Thanks. I have started a career in electrical transmission (utility control). This video is a great resource in addition to my textbooks. I am definitely going to watch some more of your content.
Hello Mr. Schett! Thank you very much for uploading the series on power systems engineering. However, as far as I can see, you've overlooked something here. The mistake you make is based on the assumtion that power line angle stability is equivalent to rotor angle stability, which it is not from my point of view. For an ideal power line & synchronous machine you'll have following formula: P = U1*U2/X*sin(theta) In case of a generator, U2 will be the more or less constant grid voltage & U1 the rotor voltage (assumed to be constant as well). Therefore, your calculations are correct. However, in case an electric line where U1 might be the voltage deep inside the grid (constant), there is no reason to assume that U2 is a constant in theta. Increasing theta will decrease U2, and at some tipping point U2 will decrease faster than sin(theta) increases -> power is at its maximum. The angle at which this happens is !not! 90° but a function of the load angle (solely). In order to calculate it, you'll have to find an equation U2=f(P,load angle). Therefore, it is convenient to use the equation of an ideal power line: U2 = U1*cos(beta*L)-j*I1*Zw*sin(beta*L)...Depending on which coordinate system you use, you'll get a different sign. Substitute I1 with some expression of P & load angle, this will yield a quadratic equation in U2. Solve it and you'll end up with a term U2=f(P, load angle) which can be plugged into the first equation P = U1*U2/X*sin(theta). This will yield a curve P(theta) similar to yours, but with the essential difference that theta(P_max) is around 50-60° , depending on the load angle. Best regards, Matthias Maier
You are correct, therefore (as far as I remember...) I said in the video that for simplification reasons I assume that the voltage and the angle include the rotor angle. In reality the synchronous reactance of the generator plays a major role for both, the angle and the voltages. Therefore as you say, the maximum angle in the system will be 30 - 60 and not 90 because the additional angle between the generator terminals and the induced generator voltage has to be considered. Thanks for your valuable comments!
Hello Schett, great video. Just wonder if there is an intuitive way to understand why active power depends on a certain phase angle between source and load?
Hello Magnus, thanks for the fb! I am working on such a video, will be published soon. You can think of two magnets. Both are mounted on the same axis in the middle of the magnet. As long as there is no torque applied on the magnets, they will be oriented in parallel however with N and S pole in opposite direction. One magnet is the rotor field, the other one is the field induced by the stator currents. As soon as there is a torque, the angle between the rotor and the stator field will increase. Applying torque on a rotating system is equal to provide active power. I hope this helps a bit.
I appreciate your video. However, regarding the formula @7:00, is V^2 the difference between the source and terminal voltage squared, or is it one of those voltages squared? If it is one of them, which one?
Good question. The assumption is, that both voltages are equeal in magnitude but different in phase angle. This is the key point. The torque establishes the phase angle and when you have torque you provide real power.
@@georgschett801 thanks for your response. So then if I understand correctly, if we took into account the voltage drop due to impedance on the transmission lines and probably source stator windings, then instead of V^2 we shall have V source multiplied by V Terminal (Vs * Vt). Is this reasoning correct?
sir, i want to ask one more question please. so what is exactly the rotor angle stability? i mean what angle exactly? angle between what and what? because i read and see another video and i am confused now. please help
It is the angle between the inner induced voltage of the synchronous generator and the outer voltage you can measure at the terminal of the generator. I am not sure where your confusion comes from. You may write me a mail: georg.schett@ecsp.ch
@@georgschett801 I think it´s a straightforward question. It may not seem that evident that the angle (in the time domain) between terminal voltage and induced e.m.f. is the same as the angle between stator and rotor magnetic fields (electrical degrees).
great explanation, but sir im still confuse, please answer. in 11:37 you say that "and you can see the phase angle between the line ends are still well below 90 degrees" pleasee tell me which one is the phase angle, helpp
you see the 3 sinusoidal voltage curves of one line end and very close by, with a slite delay the 3 other curves. The delay at the instant you are pointing to is a few miliseconds (1 or 2 ms if I remeber well). The time delay is called the phase angle (or the delay angle or angle between the phasors) between both line ends. For 90 degrees you would need a time delay of 5 ms, enough to loose stability.
@georgschett801 First, thank you very much for your interest. at the minute of 14:12, you said ( I open the breaker, and you can see now is how much this angle between the sending end and receiving end has been decreased) , you said before when adding the capacitor, the angle will decrease, so when we remove the capacitor, the angle is supposed to increase, or what? Or you may be mean when you open the breaker of the second line and not the breaker of the capacitor , is this right ?
@@user-lr6hx8bb2m I guess I see your point. When the breaker is closed, the capacitor is short circuited, i.e. not in. Only when the breaker is open, the capacitor is in. I hope this clarifies
Thanks. I have started a career in electrical transmission (utility control). This video is a great resource in addition to my textbooks. I am definitely going to watch some more of your content.
Love your explanations w/great graphic representations! Clear to the point tutorials! TX
I really liked your video, straightforward explanation and nice analogy with the horse-spring-weight example.
Hello Mr. Schett! Thank you very much for uploading the series on power systems engineering. However, as far as I can see, you've overlooked something here. The mistake you make is based on the assumtion that power line angle stability is equivalent to rotor angle stability, which it is not from my point of view.
