Symmetrical Fault Analysis

[@DrAfrozAlam]

Download Fault Analysis handwritten lecture notes (pdf) using the following links: drive.google.com/file/d/18Zvgdi3YD6iAIkWJwz5-s7dVHRgM6lwZ/view?usp=sharing drive.google.com/file/d/1sfMf8BDsZ1BlKoehLLkokwizgtWHIbrr/view?usp=sharing drive.google.com/file/d/1eYOtrSzN7VwsqiLwjnrJ79BQd2LcE_Vy/view?usp=sharing

[@omerkarar8405]

This is exactly what I expect from a teacher, being direct and unbelievably precise! Thank you very much, teacher!

[@md.mehedihasan8314]

MashALLAH sir. First of all, your teaching quality too much good and uses simple & easy word. This is a good support for me.

[@ganinitianelectricalkkr9686]

mashallah 😍.. bahut khubsurat lectures....... as like as dr. afroz alam sir. shukriya( thanks ) so much for providing these lectures as well as notes.

[@sufiyashama8877]

Sir apka bhot bhot shukriya.. Apki wajah se meku bhot ache se samaj aya

[@Yusuf-dx4hw]

Thank your for your labor Sir. Much appreciated. Greetings from Ankara,Turkey

[@YoungsProton]

Sir... great lecture.. Cleared all my doubts.. on symmetrical faults. I calculated rest of the math.. Anyone Correct me if I am wrong: Current supplied on LT side = Ifault X 100/11X (root 3)=20 KA By current divider Ig1= 11.803 A and Ig2=8.196 A.

[@DrAfrozAlam]

Thank you Asif Khan for the comment. Your calculations are almost correct. As per my calculation, Total Current supplied on LT side = - j 19.995 kA ≈ - j 20 kA. Ig1= - j 11.8003 kA and Ig2= - j 8.1947 kA.

[@DrAfrozAlam]

For more problem practices on "Symmetrical Fault Analysis", you may visit the following link: kzbin.info/www/bejne/b3TInpeZa9ZqhZI

[@sattamdutta1899]

@@DrAfrozAlam thank you sir....you are the real saviour

[@DrAfrozAlam]

@Sattam Dutta, You are most welcome dear

[@DrAfrozAlam]

Download Fault Analysis lecture notes (pdf) using the following link: drive.google.com/file/d/18Zvgdi3YD6iAIkWJwz5-s7dVHRgM6lwZ/view?usp=sharing

[@baijaltudu3002]

Sr. Your lecture is vety good & explain easy language.

[@SurajChaudhary-mp1uw]

Sir, your method of teaching is awesome..

[@ohgoditsjames94]

Excellent video, many thanks from the U.K.

[@mjmontesinos1]

Sir, thank you very much for your videos. They have taught me a lot. much Better explained than professor at my university.

[@ahmedsaleem5586]

بارك الله فيك دكتور و جزاك الله كل خير على هذا المجهود الطيب حفظكم الله

[@DrAfrozAlam]

أمين ، شكراً لك يا أخي على الثناء والاهتمام بمحاضراتي. يرجى مشاركة هذه المحاضرات في مجموعتك.

[@ahmedsaleem5586]

@@DrAfrozAlam تم نشر محاضراتك على قنواتنا في برنامج التلكرام I share it in telegram app Dr 💚 In : t.me/Electricity_Channel And t.me/ib_eng97

[@asrakhursheed664]

May Allah bless you and give you gardens of heaven.

[@caDet07]

Guys ANSWERS ARE i (G1) = -j 11.800 K ampere i (G2) = -j 8.194 K ampere ( here K= 1000) Thank you sir ❤❤

[@malakspeters6052]

hi Sir,,,thank you for your insightful tutorial videos....they helped me understand power systems...great teaching...

[@munierhussain]

THANKS a lot sir, Your teaching quality is top class ❤

[@crackbanking]

These are lectures are really helpful.thank you sir.

