
Software Technical Lead
Premium job
 Hanslope Park, Milton Keynes
 £40,722 to £53,575 + £5k Welcome Package
From the smallest details to leading the biggest decisions, your impact will be out of this world.
 Recruiter: HMGCC

Lead Planning Investment Engineer
 Aberdeen, Aberdeenshire
 up to £62,500
Lead Planning and Investment Engineer (514108) Location: Aberdeen, Perth, Inverness (flexible) Salary: up to £62,500 depending on skills and experi...
 Recruiter: SSE

Electronics Technical Engineer (Design and Operation)
 Switzerland
 £5106.81  £7292.30 per month
Founded in 1954 in Geneva, CERN ? the European Organization for Nuclear Research is the world's most advanced fundamental research institute for particle physics with currently 22 Member States.
 Recruiter: CERN
IET 
search :
help :
home


Latest News:


Topic Title: CTE Transformer Ze calculations Topic Summary: Created On: 15 May 2009 02:23 PM Status: Read Only 
Linear : Threading : Single : Branch 

Search Topic

Topic Tools

15 May 2009 02:23 PM


I have a good knowledge of Ze calculations but when it comes to a typical use of a 240/110vac Transformer with 55055 CTE Secondary, I come unstuck.
How do I calculate the impedence figures for the 110vac CTE secondary circuits. I keep getting dragged into Transformer Regulation conversations. Any help would be much appreciated. Cheers, Steve 



15 May 2009 07:22 PM


Here is a starter.
Reflect the primary supply circuit impedance (Zp) into the secondary and then obtain the phasor sum of primary and secondary impedances . Zsec = Zp *(Vs/Vp)² + Zpu (trans) * (Vs)²/VA + (R1 + R2) ohms The first bit: Zp *(Vs/Vp)² reflects the primary impedance into the secondary Next we take the Zpu (obtainable from the manufacturer) and convert the secondary impedance into a value: Zpu (trans) * (Vs)²/VA Finally R1 + R2 is the the resistance of phase and cpc for the secondary circuit (flex to appliance, etc). I was on the BS 7375 (Distribution of electricity on building sites) committee and I did a lot of work on this  I don't have it to hand right now but if the above doesn't help post again. EDIT: I have not checked my work on this yet, but I would expect that the above formula may need to be adjusted. As written it would appear to fail to take account of the fact that a fault creating an 'indirect contact' shock risk would only usually only involve half of the secondary winding, so a divisor of 2 may be required  I need to check. The formula is taken from Paul Cook's Commentary but he gives a simplification that I do not agree with  this involves the earth fault loop path impedance for the primary circuit, rather than the line neutral impedance, which IMO is the correct method. Regards Geoff Blackwell Edited: 15 May 2009 at 08:07 PM by GeoffBlackwell 



19 May 2009 10:27 AM


Geoff,
Many thanks, For your information. We use calculation software, either Amtech or CableCalc and neither deal with the 'Typical' scenarios we come across almost every day. This being a 240vac single phase supply, with known Ze/PSCC figures, supplying typically a 2kVA 240/110vac (55055)transformer, which goes on to supply a number of CTE sockets. (Portable equipment) They tell us to treat the transformer as a 'load' say 12A primary current, and then calculate Zs (Ze+R1+R2). This is where I am struggling. I simplistically thought I could either get an impedence figure fromm the transformer vendor, and add it to the previously calculated Zs figure, which would form the Ze for the secondary calculations for the final circuits. This may be '*****' and by the looks of it I would need to apply the formula you indicate. Is Zp effectively my calculated Zs figure? Is Zpu an Impedence Per Unit? Clearly I am a bit of a 'dummy'!!! Many thanks, Steve 



