure out, anyone could do it. The fact that you have to sometimes multiply by the square root of three or 1.73 confuses the heck out of most people. Once you do figure it out, you will be one of the few and set to make the big bucks!!!

When considering the voltage of a sin wave, you have to take the peak voltage divided by the square root of 2 or 1.41 to get RMS. Another thing to confuse people, perfect! More money for us!
 
I am actually the guy paying you men (worth every penny!). I am a small business so I cannot afford blown transformers/under-rating, which is why I am trying to learn and verify what is being quoted to me.

Thanks again!
 
cottora said:
I am actually the guy paying you men (worth every penny!). I am a small business so I cannot afford blown transformers/under-rating, which is why I am trying to learn and verify what is being quoted to me.

Thanks again!
Click to expand...
Another way to look at it is 1000 kVA transformer can handle 1000 kVA of load. If you have multiple 6 amp @ 208 volt loads (1248 VA) then you can have 801 of them, but with that much load they must be balanced across all three phases. In a three phase system, current on one phase that has incoming matching current from each of the other two phases increases by a factor of 1.732 because of the 120 degree phase angle, with a single phase source everything you connect involves a 180 degree phase angle and is direct additive, with use of only two lines of three phase source there is only 180 degree angle between the two points you are connecting to so same additive effect as a single phase source.

Gets a little more complex when you add loads that use the neutral of the wye system, but bottom line is you still can supply 1000 kVA but can not load any one of the three secondary windings to more than one third of the total rating.
 
powerpete69 said:
Look at it this way too. If this stuff were easy to figure out, anyone could do it. The fact that you have to sometimes multiply by the square root of three or 1.73 confuses the heck out of most people. Once you do figure it out, you will be one of the few and set to make the big bucks!!!
Click to expand...

Here's where the 1.73 comes from, and why we use it.

A three phase system, will deliver the same power as each line current throughout three 1-phase/2-wire systems, operating at the line-to-neutral voltage. Should you use the line-to-neutral voltage in your calculation, it would be P = 3*I*Vln.

From the geometry of the triangle that represents the arrangement of the phases about neutral, you can show that Vll = sqrt(3)*Vln. To replace Vln with Vll, multiply by 1 in a fancy way, i.e. sqrt(3)/sqrt(3), and regroup. We end up getting P = 3/sqrt(3) * I * Vll, which simplifies to P=sqrt(3)*I*Vll.
 
Status
Not open for further replies.
Top