The interesting thing here to see is the value of voltage and the impedance used in calculating the fault through the equivalent circuit: The voltage and the impedance are used as a pre-fault value, not the actual value, in reality during the fault the voltage decreases to zero, but in calculation we only consider the pre-fault value to have the worst case value to size the equipment
Hello Sir, one question is that, how is it possible that at source kA (10kA) is lower that at Feeder end (13.62kA)? As we go down stream, kA will reduce.
Hello, you are right with the concept if we consider fault on the same voltage level. However, the example described in video 10kA is fault level at a source of 11kV, and 13.62 kA is fault at LV side of transformer (415V). If we convert on the HV side (11kV), it will be less than 10kA.
Perfect , but only one question , transformer ratio considered as a phase voltage not line voltage as they are different in case of delta star transformer in our case , please clarify
@@railwayengineeringinsights6463 No you are wrong, Transformer Turn Ratio is calculated based on phase voltage not on based online voltage. In the example case also the impedance of cable when you are transferring on LV side of transformer and diving by turn ratio, then it will be (11/ (0.4/sqrt (3)) ², then only the value what you are writing is coming which is 0.000173+j0.00019.
@@sudhirsharma3435 please note that turn ratio is 11/0.4 and not 11/0.4 srqt 3 as stated by you. Furthermore, if done square of 11/0.4 it is 756.25 and if you substitute it in expression for impedance you will get from R= 0.131/ 756.25 ohm/km = 0.000173 and same for X.
![I.N "HALLUCINATION" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/n5B5q1Hwt_U/mqdefault.jpg)
The interesting thing here to see is the value of voltage and the impedance used in calculating the fault through the equivalent circuit: The voltage and the impedance are used as a pre-fault value, not the actual value, in reality during the fault the voltage decreases to zero, but in calculation we only consider the pre-fault value to have the worst case value to size the equipment
Very good job and very informative which we can't able to interpret from the standards.
Many thanks!
Good job!
Many thanks
Very good. Thanks 🙏
Thanks for sharing this video.
🙏🙏
How to Get X/R ratio?
Hello Sir, one question is that, how is it possible that at source kA (10kA) is lower that at Feeder end (13.62kA)? As we go down stream, kA will reduce.
Hello, you are right with the concept if we consider fault on the same voltage level. However, the example described in video 10kA is fault level at a source of 11kV, and 13.62 kA is fault at LV side of transformer (415V). If we convert on the HV side (11kV), it will be less than 10kA.
@@railwayengineeringinsights6463 okay, thanks sir.
Well explained... 👏
Many thanks..
Perfect , but only one question , transformer ratio considered as a phase voltage not line voltage as they are different in case of delta star transformer in our case , please clarify
Many thanks...Transformer Turn Ratio (TTR) is considered as ratio of line voltages... For example Dyn11 of 33kV/0.433V will have a TTR 33/0.433....
@@railwayengineeringinsights6463 No you are wrong, Transformer Turn Ratio is calculated based on phase voltage not on based online voltage. In the example case also the impedance of cable when you are transferring on LV side of transformer and diving by turn ratio, then it will be (11/ (0.4/sqrt (3)) ², then only the value what you are writing is coming which is 0.000173+j0.00019.
@@sudhirsharma3435 please note that turn ratio is 11/0.4 and not 11/0.4 srqt 3 as stated by you. Furthermore, if done square of 11/0.4 it is 756.25 and if you substitute it in expression for impedance you will get from R= 0.131/ 756.25 ohm/km = 0.000173 and same for X.
10 KA fault value is a 3 phase fault
at grid side?
Yes