Wye - Delta transformer kW calculation

[JasonCo]

I have a diagram, where my primary is a Wye 3-phase 277/480 volt system. And my Secondary is a Delta (Volts not given in the diagram). The Diagram shows all the 3 lines connected to each phase of the transformers. The diagram shows 3 ammeters connected in the middle of the lines on all 3 lines that shows the amps of each line.

The question is: If Ammeter 3 indicates 45.8 amps, the duct heater (3 phase Duct Heater in Delta) is delivering ___ kW of power.

Answer choices:
A. 26.443
B. 12.687
C. 10.480
D. 21.984
E. 38.077

Okay two ways I can think of doing this, I am just confused though..
First way is 45.8 x 1.732 x 277 = D as my answer.
Second way is 1.732 x 45.8 x 480 = E as my answer.

Very confused though on the theory of all this. Firstly, if I have a Wye system and E_line = E_coil x 1.732.. How is this possible. I understand line to line would be E_coil x 1.732 but looking at the diagram, Line 1,2,3 is coming out of the 3 end points of the wye transformer. So if I read the volts on the line (E_Line) it should be 277 not 480... If I read line to line it would be 480 though... But the equation is E_line = E_coil x 1.732.. Just very confusing to me.


Edit: The diagram shown in this thread is vaguely what I am trying to describe, besides the volts are different though. Just wanted to show the 3 lines all connected to the Wye - Delta accordingly. http://forums.mikeholt.com/showthread.php?t=122813&page=2

 

[Smart $]

Assuming your ammeters are on the primary conductors, the calculation is VA = 480V × 45.8A × sqrt(3).

 

[kwired]

I have a diagram, where my primary is a Wye 3-phase 277/480 volt system. And my Secondary is a Delta (Volts not given in the diagram). The Diagram shows all the 3 lines connected to each phase of the transformers. The diagram shows 3 ammeters connected in the middle of the lines on all 3 lines that shows the amps of each line.

The question is: If Ammeter 3 indicates 45.8 amps, the duct heater (3 phase Duct Heater in Delta) is delivering ___ kW of power.

Answer choices:
A. 26.443
B. 12.687
C. 10.480
D. 21.984
E. 38.077

Okay two ways I can think of doing this, I am just confused though..
First way is 45.8 x 1.732 x 277 = D as my answer.
Second way is 1.732 x 45.8 x 480 = E as my answer.

Very confused though on the theory of all this. Firstly, if I have a Wye system and E_line = E_coil x 1.732.. How is this possible. I understand line to line would be E_coil x 1.732 but looking at the diagram, Line 1,2,3 is coming out of the 3 end points of the wye transformer. So if I read the volts on the line (E_Line) it should be 277 not 480... If I read line to line it would be 480 though... But the equation is E_line = E_coil x 1.732.. Just very confusing to me.


Edit: The diagram shown in this thread is vaguely what I am trying to describe, besides the volts are different though. Just wanted to show the 3 lines all connected to the Wye - Delta accordingly. http://forums.mikeholt.com/showthread.php?t=122813&page=2

Reverse calculation says:38,077 VA / 45.8 A / 1.732 = 480 volts. You said the load was a three phase duct heater so that means all three phases are drawing same current to me.

If you want to calculate by phase then divide the total VA by 3 then figure each phase.

38,077 VA / 3 = 12692.333..A per phase. 12692.333../ 45.8 A = 277 volts.

 

[Smart $]

If you want to do it one ammeter at a time, it be for example VA_E-LINE = 277V × 45.8A.

Then VA_{(TOTAL) =} VA_{D-LINE +} VA_{E-LINE +} VA_F-LINE

 

[JasonCo]

Assuming your ammeters are on the primary conductors, the calculation is VA = 480V × 45.8A × sqrt(3).

So for test purposes I assume it is on primary side? Because the ammeter is literally in the middle of the Line 1,2 or 3 with the Wye and Delta on each sides of the line 1, 2 or 3. It as each phases connected together on the Wye and Delta sides. All 3 lines going across, and in the middle of the lines is an ammeter showing 45.8 amps...

