Energy Conversion Academy

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@mohammedalahmadi6238 7 күн бұрын

Thank you very much. That's really helpful. Please continue for fractional pitch winding.

@EnergyConversionAcademy 7 күн бұрын

You are very welcome. Thanks

@AhmedAhmed-uj5ck 9 күн бұрын

مساء الخير كيف اتمكن للوصول اليك

@EnergyConversionAcademy 8 күн бұрын

Hello Ahmed. If you have question, you can contact me through my LinkedIn account listed in this channel. Thanks

@electr1fy0 24 күн бұрын

Extremely underrated! How are you not big already?

@EnergyConversionAcademy 23 күн бұрын

Thank you for your nice words.

@tonypitsacota2513 27 күн бұрын

Thank you for the very nice video.

@EnergyConversionAcademy 27 күн бұрын

You are very welcome.

@user-UnkleAyasi Ай бұрын

Thanks, stay blessed brother 🙏

@EnergyConversionAcademy Ай бұрын

You are very welcome. Thanks

@noelomaolchraoibhe3911 Ай бұрын

Superb content. Thank you! I now see how the "RH Grip Rule' can be applied to the coil in two ways: (a) using your first approach based on the coil pattern itself but then also (b) by applying the current-carrying-wire approach (see 10'18") to the individual wire that makes up the coil; both give the same result. This has cleared up some confusion for me. And I assume 'current' implies 'conventional current'.direction i.e., opposite electron flow. This content is a masterclass BTW.

@EnergyConversionAcademy Ай бұрын

You are very welcome. Thanks for your nice comment.

@noelomaolchraoibhe3911 Ай бұрын

@@EnergyConversionAcademy You're very welcome! After having watched the 4th video in the playlist, I realize I was wrong in thinking the direction of current flow in the coil does not matter; the four examples you present in your Lenz's Law introduction show my error. My take-away is that the coil wrap direction AND the conventional current direction both need to be taken into consideration and just relying on the RH grip-rule around the outside of the coil can be misleading. Please don't feel the need to reply: this is really for my own future reference (and the algorithm!).

@nourhanmohsen3625 2 ай бұрын

Thanks for this well organized explanation. May you publish the notes you wrote in the videos, as a pdf file?

@EnergyConversionAcademy 2 ай бұрын

Thank you for your nice comment.

@ricardo.m.b 2 ай бұрын

excelente explicación. Extrañaba sus videos, son muy instructivos,.

@EnergyConversionAcademy 2 ай бұрын

Thank you for your encouraging words

@donberg01 2 ай бұрын

I like the graphics, it makes the working principles clearer.

@raphaelsales5767 2 ай бұрын

At 7:19 in my head the current is entering but also leaving the closed area therefore they cancel some terms. If anyone could help me understand why this is wrong I would be grateful.

@ЄгорБорса 3 ай бұрын

What is going on with dW_f when you just increase the resistance of the wire? What is the difference between increasing on the resistance of the wire and doing mechanical movement?

@ЄгорБорса 3 ай бұрын

And if I is increasing in the section A, the copper loses should also increased.

@ЄгорБорса 3 ай бұрын

Shouldn`t current in section D be higher than V / R ? I mean we need energy to maintain the position of movable part after the movement was completed. Like if you would pull the movable part by hands exept the energy you need to pull the part, you should also need the engry to maintain the position of the part. Because you would need to conteract the force of the spring

@ЄгорБорса 3 ай бұрын

Why current in section A is increasing significanly in compare to the current in section D? The maximum value of current should be determined by the applied voltage and active resistance (V / R). Howewer in section A current is much higher than V / R.

