Short Circuit Fault Level Calculation

[@pietervdberg7055]

Nice video if the questions can be explained. A number of people (including myself) asked why the factor of 1.05 comes into the equation. I still prefer the per unit method. It eliminates the need to determine what voltages should be used at different levels if you have multiple transformers. Also, all those motor loads will definitely contribute to increasing the Isc at point 3. Suggestion is that if you don’t want to bring motor contribution into example, don’t specify motor loads. It can confuse some readers thinking that you don’t need to add motor contribution.

[@ratssaesquareallaboutelect5919]

Thank you for sharing your views and thank you for the question, Answer to 1.05 question is : 1.05 is a voltage factor required in the calculation to account for Voltage variations in the system,if we consider voltage variation of +-6% in voltage , we have to use factor of 1.05 in the formula as per standard IEC 60909.

[@pietervdberg7055]

@@ratssaesquareallaboutelect5919 Thank you for your reply. Much appreciated.

[@richhall3662]

@@pietervdberg7055 Also note that motor contribution only occurs for 3 to 4 cycles for asynchronous or 6 cycles for synchronous machines, so not a huge factor when sizing equipment based on steady state conditions or fault levels that can exist for hundreds of milliseconds or more. Also the 1.05 takes into account the voltage factor or voltage rise that occurs on phase due to the raised neutral potential and normally relevant on single phase shorts

[@sumitmondal548]

@@richhall3662 nice

[@annasmohammed4365]

In short, calculate the short circuit current at nominal voltage and add 5% extra value to Isc (considering the factor for rise in voltage level).

[@hijazahmad1736]

Nice... I was searching for this type of explanation.. Excellent

[@charzanboo9940]

I was asked to perform the above short circuit calculation in an interview...in my head.

[@musabkanani2873]

Iam senior electrical engineer Your explain is soo good so thanks

[@walterlopez5465]

Excuse me Sir, can you tell which standard support this calculation, i'd really appreciate your answer. Greetings!

[@justinlow-thion77]

very good video tutorail. simple explanation and easy to understsnd

[@ratssaesquareallaboutelect5919]

Glad it was helpful!

[@VeeshadhaILLANGASOORIYA]

one of the best

[@muhammadabduljaleel1811]

Very nice with good explanation

[@ImranKhan-nu6xw]

Thanks nice vedio about short circuit study

[@vikasgupta-hl9sq]

Explained very nice

[@subhendumukhopadhyay3535]

Nice one

[@liewts1049]

Could you also comment on the assumption made on the impedance of the power supply utility? Thank you.

[@mthuthuzelimatsheke5422]

It's so meaningful and professional classes i would like to join as a mamber please admin

[@niluv3982]

nice

[@ratssaesquareallaboutelect5919]

Thanks

[@e-powersolution7225]

1:41 thanks for your video, why while calculating z transformer multiple with 10

[@ktsmurthy21]

If Generator(1.5 MVA) is running , Then How to Calculate Fault current.

[@noname-ug2qx]

Why is the backup generator not included when calculating If at point 2 (Busbar)?

[@amritthakur9712]

Nice

[@ricrobertogonzales8341]

what is the effect of the motors in the short circuit calculation?

[@ratssaesquareallaboutelect5919]

If Total Motor Current (100 + 50 + 50 + 30 HP , 415V , 0.85 PF) = 281A Total Motor current

[@atoz5950]

Many thanks for you .. please and kindly, Re-calculate the Isc for generators

[@bradencutajar3301]

At 1:43, there is no reason to multiply by 10

[@shuvanilmukherjee1595]

I THINK SHE DONT KNOW

[@abcdchanchan1]

Maybe No need to multiply by 10, she can’t explain

[@JDABD]

Why it is multiplied by 10?

[@mourshadhaider7361]

Could you please explain number "10" in the formula for z(transformer). Where did it come from.

[@itanongmokayengineer]

its contant in order not to include the thousand in the kva of transformer instead using 2000000 you will use 2000

[@umaid-nungshatpa]

Hi Informative New Subscriber

[@ratssaesquareallaboutelect5919]

Thanks and welcome

[@coopers0218]

Pls help with data sheet reference for Cable impedance value. It seems you have considered Cable resistance value instead of Impedance. Pls clarify. Thanks

[@Nour-ds3ib]

If you multiply kva by 1000 and kv by 1000 but undersquare and make per unit impedance 0.06 It makes the same number

[@devendrasharma2418]

You are right, I think data of DC resistance value has been considered here, instead of which cable impedance as per manufacturer's data should have been considered.

