I = VA / (E x 1.732 x Z)?

[JasonCo]

I have a homework question that reads:

A 240v, 3-ph, 3-wire panel is fed from a 100 KVA 3phase transformer that has Z = 1.8% on its nameplate. The transformer is supplied power from a 3-pole breaker in a 277/480v, 3-ph, 4-wire panel. If the Aphase and the Bphase feeder conductors going to the 240-volt panel are "shorted" together inside the transformer enclosure, about ___ amps of current will flow through the transformer winding.

A. 23,148
B. 7,716
C. 6,682
D. 13,365
E. 11,574

Okay so Equation is I = VA / (E x 1.732 x Z)?
So I do 100,000 / (480 x 1.732 x .018) = 6,682

My question is though, that the "shorted" part of the question is throwing me off, its talking about Aphase and Bphase shorting together, sending ___ amps down the transformer windings? So I see that answer choice D is double answer choice C. 6,682 x 2 = 13,365.

Do I have to multiply it by 2, just unsure what to do when the question throws in this "shorted" part...

 

[mivey]

My guess is they wanted secondary fault so you used 480 volts and you should have used 240 volts. That's really the formula for a three phase fault but no biggie.

 

[JasonCo]

My guess is they wanted secondary fault so you used 480 volts and you should have used 240 volts. That's really the formula for a three phase fault but no biggie.

Oh whoops, I misread the question. Thanks!

Can I ask a new question though, because I have other questions that are similar. For instance, I have exact same question but now it says that someone sabotaged the transformer during the night by installing a jumper between the X1 and X4 transformer lugs. The following morning the electrician energized the panel. How many amps of current flows through the jumper?

And another scenario for the same question is that a metal tool was lodged into the panel and shorted the B and C busbars together. About how many amps of current flows through the metal tool.

 

[mivey]

shorted, jumper, metal tool are all pointing to the same thing: a bolted fault (essentially zero fault impedance).

 

[JasonCo]

shorted, jumper, metal tool are all pointing to the same thing: a bolted fault (essentially zero fault impedance).

So if there is a jumper (wires,tools,etc...) between any 2 phases, then it would be the same equation but just don't use Z=impedance, so the equation would be I = VA/ (E x 1.732)? Without the Z (impedance)

Edit: Okay this can't be right... Because none of my answer choices come out to this.

Here let me give you entire question:


A 500 KVA 3phase transformer feeds a 120/208v, 3phase, 4-wire panelboard. The transformer has a Z= 1.2% on its nameplate. The cover and deadfront were left off the energized panel while the electrician went to his service truck. While he was gone someone stuck a metal tool into the panel and shorted the Bphase and Cphase busbars together. About ___ amps of current flows through the metal tool.

 

[mivey]

So if there is a jumper (wires,tools,etc...) between any 2 phases, then it would be the same equation but just don't use Z=impedance, so the equation would be I = VA/ (E x 1.732)? Without the Z (impedance)

No, you still use the source Z. The questions are all telling you you don't have additional Z (like a very long jumper or a semi-conducting tool).

 

[JasonCo]

I'm confused, so nothing changes.. All this information is just meant to throw me off and I just use the normal equation of I = VA / (E x 1.732 x Z)?

 

[mivey]

Here let me give you entire question:


A 500 KVA 3phase transformer feeds a 120/208v, 3phase, 4-wire panelboard. The transformer has a Z= 1.2% on its nameplate. The cover and deadfront were left off the energized panel while the electrician went to his service truck. While he was gone someone stuck a metal tool into the panel and shorted the Bphase and Cphase busbars together. About ___ amps of current flows through the metal tool.

That just means zero impedance added to the source 1.2% impedance.

You would add more impedance if the short occured at the end of the secondary feeder (add conductor impedance) or if you stuck a device in the panel of some known impedance (a wrench has essentially zero impedance but a heating element would have a significant impedance).

 

[mivey]

I'm confused, so nothing changes.. All this information is just meant to throw me off and I just use the normal equation of I = VA / (E x 1.732 x Z)?

They want to know if you realize a short is the same as a jumper is the same as a bolted fault is the same as a wrench across the bus, etc.

 

[JasonCo]

That just means zero impedance added to the source 1.2% impedance.

You would add more impedance if the short occured at the end of the secondary feeder (add conductor impedance) or if you stuck a device in the panel of some known impedance (a wrench has essentially zero impedance but a heating element would have a significant impedance).

Okay thanks, its just very hard for me to wrap my mind around all this. I just thought whatever the nameplate impedance is given, that number will never change. I guess for this particular question, the nameplate impedance of 1.2% will just stay the same.

 

[JasonCo]

They want to know if you realize a short is the same as a jumper is the same as a bolted fault is the same as a wrench across the bus, etc.

Oh yes I 100% realize that they are all the same. But I was thinking that all this information they through into these questions would change the equation a bit, I guess not?

