# Transformer Problem



## EEVA PE (Sep 21, 2011)

This is 3 phase problem. If there is a transformer with the high side (15KV) delta and low side (6.6KV) WYE. The current is 62.7 on the high side, what is the current on the low side?

My thought process is multiply the current (62.7) by (15/6.6) and divide by sqrt3 due to going from WYE to delta and adding -30deg angle for going from high WYE side to low Delta side. So (62.7)(15/6.6)(1/1.73)(angle -30 deg) = 82.3 A angle -30 deg

The answer key says it is 62.7 (15/6.6) = 143 A. I think this is wrong because they did not divide by sqrt3 to take into account going from WYE to Delta. What do others think?


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## BamaBino (Sep 21, 2011)

I get the 143A answer.

This is a really good problem. Is it from one Complex Imag sample exam books?


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## Flyer_PE (Sep 21, 2011)

If you are looking at the line currents entering and leaving a 3-phase transformer, the sqrt3 multiplier divides out so you wind up with pretty much the turns ratio regardless of how the individual windings are configured.


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## Wildsoldier PE (Sep 22, 2011)

Flyer_PE said:


> If you are looking at the line currents entering and leaving a 3-phase transformer, the sqrt3 multiplier divides out so you wind up with pretty much the turns ratio regardless of how the individual windings are configured.



This is what i do with problems like that. Power in the primary side = Power in the secondary side

S1= (15000VLine to line) (sqrt(3)) (62.7Amps)=1,628,993.785 VA (total in the system)

S1=S2=1,628,993.785 VA

I2= (1,628,993.785 VA) / (6,600VLine to line) (sqrt(3)) = 142.5 Amps round to 143

or you can take The S and divide by 3 to get the S per phase and divide the S per phase by Secondary phase voltate like this:

S1per phase = S2per phase = 1,628,993.785 / 3 = 542,997.9283 VA

Vphase secondary = (6600V line to line) / (sqrt(3)) =3810.51 Vphase

Iline =(542,997.9283 Va) / (3810.51 Vphase)= 142.5 Amps

Also the assumption on using the ratios is correct. But i prefer the KVA method since thats the way I do it at work almost every day.


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## Jonjo (Sep 22, 2011)

EEVA said:


> This is 3 phase problem. If there is a transformer with the high side (15KV) delta and low side (6.6KV) WYE. The current is 62.7 on the high side, what is the current on the low side?
> My thought process is multiply the current (62.7) by (15/6.6) and divide by sqrt3 due to going from WYE to delta and adding -30deg angle for going from high WYE side to low Delta side. So (62.7)(15/6.6)(1/1.73)(angle -30 deg) = 82.3 A angle -30 deg
> 
> The answer key says it is 62.7 (15/6.6) = 143 A. I think this is wrong because they did not divide by sqrt3 to take into account going from WYE to Delta. What do others think?



Another way to tackle this problem is ,

*Power of the primary = Power of the secondary *

1.7321 x V [SIZE=8pt]primary(Line to Line)[/SIZE] x I [SIZE=8pt]primary (Line)[/SIZE] = 1.7321 x V [SIZE=8pt]secondary ( Line to Line )[/SIZE] x I [SIZE=8pt]Primary (Line)[/SIZE]

1.7321 x 15(kV) x 62.7(Amp) = 1.7321 x 6.6 (kV) x I (Amp).

And that is your I you looking for (I line) , just solve for I (Amp) , and that will be 142.5 aprox 143 Amp .


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## EEVA PE (Sep 22, 2011)

I agree with the above posts, but after looking at the Power Reference Manual - Camara's on page 34-9, Figure 34.7 b&amp;c, I see the sqrt3 used which hurts my head and makes the solution fuzzy when compared with the above posts. If I follow the current using these diagrams, I get a turns ratio factored in and also dividing or multiplying by sqrt3 included.

For example:

In 34.7B: the delta-wye. Current (I) starts from the delta and when it is leaving the WYE it is aI/sqrt3.

In 34.7C: the wye-delta. Current (I) starts from the WYE and when it is leaving the delta it is (sqrt3)X(a)X(I)

A little confused: If I follow the logic from above posts for delta-wye or wye-delta, there should only be the turns ratio (a) as a factor.


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## Wildsoldier PE (Sep 22, 2011)

EEVA said:


> I agree with the above posts, but after looking at the Power Reference Manual - Camara's on page 34-9, Figure 34.7 b&amp;c, I see the sqrt3 used which hurts my head and makes the solution fuzzy when compared with the above posts. If I follow the current using these diagrams, I get a turns ratio factored in and also dividing or multiplying by sqrt3 included.
> For example:
> 
> In 34.7B: the delta-wye. Current (I) starts from the delta and when it is leaving the WYE it is aI/sqrt3.
> ...


This formula always get you the *LINE CURRENT* regardless if it is Y - delta or Delta - Y or Y-Y or Delta delta...but this is as i said only Line Current NOT phase current because phase current is Line current divided by sqrt(3) in delta. The formula is

*Iline = S(total NOT Sper-phase) / [(Vline-to line)(sqrt(3)]*

This is how i see it in Y you have 3 voltages, Vphase, Vline to line, and V 3phase voltage. Lets say for example comercial voltages for small buildings you can get Vphase=120V, Vline to line=120(sqrt(3))=208V and you have 3 phase Voltage, V3phase=120 x square(3) x Square (3) = 360Volts

So for example when you are calculating Iline but they give you phase voltage of 120V and *total* power is 180KVA ...to get the I line you need to divide both total numbers 180kva/(120 x square(3) x square (3)) = 500Amps this 500 amps will be in phase A, B and C, but remember they are out of phase by 120 degrees depending in the sequence

I hope this helps.


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## EEVA PE (Sep 23, 2011)

Wildsoldier said:


> EEVA said:
> 
> 
> > I agree with the above posts, but after looking at the Power Reference Manual - Camara's on page 34-9, Figure 34.7 b&amp;c, I see the sqrt3 used which hurts my head and makes the solution fuzzy when compared with the above posts. If I follow the current using these diagrams, I get a turns ratio factored in and also dividing or multiplying by sqrt3 included.
> ...




Thanks. I found a good problem 3.2 in the 6min solution manual that makes you go through two transformers starting from the WYE-Delta, then continuing through Delta-WYE. If I stay with the line values it is easy, and sqrt of 3 does not come into the picture. If I start converting by using the phases requires a little more work, but I get the same answer.

Thanks for the tip of using Power of the Primary = Power of the Secondary.

Now I could put this one to rest.


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## Jonjo (Sep 23, 2011)

Jonjo said:


> EEVA said:
> 
> 
> > This is 3 phase problem. If there is a transformer with the high side (15KV) delta and low side (6.6KV) WYE. The current is 62.7 on the high side, what is the current on the low side?
> ...



For any config this formula is valid , after you apply this formula you have to see if your trafo side you solve for is in Delta or Wye config and then you can figure the current you need.

Because the formula above always show the Line current ( I line )

If is for the Delta side, just divided by √3 , =&gt; phase current, = line current /√3

If it is for Wye side, just leave the current as it is =&gt; line current = phase current.


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