19 May 2009 04:51 PM


Zp is the impedance of the supply to the transformer primary. So it is the line  neutral impedance of the supply circuit rather than the line  earth impedance.
Zpu is the quoted per unit impedance (it might be given as a percentage). The figure I used in my research was 0.03 (or 3%) for a 10kVA transformer. This figure was given to me by a manufacturer. Before you can apply all of this you need to determine the form of the output circuit. The most likely fault would be one line to earth and this would involve just one half on the 110 volt transformer secondary (55 volts to earth). So IMO the Zpu should be divided by 2. Note that a fault involving a short between both lines and earth would not produce any fault current in the cpc and, therefore, no touch voltage (I can prove this but that takes pages of calculations ). That should get you started  I will post a sample calculation if I get time. Regards Geoff Blackwell 



19 May 2009 05:00 PM


Geoff,
Many thanks. A sample calc would be greatly appreciated. Steve 



22 May 2009 01:14 PM


The first question I must ask is why do you want to know?
If it is because you think you must meet a 5 second disconnection time because the boys who wrote BS 7671 say so, I fear you have been misguided . Those boys never did understand the principles of the shock protection provided by a Reduced Low Voltage System (RLVS). They also seem to ignore the work of Biegelmeier and Lee, and their own regulations . They continue to insist on a 5 second disconnection time for the RLVS. They even produce a nice table 41.6 backed by 411.8.3. They tell you that if you cannot engineer an impedance that is low enough to meet their nice table then you must use an RCD. Think about it  that is use an RCD in a site transformer that gets dropped and kicked all over the place  yer that will work for a long time won't it. Still I expect the boys on site will test it often to make sure it works . Now according to Biegelmeier and Lee (the touch voltage curve) if the touch voltage (indirect contact shock voltage) is less than 50 volts in dry conditions 95% of the population could tolerate the shock without injury (due to the shock). In wet conditions this would reduce to 25 volts. According to BS 7671  411.3.2.5  For a system with Uo > 50 volts  automatic disconnection is not required in the event of a fault to a protective or earth provide the output voltage of the source reduces to 50 volts or less within what would have been the required disconnection time. Enter the RLVS  it can be shown (by some reasonably tough calculations involving such delights as superposition theorem) that under fault conditions the touch voltage on a RLVS instantly reduces to around 30 volts for some faults and zero for others. This even applies to the three phase version that has a Uo of 63.5 volts. Given this you only need to disconnect to prevent fire and because it is good practice not leave a system in fault. Now the boys who wrote BS 7375 understood this and that standard does not specify any disconnection time. The final joke is that BS 7671 Section 704  704.0 says that this section should be read in conjunction with BS 7375  its a pity they didn't follow their own guidance . Now why I am bringing all this up  well because it is at the heart of why you think you need to get a value for Zs. You will use this value to determine if a circuit meets table 41.6. This will impose restrictions on the maximum length of the circuit and in particular on the length of flex that can be used between . If you fit RCDs to compensate how will you ensure that they remain operational? If you do not impose such limits you will find that all but the most ridiculously long flexes will cause disconnection in a 'reasonable' time. I will assume that you still feel the need to work out Zs so  to the calculations I will divide the circuit into three sections. The calculations will be for a single phase system. 1) the supply circuit i.e the mains and the cable feeding the 110 volt transformer. 2) the 110 volt transformer. 3) the final circuit i.e the flex from the 110 volt transformer to the load (probably a power tool). 1) The supply circuit This supplies the energy to the 110 volt transformer. It is this energy that would feed into a fault on the output side of the 110 volt transformer. We are not concerned with the supply circuit earth fault loop path as it is not involved in the system we are evaluating. We are concerned with the line to neutral impedance of the supply circuit as this will limit the flow of energy into the system. We could determine this line to neutral impedance for the whole supply circuit if we measure it or calculate it, at the input terminals to the 110 volt transformer. This impedance consists of the line to neutral impedance of the DNO supply and the line and neutral resistance of the cable from the supply to the 110 volt transformer. Lets call this Zp. The value of Zp could be used directly in the complete calculation or, more commonly, it could be converted into an equivalent value that can be placed in the transformer secondary. This is a so called reflected value (reflected from primary to secondary) and we do this by multiply Zp by (Vs/Vp)². Note that (Vs/Vp)² = (110/230)² = 0.23. So any primary circuit impedance only causes an effect equivalent to 23% of its value when reflected into the secondary. This tells us that the length of the supply circuit may not be all that significant in the overall results. 