Відео

Hi professor, question: At 7:14, should the angle be negative because it is a leading load?

Thank you for you help. Please keep posting more videos.

Thank you very much, indeed!

So basically the questions are wrong, is there anyway PEO is taking responsibilities for them? I believe they stressed the applicants a lot when encountering these mistakes?

Great explanation at the end of the video Prof

In todays exam there were not even 1 question on Electrical machines. No magnetic circuit no dc machine no induction motor no Synchronous Motor and Ge no Single phase trasnformer. Only power sysyem single line diagram per unit system and 1 single phase question. That is soo disappointing. Non of previous exams had similar questions not even in the reference books are there.

This is a great video~ Thank you!

Hi Professor, thanks for the videos! this topic in not in the A6 textbook. Why did they create and give this problem?

Hello professor, why did they give this type of question? It is not in the text book.

Thanks a lot! Tomorrow is A6 Exam from PEO! This is super helpful!

This is really helpful. Thank you so much professor

Thanks a lot for your videos!

Hi Professor, for part b, the power factor increased from 0.8 to 0.96 by opening the steam valve seems quite a lot, or can we calculate the new power angle by using the equation P=Ef x Vt x Sin (power angle) / Xs? For part c, would the Vt be kept constant if Ef is increased? Sometimes I am confused about which unknown is the constant after the changes. Thank you so much!

Could you explain why a negative sign was added for a leading power factor when it was stated that the factory operates at 0.8 leading power factor? (Q- d)

I have an exam on Monday. Thank you professor for updating with latest question

Don't know how much I can say thanks to your channel~ Thank you!

Thank you so much professor for your support and effort. We have one week left for the exam (on Oct 28) so if you can solve couple more questions next days, much appreciated.

Thank you so much professor

Thanks professsor. Really appreciate it. Exam of B7 is on 28 October.

these videos are very helpful

Could we calculate zth by Voltage source SC and current sources OC in this question?

Hello there I like your videos Thank you for uploading in utube I am preparing for 16-Ele-A1 and looking for paper solutions from last couple of years. I would like to learn from you. Are you providing classes online or offline ?

Thank you

Dear Professor, I have a question regarding calculations in part b and c: in part b, we used line-line voltage of 460V at 0 degree angle to calculate current but then in part c, why did we use the line to neutral voltage 460/root 3, also at 0 degrees when calculating Ea? Shouldn't we account for the -30-degree phase shift that occurs when transitioning from line-to-line to line-to-neutral? Since we are using the current 𝐼 calculated in part b for the calculation of 𝐸𝑎 in part c, wouldn’t using the incorrect phase angle for the voltage 𝑉 affect our results? Thank you for your guidance!

Hi Professor Ayman, I've been watching your A6 video playlist and I greatly appreciate the effort you've put in. Much love! <3 I'd like to propose an alternate way to solve step 2 instead of the triangles because i find them complicated. Write out Ea∠? = VT∠0 + Ia * jXs as you did. (note Ia *jXs = -18.11 +j24.12) Then expand Ea into Ea*cos? + jEa*sin? = VT∠0 -18.11 +j24.12. This can be separated into two equations. Real: Ea*cos? = VT∠0 -18.11 and imaginary: Ea*sin? = 24.12 From here we can solve the imaginary side 480*sin? = 24.12 > ? = sin-1 (24.12/480) = 2.88° VT = 497.5V same as yours but I find the separation of real and imaginary to be an easier concept. Would love to hear your thoughts!

in some videos finding core lenght they make mean lenght- gap lenght in your case you did core lenght = mean core lenght which one is correct? dear Ayman El-Hag

very helpful

A 50 hp, 440 V, 60 Hz, three-phase, Y-connected induction motor has the following equivalent circuit parameters: R1=0.1Ω R2′=0.12Ω X1 =0.35Ω X2′=0.4Ω 𝑃rot=2150W ​With no load, the line current is 18 A at 0.089 p.f. lagging. If the motor speed is 1755 rpm, determine: a. the line current; b. the developed torque; c. the output power (may be different than rated output); and, d. the efficiency. dear prof, I am unable tio figure out this question. I tired, but the answers aren't making sense. i have never seen a question with excitation branch elements not given. is the question itself wrong or is there a way to tackle this? Really need your guidance on this one. I shall be grateful to you.

