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Topic Title: Amtech - Ze for Transformers
Topic Summary: Different Ze for different transformers
Created On: 12 October 2017 12:02 PM
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 12 October 2017 12:02 PM
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Apostolos1983

Posts: 133
Joined: 03 December 2012

Good morning everyone.
When you select a 500kVA transformer, Amtech automatically sets Ze to 0.0166 giving an Earth Fault of 13.881kA.
When you select a 1000kVA transformer, the respective values set by the program are Ze=0.0086 and prospective Earth Fault of 26.739kA.
When you select a 2000kVA transformer, the program sets Ze=0.0046 and Earth Fault 49.802kA.

So there is a patern here. Every time you double the kVA of the transformer, the Ze halves and the prospective Earth Fault doubles.

However, excuse my question:
The Ze should be related firstly to the Earth Mat and the Earth Electrodes we install around the substation.
Isn't this the case?
Or the winding of the transformer is the most important criterion for the Ze?

If I seam a bit confused is because I am... :-)
Any help is welcome!!!
 12 October 2017 12:31 PM
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AJJewsbury

Posts: 16114
Joined: 13 August 2003

As long as you have a TN system (rather than TT) all the earth fault current may be assumed to return through the metallic protective conductors - the earth electrode isn't involved in a TN earth fault loop.
- Andy.
 12 October 2017 12:34 PM
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Apostolos1983

Posts: 133
Joined: 03 December 2012

The earth electrode (earth mat in the substation plus additional electrodes) does not distribute the fault current inside the ground??? How can you say that the earthing system is not involved?
???
 12 October 2017 02:47 PM
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mawry

Posts: 324
Joined: 26 April 2004

Draw out the fault loop.
Assuming you have a PNB arrangement, starting from the transformer through to your main LV switchboard the distribution system, through the fault, back through the cpc(s), to the LV switchboard, through the N-E link and back to the transformer star point,

That should demonstrate why the earth electrode isn't involved (if the description above doesn't).
 12 October 2017 02:50 PM
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Apostolos1983

Posts: 133
Joined: 03 December 2012

Mawry, then why do we need the earth electrode if it does not diffuse the fault current to earth? I understand what you say. The way you describe it only the winding of the transformer participates. What is then the use of the earth electrode?
 12 October 2017 03:03 PM
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gkenyon

Posts: 4982
Joined: 06 May 2002

Originally posted by: Apostolos1983

The earth electrode (earth mat in the substation plus additional electrodes) does not distribute the fault current inside the ground??? How can you say that the earthing system is not involved?

???
It is not involved in the intended earth fault path in a TN system for a fault of negligible impedance to an exposed conductive part.

For example, in a TN-S system, the earth fault path includes:
- the transformer
- the line conductor to the point of fault.
- the exposed conductive part at the point of fault.
- the circuit protective conductors
- the main earthing conductor / busbar / the main earth terminal
- the supply protective conductor back to the star point of the transformer

-------------------------
EUR ING Graham Kenyon CEng MIET TechIOSH
G Kenyon Technology Ltd

Web-Site: www.gkenyontech.com
 12 October 2017 03:18 PM
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Apostolos1983

Posts: 133
Joined: 03 December 2012

Owow this freaks me a bit.
Then why do we connect the earth bar to an earth electrode/earth mat??? What is the usefulness in this if the current from the earth bar does not go to the mass of earth but continues straight to the star point of the Tx???
 12 October 2017 04:51 PM
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AJJewsbury

Posts: 16114
Joined: 13 August 2003

Then why do we connect the earth bar to an earth electrode/earth mat???

Mostly to provide a reference for the LV system - without that the whole system could be raised to a significant voltage above true earth by capacitive coupling from the HV side (possibly several kV). As people could touch true earth and things connected to the LV protective conductors at the same time, it's generally good to have them at about the same potential. It would also act as part of the earth fault path for any TT installations supplied. Generally the resistance to earth doesn't have to be particularly low if the electrode is only serving the LV side - often anything below 20 Ohms is acceptable.

If the electrode serves the HV as well as LV side then things get a lot more interesting - especially if there's no metallic path back to the HV star point (e.g. because there an overhead line on the HV side) - then the electrode is part of the HV earth fault path and much much lower resistances are needed (traditionally below 1 Ohm, but possibly even lower by modern calculation methods).

Any electrode's resistance to earth (i.e. the resistance of the soil around the bit of metal you bury) will by many many times the resistance of any metallic protective conductor - so if you're talking of Ze in the order of a small faction of an Ohm or PFC in the kA range you're talking about a fully metallic earth fault loop - and nowhere relying on passing a current though the general mass of the earth.

Sometimes you need an extensive electrode system not only to reduce the overall resistance to earth, but to limit the voltage that appears across (each metre say) of the surface - i.e.to limit the voltage someone walking in the vicinity might have but between their feet, or the voltage between bonded metal parts and the adjacent soil surface.

- Andy.
 14 October 2017 08:06 PM
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Nedryerson

Posts: 109
Joined: 12 December 2009

Hi,

The fault level is at the LV terminals of the transformer only.
Nothing to do with any external circuits.

Ned
 14 October 2017 08:25 PM
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mapj1

Posts: 9707
Joined: 22 July 2004

Please be very clear when you talk about fault currents, or prospective faults, to 'earth', you need to be careful to distinguish

1) 'Terra Firma earth', the muddy stuff under your shoes, where any real fault will be ohms, if not tens of ohms.

2) CPC 'earth' a copper path back to the transformer windings, which will be similar to the neutral current path, and therefore will scale with transformer ratings.

