Ohm value of transformer

[Cookie]

For those of you who work with onsite MV-LV transformers, does anyone know the line to earth ohm value of a transformer? I'd like to include that in the loop impedance path. Similarly when stepping down LV to LV.

https://imgur.com/a/PIsy0nt

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https://imgur.com/JwUM0fv

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[pc1966]

Do you mean the winding / cable resistance, or the source (Thévenin equivalent) impedance?

 

[Julie.]

Obviously this changes from size to size, and will be specific to the manufacturer.

However, it isn't used outwith the check for solid connections, as it is not really relevant for anything, if you are calculating fault level/currents then the voltage impedance and winding group are the deciding factors (along with core type if there's no zero sequence path from the windings)

 

[Cookie]

Do you mean the winding / cable resistance, or the source (Thévenin equivalent) impedance?

Yes, transformer P-N impedance + (R1 +R2) for the max loop impedance. If my thinking is correct.

 

[Cookie]

Obviously this changes from size to size, and will be specific to the manufacturer.

However, it isn't used outwith the check for solid connections, as it is not really relevant for anything, if you are calculating fault level/currents then the voltage impedance and winding group are the deciding factors (along with core type if there's no zero sequence path from the windings)

Delta-wye, and I have the voltage impedance listed on the unit.

Basically the MV-LV transformer will be my Ze, however I have no clue how to turn that information into a Ze. Typically for residential guys they can get a Ze via PFC/loop tester.

 

[Julie.]

If you assume an infinite bus on the primary, the maximum current that can flow for a dead short on the secondary will be restricted only by the voltage impedance of the transformer, so for say a typical 1 MVA Tx 5.75% Vimpedance then the current will be limited to 17.4x full load. so if the FLC is 1200A, the fault level is 20.8kA.

You can convert this to Ze' if you want - including any secondary network cable resistance as well of course to get Ze.

In practice the transformer isn't connected to an infinite bus, so whatever the primary impedances are (such as an upstream 20MVA 10.7% transformer) then you can assume the infinite bus is above this, and include both the voltage impedances in series (once corrected to a common base such as the 1MVA in this example) , you could also include the cable/line impedances again transforming to a common base before including.

Typically the base used is 1MVA, 10MVA, 100MVA, or 1000MVA as a nice round figure, but you can use any base at all as long as you convert all impedances to the same base

 

[Cookie]

If you assume an infinite bus on the primary, the maximum current that can flow for a dead short on the secondary will be restricted only by the voltage impedance of the transformer, so for say a typical 1 MVA Tx 5.75% Vimpedance then the current will be limited to 17.4x full load. so if the FLC is 1200A, the fault level is 20.8kA.

You can convert this to Ze' if you want - including any secondary network cable resistance as well of course to get Ze.

In practice the transformer isn't connected to an infinite bus, so whatever the primary impedances are (such as an upstream 20MVA 10.7% transformer) then you can assume the infinite bus is above this, and include both the voltage impedances in series (once corrected to a common base such as the 1MVA in this example) , you could also include the cable/line impedances again transforming to a common base before including.

Typically the base used is 1MVA, 10MVA, 100MVA, or 1000MVA as a nice round figure, but you can use any base at all as long as you convert all impedances to the same base

Correct me if I'm wrong, but impedance volts is for a 3 phase bolted fault? What about a line neutral fault, does the transformer's impedance change?

Worse case is good for PFC, but loop impedance needs to sum the "reciprocal" (lowest fault current) if the math is not perfect.

 

[Julie.]

Correct me if I'm wrong, but impedance volts is for a 3 phase bolted fault? What about a line neutral fault, does the transformer's impedance change?

Worse case is good for PFC, but loop impedance needs to sum the "reciprocal" (lowest fault current) if the math is not perfect.

Impedance volts, is the voltage required on the primary in order to obtain flc on the secondary whilst shorted.

For a delta-star transformer connection with solid earthed star point, the current would be the same as a three phase bolted fault per phase.

You are correct for other faults, such as line-line, in this case you would split the components into positive and negative phase sequence, then perform the fault calculation using the +ve & -ve components.

For line-line-earth, then you need to use the zero phase component as well - this would be dependent on the winding group.

 

[Julie.]

Ok I was so bored - apparently 'we' are watching all the star wars films tonight, so it was either watching them with my husband or cutting my feet and bathing them in a bowl of vinegar!

Since we didn't have enough vinegar I decided to run calculations for three phase, phase to phase, and single phase faults based on example figures - say 1MVA 5.75% with a 20MVA 10.7% upstream, and 477V final circuit.

Here they are attached.

I used symmetrical components to calculate the unbalanced fault conditions.

Note, if you just perform the same calculations using only the one end transformer, then the phase to earth fault current is the same as the three phase balanced fault current, however when the upstream transformer is included, the phase earth current is slightly more than the three phase fault current - as one would expect.

