Transformer primary to secondary short?

  • Thread starter Evil Bunny
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  • #1
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Main Question or Discussion Point

Let's assume a 3 phase wye distribution system with a single leg supplying a typical residential center tapped transformer (like we have in the US)... For simple math, let's say it's a 10:1 transformer with 2400V on the primary and 240V on the secondary.

The center tap on the secondary is tied in with the "return" on the primary and grounded to earth, correct?

If a turn to turn short occurred at a point 1/4 of the way down on the primary (let's say that it shorted between the start of the first winding to a point 1/4 of the way down) we would see only 180V on the secondary and we would see 60V on one side of the center tap and the normal 120V on the other side of the center tap, correct?

If the same type of turn to turn short occurred on the secondary in the same location, we would see the same symptoms, correct?

Assuming all of that is correct, here's my question... If a primary to secondary short occured, at this same location, what would we see on the secondary? Would we see 1800V on the secondary with 1200V on one side of the center tap and 600V on the other side of the center tap? Something tells me it's not quite that simple... We now have the same source current running through two separate windings... What effect does this current running through both windings at the same time have on the output? What role would inductance play in this scenario?
 

Answers and Replies

  • #2
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If a turn to turn short occurred at a point 1/4 of the way down on the primary (let's say that it shorted between the start of the first winding to a point 1/4 of the way down) we would see only 180V on the secondary and we would see 60V on one side of the center tap and the normal 120V on the other side of the center tap, correct?
If I understand you correctly, you mean to say that you shorted out 1/4 of the primary winding? (it doesn't matter which 1/4 portion is shorted out, top, bottom, middle etc)
Then, of course the primary no. of turns become 3/4 of original, hence the secondary voltage
will be 4/3 of original, i.e. 240*4/3 = 320V.
You will see 160V on both sides of the Center-taps. You can never induce different voltage, on the two sides of center tap. Center-tap, by virtue of tapping from the CENTER, divides the voltage equally.

I think you are missing something over here. The voltage induced in the secondary isn't 1-to-1 corresponding with the respective part of primary (i.e. top coil induces voltage in top part, etc). Ask, if something confuses.
Cheers
 
  • #3
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I think you are missing something over here. The voltage induced in the secondary isn't 1-to-1 corresponding with the respective part of primary (i.e. top coil induces voltage in top part, etc). Ask, if something confuses.
Cheers
I think with some of the primary windings out of the circuit, the inductive coupling with the secondary in that section of the winding wouldn't exist, or would be very weak... You're right, I'm missing something here. Maybe it's that I'm envisioning a length of a single coil of wire (on the primary) coupled with another single coil of wire on the secondary side. I've seen the inside of some pad mounted transformers with very thick copper windings and the windings didn't appear to "overlap" each other (meaning there weren't 2 rows deep on the primary side or the secondary side, as far as I could tell), but maybe most transformers aren't set up this way? Or maybe I'm not remembering what I saw correctly.

And we haven't even discussed the main question of the primary to secondary short...
 
  • #4
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First of all, you need to understand, that after you shorted out 1/4 of the primary winding, the total secondary voltage will be 320V (not 180 Volts) as you suggested.
Vsecondary = Vprimary * Turns_secondary/ Turns_primary.

Can you understand this?

So, there you started wrong. I will get to primary to secondary short.
 
  • #5
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Yes, I see I had that part wrong... 7.5:1 instead of 10:1 would give us a greater number on the secondary. I did have that part screwed up. Thanks for correcting me on that.

I suppose now I'm questioning if 3/4 of the primary would couple with the entire secondary or if there would be a "missing" section... maybe I'm not asking it correctly. I hope that makes sense.
 
  • #6
Zryn
Gold Member
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I think with some of the primary windings out of the circuit, the inductive coupling with the secondary in that section of the winding wouldn't exist, or would be very weak
The coils are designed to sit close together to prevent flux leakage. If you lose the first 1/4 of your primary, the resulting field will interact with the entire secondary. Its not like the first 1/4 of the secondary now misses out on flux linkage and doesn't produce a voltage.
 
  • #7
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:smile: Okay... thanks. I get that part now.
 
  • #8
677
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The transformer primary to secondary short was quite challenging and interesting.
Look at these figures.
Tshort.jpg


In Figure 1. There is a short between 1/4 of the primary to 1/4 of the secondary. In this case, the grounds aren't connected. Nothing happens here. Nothing.

In Figure 2. There is same short, but ground is connected. The voltages are shown at the instant of shorting. Since there is a huge voltage difference, heavy current will flow for some time, sparks may arise. But the condition won't las long.
The red lines shows the path of steady state primary currents, Because the path CD and AB are parallel, current will divide. But CD is made of up few turns of heavy secondary, almost all currents will flow through CD effectively shorting out AB.

Figure 3. (This is actually Figure 2 twisted and turned) Since AB is irrelevant as it shorted out by CD, it is shaded gray. Now we effectively have a auto-transformer. From the original spec, we can say that the primary has now nearly 1/4 of turns of original. So, we can calculate that the secondary voltage of the on-shorted part DG.
I have calculated V(dg) = 480 V (approx).

Ask, if you are interested.
 
