single phase transformer calculation

bee · Jun 3, 2004

100% eff. transformer.

primary is 7200 volt.

2-120 volt loads on leg 1.
first load has 15 ohms of resistance
second load pulls 10 amps

2-120 volt loads on leg 2.
first load pulls 12 amps
second load has 8 ohms of resistance

2-240 volt loads
first load uses 9600 watts
second load uses 4500 watts of power.

I have 3 questions to start out.

first question is what is the current flowing in line 1 with all loads on?

2. what is the current flowing in line 2 with all loads on?

3. what is the current flowing in the secondary with all loads on?

book answers are:
1. 76.75 amps
2. 85.75 amps
3. 81.25 amps

i understand the answers to 1 and 2. the answer to question 3 is my main problem.

to sum all this up my main question is:
answer to question 2 says In line 2 85.75 amps are being pulled with all loads on.
answer to question 3 says 81.25 amps are being pulled in the secondary with all loads on.

how can the whole secondary pull less amps than just line 2 with all loads on?

paul32 · Jun 3, 2004

Re: single phase transformer calculation

I would wonder what they mean by current through the secondary if I were taking the test, since the two halves of the secondary have different currents. It looks like the answer they have is the average of the two, which seems reasonable, but I think the question is a bit strange.

charlie b · Jun 4, 2004

Re: single phase transformer calculation

Paul is right. The currents in the two halves of the secondary are different. The #2 half has 9 more amps than the #1 half, and those 9 amps will flow in the neutral.

You calculate the ?secondary current? of a transformer by looking at the power (VA) it is providing to the system, and dividing by its nominal voltage.

The VA in the #1 half is (76.75) x (120)

The VA in the #2 half is (85.75) x (120)

The total VA is (76.75) x (120) plus (85.75) x (120)

The secondary current is this value divided by 240 volts

But let?s look at how that turns out:

I(secondary) = { (76.75) x (120) + (85.75) x (120) } / (240)

Without actually crunching the numbers, you can factor out the 120/240 (or one half) from this equation, and you will get

I(secondary) = (1/2) x { (76.75) + (85.75) }

This formula is, in fact, the average of the two currents in the two halves of the secondary.

bee · Jun 4, 2004

Re: single phase transformer calculation

Thank you very much Paul and Charlie. Figuring VA every answer works out perfect. I still do not understand how leg 2 can pull 85.75 amps and then the whole secondary just pull 81.25 amps.

the numbers work fine. I get the right answers using the formula. In my head it seems 9 amps should go back through the neutral and 76.75 amps in line 1 so there should be 85.75 amps in secondary. I know the formula is right but I was just having trouble understanding how there is 85.75 amps in leg 2 and just 81.25 in whole secondary. Why do we take the average? where did the 4.5 amps go? I am learning so please hang with me.
Thanks.

charlie b · Jun 5, 2004

Re: single phase transformer calculation

OK. Since you were kind enough to ask, here is THE TRUTH!!

It is, in fact, absolutely meaningless to speak of a value of current in THE secondary. The truth is that there are TWO secondaries. Secondary #2 has a current of 85.75 amps. Of that value, 9 amps goes through the neutral, and 76.75 amps becomes the current in Secondary #1. So if you wish to provide a truthful answer to the question, ?what is the secondary current,? then the only truthful response is to answer that question with a question of your own, namely, ?Which secondary??

That said, it is common enough, and there is some value in it, to speak of average values as though they represented the component as a whole. For example, the average value can tell the mechanical engineer how much ventilation to provide for the space occupied by the transformer, and the average value can be used by the electrical engineer to evaluate whether the transformer has been overloaded. So you ask, ?what happened to the 4.5 amps?? The answer is that nothing happened to it. It is a fictitious, albeit useful, representation of a physical situation.

charlie · Jun 5, 2004

Re: single phase transformer calculation

If this confuses the situation, please disregard since this is just an explanation of one of the Code rules.

If you have a single winding primary and single winding secondary transformer, you are permitted to protect the secondary winding with the primary overcurrent device. However you are not permitted to protect the secondary winding with the primary overcurrent device if you have a two winding secondary. It is possible to have all the load on one of the secondary coils (phase and neutral) and not of the load on the other coil (the other phase and neutral.

That secondary load may be twice as much load as that half of the secondary winding is permitted to have. However, the current on the primary coil will just show as 100% loaded. The bottom line is that a transformer that is protected in that manner is not protected.

bee · Jun 5, 2004

Re: single phase transformer calculation

Thank you Charlie
You have been tremendous help. I can move on now. You got me unstuck. I passed my journeymans test through experior and am taking the masters 10/9.
I have a lot of studying to do and am just trying to make sure I understand the answers when I get them.

single phase transformer calculation

bee

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bee

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