 

[pc1966]

Interesting. Power transformers are a bit out of my field but maybe you can cover the 'regulation' value of 5.75%, 10.7%, etc.

I guess that is mostly leakage inductance (or the equivalent, not resistive loss) but somewhere I remember reading that in the USA some are closer to half that value. Is the larger value a deliberate design aspect to limit the fault MVA, or is it simply a by product of the style of construction?

 

[Julie.]

It's not a regulation value, it's merely a method of measuring the actual impedance of a transformer, do a search on 'transformer voltage impedance' - there will be much better descriptions than I could do!

As for values, yes it depends on the construction, I don't think they are too different between the US and us - I just went for typical values from memory in the example, but here is an extract from the IEEE buff book - it's their recommended practice guide for industrial and distribution systems, although my copy is from 1986 I doubt much has changed

This is a similar extract from the j&P transformer book, so it looks like I over egged the 1MVA value, should have been more 4.75 - 5% rather than 5.75%

 

[Cookie]

Impedance volts, is the voltage required on the primary in order to obtain flc on the secondary whilst shorted.

For a delta-star transformer connection with solid earthed star point, the current would be the same as a three phase bolted fault per phase.

You are correct for other faults, such as line-line, in this case you would split the components into positive and negative phase sequence, then perform the fault calculation using the +ve & -ve components.

For line-line-earth, then you need to use the zero phase component as well - this would be dependent on the winding group.

I 100% agree with everything you said, like given

Two Questions:

1) Will there ever be a time when the phase - ground value is higher in actual reality? I ask because this popular document says it could as high as 125%, but as you say tells me to assume 100% of the L-L-L short circuit value.

Page 7: https://www.mikeholt.com/files/PDF/Short_Circuit_Calculatons_Book.pdf

https://imgur.com/hRaEcB0

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2) If I obtain 66,902 amps for a 2000kva 277/480 volt Dyn11 4.00% Z transformer, how do I turn the L-N short circuit value into a Ze ohm value?

 

[Cookie]

Speaking of regulation, is this a legit calculator?

Voltage Regulation – Voltage Disturbance

voltage-disturbance.com

I'd like to use this daily.

Beyond words I am grateful for your sequence competent calcs. I am terrible with math, but I will try my best and brush up on this subject as I really ought to learn this to heart.

 

[Julie.]

I 100% agree with everything you said, like given

Two Questions:

1) Will there ever be a time when the phase - ground value is higher in actual reality? I ask because this popular document says it could as high as 125%, but as you say tells me to assume 100% of the L-L-L short circuit value.

Page 7: https://www.mikeholt.com/files/PDF/Short_Circuit_Calculatons_Book.pdf

View: https://Upload the image directly to the thread.com/hRaEcB0


2) If I obtain 66,902 amps for a 2000kva 277/480 volt Dyn11 4.00% Z transformer, how do I turn the L-N short circuit value into a Ze ohm value?

If you only have the details of the transformer itself, then yes the 3phase fault current would be identical to the phase earth (or phase neutral) current.

However if you are able to include any upstream impedance, such as the known fault level on the hv bus, or the upstream transformer, or the cable/ohl then it would be different.

This is because the path for pps and nps currents would include the upstream network, whilst the path for the zero phase sequence current depends on the winding group of the transformer.

So if it's unearthed star-star this would give an open circuit for zps - so no earth fault current.

If the winding is delta-star then the path for the zps is around the delta - within the transformer, hence the additional impedance of the upstream network is not included in the zps network, and the earth fault current is therefore higher than the 3phase fault current - the higher the upstream network impedance, the bigger the difference.

In the UK it is often the case that the earth fault current is higher than the 3phase fault current.

This would give the current for a fault right up at the transformer terminals you could work out the equivalent Thevenin impedance as just v/I at this point to give you Ze'

The Ze at the subsequent 'customer intake' is this Ze' + Line resistance + neutral (or earth) resistance

So depending on conductor sizes as the neutral or earth wires are usually different sizes to each other and to the phase size, then you could end up with lower levels the further you get from the transformer.

 

[Cookie]

Alright, here is an example:

https://imgur.com/iC6K9mj

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Using Table 9:

https://imgur.com/ADFOJDS

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75 kva transformer, 1% Z, 23,132 amps 3 phase fault.

Using ohms law I get a transformer value of: 0.00519 ohms

.................................

For the service:

Plastic PVC conduit, 600MCM copper-

(0.039x0.039)+(0.023x0.023)=0.00205 root= 0.0452769256906871

/1000 x 25 feet = 0.0011319231422672 ohm

400MCM copper-

(0.040x0.040) + (0.033x0.033)= 0.002689 root = 0.0518555686498567‬ /1000 x 25

feet= 0.0012963892162464‬ ohm

...................