  • #9
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In Figure 1. There is a short between 1/4 of the primary to 1/4 of the secondary. In this case, the grounds aren't connected. Nothing happens here. Nothing.
I agree with this... no complete circuit. Nothing happens.

In Figure 2. There is same short, but ground is connected. The voltages are shown at the instant of shorting. Since there is a huge voltage difference, heavy current will flow for some time, sparks may arise. But the condition won't las long.
The red lines shows the path of steady state primary currents, Because the path CD and AB are parallel, current will divide. But CD is made of up few turns of heavy secondary, almost all currents will flow through CD effectively shorting out AB.
Good point... I hadn't thought of the conductor sizes of the windings.

Figure 3. (This is actually Figure 2 twisted and turned) Since AB is irrelevant as it shorted out by CD, it is shaded gray. Now we effectively have a auto-transformer. From the original spec, we can say that the primary has now nearly 1/4 of turns of original. So, we can calculate that the secondary voltage of the on-shorted part DG.
I have calculated V(dg) = 480 V (approx).

Ask, if you are interested.
Okay. You're saying that we now have 2400V between C and D, since this is now the path that the primary takes. This part makes sense.

So the DG voltage is now 1/2 of the remaining 2.5:1 (2400/2.5=960) transformer that is still in place. Since we'd only have half of this voltage (960/2=) we see 480V.

Do I have it right?

EDIT: What about V(de)?
 
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  • #10
677
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I agree with this... no complete circuit. Nothing happens.



Good point... I hadn't thought of the conductor sizes of the windings.



Okay. You're saying that we now have 2400V between C and D, since this is now the path that the primary takes. This part makes sense.

So the DG voltage is now 1/2 of the remaining 2.5:1 (2400/2.5=960) transformer that is still in place. Since we'd only have half of this voltage (960/2=) we see 480V.

Do I have it right?

EDIT: What about V(de)?
Hmm, not quite, but it was interesting that you arrived at 480V somehow! :)

We don't have 2400V between C and D.
We have 2400 V between F and D.

So we have 2400 V as primary. How many turns does the primary have?
Primary Turns = Turns in FA + Turns in CD == Turns in FA approx. (Turns in CD << Turns in FC)
Turns in FA is 1/4 of the original turns.
So, effectively, the Turns in primary has been decreased to about 1/4 of original

So, we get 4 times the original voltage in DG = 4*120 = 480V.

(You can do more accurate calculation by taking into account the turns in CD as well)

However remember that, you can't run a the transformer like that and obtain and use 480V in secondary, because, the core in Transformer is designed to used for the original no. of turns. You can't change (decrease) the no. of turns from the original design. It will result in excessive core losses.

Edit: Voltage across CD = 1/2 Voltage across DG = 240 V.
 
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  • #11
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So, effectively, the Turns in primary has been decreased to about 1/4 of original

So, we get 4 times the original voltage in DG = 4*120 = 480V.
This is just a different way of stating what I did... 1/4 of the original turns changes the ratio from 10:1 to a new ratio of 2.5:1 across the entire secondary. This gives us a secondary total voltage of (2400/2.5=) 960V. Since DG is being tapped at the center of this total voltage, then we get half of it... which is 480V.

It wasn't fuzzy math :smile:

So anyway... with all of that out of the way... What would happen in real life if this happened? Would there be an explosion? smoke?
 
  • #12
677
16
This is just a different way of stating what I did... 1/4 of the original turns changes the ratio from 10:1 to a new ratio of 2.5:1 across the entire secondary. This gives us a secondary total voltage of (2400/2.5=) 960V. Since DG is being tapped at the center of this total voltage, then we get half of it... which is 480V.

It wasn't fuzzy math :smile:

So anyway... with all of that out of the way... What would happen in real life if this happened? Would there be an explosion? smoke?
Yeah, you were right (I couldn't understand what you told then. I understand it now. Yours and mine explanation is the one and the same :) ) So, no surprise you arrived at 480V.

If that happens here is a brief explanation of what will happen!

Since the turns in Primary, is 1/4 th the design turns
E = N d(phi)/dt
The flux in the core will be 4 times as much and so will be the Flux density.

This will lead to more than 4 times in increasing in Iron Loss, may be resulting in Red-hot Core! :P

Also, as the core is pushed deep into saturation, it will consume huge magnetizing currents .In this case, I guess more than 10x increase. (No-Load Currents). This will result in excessive copper loss. May be resulting in red-hot copper. Infact due to huge magnetizing currents, there will be large IR drop due to copper resistance in Primary, so, ineffect, we won't get that 480V in secondary (which was obtained by ignoring no load current, which can't be done in our drastic condition!). I guess the secondary voltage will be around 400 V

So, we see that the transformer will be heated excessively so sure the transformer will be completely damaged if left like that for some time. So, may be after some time the enamel of the conductor of primary melts away and direct short may result, deteriorating the condition even more. I don't know whether there will be an explosion, though!
 
  • #13
677
16
These Curves will help Estimate the Increase In Iron Loss (W/Kg) and Increase in No-Load Currents, when the Flux densitiy is increase 4x. Take original Flux Density to be 1.4 T and New Flux Density = 4*1.4 T
fig%201.jpg
fig%202.jpg
 

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