2) The 110 volt transformer The windings of the transformer (primary and secondary) both have impedance and this will limit energy flow. Note that in calculations on transformers it is usually best to place all of the variables in either the primary winding or the secondary winding. The other winding then being assumed to be 'perfect' i.e to have zero impedance etc. I will place values in the secondary winding. This means that primary values must be reflected into the secondary. A transformer manufacturer should be able to supply a value for the winding impedances and they will usually give this as a per unit quantity. This can be converted into a value for use in primary or the secondary  we will use it in the secondary. The transformer impedance is given by Zpu * Vs²/VA ohms. Zpu comes from the manufacturer and VA is the power rating of the transformer in voltamperes. 3) The final circuit This is the R1 + R2 of the flex from the transformer to the load. It should really be adjusted for the conductor temperature under fault conditions. An example  we will use a 10kVA 230/110 transformer with a quoted Zpu of 0.03 and a CT secondary winding with a Uo of 55 volts. The full load primary circuit current would be 10000/230 = 43.5 A. This gives us an idea of the size needed for the supply cable etc. Lets assume that the line to neutral supply impedance is 0.2 ohms and that the cable from the mains to the transformer also has a line to neutral impedance of 0.2 ohms at full load temperature. We could simply use the arithmetic sum of these impedances without too much error (the circuit will usually be mainly resistive) so we have: Zp = 0.2 + 0.2 = 0.4 ohms When we reflect this into the secondary circuit of the transformer we have: Zp' = 0.4 *(Vs/Vp)² = 0.4 * 0.23 = 0.09 ohms Next the transformer Z(transformer) = Zpu * (Vs)²/VA = 0.03 * 110²/10000 = 0.03 * 1.21 = 0.04 ohms Now this calculation includes the whole of the secondary winding but, in fact, if we consider a fault between a line conductor and earth on the secondary we would only be using half of the winding (55 volts to earth). So using a value for the whole winding will produce a higher impedance. The trouble is that the Zpu quantity also includes the whole primary impedance and this is required. Now we could eliminate this error but it is probably not worth the effort. Leaving it will produce a higher figure and this error is on the right side for caution. Bringing things together we have an impedance at the output terminals of the transformer of Zp + Z(transformer) and this is approximately: Z (at output) = 0.09 + 0.04 (again just adding arithmetically) = 0.13 ohms. Now bear in mind that this is for a 10kVA transformer  probably supplying a number of socket outlets. If we used a smaller transformer the impedance would be higher. As I mentioned in the last post  Paul Cook simplifies the calculation a bit in his commentary  basically he ignores the transformer impedance. This is a valid approach and I used the same idea back in the mid nineties when I was on the BS 7375 committee. The error is not great as the transformer impedance will be mainly inductive and if we did the calculation correctly (i.e not taking simple arithmetic sums) it would not make much difference. The bit of his approach I disagree with is the use of R1 + R2 on the supply side of the transformer  IMO this should be R1 + Rn. So now we get to the fun part  we are going to use our results to determine that maximum flex length between the transformer and the load (or you could say socket outlet close to the load). Referring to BS 7671 table 41.6  lets assume we have a 16A circuit protected by a 16A type C mcb. The table gives a maximum Zs of 0.34 ohms. We have 0.13 ohms at the transformer so the flex can be R1 + R2 can be a maximum of: (R1 + R2)flex = 0.34  0.13 = 0.21 ohms Now using the data from GN3 for copper conductors We have 12.10 milliohms per metre 20 degrees for 1.5mm² and 7.41 milliohms per metre for 2.5mm² now these would need to be multiplied by 1.2 to take account of conductor temperature and to match them to table 41.6. Note that these are just for one conductor so we need to multiply by 2 to get R1 + R2. Maximum flex length will be: Lmax = 0.21/((12.10/1000) * 1.2 * 2) = 0.21/0.029 = 7.24 metres Oh dear, not very long is it Well lets try a type B (R1 + R2)flex = 0.69  0.13 = 0.56 ohms Lmax = 0.56/((12.10/1000) * 1.2 * 2) = 0.56/0.029 = 19.3 metres Better, but still not as long as some site flexes I have seen and we are using a type B  that could be a problem with motor loads Remember we have used a fairly large transformer  it would be worse with a small one  still you could use RCDs but how long would they last Better to bin your copy of BS 7671 and use the far superior BS 7375 . Regards Geoff Blackwell 