 

[JasonCo]

Reverse calculation says:38,077 VA / 45.8 A / 1.732 = 480 volts. You said the load was a three phase duct heater so that means all three phases are drawing same current to me.

If you want to calculate by phase then divide the total VA by 3 then figure each phase.

38,077 VA / 3 = 12692.333..A per phase. 12692.333../ 45.8 A = 277 volts.

So each phase of the delta is 277, I don't think that can be. Its asking what the total kW of power the 3-phase duct heater is delivering. The duct heater is a Delta. This is exactly what I am really confused about actually. if the Wye side is 277/480 and line to neutral is 277, and each line is going across the screen to the Delta side. Then how is line voltage 480... Its not it should be 277. Literally Line 1 is connected to one of the windings of the Wye transformer. Line to Neutral on the Wye is 277, but the equation to find E_Line is E_Line = I_Coil x 1.732... So how is this possible!!!!!! If the diagram has Line 1,2 or 3 connected to each winding on a Wye transformer. And Line to Neutral on the Wye is 277, then why would the line not also be 277... Hopefully I am explaining in a way that you can understand.

 

[jumper]

So for test purposes I assume it is on primary side? Because the ammeter is literally in the middle of the Line 1,2 or 3 with the Wye and Delta on each sides of the line 1, 2 or 3. It as each phases connected together on the Wye and Delta sides. All 3 lines going across, and in the middle of the lines is an ammeter showing 45.8 amps...

You would have to assume the ammeters are on the primary since you said secondary voltage was not given and as such you could not calculate kW without it.

 

[Smart $]

So for test purposes I assume it is on primary side? Because the ammeter is literally in the middle of the Line 1,2 or 3 with the Wye and Delta on each sides of the line 1, 2 or 3. It as each phases connected together on the Wye and Delta sides. All 3 lines going across, and in the middle of the lines is an ammeter showing 45.8 amps...

Hmmm... it sounds as though the diagram is the transformer secondary (wye) and the load (delta). The transformer primary is not shown.

I was wondering about this when you said same as diagram in linked thread, so I made the assumption the ammeters were in the 480/277 lines, so it doesn't really matter as no other voltage is provided.

 

[kwired]

So each phase of the delta is 277, I don't think that can be. Its asking what the total kW of power the 3-phase duct heater is delivering. The duct heater is a Delta. This is exactly what I am really confused about actually. if the Wye side is 277/480 and line to neutral is 277, and each line is going across the screen to the Delta side. Then how is line voltage 480... Its not it should be 277. Literally Line 1 is connected to one of the windings of the Wye transformer. Line to Neutral on the Wye is 277, but the equation to find E_Line is E_Line = I_Coil x 1.732... So how is this possible!!!!!! If the diagram has Line 1,2 or 3 connected to each winding on a Wye transformer. And Line to Neutral on the Wye is 277, then why would the line not also be 277... Hopefully I am explaining in a way that you can understand.

As mentioned secondary voltage was not specified - this problem is not possible to solve if the ammeter reading in question is showing secondary current as we don't know what secondary voltage is, so that reading is assumed to be one leg of the primary current. It is also assumed that since the load was stated to be a three phase heater, with no other detail we must also assume it draws same current on each phase.

Total VA will be same on each side of transformer, and if secondary is balanced current then primary is balanced as well.

So if you have a balanced three phase load each phase is carrying one third of the total VA. That one phase is just 277 volts but is also one third of the total load which in OP you tried to apply 277 to the total load.

Simpler round numbers : if you have three 10 kVA 277V heaters and balance them across 480 volt three phase, you still have 30 kVA of load, but one third of it is supplied by each phase and the neutral sees no current so the source sees it no different then a 30 kVA three phase load.
You could connect this setup to a delta supply with one phase conductor to each heater and other end of each heater tied together to make a wye network out of the heaters and each still sees 277 volts (they all have to be same rating or it will throw the balance off) even though there is no source neutral.