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Even so, a short circuit at the end of a 137m run on a 20A cable with a 20A overcurrent device is likely to take 20+ seconds to operate.

Just using the loop resistance of the final circuit would demonstrate it.

It's a big factor out.
 
Even so, a short circuit at the end of a 137m run on a 20A cable with a 20A overcurrent device is likely to take 20+ seconds to operate.

Just using the loop resistance of the final circuit would demonstrate it.

It's a big factor out.

It is, but remember its not in the NEC. Hence why I'm trying to wrap my mind around it.
 
Err not really.

I am not familiar with the tables, or figures, but if I understand it, you are talking the R + X values for each cable section, converting each to a Z value (ignoring the phase angle) and adding all the Z's.

Yes, simply adding resistance and reactance together to get Z. Then using Z for R1+R2 in addition to Ze.

I don't know how to work phase angles. But I'm willing to learn.



Technically this is incorrect, if you are adding phasor values (vectors) then you must add including the phase angle - imagine one part of the cable is pure inductance R=0, X=1 ohm (Z=1), and one is pure capacitance again R=0, X=1 ohm (Z=1).

Using your method this would give a total of 2 ohms, however if you remember the phase angle, this is 90degree leading for 1 and 90degree lagging for the other, giving a total answer of 0 ohms.

In practice all the cables are likely to have a similar x/r ratio - so won't actually be radically different like my extreme example.


How much of a difference? Does the short circuit power factor of the transformer make any difference? How about large cables such as over 4/0 (107.2 mm2) where X begins to dominate?


Add all the resistances and all the reactances first, then convert to Z

As an aside, you appear to have a shower load ~20A at 120V at the end of a cable 137 metres long - is this realistic?

Sadly, in the states it is often the case when dealing with parking lot lighting, ball field lighting, caravans, rural properties, irrigation and oil fields, ect.

Things are becoming live, breakers aren't opening and the NEC still refuses to add disconnect times are at least a limit on feeder/final circuit runs.


We would usually have the dist board much closer so it is supplied with larger cables, in your case, you are likely to drop ~18V or so. (obviously I don't know what the actual current is, but even so it looks a bit wrong)
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Here is how I would use the figures:
View attachment 58222

Note, I think the layout is important as it helps, also to signify the reactive component I used the prefix j in the layout.

I haven't checked the values are correct from the table as I wouldn't know which ones are correct or not, so I just used what I understand are the correct R & X from your calcs.

Btw in the UK for normal circuits - basically anything below ~25mm^2 we only use the resistance, as the X/R ratio is ~10 and sqrt(10^2 + 1^2) is pretty much 10.

In this case, we would either use the standard, or the on-site guide, this shows for say a 2.5mm^2 cable with the same size earth/cpc a resistance of 14.82 ohm per kilometre so 137M would be 2 ohm amd then use that (obviously this is different as your choice of standard cable is slightly bigger than our standard cable here)


I'll look at your calcs. I wish I could do them in vector, polar and cartisan forum.
 
It depends on the situation.

If it's normal domestic / commercial / industrial situations the cables tend to be small in comparison to the distance between them, i.e. the insulation thickness is comparable to wire diameter, then the X/R ratio is >>10 (actually it's the R/X that's greater!) and therefore you could just use the resistance value alone, the error is very small.

If however you are dealing with specialist situations, where the X/R ratio may be closer to 1, then the reactance must be taken into account, in these conditions the X/R ratio can be radically different to each other over various parts of the circuit so it would make a significant difference.

The situations where this occurs is where the insulation/gap between conductors is small compared to the size of the conductor, and where the resistance is low, all conductors have a self-inductance and this becomes more significant as the resistance falls.

You can see this in your table, smaller cables have a higher reactance, but a much higher resistance when compared with the thicker cables.

These sort of large cables are usually used only at the biggest of transformers, and usually for such short lengths, that they are so low an impact overall - your example is like this, without including way to many significant places, the resistance and reactance of the big cables are lost in the overall picture as the long small cable characteristics dominate.

If you are dealing with large generating stations, or heavy industrial plants however, then you can have very large cables over respectable lengths, and no small long cables to dominate the calculation - in these cases it is very important to use the self and mutual inductances to get the reactance, and include it in the calculations
 
Makes sense. With larger sizes, does adding X and R together still give acceptable results?
No, you have to add up the Rs and Xs terms as two separate values, then finally covert to a magnitude using:

Z = sqrt(R**2 + X**2)

Then from V/Z you get the current magnitude (which is usually the most important factor, as phase angle is unlikely to be too crazy).