For an ideal power line & synchronous machine you'll have following formula:
P = U1*U2/X*sin(theta)
In case of a generator, U2 will be the more or less constant grid voltage & U1 the rotor voltage (assumed to be constant as well). Therefore, your calculations are correct. However, in case an electric line where U1 might be the voltage deep inside the grid (constant), there is no reason to assume that U2 is a constant in theta. Increasing theta will decrease U2, and at some tipping point U2 will decrease faster than sin(theta) increases -> power is at its maximum. The angle at which this happens is !not! 90° but a function of the load angle (solely). In order to calculate it, you'll have to find an equation U2=f(P,load angle). Therefore, it is convenient to use the equation of an ideal power line: U2 = U1*cos(beta*L)-j*I1*Zw*sin(beta*L)...Depending on which coordinate system you use, you'll get a different sign. Substitute I1 with some expression of P & load angle, this will yield a quadratic equation in U2. Solve it and you'll end up with a term U2=f(P, load angle) which can be plugged into the first equation P = U1*U2/X*sin(theta). This will yield a curve P(theta) similar to yours, but with the essential difference that theta(P_max) is around 50-60° , depending on the load angle.
Best regards,
Matthias Maier
You are correct, therefore (as far as I remember...) I said in the video that for simplification reasons I assume that the voltage and the angle include the rotor angle. In reality the synchronous reactance of the generator plays a major role for both, the angle and the voltages. Therefore as you say, the maximum angle in the system will be 30 - 60 and not 90 because the additional angle between the generator terminals and the induced generator voltage has to be considered. Thanks for your valuable comments!
Excellent work THANK YOU... I paid for degree and I STILL looking for information to understand!!!
Thanks for your feedback. You could support by sharing the videos in your community.
Great explanation, please keep making this kind of videos.
Sure, Ahmed, thanks for your feedback.
Excellent work! Regards from Argentina.
Thanks for your kind feedback. Please share it with your connections for me to get visibility in the community. All the best!
@@georgschett801 It was the first I did when a finished the video.
Hello Schett, great video. Just wonder if there is an intuitive way to understand why active power depends on a certain phase angle between source and load?
Hello Magnus, thanks for the fb! I am working on such a video, will be published soon. You can think of two magnets. Both are mounted on the same axis in the middle of the magnet. As long as there is no torque applied on the magnets, they will be oriented in parallel however with N and S pole in opposite direction. One magnet is the rotor field, the other one is the field induced by the stator currents. As soon as there is a torque, the angle between the rotor and the stator field will increase. Applying torque on a rotating system is equal to provide active power. I hope this helps a bit.
@@georgschett801 Thanks, yes absolutely, cheers :)
I appreciate your video. However, regarding the formula @7:00, is V^2 the difference between the source and terminal voltage squared, or is it one of those voltages squared? If it is one of them, which one?
Good question. The assumption is, that both voltages are equeal in magnitude but different in phase angle. This is the key point. The torque establishes the phase angle and when you have torque you provide real power.
@@georgschett801 thanks for your response. So then if I understand correctly, if we took into account the voltage drop due to impedance on the transmission lines and probably source stator windings, then instead of V^2 we shall have V source multiplied by V Terminal (Vs * Vt). Is this reasoning correct?
@@dubol07 Perfect!!
Thank you sir. What is the direction of torque. Same with rotation of rotor or opposite it
It depends on the power delivery: With power out, the torque has the same direction as the rotation and vice versa.
sir, i want to ask one more question please. so what is exactly the rotor angle stability? i mean what angle exactly? angle between what and what? because i read and see another video and i am confused now. please help
It is the angle between the inner induced voltage of the synchronous generator and the outer voltage you can measure at the terminal of the generator. I am not sure where your confusion comes from. You may write me a mail: georg.schett@ecsp.ch
@@georgschett801 I think it´s a straightforward question. It may not seem that evident that the angle (in the time domain) between terminal voltage and induced e.m.f. is the same as the angle between stator and rotor magnetic fields (electrical degrees).
great explanation, but sir im still confuse, please answer. in 11:37 you say that "and you can see the phase angle between the line ends are still well below 90 degrees"
pleasee tell me which one is the phase angle, helpp
you see the 3 sinusoidal voltage curves of one line end and very close by, with a slite delay the 3 other curves. The delay at the instant you are pointing to is a few miliseconds (1 or 2 ms if I remeber well). The time delay is called the phase angle (or the delay angle or angle between the phasors) between both line ends. For 90 degrees you would need a time delay of 5 ms, enough to loose stability.
@@georgschett801 thankyou sir!
Excellent I required a little more explanation and exposure could you help me...Electrical Safety Inspector
Sure, let me know your questions / suggestions. You can contact me by mail on georg.schett@ecsp.ch
Please , I think there is a mistake in the last two minutes, or may be the mistake in my understanding.
OK, thanks! But I am not aware of a mistake, can you please let me know?
@georgschett801 First, thank you very much for your interest. at the minute of 14:12, you said ( I open the breaker, and you can see now is how much this angle between the sending end and receiving end has been decreased) ,
you said before when adding the capacitor, the angle will decrease, so when we remove the capacitor, the angle is supposed to increase, or what?
Or you may be mean when you open the breaker of the second line and not the breaker of the capacitor , is this right ?
@@user-lr6hx8bb2m I guess I see your point. When the breaker is closed, the capacitor is short circuited, i.e. not in. Only when the breaker is open, the capacitor is in. I hope this clarifies
@georgschett801
You really added valuable information to me. Thank you very much 😊