Line current. Wye inverter to Delta service

McMac

Member
A 480Y inverter states 40A per phase. In a wye config the phase current is equal to line current (what I size my conductors to).

In a Delta config the line current is phase current x √3 (correct?).

My question is for calculating line current on the delta side (240v).

Is it
(a) the 40A line current (which is the same and stated as phase current on the spec sheets) from the 480Y is a simple and direct calculation. At 240D my line current is now 80A and I size my conductor to 125% of that (100A). Leaving the phase current on the delta is then less by √3. But phase current is more applicable to the transformer and not where my concern lies.

Or
(B) the 40A phase current from the 480Y is the simple and direct calculation. At 240D the phase current is 80A. Leaving that my line current is then greater by √3. Which means I would need to size conductors even larger. In this example 80x√3=138.6Ax1.25=173.3A. Wow I hope this a silly concern based on incorrect application of the knowledge.

Full disclosure, I hope (a) is right because this is how I have been designing, and I haven't had any issues. But when I was forced to put three single phase inverters on a 240 delta, I did my research and learned that at least for that application I did needed to factor the line current to be the inverter max x √3 to properly size my conductors. But should I have been applying this same consideration for my 3P 480wye inverters when xfrming to 240delta? I hope not.

I appreciate your help in adressing this concern of mine. Thanks in advance for your time.

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electrofelon

Senior Member
As you state, in a Wye, the line current is the same as the phase current. In a delta, the line current is phase current X 1.732, however you dont really ever come across the phase current in a delta, except if you are building a three phase bank from single phase components, like may be done with single phase inverters. In those cases, you could think of it in terms of phase vs line currents, but I personally find it simpler to work with KVA. For Example:

Say I have 3 10 KW inverters connected A-B,B-C,C-A (which is a delta configuration). IF I want to know the total current on that feeder or service, I would use 30,000/(240*1.732).

For transformers, dont worry about what is going on inside (yes there is some L-P dosey doe going on if the primary and secondary have different winding configurations) just take the current in vs current out as inversely proportional to the voltage change.
 

McMac

Member
As you state, in a Wye, the line current is the same as the phase current. In a delta, the line current is phase current X 1.732, however you dont really ever come across the phase current in a delta, except if you are building a three phase bank from single phase components, like may be done with single phase inverters. In those cases, you could think of it in terms of phase vs line currents, but I personally find it simpler to work with KVA. For Example:

Say I have 3 10 KW inverters connected A-B,B-C,C-A (which is a delta configuration). IF I want to know the total current on that feeder or service, I would use 30,000/(240*1.732).

For transformers, dont worry about what is going on inside (yes there is some L-P dosey doe going on if the primary and secondary have different winding configurations) just take the current in vs current out as inversely proportional to the voltage change.
Thanks for confirming.

You're correct that it much simpler to think in kva, and thinking in those terms makes the answer to my concern a little more obvious.

I have done the 3 single phase in Delta a couple times before because sales figured out on accident that they could get more incentive money for the customer with single phase inverter due to size options in each class. But that is what made me second guess myself.

With a couple 100kva wye to Delta projects on my desk, I just needed to be sure.

Thanks again.

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synchro

Senior Member
Or
(B) the 40A phase current from the 480Y is the simple and direct calculation. At 240D the phase current is 80A. Leaving that my line current is then greater by √3.
This is where version (B) has a discrepancy. The L-N voltage of the 480Y is 480V/√3 = 277V. So the phase current in a winding of the 240 delta output is 277/240 x 40A = 46.2A.
Then the line current out of the delta is √3 x 46.2 = 80A. So the √3 's get divided out in version (B) and you come up with the same answer as version (A), which is correct.