[@joegellido2596]

The presentation will be complete if motor contribution to fault is considered which is very important factor..

[@mohammedsaifullah6437]

Motor and Generator as well

[@ratssaesquareallaboutelect5919]

If Total Motor Current (100 + 50 + 50 + 30 HP , 415V , 0.85 PF) = 281A Total Motor current

[@satishkumar-wm9yl]

please can you share and update more videos of typical calcualtions of short ckt Transformers and RMU Swithchgears, thanks

[@sunilkumarmoharana7639]

Please explain the factor 1.05. how it comes.

[@mazmaz1981]

sunil kumar moharana 5% voltage tolerance, which is very typical.

[@ratssaesquareallaboutelect5919]

thank you for the question, Answer to 1.05 question is : 1.05 is a voltage factor required in the calculation to account for Voltage variations in the system,if we consider voltage variation of +-6% in voltage , we have to use factor of 1.05 in the formula as per standard IEC 60909.

[@KonduriMurthy]

Very good . .

[@ratssaesquareallaboutelect5919]

Thanks

[@surendrakverma555]

Good

[@ratssaesquareallaboutelect5919]

If Total Motor Current (100 + 50 + 50 + 30 HP , 415V , 0.85 PF) = 281A Total Motor current

[@RaviShankar-sj7qs]

Very Informative video admin. Please upload video of disadvantages of harmonics and ways to arrest them.

[@ratssaesquareallaboutelect5919]

Thanks for the feedback , will try to make the required video

[@tapasbarua4301]

Very nice video. But I did not see any contribution of motors into the fault point as a secondary source and alternator and also why you have included 1.05XV in the calculation?

[@shuvanilmukherjee1595]

HOPE YOU HAVE NOT RECEIVED ANY ANSWER.

[@ratssaesquareallaboutelect5919]

If Total Motor Current (100 + 50 + 50 + 30 HP , 415V , 0.85 PF) = 281A Total Motor current

[@mudasirhashmi8907]

How do you the cable impedance as .0077 and what are the formula you are using

[@trimafs]

Hi, Thanks for informative video can you please also give the reference for formulas

[@jkj1459]

BEAUTIFUL

[@velusamypothiraj2300]

Where it's coming 10, in impedance of transformer, what is the fault current in 11kv side

[@ratssaesquareallaboutelect5919]

it can be obtained from transformer rating plate.

[@amarsrivastav1167]

@@ratssaesquareallaboutelect5919 sir pls explain from where it's coming 10

[@prakashksp2918]

Please upload videos on instantaneous current formula

[@mazmaz1981]

What about motors contribution to the fault?

[@Keepit-nq3id]

Good question the reverse currents shall be calculated

[@ratssaesquareallaboutelect5919]

If Total Motor Current (100 + 50 + 50 + 30 HP , 415V , 0.85 PF) = 281A Total Motor current

[@jkmultimediasolutions3935]

Thanks for the video. Please when you mentioned number of runs; did you mean the number of cables or what ? Please, explain it further. Thanks.

[@ratssaesquareallaboutelect5919]

No of run means no of parallel cables

[@poojap2508]

Hw did u get 1.05 while calculating Isc

[@ratssaesquareallaboutelect5919]

thank you for the question, Answer to 1.05 question is : 1.05 is a voltage factor required in the calculation to account for Voltage variations in the system,if we consider voltage variation of +-6% in voltage , we have to use factor of 1.05 in the formula as per standard IEC 60909.

[@manojshukla3598]

how 1.5 comes in this equation

[@loubensoctave3986]

Nice video. How can I contact you? I have many questions to ask.

[@joelcastillo3902]

No way, she's Siri the robot!

[@sakirdag2976]

thx

[@lanzsolissiargao8477]

What do you mean of number of runs?

[@ratssaesquareallaboutelect5919]

no of parallel paths are termed as no of run.

[@TheSuperAboody]

What do you mean by Number of Runs?

