Complex Imaginary Volum 3 question 6

[saw]

Hi,

I am trying to solve the problem 6, Vol 2 of the imaginary book. I blieve the answer is wrong.

Here is the question. Hope some one give me the right answer:

A delta-wye transformer is rated for: 34.5kv/480v, 125 A primary current (L-L). What is the transformer's power rating (KVA) and the secondary current?

The answer in the book is: 4300KVA, 8960A

Thanks

 

[Flyer_PE]

For transformer rating, I get:

34.5kV*125A*sqrt3 = 7469 kVA

Secondary Current:

7469kVA/(sqrt3*0.48kV)=8984A

 

[gte636i]

For transformer rating, I get:

34.5kV*125A*sqrt3 = 7469 kVA

Secondary Current:

7469kVA/(sqrt3*0.48kV)=8984A

I think the current they are giving is phase current in a delta so the rating for the three phase transformer would be...

3*125A*34500V = 12.9MVA

The rating on each single phase transformer is... (not sure if that's actually what they're asking or not)

12.9 MVA / 3 = 4300 KVA

The secondary current would be...

((34500/(480/sqrt3))*125A) / sqrt3 = 8984A

 

[Flyer_PE]

I think they are using one to may sqrt 3's. 8984A at a line voltage of 480 gives you 7.469 MVA.

 

[saw]

I agree with Flyer and company pilot. The only issue that I have is on secondary Wye why don't we divid the voltage by sqrt 3 since wye phase voltage is differ by squrt 3.

 

[Flyer_PE]

In a balanced system with all phases operating, the internal transformer connection doesn't matter. The magnitude of the 3-phase apparent power will be sqrt(3)*VL-L*ILine The only difference between a wye and a delta is whether the sqrt(3) term originates with the voltage or the current. Mathematically, it works out the same either way.

 

[gte636i]

I agree with Flyer and company pilot. The only issue that I have is on secondary Wye why don't we divid the voltage by sqrt 3 since wye phase voltage is differ by squrt 3.

You could divide it by the sqrt of 3, but then you couldn't use the equation Flyer_PE is using. You would need to use the equation 3*VL-N*IL which uses the phase voltage rather than the line to line voltage, both methods work out the same. On the primary side they specified the current as line to line (i.e. phase current in the delta winding) you would need to multiply that by sqrt 3 to get line current on the primary side and use the sqrt(3)*VL-L*ILine equation or alternatively work in per phase and multiply by 3 which is the method I mentioned above in my previous post.

On the secondary side I think Flyer is correct. They used one to many sqrt 3's. Working in reverse with their own solution doesn't come out to the right rating.

 

[saw]

Thanks for all the help. I got it. The hardest part is that I get confuse when to use squrt 3 and when not to use it. Thanks again.