Two 600-kW, 60-Hz, diesel-driven synchronous generators A and B have governor speed regulations of 2.0 and 5.0 percent, respectively. Both machines are in parallel and supplying equal shares of a 1000-kW bus load at 57 Hz. a. Sketch the approximate governor characteristics for both machines on one set ofcoordinate axes, and indicate the operating frequency. Label both curves. b. On the same diagram, approximate a new operating condition that assumes theload on the bus decreases a total of 400kW. c. Determine the new frequency and the new load distribution for the conditions in part (b). This question appeared in May 2016, Sir? To solve such questions, we need to have slope in MW/Hz that can be calculated from speed drop. However, I am having confusion, since no-load frequency is not given explicitly. Should i suppose the rated frequency of 60 Hz as Full load frquency and calculate no-load frequency , or shall i assume it as no-load frequency and calculate the full load frequency? secondly, with these dropping percents and given freuqnecy, the two generators can't share load of 1 MW at 57 hz. This system frequency isn't correct, in my opinion. Can u please guide me on this? Thanks

it can also be said that at lower slip, rotor losses are less (since P,rcl = s*P,ag) and and more airgap power transforms into converted power, hence efficinecy increases. Similary, if rotor is blocked, slip becomes 1, and all airgap power is consumed in the rotor (P,conv = P,ag- P,rcl). and output power becomes zero.

Sir, what's the name of the software?

Thank you professor, Can you please solve one question about Salient generator. Can you please explain how to deal with Xd, Xd’ and Xd” and what is the difference between steady state condition , subtransient and transient ?

Dr, are there synchronous Generator questions in the exam?. I saw generator questions being in exams from 2013, 2012 and 2016

thank you!

Thank you so much for elaborating on the steps to solve this. Could you share how to solve question no. 3 of Dec 2017? The question has the same structure, but instead of asking for the current for the inductor, the question asks for the voltage for a capacitor. I am wondering how the switch affects the capacitor circuit when it is in a steady state. Thanks a lot.

Hi professor there is a video in this playlist on unsymmeterical faults that is hidden it it possible to unhide it.

I believe the open circuit test should be performed on the low voltage side to reduce the maximum voltage to be measured. It is typically done on the low voltage side, but the data in this question might be incorrect. The open circuit test voltage should be 230V, not 2300V.

Hi Professor, would you mind helping us to solve the Dec 2020 Question 5? I am a bit confused with the use of sub-transient reactance. Or any Hints for us? Thank you so much~

Thank you so much! 🙏🙏🙏🙏🙏

Thank yo Dr for your helpful videos.

A 600V, 60Hz, 8-pole, Y -connected, three-phase synchronous motor has a synchronous reactance of 9 ohm per phase and negligible armature resistance. When it runs at no load, the motor draws an armature current of 8 A at a 0.08 pf leading. For the no-load condition, calculate: a. the induced voltage; b. the power angle, and, c. the rotational losses of the motor. This motor is used as a synchronous condenser to improve the overall power factor of a small factory, whose other loads include: a three-phase induction motor delivering an output power of 15 kW at an efficiency of 90% and a power factor of 0.85 lagging. a combined heating and lighting load at a power factor of 0.95 lagging. Calculate: d. the synchronous motor active and reactive power required to operate the plant at unity power factor; e. the synchronous motor armature current for the conditions in (d); and, f. the excitation emf required for the conditions in (d).

Thank you so much!

in part e, how do we know that it wants us to compare torque angle and Ea with that of what is calculated in d (iii) , and not Ea from part (c)?

Dear Sir, In part d (i), it should be cosine inverse of 0.8, not tan inverse.

Hi professor! At 5:06, you put IA=55 in the equivalent circuit diagram, or should this be IA= 55/sq rt 3 as a Delta connection? Thank you.

Thank you Prof!

Thank you for this video!

Thank you so much for your help!

Very down to earth explanation

Dear Professor, Thank you once again. Your explanation is so clear, which helps us a lot. Regards, Jerry.