[@DrAfrozAlam]

It's my pleasure

[@surendrakverma555]

Excellent lecture Sir. Thanks 🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏

[@1973jdmc]

Thank you so much for these great lecturers- truly appreciated-

[@vindicatress_]

Thank you so much sir, you explained it so well🔥👏👏👏👏

[@HARVINDERSINGH-lx9vy]

Your lectures are awesome ...Thankyou sir🥰

[@DrAfrozAlam]

Most welcome Harvinder Bajwa

[@abuanas8569]

Thank you Dr may Allah reward you greatly

[@komalyadav4051]

Very nice lecture sir.thank u so much

[@neminathnarule2257]

A huge respect and hatts off u sir.. Really ur teaching is awesome👏🙏👌tq sir

[@DrAfrozAlam]

You are most welcome dear

[@fioricoziuc1792]

Good explanation! Thank you!

[@stanleypete1807]

...this is just perfect

[@kashifrehman2060]

MashALLAH Sir, your lectures are helping me a lot in PSA Subject. which book you are following for these lectures???

[@DrAfrozAlam]

I m glad to know this..... Some times, I follow Kothari and Nagrath.....You can see Hadi Saadat also. BTW, I use to prepare my own lectures as per my convenience.

[@mohammadalisina8670]

thank you very much for your nice videos. may Allah save you

[@Aspirant-lp8cw]

Great lecture sir.. Thank you.

[@shubhampuri8865]

Sir if possible could you deliver lecture of power system protection as I can't understand the relays perfectly especially mho relay... Your lectures are really helpful for competitions

[@Haider_Waseem_445]

Mashallah Sri G thank you so much for such kind effort.

[@DrAfrozAlam]

Thank you Waseem for the comment.

[@utsavrai7587]

Love the way you teach

[@DrAfrozAlam]

Thank you Pratap

[@newindia294]

Awesome lecture Thanks lot Sir

[@DrAfrozAlam]

Most welcome

[@sathwik__goud]

classic and neat explanation

[@DrAfrozAlam]

Thank you Sathwik

[@shahabrar1814]

Shukriya janab for the wonderful lectures..👍. One more request to you kindly arrange the uploaded lecture in sequence.it will be helpful to us🙂

[@DrAfrozAlam]

Thank you for the comment. The uploaded lectures are already arranged in the following Playlist: kzbin.info/aero/PLEVP-S4VFKO9skYbAegYfSQNXNJajL3oD

[@amarnishant7464]

Maulvi and Mathematics... great... actual requirement...

[@amaan-123]

Thank you sir!

[@prajwalraut4695]

Really good explanation

[@DrAfrozAlam]

Thank you Prajwal

[@monicaraju903]

Master class sir..please post more videos

[@RAHULNIGAM1985]

Sir. When calculating XT for transformer in pu terms why take voltage on HV side ..why not LV side voltage as old voltage...

[@pranaymeshram8888]

Thank you 😊😊

[@roshandahal8344]

hello sir thank you for this lecture

[@KAFKUBA]

Why isn't line impedance converted to new p.u.??

[@pavanvenkatasai5767]

Thank you sir

[@MahadKhaliq]

Thank you so much Sir, it was so helpful!

[@sajadkhan887]

very well explained

[@DrAfrozAlam]

Thank you Sajad

[@theyab65]

jzak allah khair

[@almoraldeme8664]

Where did you get VL and why is S=100?

[@swaroopapte3440]

Sir i just have one doubt, sir आपने पहले numerical के पहले एक प्वाइंट लिया था की "if some fault impedance is given, then how to calculate" करके। सर मेरा सवाल ये है, की वहा पे fault impedance क्यों आता है??

[@MrOptimusPrime.]

The reason why education is boring without practical classes is the difference between theory and reality. Most engineering students do not know what is motor, transformer, transmission line, circuit breaker, current transformer even in their advanced studies. But they have to make all these calculations theoretically in their engineering studies only on their mind....... This works only when you have hands on experience on substations and power stations........Otherwise this session is just an asmr.

[@maheboobmula1166]

thanks chacha jan.

[@dganesh9391]

Good explanation

[@DrAfrozAlam]

Thank you D Ganesh

[@zabehullah1033]

Sir ! Plz make some lectures on power system protection

[@itsabhilasa]

Sir, can you please make a video on pre fault loaded systems ?