For the feeder:

# 3 copper in steal conduit-

(0.059x0.059)+(0.25x0.25)= 0.065981 root= 0.2568676702117259 /1000 x 100 =

0.0256867670211726‬ ohm

# 8 copper EGC-

(0.065x0.065)+(0.78x0.78)= 0.612625‬ root = 0.7827036476214992 ‬ /1000 x100


0.0782703647621499 ohm

.................................

The branch circuit, #12 copper, PVC conduit-

live- (0.054x0.054)+(2.0x2.0)= 4.002916 root= 2.000728867188156‬ /1000 x 450 =

0.90032799023467 ohm


ground- 0.90032799023467 ohm






Grand total= 1.912231424058514‬ ohms

So using ohms law at 120 volts I get 62.75396 amps.

Did I do my math right?

 

[pc1966]

Did I do my math right?

Are you sure you got enough significant digits in?

 

[Julie.]

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.

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.

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?

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)
[automerge]1589880502[/automerge]
Here is how I would use the figures:

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)

 

[pc1966]

As Julie already pointed out, it looks a bit odd as a PFC of 63A or so on a 20A breaker is not going to trip fast enough to be acceptably safe (at least by our regs)?

The "rule of tumb" is usually minimum PFC > 5*In so a B-curve MCB hits the instantaneous trip, but obviously that varies by OCPD, and if there is an RCD to cover earth faults.

 

[Cookie]

Its actually an outdoor light in a parking lot to demonstrate one example for where loop impedance may be questionable.

 

[pc1966]

Its actually an outdoor light in a parking lot to demonstrate one example for where loop impedance may be questionable.

Ah, OK that does then make sense.

 

[Julie.]

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.

 

[Cookie]

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.

 

[Cookie]

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)
[automerge]1589880502[/automerge]
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.

 

[Julie.]

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

 

[Cookie]

Makes sense. With larger sizes, does adding X and R together still give acceptable results?

 

[pc1966]

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).

 

[pc1966]

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!

 

[Cookie]

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.

 

[Cookie]

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?

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?

 

[Julie.]

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:

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.

 

[pc1966]

bit-shift? over flow? instruction cycles?

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:

Hager Brochures, Catalogues & Documentations | Hager UK

Discover all the product documentation you need in our brochures & catalogues! Download what you need in one click straight from Hager Official website!

hager.com

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).

 

[Cookie]

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:

Hager Brochures, Catalogues & Documentations | Hager UK

Discover all the product documentation you need in our brochures & catalogues! Download what you need in one click straight from Hager Official website!

hager.com

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.
[automerge]1590141316[/automerge]

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

https://imgur.com/a/PAkh5XJ

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!

 

[Julie.]

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!

 

[Cookie]

Can I ask about conductors in parallel? What is the proper way to calc impedance for them?

 

[pc1966]

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.

 

[Cookie]

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.

 

[Julie.]

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!

 

[Vortigern]

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.

 

[Julie.]

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:

 

[pc1966]

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!

 

[Cookie]

Transposition towers are semi common in the US, so I am familiar with that concept.

But going back- 6-500 kcmils in a 4 inch steal conduit. Or 32 600kcmil sets in 8 separate PVC conduits.

If anyone is curious, here are US cable sizes in mm2:

https://imgur.com/a/lZvLk2D

View: https://Upload the image directly to the thread.com/a/lZvLk2D


500kcmil = 253mm2

600kcmil= 304mm2

 

[pc1966]

Transposition towers are semi common in the US, so I am familiar with that concept.

But going back- 6-500 kcmils in a 4 inch steal conduit. Or 32 600kcmil sets in 8 separate PVC conduits.

The sort of places I have seen conductors that large tend to have them on open trays, probably for ease of installation as much as for air cooling. But equally it is not really my area.

 

[Cookie]

The sort of places I have seen conductors that large tend to have them on open trays, probably for ease of installation as much as for air cooling. But equally it is not really my area.

In the EU I've heard that being the norm, but in North America conduit- (either steal or PVC)- rains supreme.

Here is an example of a chiller breaker feeding two parallel runs in separate conduit:

https://imgur.com/JPeRNMq

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US power companies deliver 120/208Y to large buildings as much as 277/480Y, so high current circuits (and their up-size for voltage drop) is very common. That and NEC article 220 load calcs tend to make feeders and services near double their actual peak current draw.

 

[pc1966]

It looks like they keep the 3 phases in each conduit, presumably to reduce magnetic interaction a bit.

However the earth wires look astonishingly thin in comparison so I presume they are using a RCD for earth fault protection!?

 

[Cookie]

It looks like they keep the 3 phases in each conduit, presumably to reduce magnetic interaction a bit.

Good eyes- correct. IF the conduit is steal it must be done that way.