@Captaincodeine 3 ай бұрын

25:58 I think this point is a little confusing. For a given load the current will generate a constant rotating magnetic field but if you change load the current will change causing more or less magnetic force

@EnergyConversionAcademy 3 ай бұрын

Thank you for your question. Sorry for the confusion. In the lecture, I meant even when the current is changing sinusoidally with the time, the rotating magnetic field will have constant magnitude. For example, if phase a current is equal to "ia=Im sin wt" , the magnitude of phase a current is constant and equal to Im. However, we can say phase a current is changing sinusoidally with time. This is what I meant in the lecture. At 22:34 I explained how the standing magnetic field is produced when the current is changing sinusoidally with time. Hope this clear your confusion. Please let me know if you need more explanation. Thanks

@Captaincodeine 3 ай бұрын

@@EnergyConversionAcademy I actually have a different question. If I increase reactive load on a synchronous generator operating alone, I assert that nothing should happen to the counter torque induced in the machine. Yet the stator field should become stronger with more current, and thus the net field should become weaker. According to the induced torque equation the induced torque should change yet we are not pulling more power out of the machine

@EnergyConversionAcademy 3 ай бұрын

The best way to understand the load change is to model the standalone generator as an A.C voltage source connected to synchronous reactance and the load impedance in series. The load can be modeled as a resistance in series with inductance. The load impedance is Z with angle of Theta. Theta represent the power factor. We can assume that the back EMF of the generator or the A.C voltage source is constant as the excitation is constant. Now As the load increase (impedance Z is reduced), the drawn current is increase and the voltage terminal will drop ( drop across the synchronous reactance) and the load angle (delta between the voltage source and the terminal voltage) will increase. This will cause increase of the active power. Basically, if we change both the resistance and the inductance the active power will increase. I highly recommend to simulate the circuit using MATLAB simulink and perform different load change. That means try to change only change the resistance, or try to change only the load inductance or change both to keep the power factor same. With all these changes observe the power which also give you idea about the torque as the speed assumed constant. Hope the explanation is clear. Please simulate this scenario and let me know if the above make sense or not. Thanks again for your question.

@sisir22 3 ай бұрын

very structured elaboration of magnetic ckt.

@EnergyConversionAcademy 3 ай бұрын

Thank you for your feedback.

@samuelenjebo4054 4 ай бұрын

Very interesting video, thank you sir

@EnergyConversionAcademy 4 ай бұрын

You are very welcome

@johnsonblakra9214 5 ай бұрын

Sir are there more lectures to come

@EnergyConversionAcademy 5 ай бұрын

Yes. I am working on one lecture and hopefully I can finish within a month.

@Kaq-k5d 5 ай бұрын

Thank youu so much it's really good to understand! I have a question, if there is an AC source instead of a voltage source, how do I get a phi function? Should I get the voltage with v= Ldi/dt and apply Faraday's law?

@EnergyConversionAcademy 5 ай бұрын

Thank you for your question. I will assume that the question is how to calculate phi if we have current signal waveform. Answer: If we assume the magnetic circuit is linear magnetic circuit, the following relationship can be used: Bsi = L x i where Bsi = N x phi . Therefore, the magnetic field phi can be calculated as follows: phi = (L/N) x i ( Note: the magnetic field and the current are in phase if the current sinusoidal waveform) Hope this answer your question

@Kaq-k5d 5 ай бұрын

@@EnergyConversionAcademy oh i got it.., thank you for answering my question!!

@JACARTISTEOFFICIEL 5 ай бұрын

Thank you sir for everything. Your courses helped me a lot. Clear and great courses. But I need your help please sir. I'm trying to do modeling of induction motor based on magnetic equivalent circuit using reluctance network but I have problem with air gap because in dynamic regim, the reluctance of air gap is not constant. And I'm stack. Please is it possible to enter in contact with you so that you can help me ? I looked for you on linkedin and found you over there but it wasn't possible to leave a message for you there. Please help me. Thanks in advance.

@EnergyConversionAcademy 5 ай бұрын

I sent you a connect invitation through Linkedin. Hope this will help to communicate

@JACARTISTEOFFICIEL 5 ай бұрын

I Hope you will see my comment ! Your courses are absolutely wonderful. Thanks a lot

@EnergyConversionAcademy 5 ай бұрын

Thank you for your nice feedback