[@KAFKUBA]

Shouldn't 132 squared be divided by 100,000 since it's MVA?

[@mdnurulislamnahid5424]

Thanks sir

[@DrAfrozAlam]

My pleasure

[@abhishekrajwanshi5265]

vaibhav kuril was here

[@mubarakali-cj5gy]

Sir at time 42:03 .100/root 3*line voltage on h.v side you get the value (-j437.4) but I get the value (j0.43738 )... When divided by 100/root 3*132=(j0.43738)....sir how you find the value of (-j437.4) ?

[@DrAfrozAlam]

As 'S' is in MVA, you need to calculate as, (100*1000)/root 3*132 = 437.4

[@amadelfarooq]

Asalamu alikum , please help sir in this Q. A 10KVA,400/200 V single phase transformer with 10% impedence draws a steady state short circuit current of ?

[@DrAfrozAlam]

Wa Alaikum Assalam. Use the same procedure explained in this video lecture. Isc(p.u.) = 1/0.1 = 10, Isc(base) = (10×1000)/200 = 50 A, Isc(actual) = Isc(p.u.)×Isc(base) = 10×50 A = 500 A (Answer). Also, in case of 3-phase transformer, Isc(p.u.) = 1/0.1 = 10, Isc(base) = (10×1000)/(√3×200) = 28.868 A, Isc(actual) = Isc(p.u.)×Isc(base) = 10×28.868 A= 288.68 A

[@Rashed_007]

Thanks..

[@hossaintarik7904]

Mashallah very interesting and simple way to do fault analysis.. But Sir, would you please explain kindly the reason of calculating value of XG2 whereas the value already given as 10% like you put the value of Xg1. wish to get more practice materials.thanks.

[@AryanBKumar]

Even if u apply the formula in xg1 u get 0.15*(100/100)*(11/11)^2 soo he directly wrote it , hope this answer helps you sir 🙏🏻

[@jaydipprajapati5763]

Nice video

[@DrAfrozAlam]

Thank you Jaydip

[@JkFilm965]

Hi sir, I understand that the fault current return to the generator through the earth if generator is earthed at the neutral. But if the generator is ungrounded Wye, how does the fault current return to the generator? Do we still calculate the fault current the same way? Really appreciate your help, thanks!

[@sarvjeetprajapati7257]

Thank you so much sir you are great sir

[@DrAfrozAlam]

Most welcome dear

[@Phoenix16422]

sir can you please explain again how to find out the current through G1 and G2... you were explaining at the end but I couldn't get it...Thankyou

[@DrAfrozAlam]

Total Fault Current supplied by LT side = - j 19.995 kA ≈ - j 20 kA. Now applying current division rule, we get, IG1= - j 11.8003 kA and IG2= - j 8.1947 kA.

[@HaiderAli-pp9go]

@@DrAfrozAlam Sir why are we considering the LT side? Really confused about this

[@DrAfrozAlam]

@ Haider Ali, If it has been asked in the problem, you have to calculate it.

[@HaiderAli-pp9go]

@@DrAfrozAlam I mean why is the current division used by taking LV current? Why can we use HV current we calculated to find the current supplied?

[@DrAfrozAlam]

In Fault analysis, we can evaluate Fault current at any point

[@KAFKUBA]

I don't understand the last step with thevenin simplification

[@atibali1087]

Sir, What will be the effect on fault current in the case of LLL or LLLG fault if neutral of synchronous generator is ungrounded??

[@DrAfrozAlam]

In such faults, negative and zero sequence components of fault current are absent. i.e. only positive sequence component of fault current exists which does not require path through neutral. Note: Only zero sequence component of current requires its path through the neutral. Recall, neutral current In = 3*Io

[@leesadik]

what happens if the two generators are having different magnitude of voltage

[@DrAfrozAlam]

The voltages of parallel operated generators are always kept same.

[@leesadik]

iam from Hyderabad sir iam a btech student at jntuh

[@leesadik]

oh yes sir right i just forgot thank you sir

[@uknown9633]

Fault current is -j19994.6

[@pritchardmabutho8018]

Thank you Dr, your video helped alot. But you didn't finish the last part, What are the values for the actual currents supplied by the two generators?

