sir because of you i will have to make my IES notes again for stability, load dispatch and fault analysis part . 😭 old ones too complicated . Thanks bdw 😂
Dear Pradeep, Once again excellent explanation and teaching methods in your video. I think one helpful addition to your channel would be if you made some lectures about developing the capability diagram of synchronous generator
for equation at 24:55, to solve for critical clearing angle, do we use Pmax @ fault? Or before fault? Or once fault is cleared? Pmax changes based on which condition.
Hye Sir. The area A1 ( or A2) is said to be the stored energy. How does the P axis vs delta axis or what is the relationship between P and delta that we can say the area is equals to the energy?
The areas A1 and A2 haven't been said to be "equal to the energy". Due to the discrepancy between the mechanical power input and the electrical power output the latter being zero on account of fault (V=0), the entire power input to the rotor manifests itself as kinetic energy, to accelerate the rotor. This accelerating power acting on the rotor, makes it's speed which was synchronous prior to the fault (on account of stable operating conditions), to go beyond synchronous now. The rotor angle delta increasing along with it. Following the fault clearance, the electrical power output becomes greater than the constant mechanical power input, as is evident from the power angle characteristic. Therefore the accelerating power goes negative now. This makes the rotor decelerate, it's speed gradually decreasing from super-synchronous to synchronous, until the kinetic energy which it had acquired during the initial accelerating period is completely given up as per the "energy conservation principle". Thus areas A1 and A2 are actually "representatives" of the Kinetic Energy gained or lost by the rotor and not mathematically equal to the kinetic energy gained or lost by the rotor. We equate A1 with A2 in equal area criteria just to comply with the "energy conservation principle".
Debanjan Datta : whats the physical reason of heigher electrical output power just after the fault clearance ? Is it , because of consistent mechanical input + the kinetic energy stored both will produce electrical power during the period it returns to sycn speed and hence the stored kinetic energy being consumed in output power generation ?
Thanks a lot Pradeep!! I’m doing Masters and I must say your are more wonderful than my professor in explaining the concept
Really great. Thank you for sharing this knowledge.
Thanks a lot for your dedicated video. Much better than my lecturer.
Good explanation sir
Makes me understand this topic even more!
sir because of you i will have to make my IES notes again for stability, load dispatch and fault analysis part . 😭 old ones too complicated . Thanks bdw 😂
Very helpful video, keep up the good work!
Excellent lecture
Dear Pradeep, Once again excellent explanation and teaching methods in your video. I think one helpful addition to your channel would be if you made some lectures about developing the capability diagram of synchronous generator
Superb......
for equation at 24:55, to solve for critical clearing angle, do we use Pmax @ fault? Or before fault? Or once fault is cleared? Pmax changes based on which condition.
superb....sir...
very good lecture
Very good, understood perfectly
Perfect video ... Best video I have ever seen... Thank you..
Excellent tutorial sir Thank u
Excelent !! , From Perú
at 15:00 you wrote anglemax = pi - intial angle, but thats wrong because anglemax is pi/2 (90 deg)
keep up the good work! very useful and easily understood, way better than my grumpy prof. thank you! meru teluga?
Why load angle increases with increase in load... Please explain mathematicallye.
Great.
nice video
Sir , can the fault condition in Generator be considered as a change in Load for the motor?
nice sir
Sir how pmech is less than pelec
Point to point swing curve means
Plz tell me 2H/w=M hota yaa fir M=SH/w hota hai sala confusion ho rha hai
Any Matlab program for multimachine
please put subtitle on this, so it can be auto translated. thanks.
❤
Excellent but try to keep ur phn in silent mode...its distracting you
Hye Sir.
The area A1 ( or A2) is said to be the stored energy. How does the P axis vs delta axis or what is the relationship between P and delta that we can say the area is equals to the energy?
The areas A1 and A2 haven't been said to be "equal to the energy". Due to the discrepancy between the mechanical power input and the electrical power output the latter being zero on account of fault (V=0), the entire power input to the rotor manifests itself as kinetic energy, to accelerate the rotor. This accelerating power acting on the rotor, makes it's speed which was synchronous prior to the fault (on account of stable operating conditions), to go beyond synchronous now. The rotor angle delta increasing along with it. Following the fault clearance, the electrical power output becomes greater than the constant mechanical power input, as is evident from the power angle characteristic. Therefore the accelerating power goes negative now. This makes the rotor decelerate, it's speed gradually decreasing from super-synchronous to synchronous, until the kinetic energy which it had acquired during the initial accelerating period is completely given up as per the "energy conservation principle". Thus areas A1 and A2 are actually "representatives" of the Kinetic Energy gained or lost by the rotor and not mathematically equal to the kinetic energy gained or lost by the rotor. We equate A1 with A2 in equal area criteria just to comply with the "energy conservation principle".
Debanjan Datta : whats the physical reason of heigher electrical output power just after the fault clearance ? Is it , because of consistent mechanical input + the kinetic energy stored both will produce electrical power during the period it returns to sycn speed and hence the stored kinetic energy being consumed in output power generation ?
it will help ua-cam.com/video/-bJ_X-5Lyfk/v-deo.html
2H/w hota hai yaa fir SH/w hota hai yaar
Sir M.E &E.E POWER BOTH ON X -axis but u say me power is in y -axis&ee power is in x-axis
It will help ua-cam.com/video/-bJ_X-5Lyfk/v-deo.html
Excellent, very much helpful. far better understandable than bookish shit and professors dull speech.
thanku sir