available 120V amps in hi-leg system

Status
Not open for further replies.
I am sizing a Delta secondary Hi-leg system for two 42 amp at 120 volt loads. I have reviewed the theory and I believe I will require a 75,Kva, Delta/Delta (due to customer requires 240/120 volt). I believe the math is; 75 Kva/3= 25 Kva. I am treating each 25 Kva winding or transformer as single phase therefore 25 Kva/240V=104.17A.
Am I correct that the available amps for 120 volt loads would be 52 amps from C1 to N and 52 amps from C2 to N since I am using half of the winding?
This is being provided for a single piece of equipment. The load requires two, single phase, 120 volt circuits of 42 amps each.
Thanks in advance.
 
If I may ask, why bother with the high leg? Just use a 12.5kva 1ph 120/240v transformer.

And, no, your 3ph math is incorrect, but your 1ph math works. 242 x 40 = 10.08kva.
 
The customer, US Army, is requesting the option to have 3 phase available. The equipment they will power immediately requires 240/120V single phase. I reviewed information on Mike's site and on ECM where I understood them to say the 3 phase delta as essentially 3 separate transformers for determining I phase would be 25 Kva divided by 240 for the 104.17 amps while I Line is 75 Kva/(240 X 1.732) = 180.427 amps. I was trying to determine what the available current for the 120 volt loads would be. I thought if the allowable current for one winding was 104 amps, then since the winding was center tapped, the current for each 1/2 winding would be 52 amps. I haven't been able to find anything yet that ether explains that particular situation or perhaps puts it into words that I have understood. I have taken your suggestion to use a single phase unit and I am submitting a request for a change from 3 phase to single phase be allowed however, the smallest transformer I find on the Eaton site is a 37.5 Kva with my required 400 Volt primary and 240/120 volt secondary. Prime power on the base is 230/400V. I appreciate your input and welcome further constructive comments. Thank you
 
I was trying to determine what the available current for the 120 volt loads would be. I thought if the allowable current for one winding was 104 amps, then since the winding was center tapped, the current for each 1/2 winding would be 52 amps. I haven't been able to find anything yet that ether explains that particular situation or perhaps puts it into words that I have understood.
A single phase transformer with a 240V/120V secondary has two 120V "half windings" where each one contributes 1/2 of the rated transformer KVA. Since the windings are in series, the output voltage across both of them doubles to 240V but the current capability is unchanged. So a 12.5 KVA single phase transformer would be adequate for two 42A 120V loads (one on each leg) as Larry suggested. If the two 120V 42A loads are both present and exactly balanced then the 42A will flow from one half-winding through the other half-winding, and no current will flow through the center tap (neutral).
 
Last edited:
This is sort of the 'amps per phase' question, but in reverse.

1) If a transformer coil is rated to produce 104A at 240V, then it can provide 104A. If you have two perfectly balanced 120V loads electrically in series, then this coil could provide the full 104A; dividing by 2 is not necessary.

2) Delta transformers often have specific ratings for the neutral terminal, which _may_ be considerably less than the full rating of the main terminals. So if you have 120V loads that are not perfectly balanced, the maximum allowed for these loads may be much less than the 104A, possibly less than 1/2 the full rating. It depends on the design of the transformer.

3) In a delta transformer, each terminal is connected to two phase coils, and both supply output current. So even though you are only connecting a single phase load, you get contribution from the other two phase coils.

-Jon
 
Status
Not open for further replies.
Top