Fault Analysis in Power Systems Part 2b
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- Опубликовано: 6 фев 2025
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In this series, we will be going over the analysis of various types of faults that occur in power systems and at the same time intuitively understanding the hand calculations involved.
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Really well done and easy to comprehend. Thank you.
I may have missed it, but is there an assumption regarding negative sequence impedance for the transformer and would this apply to all unbalanced fault studies?
Thank you for the videos you offered. I have a question about a program you used as a black board when you teaching. what is the program name?
Hi, Thank you for the informative video. When you made the assumption where +ve, -ve & 0 seq impedance to be the same, does that mean that the generator & the transformer are assumed to be solidly grounded star windings only?
Thank you for all your videos. Just one question, is there any explaination to why Kilo and Mega are being used without any form of conversion, First was when calculating the Base impendance (Yes! just as a base, I could understand, cause it is just being used as a refference), but when calculating the short circuit impedance of the Generator, the voltage is in KV while the Apparent power is in Mva, here I would need a good reson to understand. Thank you in advance.
Kilo = 10^3 ; M = 10^6
Z(or)R = V^2/P ;
(10^3)^2/10^6 = 10^6/10^6 = 1
Voltage is in (Kilo) Volts; Power is in (Mega) Volt Amperes
V^2/P = (Volt)^2/(Volt*Amp) = volt/Amp = ohm
try this: (115*1000)^2V / (30*1000000) VA = 440.83 OHM
how did we get j0.10 pu, we nvr solved anything
Isn't it because pu is 10%?
Why Transformer P.U takes as 0.10. Why not 0.10/440.83 ??
Hi Farhan, the transformer was already provided in the question so there was no need to divide it with base value.
Hope you find it useful.
Best Regards,
GeneralPAC by AllumiaX
how p.u values for transformer taken as 0.1 i dont get it
never mind i got it.
Strange network; generator 115 kV nominal voltage and line at 13 kV.