[@DrAfrozAlam]

Please see the pinned comment

[@pritchardmabutho8018]

Thank you Dr. If you find time please make a video on symmetrical fault analysis with a Load connected.

[@adnanrafiq4507]

Couldn't understand a little portion how to find G1 and G2 current Kindly tell me how to find these current after finding If(actual) on L.V & H.V side

[@DrAfrozAlam]

Apply current division rule

[@ermuhanmaikhanov8259]

Thank you sir for this lecture! Can you just clarify 1 point for me? When making pu system in your example. We assume 100 MVA as a Base. Then Gen1 is 1 pu, but shouldn't Gen2 be 0.5pu?

[@DrAfrozAlam]

Both the generators are operating at 11 kV, hence at 11 kV base, operating voltages in pu for Gen 1 and Gen 2 are shown as 1 pu.

[@kalyanreddy4820]

Sir why in symmetrical fault calculations shunt elements are neglected. Please clear my doubt sir.

[@DrAfrozAlam]

To make the calculations simpler for undergraduate level students

[@DrAfrozAlam]

This assumption doesn't make significant difference in the accuracy of the results

[@MANOJKUMAR-mb2uw]

Sir why we take base voltages different sir on hv and lv side Can we take single base voltage for both sides

[@DrAfrozAlam]

As we know that actual value of voltage at any point of the system is equal to per unit value multiplied by its base value. Hence, base value is changed whenever actual value of voltage is changed due to presence of a transformer.

[@electricalengineeringmat6827]

Thankyou sir

[@technicalzargar8138]

Assalam ali kum Sir S/√3 VL = 437 .. how......? Sir what we take S 100* 10^6 = 100 MVA or If we take 100*10^3 the looks 100 KVA

[@DrAfrozAlam]

Wa Alaikum Assalam dear. S/√3 VL = (100*10^6)/(√3*132*10^3) = 437

[@technicalzargar8138]

@@DrAfrozAlam ok thanku sir

[@DrAfrozAlam]

Most welcome

[@ismailpinjar]

Sir I think base current @hv side is -j3.8095*(S/(√3V)^2)=728159

[@DrAfrozAlam]

I think you should watch the following lectures on p.u. Systems: Per Unit System (In Hindi): kzbin.info/aero/PLEVP-S4VFKO85zhnjJYg_ZCVh1_Wqpvqi

[@honarfarhad]

Sir ican i get a notebook of fault analysis

[@DrAfrozAlam]

You can download Fault Analysis lecture notes (pdf) using the following link: drive.google.com/file/d/18Zvgdi3YD6iAIkWJwz5-s7dVHRgM6lwZ/view?usp=sharing

[@honarfarhad]

@@DrAfrozAlam thanks a lot

[@leesadik]

wonderful class sir👌 may i konw where r u from

[@DrAfrozAlam]

Thank you Sadik. I teach in AMU, Aligarh, India. Where are you from?

[@ranjeetmaurya7913]

Makwana sir suggested me for this lecture

[@DrAfrozAlam]

Yes, Dr. Makwana sir has informed me about this. Feel free to ask me in case of any difficulty.

[@ranjeetmaurya7913]

Thank you sir

[@asror94]

what formulas calculation for LLLG sir?

[@DrAfrozAlam]

kzbin.info/www/bejne/f2qbeHSOnMmriJY

[@darmavarapugovindarao8386]

Sir please make a vedio psa first chepter ( chapter name is per unit representation and topology )

[@DrAfrozAlam]

Per Unit System (In Hindi): kzbin.info/aero/PLEVP-S4VFKO85zhnjJYg_ZCVh1_Wqpvqi

[@darmavarapugovindarao8386]

Sir I want to English language psa lectures for the purpose of 3-2 semister examinations

[@gustavomarcelo7250]

Hi sir. Not clear on the source; how can we say it is 1pu?

[@DrAfrozAlam]

Source voltage will be taken as base voltage (Base kV). Hence, its per unit value will be 1 p.u.