[@Keepit-nq3id]

For example 3 cable runs means 3 number of cables

[@TheSuperAboody]

Thank you

[@zainliaquat9537]

Can you please explain why you are dividing the line voltage by the square root of 3 in Isc formula (At Point 1) ? Your response will be very helpful.

[@jay403]

it is incorrect. There is no point dividing by root3. Any current is simply V/Z. 1.05 is form factor.

[@MEOVLOGz]

what method is this?

[@rolandomullasgo48]

Where did you get 1.05 shown in your cakculation.

[@ratssaesquareallaboutelect5919]

As per standard IEC 60909 - 2001 , We have to consider the voltage factor of 1.05 in LV fault calculation (100V to 1000V).

[@shelfrobthomas449]

is 1.05 in the equation a constant value and how about the impedance value of cable ohms per km is it also constant?

[@ratssaesquareallaboutelect5919]

1.05 is part of equation and impedance of cable can be obtained from the cable manufacturer . Thanks for writing

[@rasengan9743]

1.05 is a voltage correction factor Voltage maximum/voltage minimum 440÷410

[@ranjan1436]

please explain, how? to size the KVA of a transformer.

[@mdshamsealam4024]

How number of runs between point 1 and 2 are 7 ?

[@mdtarique6305]

Provided in Ts

[@sungmopark0405]

Could you explain why symmetrical fault L-L-L-G and L-L-L is same?

[@ratssaesquareallaboutelect5919]

The behaviour of LLLG fault and LLL fault is identical due to the balanced nature of the fault Three phases are equally affected in these type of faults and the system remains balanced. Hope your query is resolved and pl keep watching the channel for more useful videos. thanks.....

[@jinishcrajan]

can explain how 1.05 has been considered

[@ratssaesquareallaboutelect5919]

thank you for the question, Answer to 1.05 question is : 1.05 is a voltage factor required in the calculation to account for Voltage variations in the system,if we consider voltage variation of +-6% in voltage , we have to use factor of 1.05 in the formula as per standard IEC 60909.

[@crisrobersonsantos9192]

Is the short circuit at fault 3 = 32.58 KA?

[@ratssaesquareallaboutelect5919]

thanks for writing to the channel , I Sc at point number 3 is : 33.5KA Pl keep on watching the Channel for other useful videos.

[@rkthemgil]

What is 1-05 in the formula?

[@ratssaesquareallaboutelect5919]

thank you for the question, Answer to 1.05 question is : 1.05 is a voltage factor required in the calculation to account for Voltage variations in the system,if we consider voltage variation of +-6% in voltage , we have to use factor of 1.05 in the formula as per standard IEC 60909.

[@mohammedsaifullah6437]

But what about generator current

[@niksontalan5803]

How to calculate MVAsc on transformator and feeder

[@iswarmahanta6905]

What is the value of 1.05

[@wirotebua6971]

As I guess, the prefault voltage is 105% of norminal voltage thus the factor 1.05 is used?

[@ratssaesquareallaboutelect5919]

thank you for the question, Answer to 1.05 question is : 1.05 is a voltage factor required in the calculation to account for Voltage variations in the system,if we consider voltage variation of +-6% in voltage , we have to use factor of 1.05 in the formula as per standard IEC 60909.

[@madhusudan6395]

What is 1.05? Pl.describe

[@soumya2221]

What is the fault current contribution from motor

[@ratssaesquareallaboutelect5919]

If Total Motor Current (100 + 50 + 50 + 30 HP , 415V , 0.85 PF) = 281A Total Motor current

[@malam4883]

What should be breaking capacity of cb at point 3.

[@kilumumahesh935]

Mr M.salam The breaking capacity of the breaker purely depends on load nature. Ex: if your load 80% const power & 20% const impedance. Then the SC contribution will be some value.. Or If you load purely 100% constant power load, then the SC Current from the load will be some other value.. So as per above examples the SC Current contribution from the load side may very as per the load type.. Therefore = SC Current contribution from upstream + SC Current contribution from load end side = breaking capacity of the breaker nothing but (Ics or Icu)..