22 May 2009 01:27 PM


Doesn't it just come out in the wash if you divide Vs by 2?  i.e. use Vs = 55V (or 63.5V) rather than 110V.  Andy. 



22 May 2009 01:34 PM


Geoff,
Excellent reply and I very much appreciate the time you have taken on my behalf. We will be using BS88 fuses and the sockets will be hard wired for portable tools, but the method is clear now. Regards, Steve 



22 May 2009 01:44 PM


Doesn't it just come out in the wash if you divide Vs by 2?  i.e. use Vs = 55V (or 63.5V) rather than 110V.  Andy. We are ultimately using the calculation against table 41.6 and this already uses the impedance divided into 55 volts. The value without a divisor contains an error because only half the secondary winding is actually in circuit. That said, any error will be small as we would only have to adjust the secondary, and the winding is mainly reactive meaning that it has little effect on the overall result. Regards Geoff Blackwell 



22 May 2009 05:06 PM


This is a load of old tosh.
I do not know of anyone who has been electrocuted by a 55055V system despite being used in some of the most appalling environmental conditions. My advice is to go outside and enjoy the sun if you have nothing better to do. Eric 



22 May 2009 05:55 PM


Really  well its a good thing that some people spend some of their time making sure these systems are safe for you then isn't it.
Regards Geoff Blackwell 



22 May 2009 06:11 PM


This is a load of old tosh. I do not know of anyone who has been electrocuted by a 55055V system despite being used in some of the most appalling environmental conditions. My advice is to go outside and enjoy the sun if you have nothing better to do. Eric I am outside, enjoying the sun. 



22 May 2009 08:59 PM


Geoff
Thanks that was most informative. I takes me back to my college days talking about reflected impedances. What ashame most of that knowledge is locked away on inaccesible areas of my internal cranium hard drive.  John Peckham http://www.astutetechnicalservices.co.uk/ 



23 May 2009 11:35 PM


Geoff
What a contribution, Cheers.  Regards Martyn. Only a mediocre person is always at their best 



25 May 2009 06:51 PM


A couple of years ago after a request by a member who was taking an exam I posted a pdf on SH on how to calculate the loop impedance at the end of a circuit fed from the sec winding of a transformer, I included a step by step worked example, changing my computer to a Mac I downloaded all my files onto a load of DVD's but will look to see if I still have a copy over the next couple of days.
regards  "Take nothing but a picture, leave nothing but footprints!"  "Oh! The drama of it all."  "You can throw all the philosophy you like at the problem, but at the end of the day it's just basic electrical theory!"  



