Discuss Ohm value of transformer in the Electrical Forum area at ElectriciansForums.net
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.
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:
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)
No, you have to add up the Rs and Xs terms as two separate values, then finally covert to a magnitude using: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!
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
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.
Unless you write DSP software (or need to approximate things on a very small integer microprocessor) you don't need to worry about it!bit-shift? over flow? instruction cycles?
In the vast majority of cases the "wiring regulations" only deal with two effects:How does BS7671 tackle the issue of say of including a transformer in loop impedance calculations?
No! Just compute it the obvious way: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?
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:
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).
- Voltage drop (L-N)
- Fault current (L-N and L-E)
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:
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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).
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
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: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.
Agree. And a DC off set during a fault if I'm on the right tack?
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.
But still struggling to find the X value of trafos- I guess this is my main concern now.
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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!
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.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.
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.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.
A broadly similar idea to Litz wire.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.
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