[@kilumumahesh935]

To each and everyone .. First go through the video carefully .. Why everyone asking regarding what about load SC Current contribution..?? Channel was created fault at the Trafo LV bus. From the upstream we can find Trafo SC current contribution and from Downstream the load SC current contribution. The upstream and downstream SC currents will get added up and visible to us at Bus right side.. So channel was trying to stress the point is how to calculate the fault current contribution from the upstreams not from the load end side or downstream side.. The channel was not stressing any point like.. this is the RMS SC current so based upon the this SC current only you have to choose breaker Ics or Icu.. Hope everyone understood.. Thank you...

[@thajmulhussain4144]

HOW YOU DETERMINED OHMS(0.0778) AT BETWEEN POINT 1 TO2)AT3.41?

[@wescointernationalgroup6494]

U have to take transformer impedance x10xline volt x line volt

[@vijaymaurya9070]

Z of cable where this value ,how to find catalogue or other method pl explain

[@ratssaesquareallaboutelect5919]

Z of cable can be obtained from the manufacturer. Thanks for writing.

[@lov7587]

Hi Sir, How did you get the figure of 1.05 in the formula? Please advice.....

[@ratssaesquareallaboutelect5919]

thank you for the question, Answer to 1.05 question is : 1.05 is a voltage factor required in the calculation to account for Voltage variations in the system,if we consider voltage variation of +-6% in voltage , we have to use factor of 1.05 in the formula as per standard IEC 60909.

[@KonduriMurthy]

IF the Voltagebminus side ,variation is 10% on plus sideand,what. is the factor.

[@Nour-ds3ib]

Any one please can tell me what's meant by number of Runs????

[@ratssaesquareallaboutelect5919]

Number of runs mean number of parallel paths.

[@РоманАпостолюк-в1л]

how do motors affect short-circuit currents?

[@berniesayson6612]

for solving of short circuit current of electrical system all connected equipment's and load must be consider, in your calculation load is not considered.

[@ratssaesquareallaboutelect5919]

If Total Motor Current (100 + 50 + 50 + 30 HP , 415V , 0.85 PF) = 281A Total Motor current

[@ratssaesquareallaboutelect5919]

If Total Motor Current (100 + 50 + 50 + 30 HP , 415V , 0.85 PF) = 281A Total Motor current

[@anugarg4864]

Dear Mam, Please tell me about the number of run ,What is this ?

[@ratssaesquareallaboutelect5919]

Number of run means , number of parallel cables, Thanks for watching.

[@Maharana_Rahul]

Ex:- each phase four runs

[@jcazur70]

Why did you not consider the contribution of the motors below and the generator above whose ratings are significant????

[@ratssaesquareallaboutelect5919]

If Total Motor Current (100 + 50 + 50 + 30 HP , 415V , 0.85 PF) = 281A Total Motor current

[@keshabchandratripathy7300]

Please find explain the formula used

[@DJLMGT]

what means 7 runs?

[@ratssaesquareallaboutelect5919]

7 runs means 7 numbers of parallel connection

[@dilermehndishirazee8077]

Four 20KVA generators are connected to 11 kV bus what is the Fault MVA at bus?

[@aspiringenggs1]

Question is incomplete...Actual Impedance of generator is required for Fault MVA

[@atmsc4nd4l]

why do you have 1.05? does that mean you have increased the voltage to 5%?

[@anilkumarmehta1923]

But you have not considered the incoming grid and generator connected which these will contribute

[@kritabarma]

What about motor contributions??

[@ratssaesquareallaboutelect5919]

If Total Motor Current (100 + 50 + 50 + 30 HP , 415V , 0.85 PF) = 281A Total Motor current

[@sohrabalmajose4082]

what happen to the current contribution of the motor ? you have not consiger it in your calculation. for your info tnx

[@ratssaesquareallaboutelect5919]

If Total Motor Current (100 + 50 + 50 + 30 HP , 415V , 0.85 PF) = 281A Total Motor current

[@atmsc4nd4l]

and why is that motor contribution is neglected?

[@ratssaesquareallaboutelect5919]

If Total Motor Current (100 + 50 + 50 + 30 HP , 415V , 0.85 PF) = 281A Total Motor current

[@liewts1049]

Thank you for the nice video, this considered single phase to ground fault. Would you also make a fault current calculation video for ph-ph, ph-ph-gnd, 3ph fault?