Like Ethan said, you don't normally have to consider what's going on inside the transformer.
 

ggunn

PE (Electrical), NABCEP certified
As you state, in a Wye, the line current is the same as the phase current. In a delta, the line current is phase current X 1.732,
On the face of it and to a cursory glance, that doesn't make sense to me. Consider a balanced wye circuit, where the current in the neutral is zero. How could the current in the conductors (phase current) be different from that in a delta circuit with the same line current?
 

electrofelon

Senior Member
On the face of it and to a cursory glance, that doesn't make sense to me. Consider a balanced wye circuit, where the current in the neutral is zero. How could the current in the conductors (phase current) be different from that in a delta circuit with the same line current?
The phase current in a Delta is the current on one of the windings, BEFORE it connects to the other windings. The line current is the "messy vector stuff" that results when two windings are connected at a common point.
 

ggunn

PE (Electrical), NABCEP certified
The phase current in a Delta is the current on one of the windings, BEFORE it connects to the other windings. The line current is the "messy vector stuff" that results when two windings are connected at a common point.
But the current in the conductors, it seems to me, has to be the same. We see discussed in here pretty often the two ways of connecting a transformer from a PV system to a three phase wye service. One is wye where you run a neutral back to the service and the other is delta where you don't. In the wye case the neutral current is zero. The current in the conductors must be the same in either case for the same power output. Likewise, the line current is the same.
 

electrofelon

Senior Member
But the current in the conductors, it seems to me, has to be the same. We see discussed in here pretty often the two ways of connecting a transformer from a PV system to a three phase wye service. One is wye where you run a neutral back to the service and the other is delta where you don't. In the wye case the neutral current is zero. The current in the conductors must be the same in either case for the same power output. Likewise, the line current is the same.
Yes the LINE current is the same for both. But "phase current" is different in a delta. Phase current is the current on one of the single phase components. Wye -> line current phase current same, line voltage phase voltage different. Delta is opposite -> line current phase current different, line voltage phase voltage same.
 

ggunn

PE (Electrical), NABCEP certified
Yes the LINE current is the same for both. But "phase current" is different in a delta. Phase current is the current on one of the single phase components. Wye -> line current phase current same, line voltage phase voltage different. Delta is opposite -> line current phase current different, line voltage phase voltage same.
I understand what you are saying but what I am not understanding is how the current in wye connected conductors in a balanced system (no neutral current) could be different from a delta connected system where the line current is the same. It's the same amount of current through three wires. Don't just tell me they are different; explain how/why they are different. I know a lot but I don't know everything. :D
 

electrofelon

Senior Member
I understand what you are saying but what I am not understanding is how the current in wye connected conductors in a balanced system (no neutral current) could be different from a delta connected system where the line current is the same. It's the same amount of current through three wires. Don't just tell me they are different; explain how/why they are different. I know a lot but I don't know everything. :D
I know you know this, I think there is just some confusion about terms, specifically (I think) "phase current" vs "line current". Line voltage and line current are the "normal" values we associate with a three phase system. Phase current and phase voltage are the values of the single phase components of a delta or wye connection. Look at the voltage and current of the single phase components in both a delta and a wye with the same line current and voltage. An example: POCOS use 3- 240V transformers in a wye connection to get 208V three phase line current. Phase voltage is 240, line voltage is 208. For the delta example, where its the opposite and voltage stays the same and current changes, Consider what I know you have done, is connect three single phase inverters AB,BC,CA to a three phase system. The current on each inverter is different than the resulting three phase current.
 

synchro

Senior Member
I understand what you are saying but what I am not understanding is how the current in wye connected conductors in a balanced system (no neutral current) could be different from a delta connected system where the line current is the same. It's the same amount of current through three wires. Don't just tell me they are different; explain how/why they are different. I know a lot but I don't know everything. :D
I personally think the common nomenclature "phase current", "phase conductor", etc. may be adding confusion because each line output conductor also has an associated phase relationship with the other two.
Be that as it may, as Ethan stated the phase current is the current flowing through each individual 2-terminal winding (or inverter). In a delta two of the three windings are connected to form each line output, and therefore because of Kirchoff's law each line output current is the sum of the currents on these two connected windings. On a delta, a 180 degree reversal (polarity flip) is effectively made on successive phase windings so the currents are 60 degrees apart from each other (instead of 120 as in a wye) in order to make the resulting vectors form a triangle (i.e. delta). Because the two phase currents are 60 degrees apart from each other the resulting line current has a phase right in between these two currents, i.e. +-30 degrees from each one. Therefore the component of each phase current that contributes to the line current is scaled by cos(30) = 0.866 because that's how much is "in phase" with the resulting line current. And so total line current is 2 x 0.866 = 1.732 times the current in one of the windings, instead of 2 times if the two windings were in phase with each other.


