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Prospective Short Circuit Currents 

Greenwood Solutions | Renewables Educator & EPC
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We look at the basics of determining the prospective short circuit currents from the transformer to a distribution board.
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27 июл 2024

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Комментарии : 25   
@WEEMAN240496
@WEEMAN240496 3 года назад
This was better than I expected it to be given I've never heard of the channel. Also being Australian it really helped to be using language and references I understand. Subscribed.
@GreenwoodSolutions
@GreenwoodSolutions 3 года назад
Thanks Joshua! :)
@nickserafin76
@nickserafin76 3 месяца назад
Awsome ausie info. thanks for taking the time to carefully explain.
@GreenwoodSolutions
@GreenwoodSolutions 2 месяца назад
No worries. Always learning and as we learn so do you. Cheers
@brockturner8430
@brockturner8430 2 года назад
Great video, thank you! It was so good to hear it from an Australian, it made it so easy!
@GreenwoodSolutions
@GreenwoodSolutions 2 года назад
Cheers Brock!!
@user-mu2qu8uo4z
@user-mu2qu8uo4z 8 месяцев назад
LOVE IT
@Chrissrihcify
@Chrissrihcify 3 месяца назад
I think there is something forgotten....(Great Aussie Content . Cheers ) In a MEN System (TNC-S) the 3 Phase Prospective Short Circuit is significantly higher than the Per phase Short circuit Current. Ipsc = Root 3 x Isc. The impedance per phase for a Phase to Earth Fault , must include the Neutral return path (Via the MEN System aka the P.E.N) A single phase to earth fault has a much higher Loop impedance for a Phase to Earth fault loop path For a Per Phase Short Circuit to earth the impedance is Zs= Ztx + Zcable +Zpen For the purpose of a Bolted Symmetrical fault (3Phase Ipsc) the fault loop Path Ignores the Neutral Return path, as the out of phase currents cancel each other, reducing the neutral fault current to zero. Therefor the impedance path for Ipsc is much Less Zs = Ztx +Z cable (with no Zpen) Therefor the analysis of Ipsc for a 3Ph Ipsc must be 1.73(Root 3) higher than the single phase short circuit current, after ignoring the neutral Impedance (Balanced Fault). 3Phase Ipsc = 1.73 x Is/c 1Phase I psc = Is/c Perhaps another way to look at it, the heat energy in fault current per Phase is 1/3 of the total energy in the fault. You can't add impedance across phases so you have to analyse the effective impedance per phase The Fault Path to earth (MEN via PEN) is twice the resistance , (Half the fault loop current per phase) You could compare the impedance Values as 2 divided by root three , Hence the three phase to single phase conversion of cable resistance in Table within AS 3008.1.1 (1.155 or 1/0.866) Perspective Short Circuit current is Line Current for a Phase to Phase to Phase fault (Worst case) in a balanced fault with no neutral impedance restricting current.. Remembering that Phase currents in a Star connecting are the same as line current Ipsc = Vline / Zs Ipsc = 1.73 x Vph / Zs when I s/c = Vph / Zs Ipsc = 1.73 x is/c Thanks again for the great content!!!
@GreenwoodSolutions
@GreenwoodSolutions 2 месяца назад
Cheers for that.
@al_loi
@al_loi 2 года назад
The best explanation that I have heard so far in Australia. I hope you can point me in the right direction as to where can I find the conductor impedance per phase value? Thank you :)
@GreenwoodSolutions
@GreenwoodSolutions 2 месяца назад
In Australia we reference AS3008.
@angelofranklin1
@angelofranklin1 3 года назад
Thank you for the vid, vry much helped me. I wish everything can be explained like this. The very best wishes to you. I have struggled to find stuff like this with the clarity you have shown.
@GreenwoodSolutions
@GreenwoodSolutions 3 года назад
Thanks Angelo. Hope these video help in other areas too!
@angelofranklin1
@angelofranklin1 3 года назад
Only part if you could explain, I was under the conception the highest fault current would be at 400 volts and not 230, can you kindly update when you can thanks. Its when you devide the fault current at the Consumer mains you have considered 230 volts devided by the impedance, I thought that would be 400 volts and not 230 volts. If you or anyone can advise the reason thanks.
@GreenwoodSolutions
@GreenwoodSolutions 3 года назад
@@angelofranklin1 Faults are per phase hence the 230 volt reference. Hope this answered your question!
@angelofranklin1
@angelofranklin1 3 года назад
@@GreenwoodSolutions Thanks appreciated.