Also for transformers you typically find X >> R so treating it as a simple resistance is going to give very wrong answers for the faults that matter to the first DB or so.

By time you get to the final circuits (like your street light example) it is going to be dominated by the final wire, and that is almost certainly going to be dominated by R if it is a circuit in the tens of amps or less region (as already well explained by Julie).
 
As an aside a quick and dirty approximation for the complex magnitude is to add the biggest magnitude and half of the smallest magnitude.

For example, if you had 1 + j0.7 the correct answer is sqrt(1*1 + 0.7*0.7) = 1.2206... or the approximation is 1 + 0.7/2 = 1.35 which is around 10% different. In fact the largest error is 11.8% so it is handy as a mental check on the results of such calculations.

Why do I know that? Well it is a very fast approximation on a computer where you can take absolute values, compare, and then divide smaller value by 2 using a bit-shift, and finally add. And all works fine in integer arithmetic (assuming no overflow) and can be done in around 6 instruction cycles!
 
No, you have to add up the Rs and Xs terms as two separate values, then finally covert to a magnitude using:

Z = sqrt(R**2 + X**2)

Yup thats what I did.

but I was wondering of doing it a different way mentioned in this thread. Adding all the Rs together and adding all the Xs together separately. Then using a2+b2=c2. Never done it that way, but was interesting that it was mentioned.



Then from V/Z you get the current magnitude (which is usually the most important factor, as phase angle is unlikely to be too crazy).

Also for transformers you typically find X >> R so treating it as a simple resistance is going to give very wrong answers for the faults that matter to the first DB or so.

This is where I need your help. How do I do the math? Or must I do it via sequence components?

I don't know how to get the positive, negative and zero sequnce of typical pad-mount and pole top transformers. Especially T-T connected units which are common in the States.

Although the ones Julie posted for 50 cycle units are good enough?

By time you get to the final circuits (like your street light example) it is going to be dominated by the final wire, and that is almost certainly going to be dominated by R if it is a circuit in the tens of amps or less region (as already well explained by Julie).

In agreement on this one. :)
 
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As an aside a quick and dirty approximation for the complex magnitude is to add the biggest magnitude and half of the smallest magnitude.

For example, if you had 1 + j0.7 the correct answer is sqrt(1*1 + 0.7*0.7) = 1.2206... or the approximation is 1 + 0.7/2 = 1.35 which is around 10% different. In fact the largest error is 11.8% so it is handy as a mental check on the results of such calculations.

Why do I know that? Well it is a very fast approximation on a computer where you can take absolute values, compare, and then divide smaller value by 2 using a bit-shift, and finally add. And all works fine in integer arithmetic (assuming no overflow) and can be done in around 6 instruction cycles!

bit-shift? over flow? instruction cycles? :flushed::flushed:

How does BS7671 tackle the issue of say of including a transformer in loop impedance calculations?

11.8% is good enough for me.

Increase the final number by 11.8% for PSC, decrease the final number by 11.8% for max disconnection times.

Or does assuming 70*C-75*C conductor temp for R work out enough for loop impedance?
 
Yup thats what I did.

but I was wondering of doing it a different way mentioned in this thread. Adding all the Rs together and adding all the Xs together separately. Then using a2+b2=c2. Never done it that way, but was interesting that it was mentioned

You didn't, you converted all the sections from R and X into individual Z values, and then added the vector/phasor values using their magnitudes only.

Pc is stating the same as me, add all the Resistances to each other, and all the Reactances to each other and THEN convert to the Z value as sqrt(R^2 + X^2).

The resistance gives a voltage drop in phase with the current, but the reactance gives a voltage drop at 90degree to the current
This is where I need your help. How do I do the math? Or must I do it via sequence components?

I don't know how to get the positive, negative and zero sequnce of typical pad-mount and pole top transformers. Especially T-T connected units which are common in the States.

The maths for the most part is just simple stuff, for the lv side of transformers you would just use normal resistance and reactance addition that is basic ac theory you would have already done during basic training.

You only need to use symmetrical components if you want to calculate the unbalanced fault levels across transformers including the high voltage network.

They would not normally be used. Typically you would just accept the fault current for an earth fault direct on a transformer is the same as the three phase fault current, however you specifically included a network with the HV included.

For a transformer, all the sequence components are fundamentally the same as the voltage impedance. The only difference is the zero phase sequence one, and the inclusion of this along with how it should be included in the calculation is dependent upon the winding group and connection.