So back to the orginal question (I think): Consider a wye and a delta connected device (transformers or inverters) that both produce the same line currents and L-L voltage on the three output wires, and with a balanced load. How do the currents compare on the "phase conductors" that that are connected to make the wye configuration vs. the delta?
On a wye, only one phase conductor is connected to each line output and so the line current is the same as the phase current. On a delta, two phase conductors are connected to each line conductor and therefore (as above) each contributes 0.866 times its phase current, for a total line current of 1.732 times one of the phase currents.
However on a wye, because the voltage across each phase winding is 120 degrees apart with a common connection at the neutral, the L-L voltage is the vector sum of two phase voltages and therefore 1.732 times the phase voltage. On a delta the L-L voltage is of course the same as the phase voltage.

So in order to make the wye and delta output have the same line output voltages and currents using the same transformer primary circuit, the delta windings would need to have 1.732 more turns than the wye windings (so they produce 1.732 times the voltage and 1/1.732 times the current as the wye ones). This would result in the same total output power for the same balanced load.
 
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ggunn

PE (Electrical), NABCEP certified
I know you know this, I think there is just some confusion about terms, specifically (I think) "phase current" vs "line current". Line voltage and line current are the "normal" values we associate with a three phase system. Phase current and phase voltage are the values of the single phase components of a delta or wye connection. Look at the voltage and current of the single phase components in both a delta and a wye with the same line current and voltage. An example: POCOS use 3- 240V transformers in a wye connection to get 208V three phase line current. Phase voltage is 240, line voltage is 208. For the delta example, where its the opposite and voltage stays the same and current changes, Consider what I know you have done, is connect three single phase inverters AB,BC,CA to a three phase system. The current on each inverter is different than the resulting three phase current.
Maybe what we have here is a semantic issue. I agree that within components connected phase to phase, i.e., delta, be they inverters or transformer windings, currents are different from those connected phase to neutral, i.e. wye, for a given line current. For the conductors between components, however, the current in the individual conductors, assuming a balanced system, are the same as the line current whether there is a neutral (wye) or not (delta).
 

electrofelon

Senior Member
Maybe what we have here is a semantic issue. I agree that within components connected phase to phase, i.e., delta, be they inverters or transformer windings, currents are different from those connected phase to neutral, i.e. wye, for a given line current. For the conductors between components, however, the current in the individual conductors, assuming a balanced system, are the same as the line current whether there is a neutral (wye) or not (delta).
I dont see the need, nor the correctness for that matter, to break a "component" into smaller parts to make this distinction. The conductors "between components" will be different than the three phase line current also. The "change" doesnt happen until that component connects to another component that has different phase angles - at the "corners" of the delta. Since this is the PV forum, lets use the three single phase inverter example. They are connected AB,BC,CA to a three phase system which is equivalent to making a delta connection out of the three inverters. Each inverter is 32 amps. The 32 amps is the current of that component. This would also be the "phase current". Its 32 amps beginning at the magic box inside which no one understands, through the conductors until it hits the busbar in the AC combiner panel where another component is also sending current. The conductor connected to that bus bar sees (1.732)(32) or 55.4 amps. 55.4 is the "line current".

