Transformer sufficient for continuous load?

[EE_FSU]

Greetings!

I've been delving into the NEC and trying to increase my understanding of power here recently. Hopefully in the future, I will be able to answer your questions! So let me start:

I have a delta-to-wye transformer rated @ 20kVA. It is converting 115VAC, Delta (3PH) to 200/115VAC, Wye (3PH), and is to deliver a total of 80A (continuous). From my calculations I get the following:

I = apparent power/Voltage (line-to-line) = 20kVA/200 = 100A
since 3PH: 100A/sqrt(3) = 57.7A

Questions: Will a 20kVA xfrmr be enough to deliver a max load of 80A?
Is the 57.7A the total amperage delivered per phase/line or is it the total of all three phases? In other words, is each phase delivering 19.3A (57.7A/3)???? (or perhaps the 100A is the total and it's 57.7A per phase).

Thanks in advance for your help! Of course, let me know if you require more information/clarification.

 

[Smart $]

It's 57.7A per line... and that is its [continuous] secondary rated current. Doesn't matter what load it will be powering, that is the max' continuous load current per line.


Welcome :thumbsup:

 

[gar]

150705-0858 EDT

EE_FSU:

I found your question quite confusing.

There are various ways of using a single transformer (a true 3 phase transformer), or multiple transformers ( 2, 3 or more single phase transformers) to change voltage levels in a 3 phase system.

I have a delta-to-wye transformer rated @ 20kVA. It is converting 115VAC, Delta (3PH) to 200/115VAC, Wye (3PH), and is to deliver a total of 80A (continuous). From my calculations I get the following:

Assume this is a true 3 phase transformer rated at 20 kVA, then this could be replaced by 3 single phase transformers rated at 6.67 kVA each. At 115 V secondary voltage each transformer can supply 6667/115 = 58 A. This is a simple way to easily see what is the secondary current capability of a single phase leg of a 3 phase system without being confused as to whether or not to use the sq-root of 3.

What do you mean by "to deliver a total of 80 A". Is this per leg or the sum of all three legs of the secondary? If it is the sum of all three legs, then clearly 80 is less than 3*58 = 174. If it is the load per leg, then you can not load to 80 A.

For a true 3 phase transformer rated at 20 kVA you can not put a 20 kVA load on one phase of the secondary.

Is this some sort of class problem?

.

 

[charlie b]

The sooner you drop, once and forever, the notion of "amps per phase" from your vocabulary, the happier your career will become. That notion is meaningless. In a three phase system, at the moment when the A phase voltage is at its positive peak, the amps that leave the source from the A phase will return to the source via the B and C phases. It is the same amps, going opposite directions. You don't add the Phase A amps to the Phase B or C amps. Rather, the Phase A amps becomes the Phase B and C amps. At a later moment in time, when it is the Phase B voltage that is at its positive peak, the amps leaving the source on Phase B will return to the source on phases A and C. Put another way, at every instant in time, the total sum of the Phase A, B, and C currents is zero.

In your example, the transformer is rated to carry a maximum secondary current of 57.7 amps. If you fully load that transformer and keep the loads balanced, an ammeter set up to read all three phase currents will read 57.7 on each phase.

Welcome to the forum.

p.s. Am I to infer that you went to school in Florida and then moved to Alaska? I don't think I could handle that extreme a change in weather. :happyno:

 

[Smart $]

The sooner you drop, once and forever, the notion of "amps per phase" from your vocabulary, the happier your career will become. That notion is meaningless. In a three phase system, at the moment when the A phase voltage is at its positive peak, the amps that leave the source from the A phase will return to the source via the B and C phases. It is the same amps, going opposite directions. You don't add the Phase A amps to the Phase B or C amps. Rather, the Phase A amps becomes the Phase B and C amps. At a later moment in time, when it is the Phase B voltage that is at its positive peak, the amps leaving the source on Phase B will return to the source on phases A and C. Put another way, at every instant in time, the total sum of the Phase A, B, and C currents is zero.

In your example, the transformer is rated to carry a maximum secondary current of 57.7 amps. If you fully load that transformer and keep the loads balanced, an ammeter set up to read all three phase currents will read 57.7 on each phase.

Welcome to the forum.

p.s. Am I to infer that you went to school in Florida and then moved to Alaska? I don't think I could handle that extreme a change in weather. :happyno:

It would help if you also adapted to using Line A rather than Phase A, etc. IIRC, the IEEE convention for 3Ø parlance is that a "phase" is line-to-line for both voltage and current. I realize it is difficult when you have done so for years, but that is part of what perpetuates the use of "amps per phase".

JMO

 

[EE_FSU]

Charlie, of course!:slaphead: Each line will peak at X amps, but at different times since they are 120 deg out of phase. The summation of these phasors are equal to zero (as you stated). Thanks!