 

[mivey]

Oh yes I 100% realize that they are all the same. But I was thinking that all this information they through into these questions would change the equation a bit, I guess not?

Not a bit. Sounds like someone was tired of the same old boring questions so wanted to spice it up a bit. You will also see a lot of un-needed info in questions: supposedly to weed out the ones that are struggling. I'm not a huge fan of that method but have been guilty of being cutsie like that on quizzes also.

 

[JasonCo]

Not a bit. Sounds like someone was tired of the same old boring questions so wanted to spice it up a bit. You will also see a lot of un-needed info in questions: supposedly to weed out the ones that are struggling. I'm not a huge fan of that method but have been guilty of being cutsie like that on quizzes also.

Can't say that I don't disagree with those tactics, its always good to spice the questions up a bit, and get the apprentices thinking outside the box. So I'm all for it! Now I just need to understand the theory behind why the amp load doesn't change when 2 phases are touching.. Is it because its a Wye system and all the phases are already jointed together with the Neutral so if you have a joint between 2 phases it makes no difference? Its this theory correct?

If so, what if it is a 120/240 Delta (Because I have this same question but its secondary is a Delta and not a Wye)

 

[Phil Corso]

JasonCo...

A phase-phase fault produces a current just 86.6 % that of a 3-phase fault!

Regards, Phil Corso

 

[mivey]

Can't say that I don't disagree with those tactics, its always good to spice the questions up a bit, and get the apprentices thinking outside the box. So I'm all for it! Now I just need to understand the theory behind why the amp load doesn't change when 2 phases are touching.. Is it because its a Wye system and all the phases are already jointed together with the Neutral so if you have a joint between 2 phases it makes no difference? Its this theory correct?

If so, what if it is a 120/240 Delta (Because I have this same question but its secondary is a Delta and not a Wye)

Are you still asking about the original question or are you on to something else? For "2 phases touching" you have a line-to-line fault (answer E). The amp load will most definitely change: a lot!

If that is not the question, then please re-phrase the question.

If you are talking about windings, the windings are electrically isolated and are excited by magnetic flux. You can connect any lead from one winding to any lead from a second winding. Any subsequent connection between the windings is where it can get ugly.

 

[kwired]

Are you still asking about the original question or are you on to something else? For "2 phases touching" you have a line-to-line fault (answer E). The amp load will most definitely change: a lot!

If that is not the question, then please re-phrase the question.

If you are talking about windings, the windings are electrically isolated and are excited by magnetic flux. You can connect any lead from one winding to any lead from a second winding. Any subsequent connection between the windings is where it can get ugly.

He only had two lines involved in the fault ever since post 1, yet everyone's answers until phil corso a couple posts back was based on three phase conductors being faulted.

 

[mivey]

He only had two lines involved in the fault ever since post 1, yet everyone's answers until phil corso a couple posts back was based on three phase conductors being faulted.

That did not appear to be the question. However, I did point out in post #2 that the OP was using the formula for a three-phase fault.

The answer D was double his three-phase calc not because it had anything to do with the term "shorted" but because he used the wrong voltage. The term "shorted" had nothing to do with a 2X multiplier.

The issue the OP was having was with the terminology of "shorted" and what it meant. He then questioned other similar terminology which was also answered. See post #4.

 

[Ingenieur]

3 phase fault
VA/(1.732 V pu Z)

line-lone
VA/(2 V pu Z)

neglecting line Z

actual xfmr Z (ohms) = (Z pu V^2)/ VA

Line-line fault i = V/Z

 

[JasonCo]

Alright I better understand all this much more now. Before I was being mislead by my own wrong terminologies of what I thought things meant. So that's my fault! I get it now. Only thing though I don't get is why would we ever want to calculate a faulty current? If there is a problem and 2 phases are grounding out or shorting out, then wouldn't we as electricians go in and just fix the problem? Why would you ever want to calculate the currents of a fault current when the problem can just be troubleshooted and fixed? This is probably a very stupid question but I had to ask haha

 

[GoldDigger]

Alright I better understand all this much more now. Before I was being mislead by my own wrong terminologies of what I thought things meant. So that's my fault! I get it now. Only thing though I don't get is why would we ever want to calculate a faulty current? If there is a problem and 2 phases are grounding out or shorting out, then wouldn't we as electricians go in and just fix the problem? Why would you ever want to calculate the currents of a fault current when the problem can just be troubleshooted and fixed? This is probably a very stupid question but I had to ask haha

Circuit breakers (and other OCPD) have a maximum interrupting capacity (AIC) and if you exceed that they will be damaged and may even fail to interrupt the fault.
Other components in series along the way have a maximum current they can safely carry while waiting for the OCPD to open (Short Circuit Current Rating = SCCR). To see whether you are within the limitations of all of these installed devices you need to be able to calculate the maximum fault current that they will be exposed to.
It also comes into play when estimating the amount of energy that might be present in an arc fault so that you can determine the right protective equipment to use.
(I.E. what should you be wearing while swinging that wrench around!)