I dont think the use of these terms "phase current" and "line current" are semantic at all. They are well defined properties of three phase sources and loads. This is different that that other use of the word "phase" that results in threads hundreds of pages long.:angel:
 

ggunn

PE (Electrical), NABCEP certified
I'm pretty sure that the current in the individual conductors from a given inverter to a service will be the same as the line current irrespective of whether the inverter is wye or delta connected; I don't know how to state it more simply than that. The OCPD on them is the same either way. Maybe that's not what you mean by phase current, hence semantics.
 

electrofelon

Senior Member
I'm pretty sure that the current in the individual conductors from a given inverter to a service will be the same as the line current irrespective of whether the inverter is wye or delta connected; I don't know how to state it more simply than that. The OCPD on them is the same either way. Maybe that's not what you mean by phase current, hence semantics.
Do we agree on the following? Three single phase inverters connected to a three phase system as in post #14. Say its a regular 208 service. Inverters are thus 208V, say 32 amps. That is 20KW total of inverters (208)(32)(3). Three phase power is W=(V)(I)(1.732) so current is 55.5 A or 1.732 times the inverter or phase current. The currents I am referring to as being different are the 32 amps and the 55.5 amps. (phase and line respectively). Are we each talking about different currents or points of measurement?
 

pv_n00b

Senior Member
I made an Excel spreadsheet that did all the calculations in vector format for deriving line currents based on phase currents of delta connected inverters. I did it just to play around with unbalanced inverters in delta connections to see how the line currents and phase relationships change. I recommend doing it, very educational.
 

electrofelon

Senior Member
I made an Excel spreadsheet that did all the calculations in vector format for deriving line currents based on phase currents of delta connected inverters. I did it just to play around with unbalanced inverters in delta connections to see how the line currents and phase relationships change. I recommend doing it, very educational.
yes it gets a lot more complicated in the unbalanced case, I wasnt going to go there ;)
 

wwhitney

Senior Member
I'm pretty sure that the current in the individual conductors from a given inverter to a service will be the same as the line current irrespective of whether the inverter is wye or delta connected
Say you have a group of 3 single phase inverters and a feeder from that group of inverters consisting of 3 ungrounded conductors and possibly a grounded conductor. The current in a feeder ungrounded conductor is the "line" current. The current through an inverter is the "phase" current.

If the inverters are wye connected, then each feeder ungrounded conductor is connected to one inverter with no other connections. That means line current = phase current.

If the inverters are delta connected, then each feeder ungrounded conductor is connected to two inverters. The line current on that feeder ungrounded conductor is the vectorial sum of the two phase currents, one from each connected inverter. If the inverters are 120 degrees out of phase, the magnitude of the line current will be sqrt(3) times the phase current.

Cheers, Wayne
 

synchro

Senior Member
If the inverters are delta connected, then each feeder ungrounded conductor is connected to two inverters. The line current on that feeder ungrounded conductor is the vectorial sum of the two phase currents, one from each connected inverter. If the inverters are 120 degrees out of phase, the magnitude of the line current will be sqrt(3) times the phase current.
And if the phase currents from two connected inverters are I[SUB]Ph1[/SUB] and I[SUB]Ph2[/SUB] with a phase angle Ø between them, then the resulting line current on the feeder conductor at that connection will be sqrt( I[SUB]Ph1[/SUB][SUP]2[/SUP] + I[SUB]Ph2[/SUB][SUP]2[/SUP] + 2 I[SUB]Ph1[/SUB] I[SUB]Ph2[/SUB] cos Ø ). If both of these two phase currents are defined as flowing into this connection, then Ø is 60 degrees (a180 flip from 120 degrees).
So if the inverters were both 10 amps then the line current output is sqrt( 10[SUP]2[/SUP] + 10[SUP]2 [/SUP]+ 2 x 10 x 10 x cos(60) ) = sqrt(100 +100 + 2 x 10 x 10 x 0.5 ) = 17.32 amps = sqrt(3) x 10 amps. If the inverter currents are different from each other then you can substitute those in the formula.
 
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