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.

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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

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.