[@fereshteh5205]

@@yaminoshinn1473 415 v is line voltage, by dividing on sqr3, it would be phase voltage

[@patelkaushik2996]

Medam send video for HT 11KV TO LT 415 V relay

[@pranayprakash3877]

No of run is 7 you said, where this come from

[@ratssaesquareallaboutelect5919]

for example, 7 parallel paths are assumed here.

[@vinnnyvinvin]

@@ratssaesquareallaboutelect5919 Hi, does this mean parallel paths for 3 cores (3 phases) and 1 core for neutral? any earth connection?

[@patricksalvacion8975]

How to obtain the number of runs?

[@Feitan0123]

it is the number of wire per phase.

[@dsmahatpure8199]

Mam in farmula 1.05 and 10 is coming from where?

[@ratssaesquareallaboutelect5919]

thank you for the question, Answer to 1.05 question is : 1.05 is a voltage factor required in the calculation to account for Voltage variations in the system,if we consider voltage variation of +-6% in voltage , we have to use factor of 1.05 in the formula as per standard IEC 60909.

[@shuvanilmukherjee1595]

MAM WHY CALCULATING Z YOU ARE USING MULTIPLY BY 10

[@emekaesenwa387]

The contribution of the electrical motors to the fault current wasn't captured. It will supply fault current to the bus which it is connected to during a fault

[@emilsify]

you are correct during the fault there are some contribution from the motor this method seems shortcut.

[@ratssaesquareallaboutelect5919]

If Total Motor Current (100 + 50 + 50 + 30 HP , 415V , 0.85 PF) = 281A Total Motor current

[@ratssaesquareallaboutelect5919]

If Total Motor Current (100 + 50 + 50 + 30 HP , 415V , 0.85 PF) = 281A Total Motor current

[@deepakgrg1408]

ok

[@anwarhussainansari13]

Transformer impedance already given %6=.06 ohm so why u are re calculate ?

[@staceyksmith1095]

The transformer impedance given in percentage is used when you convert all impedances to Per Unit. She avoided the per unit conversions by calculating the Z impedance first

[@quirinoguillepajr3979]

Not comprehensive it does not consider the available fault MVA at the utility and motors subtransient reactance

[@ratssaesquareallaboutelect5919]

This type of calculation is used for deciding the Short circuit Breaking Capacity of MCCB to be used. Other factor as mentioned can be also calculated. Thanks for writing

[@shamirdebnath7407]

@@ratssaesquareallaboutelect5919 Could you please explain how to select the required breaking capacity of individual motors?

[@patelkaushik2996]

Nice pic medam

[@haithamahmed8747]

good video but where is the motors contribution, they will raise the fault level !!!

[@ratssaesquareallaboutelect5919]

If Total Motor Current (100 + 50 + 50 + 30 HP , 415V , 0.85 PF) = 281A Total Motor current

[@senthivelgopal567]

why here u r multiplying Z% by 10 , could please explain, if anyone else also most welcome

[@keshabchandratripathy7300]

I feel for Z Tago in formula given instead of 10 we will write 1000.please clarify

[@IbrahimMA-EGY-suez]

but you ignore motor reactance how could you !!!

[@ratssaesquareallaboutelect5919]

If Total Motor Current (100 + 50 + 50 + 30 HP , 415V , 0.85 PF) = 281A Total Motor current

[@muhammadbinali6223]

too complicated , instead do this, first calculate the sec current of transformer, which is Isec= rated KVA/sec voltage, incase its 3 phase (1.73 into voltage) , u will get Isec of transformer, now next step to calculate short circuit current by using the formulae , isc= Isec*100/%imp of transfomer , which will give u the short circuit curent on busbar by transformer

[@nagendranm7322]

How to calculate point no 2?

[@patelkaushik2996]

Medam send video for HT 11KV TO LT 415 V relay coordination method

[@lemmirorissa5217]

Could you please explain number "10" in the formula for z(transformer). Where did it come from.

[@AzkadanAhmad]

Same Question

[@AbdulRahman-pm1nd]

What is value of 1.05

[@ratssaesquareallaboutelect5919]

As per standard IEC 60909 - 2001 , We have to consider the voltage factor of 1.05 in LV fault calculation (100V to 1000V).