The electrode at the substation only helps in the first case, for fallen wires in puddles etc.
The second case is to catch wires damaged in in earthed trunking or similar..

-------------------------
regards Mike
 14 October 2017 09:25 PM
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Nedryerson

Posts: 109
Joined: 12 December 2009

Hi,

I believe Apostolos1983 was confusing the 'Ze' of an 'installation' with the L-N quoted impedance of a distribution transformer.

0.0166 ohms x 13.881A = 230 Volts

Input a kVA rating of a transformer into Amtec will produce the above numbers based on standard dyn11 ONAN distribution transformer data. (Can't remember if one inputs the transformer impedance also)

This is the PSCC/impedance at the origin point before any form of earthing conductors are are made to the star point.

Ned
 16 October 2017 08:40 AM
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Apostolos1983

Posts: 133
Joined: 03 December 2012

Yes Nedryerson, however Amtech uses this Ze to start calculating the Zs further down the network adding cables etc.
So IT IS the Ze of my installation. Isn't it?
 16 October 2017 09:10 AM
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Nedryerson

Posts: 109
Joined: 12 December 2009

Yes, that's your starting point before you start adding the downstream cabling and any earthing conductors be they cable cores, supplementary external conductors or armour or both or all.

If you intend to solidly earth the neutral then forget about earth rods all L-N or L-E fault currents will go straight back to the star point.

Ned
 18 October 2017 09:26 AM
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Apostolos1983

Posts: 133
Joined: 03 December 2012

Nedryerson, I think I start to see some light.
However what I do not understand is where does Amtech find these predetermined Ze values for the transformers.
As I said in the very first post Amtech spits out
For a 500kVA: Ze=0.0166 & Ief=13.881kA
For a 1000kVA: Ze=0.0086 & Ief=26.739kA
For a 2000kVA: Ze=0.0046 & Ief=49.802kA

As you very correctly said in every case Ze x Ief = 230V
So I believe that amtech calculated the Ief for each transformer rating according to the Ze value for each transformer.
So far so good...
1) Where however did Amtech get these typical values for the Transformers?
2) And why for double the kVA the Ze halves??? Is there s correlation between the kVA of the Transformer & the Ze of the transformer?
3) As I understand, this Ze is the resistance of the winding of the transformer per phase. Am I wrong?
Kind Regards
 18 October 2017 11:17 AM
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AJJewsbury

Posts: 16114
Joined: 13 August 2003

As I understand, this Ze is the resistance of the winding of the transformer per phase. Am I wrong?

That's an approximation. Impedance rather than resistance would be a better approximation, and taking account of the L-N (or L-L) loop impedance on the primary side would be a further improvement (as the power to drive the LV fault current has to come from somewhere) - although for a transformer connected to the HV national grid that's usually so small as can often be ignored in practice.

2) And why for double the kVA the Ze halves??? Is there s correlation between the kVA of the Transformer & the Ze of the transformer?

I would imagine so - after all, all else being equal, if a transformer windings have to carry twice as much current in normal service they're likely to have about twice the c.s.a. (or have two secondary windings in parallel) and therefore half the resistance (plus or minus a bit for all the other approximations).

- Andy.
 18 October 2017 11:40 AM
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Apostolos1983

Posts: 133
Joined: 03 December 2012

Thanks Andy.
Google is a powerful thing :-)
So after research, the reason for Ze halving each time you double the kVA of your Tx is that this Ze is actually your transformer impedance in Ohms.
If you assume 5% impedance both for a 500kVA & a 1000kVA transformer, the Zbase that you multiply the 5% to get your ohms value halves every time you double your kVA rating.
Zbase = V2/S
In both cases V2=400x400, however when you double the S, the Zbase halves.
Then Ze=Zbase x Z(%), so the Ze in Ohms also halves since the Zbase halves.
Cheers
 18 October 2017 11:46 AM
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kenelmh

Posts: 92
Joined: 17 February 2012

Hi,

It would be useful to take note of the formula by which Amtech calculates the fault current (or that that they should be using). This is... Isc = 100/Z% x FLC. Where Z% is the expression of the percentage of primary voltage required to flow full load current on the secondary when shorted, usually this is around 5%, so 100/5 x FLC = Isc (single line to earth).

Once you have Isc, you can apply Ohms law to find the internal impedance of the transformer LV winding and thus you have a 'Ze' figure for any cable directly fed from the transformer.

As people have said, the earthing of the transformer and it's star point is to enable earth fault protection. Unearthed (at all) star points I think are rare as they can suffer arcing grounds if the connected cable to ground capacitance is sufficient.

I hope this helps some.
 22 October 2017 07:14 AM
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Nedryerson

Posts: 109
Joined: 12 December 2009

Apostolos1983,

I don't think there is a lot I can add to the previous two contributors.

Not sure if Amtec calculates anything or assumes typical data for standard distribution transformers.

Typical data for 415V transformer


kVA R X

200 0.012 0.0423
315 0.0071 0.027
500 0.0038 0.0173 Impedance = 0.0177
800 0.0022 0.011
1000 0.0016 0.0087
1500 0.00099 0.0068
1600 0.0012 0.00584
2000 0.00079 0.0056

Amtec gives 0.0166 for your 500kVA transformer but its near enough and remember in reality the fault level will be lower depending on the HV network impedance which is taken as zero (or assumed to be 250MVA or the rating of the HV switchgear) . So using zero you have a safety factor built in before you begin your LV network calculations.

Well. must get on,

Ned
 22 October 2017 07:16 AM
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Nedryerson

Posts: 109
Joined: 12 December 2009

Oh, table didn't come out too well

Ist column is kVA
2nd column is Resistance
3rd column is Reactance
Statistics

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