27 May 2009 08:13 AM


Hi GeoffBlackwell,
Subject discussion : Building Design  Earth loop impedence good day to you, i have face problem with Earth loop impedence calculation. Data input as Trasformer 11000 /400 volts, Trasformer imp  6.25 % (.0625 p.u.) doing with amtech prodesign given as ZE of 0.0071 ohm. i have faced problem of Earth loop impedence error as 1.Cbl_FC10 Zs=0.0706 Ohm Max.Zs=0.0625 Ohm Tef=28.68s Tefmax=5.00s The earth loop impedance Zs at the load end of the cable is 0.0706 Ohm > Max. Zs 0.0625 Ohm for the overcurrent protection. The Disconnection time Tef=28.68s exceeds the Max. time Tefmax=5.00s. Error 2. Cbl_SM3 Zs=0.0396 Ohm Max.Zs=0.0313 Ohm Tef=36.21s Tefmax=5.00s The earth loop impedance Zs at the load end of the cable is 0.0396 Ohm > Max. Zs 0.0313 Ohm for the overcurrent protection. The Disconnection time Tef=36.21s exceeds the Max. time Tefmax=5.00s. Error in the above Data 1 Error as (cbfc10) MCC panel located in Terrace . LV panel at basement Totally length shoul be 248 Mtr. Circuit protective device as 800 Amps ( 300 kw chiller .as per manufacture table starting current shall be 699 amps, reference : Trane chiller) 2.Data 2 Error as panel feeder supply from basement to 7th floor level . total length as 205 Mtr. Protective device as 1600 amps. in manual calculation of Earth loop impedence zs = ze+ ( r1+ r2) ze  earth loop imp. r1  Phase Conductor Resistance r2Circuit Protective Conductor Resistance in the above case i have connect from transformer to lv panel as bus duct of 2500 amps. length as 30 mtr.protective device as 2500 amps ACB. kindly advice me how to claculate Earth loop impedence calculation. await for your response . thanks kumaraguru Edited: 27 May 2009 at 08:16 AM by kumaraguru 



27 May 2009 08:53 AM


This is a detailed design and as such beyond what can be undertaken on a forum such as this. There are all sorts of issues such as PI insurance, etc, etc  not to mention that a designer would expect a considerable fee for all the work, and the risk .
You have not given sufficient data (transformer capacity (KVA) for a start. I assume that you are not able to meet 5 second disconnection times as required by BS 7671. There are many solutions, but a good starting point would be to investigate what earth fault protection can be fitted as standard to your ACBs / MCCBs. That is the best I can do I. BTW I don't use Amtech ProDesign. Regards Geoff Blackwell 



27 May 2009 12:10 PM


Hi Geoff Blackwell ,
Thanks for your response, i agree this is detailed design. 1. Transformer capacity shall be 1500 KVA. 11kv /400v, 50HZ, 6.25 % impedence. ( CPD rated as 2500 amps ACB ,50 ka icu, Electronic LSI) From Transformer to LV panel utilised BUs duct 2500amps rated. can you correct if i am wrong in earth loop impedence calc. In earth loop impedence calc: Zs= Ze +( R1 + R2) Ze = 0.0071 ohms ( based on input from amtech result. R= .0014 ohm, X = 0.007ohm) R1 =0.028 ohms R2 = 0.126 ohms Here R1, R2 input values taken from Schneider catelogues, Canalis KTA.( because of Busduct used here) Zs =0.1611 ohm but my software calculation used in amtech prodesign as LV panel at basement Circuit breaker rated as 1600amps.( connect to LV panel 7th floor . length as 205 Mtr) LV panel at 7th floor circuit breaker 1600 Amps MCCB. no earth fault protection can be fitted . just for amtech feed input as generic. Even i have checked with BS 7671 Fixed equipment as 5 sec. socket outlets etc., 0.4 sec. here chiller as fixed equipment. ( cable 2Runs of 300 sq.mm CPD rated as 800 amps electronic LS) but 7th floor LV panel incomer cable (4runs of 300 sq.mm ,CPD = 1600 amps MCCB electronic LS) thanks & regards, kumaraguru 



27 May 2009 12:38 PM


Kumaraguru, what are you trying to ask?
As Geoff has already said what you are asking probably goes beyond the advice you are going to get on any forum, can you not consult another engineer at your workplace? Also why can you not fit earth leakage protection to your MCCB? 



27 May 2009 01:00 PM


socket outlets etc., 0.4 se Sounds like you're looking at an outofdate copy of BS 7671 too. The "fixed equipment" category disappeared with BS 7671:2008 (17th Ed).  Andy. 



IET
» Wiring and the regulations
»
CTE Transformer Ze calculations

Topic Tools

New here?
See Also:
FuseTalk Standard Edition v3.2  © 19992017 FuseTalk Inc. All rights reserved.