4 Single Phase Transformers Current Draw

I have a 460VAC 3 phase system that I want to add 4 transformers to, Each transformer is 35KVA(76A@460V). I know in 2 of the three phases the current should be higher than the 3rd but I am not sure how to calculate it. I know if I had 3 transformers the current should be 131A(76A*SQRT3)through each leg.
I believe L3 would remain at 131A but L1 and L2 would change.

Any input is appreciated.

From the three transformers assuming they are all balanced and disregarding PF I get 132 amps per phase on the line. the remainder would add 77 to the phases it is connected to. So on two you wil have about 210 and on third 132.

Dan Bentler

With 4 transformers you have an unbalanced Delta connected load. With 3 transformers you have a delta connected load which is probably unbalanced. If the 3 transformers are a balanced load then as you have stated the line currents are (sqrt3 * 76 amps). With unbalanced loads you need to know the power factor and currents of each load. You would then use complex notation and Kirchoffs laws to determine each of the line currents. The line currents will not be 120 degrees out of phase as for a balanced load.

yes in the l1 and l2 is going 223 amp
in L3 is 2times 76 is 152 amps going.
this is when the power supply on the begin is unlimited.
However as this will be also a transformer the real is a little diffrent, but for your calcs just plain and simple.

Stuntman:
PS: You should relabel the 4th transformer from "XFMR 3" to "XFMR 4".

3-phase currents (not single-phase currents which do not use SQRT 3):
Amps through L1 = (35+35+35) x 1000 / 460 X SQRT 3 = 131.8 Amps
Amps through L2 = (35+35+35) x 1000 / 460 X SQRT 3 = 131.8 Amps
Amps through L3 = (35+35) x 1000 / 460 X SQRT 3 = 87.9 Amps

Your actual voltage is probably higher than 460 (usually at least 480).

Lancie- I am a little confused by your post.
(35+35+35) X 1000 / 460 X SQRT 3 = 395.35

Stuntman you cannot simply just add the currents together even when you assume a unity power factor to arrive at the line currents. See attached calculation which assumes unity power factor for all loads.

Lancie- I am a little confused by your post.

Click to expand...

Stumtman,

You cannot add 1-phase currents to a 3-phase system. But you can add KVAs in either system. Therefore add the 3-phase KVAs on each leg, then calculate the current for that KVA in that leg. While working with KVA values, you don't have to worry about Power Factors.

The equations for alternating-current KVA:
1-phase: KVA = (Volts X Current)/1000
3-phase: KVA = (Volts X Current X SQRT 3)/1000
Therefore 3-phase current I = (Sum of KVA in each leg X 1000) / (460 X SQRT 3)

Current I in legs L1 & L2 = [(35+35+35) X 1000] / (460 X SQRT 3) = (105 X 1000)/(460 X 1.732),
then simpifying:
Current I in legs L1 & L2 = 105,000/796.74 = 131.78 Amps (NOT 395.35 as you calculated)

The real voltage is probably 480 and the real current is most likely 105,00/(480 X 1.732) = 126.3 Amps

Sorry Lancie I don't agree. You can only add KVA's in a balanced system. This system is not balanced and so you cannot just add the KVA's and then work out the line currents from those values.

Looks complicated. It would be interesting to see the actual measured currents.

Calistodwt said:

Stuntman you cannot simply just add the currents together even when you assume a unity power factor to arrive at the line currents. See attached calculation which assumes unity power factor for all loads.

Click to expand...

Principle correct, error in calculation though:

Yes L2 phase angle should be -41 degrees

I believe a man called Kirshoff said currents add in parallel circuits.
SO if I add resistors across phase A and B of 3 phase I can directly add the currents. If I install them to balance loading across all 3 phases I cannot directly add the currents but must use vector math to do the calculations. That is where the magic 1.73 comes in.

Lancies calcs were correct. He took 3 35 kVAR transformers and added power ot get 105 then ran that thru P = I E PF 1.73. He fouled up the typing on the formula - it should have read

I = kW * 1000 / (E x 1.73) NOTE just one of many ways of writing the calc.
He then took the remaining transformer ran thru same calc and threw out 1.73 since it is single phase. He then added the amps of that on the phases to the calculated amps of the balanced loads to get total current on each of 3 line conductors.

He got reactive amps which is OK because they will be slightly higher than if he (and I) had plugged in PF. But that is OK because if you stay with that calc to select your wire size then you get slightly oversized wire which is almost always a good thing.

Dan Bentler

Dan
maybe you should read up on the calculations for un-balanced loads in your text books ? I was under the impression from Stuntman's original post that he wanted to know how to calculate the currents. Normally one first does the correct design calculations and then adds a factor of safety when considering cable sizes etc. If he is happy to make a rough guess then use the 3 phase KVA formula. The man's name was Kirchoff by the way.

Well I dont worry too much about the name of the guy who wrote out the concept I just remember the concept.

I came up with 132 and 210 -- see post #2.

Pretty close to your 132 and 201.

I agree the proper way to do it is with vector math as you did.

Dan Bentler