@jimmy2045
@jimmy2045 2 года назад
THANK YOU
@curentarul
@curentarul Год назад
Great video! Thank you! When we calculate how many amps are on a secondary side of a transformer we add current of each indivual phase (I1+I2+I3)?
@michaelriley5158
@michaelriley5158 3 года назад
Great Video Thanks mate, made it so easy to understand! Is it the same for Calculating Ipfc at the sub board but doubling the Z1,Z2 and Z3 impedance values?
@GreenwoodSolutions
@GreenwoodSolutions 3 года назад
Sort of getting out of my depth here but impedance increases the further you move away from the transformer, hence the kA rating of components tends to be lowered. Fault level at a sub-board would include the previous impedances so fault level at subcircuit connected to DB would be: 230/(0.0159 + 0.0037 + 0.026 + impedance of the cable run from DB to subcircuit) Hope this helps!
@michaelriley5158
@michaelriley5158 3 года назад
@@GreenwoodSolutions Thank you mate, is that for Ipsc? i'm curious if ipfc is the same?
@GreenwoodSolutions
@GreenwoodSolutions 3 года назад
@@michaelriley5158 If you are using a Loop Tester, you would record both PSC and PFC values and record the highest value. Due to the nature of different supply types, you would expect to find a PSC value higher than a PFC value on both TT and TN-S systems, however on a TNC-S system both the PFC and PSC value should be identical. The vast majority of installation of electrical supply throughout Australia will fall under the TT classification of BS 7671 Both PFC and PSC tests are designed to calculate the max current that will flow within a fault loop path during the event of an electrical fault as required by regulation 612.11 of BS7671:2008. A PSC test calculates the current that will flow in the event of a short circuit fault between the live conductors. That is, Line and Neutral on a single phase installation or Line to Line/ Line to Neutral on a three phase installation. A PFC test calculates the current that will flow in the event of an earth fault; i.e., Line to Earth. Hope this helps!
@Chrissrihcify
@Chrissrihcify 3 месяца назад
I think there is something forgotten....(Great Aussie Content . Cheers ) In a MEN System (TNC-S) the 3 Phase Prospective Short Circuit is significantly higher than the Per phase Short circuit Current. Ipsc = Root 3 x Isc. The impedance per phase for a Phase to Earth Fault , must include the Neutral return path (Via the MEN System aka the P.E.N) A single phase to earth fault has a much higher Loop impedance for a Phase to Earth fault loop path For a Per Phase Short Circuit to earth the impedance is Zs= Ztx + Zcable +Zpen For the purpose of a Bolted Symmetrical fault (3Phase Ipsc) the fault loop Path Ignores the Neutral Return path, as the out of phase currents cancel each other, reducing the neutral fault current to zero. Therefor the impedance path for Ipsc is much Less Zs = Ztx +Z cable (with no Zpen) Therefor the analysis of Ipsc for a 3Ph Ipsc must be 1.73(Root 3) higher than the single phase short circuit current, after ignoring the neutral Impedance (Balanced Fault). 3Phase Ipsc = 1.73 x Is/c 1Phase I psc = Is/c Perhaps another way to look at it, the heat energy in fault current per Phase is 1/3 of the total energy in the fault. You can't add impedance across phases so you have to analyse the effective impedance per phase The Fault Path to earth (MEN via PEN) is twice the resistance , (Half the fault loop current per phase) You could compare the impedance Values as 2 divided by root three , Hence the three phase to single phase conversion of cable resistance in Table within AS 3008.1.1 (1.155 or 1/0.866) Perspective Short Circuit current is Line Current for a Phase to Phase to Phase fault (Worst case) in a balanced fault with no neutral impedance restricting current.. Remembering that Phase currents in a Star connecting are the same as line current Ipsc = Vline / Zs Ipsc = 1.73 x Vph / Zs when I s/c = Vph / Zs Ipsc = 1.73 x is/c Thanks again for the great content!!! @@GreenwoodSolutions
@angelofranklin1
@angelofranklin1 10 месяцев назад
just a question. I assumed this was a single phase dsistribution board, if it was a three phase distribution board instead and the fault was accross all three phases I assume the fault calculated on your example should be multyplied by 2, or accross 2 phases them 1.73. please help.
@GreenwoodSolutions
@GreenwoodSolutions 10 месяцев назад
The board is three phase. When you are talking about a three-phase inverter etc the current mentioned is per phase. For example 200 kVA worth of inverters have a current value of 200,000 vA/( 400V nominal x 1.732 ( SQR of 3) = 289A/phase
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