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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)
 
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.
 
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
 
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.
 
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.
 
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.Ohm value of transformer DSC_0527.JPG - EletriciansForums.netOhm value of transformer DSC_0528.JPG - EletriciansForums.net

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.
 
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?
 
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 Ohm value of transformer DSC_0529.JPG - EletriciansForums.net

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%

Ohm value of transformer DSC_0530.JPG - EletriciansForums.net
 
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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

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?
 
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.
 
Alright, here is an example:

View: https://Upload the image directly to the thread.com/iC6K9mj


Using Table 9:

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



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?
 
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)
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Here is how I would use the figures:
Ohm value of transformer DSC_0531.JPG - EletriciansForums.net

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)
 
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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.
 

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