1. Calculate the relay setting of single line diagram of radial feeder shown in the fig 1. The relay used are standard IDMT phase over current relay, having rated current of 1 A. The normal range of PS (Plug Setting) is fro 50%-200% of 1 A in equal step of 25%, where as TDS range is 0.1 to 1 sec in steps of 0.05. The PS of relay R2 is givens 50% of CT Secondary rating and TDS is given as 0.2 Sec. How to solve example Sir
can we see an education system....where we can experience the same thing from today generation numerical relay, means how an over current setting is done, what are different over current setting are available and when to apply a particular setting , what are the different standards used for protection system setting, tools where we can check the relay operation with its behavior and can get a conclusion that the relay has operated as per the setting. We need our education system to change and teach the thing which are used at practical field...not just a book worm who has explored protection system just on a pen, paper and calculator.
use top=0.14*(TMS)/((fault current/settings current)^0.02-1) we can get the operating time of the relay for IDMT relay under normal standard condition top=operating time of relay tms=time multiplier setting
How the rated secondary current 5 A is taken. Is that fixed for every problem or it will be changed according to the problem. And tell me how secondary current is taken?
Sir mujhe 33/11kv substation me relay cordnetion karna hai as per power transformer, 11kv main and feeder protection my substation in rural areas and feeder connected in agriculture fault generated havey my problem is cum to feeder fault feeder trip ho jata hi leki 11kv main and Transformer vcb bhi trip ho jata hai power transformer installed in my substation caipcity 8 mva 5 mva and 3.15 mva ct connected ratio 400-200/5 200-100/5 and 300-150/5 for transformer protection 300-150/5 and 200-100/5 11kv main and feeder protection relay connected make earth fault over current Numerical relay sir mujhe relay cordnetion bata do jisse ki feeder fault aane par feeder hi trip ho transformer Tak fault na jaye
The value of time is to be taken from the time vs psm graph for the 3 sec relay for TMS = 1. Then, multiply that time with the TMS of 0.6. That will give you the answer.
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Good video for Semister exam, Thanks sir
th-cam.com/video/ISdnXvNDRvc/w-d-xo.html
1. Calculate the relay setting of single line diagram of radial feeder shown in the fig 1. The relay
used are standard IDMT phase over current relay, having rated current of 1 A. The normal
range of PS (Plug Setting) is fro 50%-200% of 1 A in equal step of 25%, where as TDS range is
0.1 to 1 sec in steps of 0.05. The PS of relay R2 is givens 50% of CT Secondary rating and TDS
is given as 0.2 Sec.
How to solve example Sir
Nice video...
can we see an education system....where we can experience the same thing from today generation numerical relay, means how an over current setting is done, what are different over current setting are available and when to apply a particular setting , what are the different standards used for protection system setting, tools where we can check the relay operation with its behavior and can get a conclusion that the relay has operated as per the setting.
We need our education system to change and teach the thing which are used at practical field...not just a book worm who has explored protection system just on a pen, paper and calculator.
Changing education system is my life time dream
S
Very good video sir. Thank you sir
th-cam.com/video/ISdnXvNDRvc/w-d-xo.html
use top=0.14*(TMS)/((fault current/settings current)^0.02-1)
we can get the operating time of the relay for IDMT relay under normal standard condition
top=operating time of relay
tms=time multiplier setting
This is an empirical formula. What's the 0.14?
Thanks you..
What ever you calculated the end result was from the given value in the question it self. 3 x 0.6 = 1.8 sec. Then why all these calculation required.
Seen
How the rated secondary current 5 A is taken. Is that fixed for every problem or it will be changed according to the problem. And tell me how secondary current is taken?
1A or 5A depend on your relay
how can we pass 50 amp secondary fault current through a relay of 5 amp relay? donot we need to change the CT ratio(1000/1)
th-cam.com/video/ISdnXvNDRvc/w-d-xo.html
Ct can carry large fault current up to some seconds
Sir mujhe 33/11kv substation me relay cordnetion karna hai as per power transformer, 11kv main and feeder protection my substation in rural areas and feeder connected in agriculture fault generated havey my problem is cum to feeder fault feeder trip ho jata hi leki 11kv main and Transformer vcb bhi trip ho jata hai power transformer installed in my substation caipcity 8 mva 5 mva and 3.15 mva ct connected ratio 400-200/5 200-100/5 and 300-150/5 for transformer protection 300-150/5 and 200-100/5 11kv main and feeder protection relay connected make earth fault over current Numerical relay sir mujhe relay cordnetion bata do jisse ki feeder fault aane par feeder hi trip ho transformer Tak fault na jaye
Sir what if the question starts in this way
A 5a ,2.2 sec .....
Sir will there be any changes in the ans ??
Psm =0.8
Time is 3sec×0.8= 2.4 sec That is ans sir.
th-cam.com/video/ISdnXvNDRvc/w-d-xo.html
Sir ji 3.5 kaha se aya samjha nahi
Definition of trip current of relay
th-cam.com/video/PLTi2z2Uz9Q/w-d-xo.html
where you get TSM=0.6?
TSM Will be given in question bro...
@@raghavms1410 🤣
125% how it's get
I have one smart excel sheet which you may see on my youtube channel. Final resultant value will be 1.97 sec using IEC SI curve.
Plz provide...
3.5 kaise aya..???
It's a mistake waha par 3 second aayega
The value of time is to be taken from the time vs psm graph for the 3 sec relay for TMS = 1. Then, multiply that time with the TMS of 0.6. That will give you the answer.
@@rahulkhatri6350exam mai diya jayega ki if psm is 8 then the time of operation will be ______
Jo exam mai diya jayega vo hi likhna hai