[@gustavomarcelo7250]

@@DrAfrozAlam How do we say two parallel source equivalent to one source?

[@DrAfrozAlam]

@@gustavomarcelo7250 Using source transformation. First convert each parallel voltage sources into current sources, then convert the equivalent current source into voltage source.

[@sohailjanjua123]

Salam , how come 437.4 ? . Please explain me. Thanks

[@DrAfrozAlam]

Wa Alaikum Assalam, On HV side, I_base=(Base MVA)/(√3*Base kV) = (100*1000000))/(√3*132*1000) = 437.4 A

[@sohailjanjua123]

@@DrAfrozAlam Salam , Thanks

[@transmissionlinetricks9760]

Assalamualikum respected sir kindly guide me about substation and transmission lines book. JazakAllah

[@shahbazalam9993]

Sir plz upload fault analysis in hindi language

[@shilpasweety641]

Srir will these lectures helpful for gate??

[@DrAfrozAlam]

Ofcourse, after this lecture, you may watch the following solved GATE questions on "Symmetrical Fault Analysis": kzbin.info/www/bejne/b3TInpeZa9ZqhZI kzbin.info/www/bejne/bpubc6d-qLpseNE

[@sarvjeetprajapati7257]

Chapter wise playlists banaiye sir

[@DrAfrozAlam]

Various playlists have now been created.

[@duttadrj]

Sir please explain source transformation theory

[@DrAfrozAlam]

If you have problem in understanding 'Source transformation', then alternatively you can use Millman's theorem which is used to compute the voltage of a circuit of parallel branches.

[@electricalengineeringmat6827]

Sir do you have whatsapp group

[@Gagandeepsingh-uf2un]

sir, what about topics like transients on transmission line, short circuit of synchronous machine ,selection of circuit breaker , sir have you covered them or not?

[@mouhammedboudjemel650]

Salamu alikum sir I want your address mail if you can لقد فهمت الدرس جيدا لكن عندي مسألة استعصت علي و اريد منك المساعدة ان امكن وشكرا

[@DrAfrozAlam]

Wa Alaikum Assalam, [e-mail]

[@mouhammedboudjemel650]

@@DrAfrozAlam 🤲👍

[@mouhammedboudjemel650]

Peace, mercy and blessings of God be upon you. I am Mohamed from Algeria, a university student majoring in electrical networks. I have sent you a picture of a topic that I could not solve in your e-mail. If you can help me in the shortest time, thank God

[@DrAfrozAlam]

The language of the questions mailed by you, is not English. So, I am unable to read those questions.

[@mouhammedboudjemel650]

@@DrAfrozAlam شكرا استاذي

[@akashpq]

sir why we use Dy11 transformers at receiving end of the transmission line and Yd1 at sending end of it?

[@atulsain6170]

Yd is used for step down (root 3 time more stepping down then yy or dd for line voltage) and dy is used for step up (more step up).

[@ijaharalam8024]

Can you speech in Hindi language? please!!!

[@DrAfrozAlam]

Per Unit System (In Hindi): kzbin.info/aero/PLEVP-S4VFKO85zhnjJYg_ZCVh1_Wqpvqi

[@DrAfrozAlam]

Bus Admittance Matrix (In Hindi): kzbin.info/aero/PLEVP-S4VFKO-BAGScDSifBQQLGXRPwjvb

[@DrAfrozAlam]

Gauss-Seidel Method of Load Flow (In Hindi): kzbin.info/aero/PLEVP-S4VFKO8MBMECe3UfdlIurue36pqa

[@MajidKhan-kq7se]

Zhcet😍

[@DrAfrozAlam]

Yes, I teach in ZHCET, AMU, Aligarh

[@KAFKUBA]

I don't buy the thevenin explanation

[@MOKRABachour]

بارك الله فيك دكتور و جزاك الله كل خير على هذا المجهود الطيب حفظكم الله

[@DrAfrozAlam]

شكرا جزيلا للتعليق. هل لي أن أعرف من أين أنت؟

[@surendrakverma555]

Very Good lecture Sir. Thanks

[@nidhichandrakar5097]

Great explanation sir