This can either be worked out by understanding the connections or by reference to sequence diagrams such as these:Ohm value of transformer DSC_0532.JPG - EletriciansForums.net
Ohm value of transformer DSC_0533.JPG - EletriciansForums.net

In the US the calculation processes are described in the IEEE red book and the IEEE buff books (at least they are from my mid '80s versions)

I would assume they have typical values for the specific equipment you use there.
 
bit-shift? over flow? instruction cycles? :flushed::flushed:
Unless you write DSP software (or need to approximate things on a very small integer microprocessor) you don't need to worry about it!

How does BS7671 tackle the issue of say of including a transformer in loop impedance calculations?
In the vast majority of cases the "wiring regulations" only deal with two effects:
  • Voltage drop (L-N)
  • Fault current (L-N and L-E)
Usually you simply have a value for PSSC & PFC at the source on an installation and accept it (or measure it if already installed) and from those you can set limits for final circuits to meed VD limits and to clear faults (and along the way not to blow up breakers).

If you get a copy of the IET's "Electrical Installation Design Guide" (ISBN 978-1-78561-471-2 for the 4th edition sitting here on my desk just now) it has some worked examples in chapter 6.

Again as this is looking at LV supply to final installations (not the HV or DNO's side) they are only really looking at an earthed star-point transformer.

An as aside, the IET book shows the "typical" let-through energy I2t of a fuse and a breaker in Figure 4.1, but the breaker's cure is somewhat over-optimistic! Compare it with the real world MCB characteristics from, say the Hager commercial catalogue here:

Page 96 for MCB let-through curves.
Page 127 for MCCB curves

11.8% is good enough for me.

Increase the final number by 11.8% for PSC, decrease the final number by 11.8% for max disconnection times.

Or does assuming 70*C-75*C conductor temp for R work out enough for loop impedance?
No! Just compute it the obvious way:
Z = sqrt(R^2 + X^2)

For all numeric range most folk work with that is perfectly good (again, not an issue for this forum, but you can have issues of the square terms overflowing/under-flowing so doing Pythagoras correctly is subtly more difficult).
 
Unless you write DSP software (or need to approximate things on a very small integer microprocessor) you don't need to worry about it!


In the vast majority of cases the "wiring regulations" only deal with two effects:
  • Voltage drop (L-N)
  • Fault current (L-N and L-E)
Usually you simply have a value for PSSC & PFC at the source on an installation and accept it (or measure it if already installed) and from those you can set limits for final circuits to meed VD limits and to clear faults (and along the way not to blow up breakers).

If you get a copy of the IET's "Electrical Installation Design Guide" (ISBN 978-1-78561-471-2 for the 4th edition sitting here on my desk just now) it has some worked examples in chapter 6.

Again as this is looking at LV supply to final installations (not the HV or DNO's side) they are only really looking at an earthed star-point transformer.

An as aside, the IET book shows the "typical" let-through energy I2t of a fuse and a breaker in Figure 4.1, but the breaker's cure is somewhat over-optimistic! Compare it with the real world MCB characteristics from, say the Hager commercial catalogue here:

Page 96 for MCB let-through curves.
Page 127 for MCCB curves


No! Just compute it the obvious way:
Z = sqrt(R^2 + X^2)

For all numeric range most folk work with that is perfectly good (again, not an issue for this forum, but you can have issues of the square terms overflowing/under-flowing so doing Pythagoras correctly is subtly more difficult).


Alright, but my understanding is that transformer reactance will change the numbers in reality, so Z = sqrt(R^2 + X^2) is not technically accurate for mains and sub board faults.

I will use your equation of 1/2 the smaller number for now.

But still struggling to find the X value of trafos- I guess this is my main concern now.
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You didn't, you converted all the sections from R and X into individual Z values, and then added the vector/phasor values using their magnitudes only.

Pc is stating the same as me, add all the Resistances to each other, and all the Reactances to each other and THEN convert to the Z value as sqrt(R^2 + X^2).

Ah! Ok, that makes more sense now! :):)


The resistance gives a voltage drop in phase with the current, but the reactance gives a voltage drop at 90degree to the current

Agree. And a DC off set during a fault if I'm on the right tack?


The maths for the most part is just simple stuff, for the lv side of transformers you would just use normal resistance and reactance addition that is basic ac theory you would have already done during basic training.

No basic training in this regard. As I said, electricians in the US do not calculate earth fault loop impedance as its not required or mentioned in the NEC. We do not carry a multi function tester and most of us have no clue what Ze, Zs or R1+R2 even means. A lot of us couldn't even tell you the per foot impedance of #12 (3.31mm2) or #10 (5.26mm2) even though we run yards of it every day.

