Armature Reaction in Synchronous Generator | Part 1

For other lectures, click the links given below:
Power System Analysis (Playlist): Click the link given below
• Power System Analysis
DC Machines (Playlist): Click the link given below
• DC Machines
Synchronous Machines: (Playlist): Click the link given below
• Synchronous Machines
Regards,
Dr. Afroz Alam,
Ph.D (IIT Roorkee),
Assistant Professor,
Department of Electrical Engineering,
Aligarh Muslim University, Aligarh, India.

[@DrAfrozAlam]

At time 24:55, rate of cutting the flux lines by conductors of coil sides "a" and "a-dash" is maximum, hence instantaneous value of EMF induced in a-phase at this moment is maximum. That's why in phasor diagram, it is being represented along vertical axis so that its vertical projection (instantaneous value) is maximum.

[@DrAfrozAlam]

Download handwritten lecture notes using the following links: drive.google.com/file/d/1ZpJCZtec-IRLmqKVqdAXtgL_-1eJhP8T/view?usp=sharing drive.google.com/file/d/1zRS-pSN6YhG0nAz7mGj-gKLjAOgEZ6Hx/view?usp=sharing drive.google.com/file/d/1i5o4YS3qseOrvZoWh1r07pJPJsunXtdX/view?usp=sharing

[@joysuryadutta2306]

U actually taught very well, in a calm manner, we need more teachers like u, Regards

[@somanathanraju847]

Very good explanation. Thank you. I learn from you a lot , that I never understand my studying.

[@nagarajvachar3076]

Nice sir.its very clear to us.

[@uttamchettri791]

You are amazing sir.

[@surendrakverma555]

Excellent lecture Sir. Thanks 🙏🙏🙏🙏

[@subhajitghosh8221]

Exceptional , and extraordinary, pls upload more topics on Synchronous machine. Thanks and regards

[@larrydurante9849]

Dear Professor, I see we are talking about armature reaction under 'balanced load' correct? If the load is unbalanced, or even different power factors per phase, is it correct to assume that the unbalance must be within a small amount for the generator to tolerate resulting unbalanced torques? I'm thinking yes of course, that there is protection for this... but I ask anyway :)

[@mdtasnim9470]

sir , Pease upload more on power system and also how these machines are simulated in matlab with closed loop

[@subhajitghosh8221]

Sir, I have a very silly doubt , but not getting cleared by any means. People say Flemming's Right hand rule is always appilcable in generator (whatever be the evaluated parameter) and left hand rule in case of motor. Sir is that statement true? Another bunch of people say that right hand rule is applicable in case of generating action ( EMF production) to find direction of induced EMF/current irrerespective of motor/generator. Menas they are saying in case of finding direction of motor Back EMF , Right hand rule is applicable. Confusing.

[@larrydurante9849]

Dear Professor, at 11:59, 'max flux linkage means max emf', I have heard other say this as well. Isn't it the max 'rate of change' of flux linkage? Therefore when passing through zero we get the maximum rate of change and thus by Faraday's Law we get the maximum 'back emf' in the stator windings? thank you

[@DrAfrozAlam]

Yes, you are right

[@larrydurante9849]

Dear Professor, I think I see it know. When flux of rotor is cutting stator windings the most that it when the flux through the stator winding is actually changing the most. When rotor flux is in quadrature with the stator windings the flux through the stator loop is the most but it's rate of change is the least as the rotor rotates. This has always been a point of confusion coming from electromagnetics back ground. Now I see that maximum cutting means maximum rate of change of field flux though stator windings and thus maximum emf.

[@dashok4749]

Sir plz answer how 3 dark lamp method and 2 bright one dark lamp method of parallel operations of alternator works

[@DrAfrozAlam]

kzbin.info/www/bejne/pZ24nJ6lmNV3r68

[@DrAfrozAlam]

kzbin.info/www/bejne/oIvNl6p5gsprmdE

[@mdtasnim9470]

Explain how rotor field and stator field rotates in same direction

[@DrAfrozAlam]

kzbin.info/www/bejne/hnXIk3l5p8trgJo

[@saimunsahriar]

Sir Assalamualaikum, i have a little confusion here. Is not magnetic direction is North to South? Is this yes. Then i think here where You write N this is S and where you write S pole this is actually N pole. ( i am not sure) please Sir explain a little..

[@DrAfrozAlam]

Wa Alaikum Assalam, It is North to South in the air i.e. outside the magnetic material, and South to North inside the magnetic material.

[@saikatmukherjee2548]

In 5:57 how can the EMF be maximum in aa' coil if the magnetic flux is parallel to the coil i.e phi = B.A = 0 as B is perpendicular to A.

[@DrAfrozAlam]

At this moment, dΦ/dt is maximum. Hence, induced EMF will be maximum.

[@saikatmukherjee2548]

@@DrAfrozAlam But sir I have a little confusion at this point the area vector is perpendicular to Magnetic field vector so phi must be 0 ?

[@DrAfrozAlam]

Please read the pinned comment that addresses the same concept.

[@saikatmukherjee2548]

@@DrAfrozAlam Ok sir got it.... Thank you 😇🙏

[@DrAfrozAlam]

Most welcome dear

[@souravmazumder1054]

30:15 sir this point not clear please clear this... we know cos 60 or theta means adjacent by hypotenuse. here adjacent memas -Im and hypotenuse means Ib so cos 60 will be Im/Ib.. so Im = -Ib x cos 60.

[@DrAfrozAlam]

Phasor length is Im. Hence, its projection along negative Y-axis is -Im*cos60°.

[@souravmazumder1054]

@@DrAfrozAlam Sir, the projection along negative Y axis is -Im and the angle between Ib and -Im is 60 degree. so cos 60 = (-Im/Ib)

[@DrAfrozAlam]

No, the projection of Im along negative Y-axis is -Im*cos60°.

[@souravmazumder1054]

@@DrAfrozAlam Sir, along with positive Y axis it is +Im according to Ia. so when we calculate Ib and taking negative of projection of Y axis that is -Im. we know cos theta = adjacent/hypotenuse. so?? not clear. how -ImCos60 came..

[@souravmazumder1054]

Sir can please you sent any u tube video link for better understanding. May be I'm unable to understand only this part.. I draw a negative axis of Y. and divide in two section between Ib and Ic, with 60 degree each. that is make a Right Triangle 📐. so put cos 60 as per the rule and i got Im=-IbCos60