The only typical exception is PVC conduit when installed between the padmount transformers and the service disconnecting means.

However the earth wires look astonishingly thin in comparison so I presume they are using a RCD for earth fault protection!?

No RCD. We struggle earth fault loop impedance lol.

Earth wires are sized based on NEC Table 250.122:

https://imgur.com/YQzWJKS

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


These values are basically calculated off the adiabatic method... the bare minimum size that won't cause the insulation to melt off during a ground fault.

So translating:

15 amp fuse or breaker requires 2.08mm2 earth wire

20 amp fuse or breaker requires 3.31mm2 earth wire

Here is where it gets rather small:

60 amp = 5.26mm2

100 amp = 8.36mm2

200 amp = 13.30mm2

300 amp = 21.15mm2

400 amp = 26.67mm2

500 amp = 33.62mm2

600 amp = 42.41mm2

800 amp = 53.49mm2

1000 amp = 67.43mm2

 

[pc1966]

These values are basically calculated off the adiabatic method... the bare minimum size that won't cause the insulation to melt off during a ground fault.

But if (as you say) they don't really check Zs, how do they know the let-through I2t for the adiabatic calculation?

 

[Cookie]

But if (as you say) they don't really check Zs, how do they know the let-through I2t for the adiabatic calculation?

Can you educate me on this concept? I have no idea how let-through in relations to Zs works in regards to EGCs.
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If curious in regards to me starting this thread I ask based on 250.4 (A) (5) and 250.4 (B) (4) as indicated in Table 250.122:

https://imgur.com/t5RbR6p

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Code mandates that impedance be low enough to trip the breaker- but does not specify the maximum time or method to determine that hence why I am turning to IEC 60364-4-41 Table 41.1 and BS7671 as a guide.

I know asking about IEC standards while simultaneously working with NFPA-70 is creating some confusion so I hope I can clear that up. Plus its fun learning new stuff

 

[pc1966]

Basically if you have a given fault, and lets assume it is a dead short at the end of your circuit, then if you know Zs you can find the lowest PFC from the minimum supply voltage as:
PFC = Umin / Zs
From this level of fault current you then look up the current-time curve of the breaker or fuse and that gives you the time to disconnect at that level. Once you know the fault current and corresponding time you have the let-through I2t "energy" and that allows you to compute what it will do to the cable.

The usual assumption is the adiabatic case - then the input heat energy is over a short period of time so the out going heat energy can be neglected, which is reasonable for a short circuit type of fault - and so if you assume a closed thermal system, then a given temperature rise depends on the material constant and input energy.

This is normally simplified even more by assuming only a couple standard conductors (Cu, Al, and Fe) and a few common thermal limits (e.g. for thermoplastic and for thermosetting insulation) as a table of a few fudge-factors to use.

It might be easier to give an example. Say you have 63A fuse (common for domestic in UK) and you find your Ze value was 0.35 ohms (typical upper limit for TN-C-S) and you wanted to size a copper earth wire. Lowest fault current would be 95% of U / Ze

PFC = 0.95 * 230 / 0.35 = 624A

Here is a typical fuse curve:

If you look up 624A prospective current for the 63A curve (as far as practical) you see the corresponding time is around 0.4 seconds, so we can compute our let-through energy as:
I2t = 624^2 * 0.4 = 156k (A2s units)

Looking at the values for copper and say 30C initial and thermoplastic (70C cable) we have k = 143 so we can size our earth using:
S >= sqrt(I2t) / k = 395 / 143 = 2.76 mm^2 CSA minimum

If we had Ze = 0.7 ohms (double the fault impedance) then our current is half at 312A but then our disconnect time is 7s hence I2t = 312^2 * 7 = 681k (A2s units) which is about 4.4 times larger and now our minimum earth conductor is given by:
S >= sqrt(681E3) / 143 = 5.77 mm^2 CSA

For a fuse (at least our BS88 ones) they limit the I2t let-through and so worst-case is at lowest fault currents when disconnection takes a long time. Of course once you get in to the ten seconds or more the adiabatic assumption no longer holds so eventually you end up with a steady-state current carrying requirement. But for a fuse the least fault energy is at max PFC.

For a breaker it is more complicated, and you have a massive difference between faults that hit the "instantaneous" trip and those that don't. Also MCB and MCCB are not fault-limiting to the same degree as fuses, so as PFC increases from that trip point you see a moderate increase in let-through energy.
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Just to add - that is why you do the lowest PFC (minimum supply voltage) for the adiabatic check as it gives the longest disconnect time and hence usually the biggest let-through energy.

 

[Cookie]

I wish I could give this reply 100 likes. Love it!

I think I might use this to compare our tables, especially with motor OCPD sizing rules.

 

[Cookie]

But how does this work out with sputtering or arcing faults?