We only size the wire based on code rules governing ampacity, terminal temps (60*C vs 75*C) and de-rating factors. We consider voltage drop after that fact, based only on our discretion. Voltage drop limits are not mandated by NFPA 70. We can legally have 20% voltage drop to any load if we deem it acceptable. Further there are no disconnection time requirements present in the NEC. If were to run 1000+ feet of 2.08mm2 wire to a swimming pool or light post we could (and can) legally do so...

In fact up until 10 years ago most US electricians mistakenly believed that a ground rod could trip a standard thermal magnetic breaker which resulted in countless tragic electrocutions and injury law suits- one example of a fence that kept becoming energized:

View: https://youtu.be/C5EiSEtxRKU?t=620


We now know better- ground rods don't clear faults- however that is only half the story. A very small minority is starting to realize that having an equipment grounding conductor isn't a guarantee. Things like railing, pools, fences, light poles, water slides, farm equipment, ect is remaining energized because the loop impedance of the circuit is to high to trip an ordinary breaker. If electricians knew about loop impedance and disconnection times these incidents would not be happening.

Lastly this was also what got AFCIs into the NEC, when it was theorized that disconnection times in excess of 6 half cycles were leading to fires. Older residential breakers had a magnetic pickup of at least 20x, some without it entirely.

Hence why I'm inquiring here with questions that are rather elementary and obvious to IEC folks. I mean no irritation hard headiness :)


You only need to use symmetrical components if you want to calculate the unbalanced fault levels across transformers including the high voltage network.

They would not normally be used. Typically you would just accept the fault current for an earth fault direct on a transformer is the same as the three phase fault current, however you specifically included a network with the HV included.

Good to know. I'm willing to assume infinite for the MV network.

However, I'd like to know if I need sequence components for a setup where 480 volts is going to 120/208Y, as is typical in buildings.

For a transformer, all the sequence components are fundamentally the same as the voltage impedance. The only difference is the zero phase sequence one, and the inclusion of this along with how it should be included in the calculation is dependent upon the winding group and connection.

This can either be worked out by understanding the connections or by reference to sequence diagrams such as these:View attachment 58306
View attachment 58307

In the US the calculation processes are described in the IEEE red book and the IEEE buff books (at least they are from my mid '80s versions)

I would assume they have typical values for the specific equipment you use there.


Thanks! But they are missing one of our most common pole top transformers :p

View: https://Upload the image directly to the thread.com/a/PAkh5XJ


T-T, which gives 3 phase power via only two core and coils- lighter and less material over delta-wye from what I'm told.

I know, we like to be different! :)
 
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Agree. And a DC off set during a fault if I'm on the right tack?

Yes, it's a DC offset - starting high and decaying down to zero - like the discharge characteristic of a capacitor.

In the US you have a simplified method of calculating the first cycle and sustained fault currents called the direct method - I assume it still exists in ANSI/IEEE Std 242-xxxx - it was in 242-1986 when I was last forced to use it!

No basic training in this regard. As I said, electricians in the US do not calculate earth fault loop impedance as its not required or mentioned in the NEC.

In terms of basic maths I was referring to standard circuit theory - you would have done something along the lines of calculating the impedance of any circuit containing R, X(L), & X(c).
But still struggling to find the X value of trafos- I guess this is my main concern now.

…...

T-T, which gives 3 phase power via only two core and coils- lighter and less material over delta-wye from what I'm told.

I know, we like to be different! :)

From your photo the impedance is 1.9%, to be sure you would need to obtain the X/R from the manufacturer, in lieu of this I would just take the value from ANSI/IEEE Std 242 - which is around 1.8; given Z and the X/R ratio you can calculate both the X and R for the transformer.

Yeah, we don't use the Tee - Tee connection over here - as a connection it has some advantages, lower iron loss, and smaller construction, the disadvantages are internal due to the interleaving required on the windings, but for us the main issue is how it fits into the greater network - a single phase fault translates to an unbalanced fault across all three phases on the primary, we tend to use Delta - Star, in this case a single phase fault this translates to a circulating current in the primary winding and a balanced fault on the HV.

We tend to run more balanced HV networks - no distributed neutral, so the Tee-Tee just isn't popular
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Oh, forgot to mention:

With a Tee-Tee type connection when you have a phase to earth fault, it isn't the whole of the winding that's in the circuit, so it has lower values of X & R so for example in your case 1.9% with an X/R of 1.8 would give X=1.66% and R=0.923% which you would use for three phase faults.

However for single phase faults then ANSI/IEEE Std 242 indicates typical values are 0.6x X and 0.75x R for a 75kVA transformer

So for a single phase earth fault - you ought to use
1.66% for X(Three phase)
0.923% for R(Three phase)
0.6 x 1.66% = 0.996% for X(single phase)
0.75 x 0.923% = 0.692% for R(single phase)

I would suggest investing in the latest buff book if you are thinking about doing these calculations!
 
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Can I ask about conductors in parallel? What is the proper way to calc impedance for them?
For resistance it is simple - just the usual parallel connection as as you would make them identical runs as far as practical (to share current evenly) the R term is going to be R/2 or R/3 for 2 or 3 in parallel, etc.

However, the inductive term depends on the layout so it is not going to be so simple. For example if you have a bunch of conductors in parallel and significantly separated from another bunch then the inductance won't change by much (i.e. no parallel reduction in X).

But more generally I don't know of any simple formulae for those cases, but probably @Julie. will know from her past work on high power stuff.
 
For resistance it is simple - just the usual parallel connection as as you would make them identical runs as far as practical (to share current evenly) the R term is going to be R/2 or R/3 for 2 or 3 in parallel, etc.

However, the inductive term depends on the layout so it is not going to be so simple. For example if you have a bunch of conductors in parallel and significantly separated from another bunch then the inductance won't change by much (i.e. no parallel reduction in X).

But more generally I don't know of any simple formulae for those cases, but probably @Julie. will know from her past work on high power stuff.


Reactance has me again. Some parallel runs will be in the same conduit.
 
No there isn't an easy answer unfortunately if you have two trifoil cables in parallel then you would treat the whole lot as two parallel paths of Resistance-Reactance in parallel.

If you can calculate using complex numbers it would be the normal equation R+JX = 1/((/R1+jX1)+(1/R2+jX2)) - in the same way you would just use R=1/((1/R1)+(1/R2)) for two resistors in parallel - this is the same thing, only the maths required isn't as simple as it is for resistance alone.

You cannot work out the resistance and the reactance separately.

If you are running single conductors then you can't do this as they influence each other.

The IEEE buff book has a simplified "guesstimate" method using a Reactance factor depending on the size of conductor and Conduit which allows you firstly to calculate each cable's Reactance based on the non-conduit value so for a conductor 250MCM (or kcmil) this would change the overall reactance by a factor of 1.149 , the Reactance itself though is made up of two parts - the self reactance Xa - which is available in a set of tables - around 11.45mOhm per 100ft for 250MCM - and the mutual reactance Xb which depends on the distance between them - say -7.95 for 0.4" spacing so Xa + Xb would be ~ 3.5 x 1.149 .

We had to do this once, a rather large generating station in India had a long run between the generator and the GSU transformer - I think it was around 800 MVA at ~22kV, so we had to run multiple parallel single core cables - the problem is if you think about it it would go something like L1-L1-L1-L1-L2-L2-L2-L2-L3-L3-L3-L3 for the four cables per phase over three phases - the problem is that only the centre two cables per phase are the same (one same phase to the left and one to the right) all the others either have no cable to one side, or have a cable of a different phase - so they would have a different reactance to each other! and therefore would not share the current equally!

We had to calculate the reactance from first principles - it gave such an unbalance that we had to transpose the conductors per phase - four times such that each one cable of each phase was next to the alongside phase once, the free air once (or the other phase) and one of the centre two twice - this caused a balanced reactance between conductors and therefore balanced sharing of currents!

I was much younger I am not sure I could do it now!
 
Wow something just went over my head! Reading (most) of the above reminds of theory we did but of course have never had to use much in the normal work environment hence such understanding of the kind of thing you are talking about tends to atrophy as the years go by. Interesting though about the answer to the power station in India, transposing the phase cores to achieve phase balance. Who thought of that Julie? Ingenious.
 
Wow something just went over my head! Reading (most) of the above reminds of theory we did but of course have never had to use much in the normal work environment hence such understanding of the kind of thing you are talking about tends to atrophy as the years go by. Interesting though about the answer to the power station in India, transposing the phase cores to achieve phase balance. Who thought of that Julie? Ingenious.
It's actually a fairly standard technique, if you look along very long transmission lines, you will find odd towers where the conductors are transposed so the phases end up of similar impedance.

Long before I was a puppy!

Transposition tower:
Ohm value of transformer tpose - EletriciansForums.net
 
It's actually a fairly standard technique, if you look along very long transmission lines, you will find odd towers where the conductors are transposed so the phases end up of similar impedance.
A broadly similar idea to Litz wire.

I have not see it since radios built pre-60s but a quick search shows there are still folk making it!
 

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