Per Unit Analysis - how does it work? (with examples) || Basics of Power Systems Analysis
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- Опубліковано 3 сер 2024
- Per-Unit analysis is still an essential tool for power systems engineers. This video looks at what per unit analysis is and how it can be used to solve power systems.
0:00 Introduction
2:37 High level intuitive overview
5:11 Step by step description of the method with simple example
13:54 Review of simple example - what can we conclude?
16:27 Dealing with complex impedances and transformers
19:19 Example single phase system
21:20 Dealing with transformers mismatched to our system bases
22:16 Three phase systems with an example - Наука та технологія
we lack such level of clearity in our educational system. Thanks for clearing the most fundamental concept!
This should have more likes. You do not rush through stuff and assume that the listener is with you. I like that you take the time and give a small example after each statement. It might be a slower pace for a lot of us impatient engineers, but if we actually listen, we can see that all nuances and bases are covered very well.
The best vid on youtube explaining per unit systems, do not expand time elsewhere. gonna share it
Very helpful! Very precisely explained! Thanks alot for your time and efforts. Much needed video!
You explained the whole per unit concept in a much more clear and understandable way than my university professor. Great Thanks!
Well explained thank you very much now I have a full intuitive understanding of per unit systems
Thank you, this is the best video on per unit Ive come across. Keep up the great work
Great explanation of this subject. Watched it a few times and it helped me a lot. Thank you 👍
U have saved my semester. What my teacher couldnt explain in an hour was better expalined here in 10min
Thank you
Excellent concept
superb explanation, Thank you.
you are amazing i wish i can watch all the adds here again and again it was really helpful thank you again
Thank you so much for the clear explanation.
Clear as crystal clear, thanks,
Thanks a lot, this was really helpful!
Sir your video is Gold. Thank you ❤
wonderful video about pu in power systems
i have many vids regarding this topic and let me tell you that your vid is the most beneficial vid
thanks a lot
This was very helpful, thank you
the bane of the current education system is the tendency to make things complicated. I wish you could be my uni prof
Thanks!
Very helpful, thanks!!
wonderful. thank you so much
Man... thank you so much🙏
Very good. Thanks 🙏🙏🙏
Great video!
Great video but I thought you changed each zone to the new base... ah you covered it... but I also thought you have to be in same voltage base and consider turns ratio
great job !
Can I comment and check with the author/publisher (ie. at video 16:22) where the Ipu = 2.39+j0.275p.u. Should it be a negative instead as it should be Ipu = 2.39 - j0.275. I think the source voltage should in polar form is 1pu phase angle of 0. When we divide this source voltage by the total impedance, the current will have a negative j term. Please advise and comment. many thanks
Oops, yes you are right, thanks for spotting.
Thanks for the great video ! In the last video, I am just wondering if the first transformer ratio is 0.415/11 V instead of 0.415/11 kV ?
Legend ❤
My comment is a bit of a side-track, and doesn't involve the 3-phase example at the end of the video. But following along and practicing the technique presented, I'm wondering if some of the examples, if one saw them in the real world, would represent overloads (for the transformers, at least).
At 6:39, the presenter explains that choice of S-base (Sb), the apparent power, can be anything, but good practice says you should choose a value that matches (as closely as possible) the apparent power base of your equipment. With that then in mind…
In the example at 22:48, the math of course all works, but in that result, is it true that the apparent power is 23 MVA, noting the chosen Sb was only 10? Doesn't that mean we've blown out both of the transformers, one of which was rated 8 MVA and the other 12 MVA? If the example represented a real system, wouldn't you need to ensure the apparent power never exceeded 8 MVA, the (I'm assuming) apparent power handling rating of the lower of the two transformers?
I was led to think about this from the prior example showing the nameplate ratings of a transformer at 17:20, which showed it was rated 50 kVA and 3.39%, and the statement following that the percent impedance only makes sense relative to the transformer's base values. Those base values influence your choice of system base (Sb).
In the example at 21:15, both transformers were given as 3% and 5% on 10 MVA, and 10 MVA was chosen as the system base. Working the example, the Ipu came out to 2.31∠-17.1° p.u. (assuming I did it right). This is also 23+ MVA. But we chose 10 MVA as system base, because…somewhere this was a rated limit (came from one or both of the transformers, presumably). It seems like only the first example, at 13:13, had p.u. that were sensible for the apparent power capacity of the system (0.9901 p.u.). The video says that power system engineers recognize that values up to 1 p.u. are immediately recognized as a well behaved system. But in the examples presented in the middle of this video, we get p.u. currents around 2.3, suggesting (if the base values were chosen as the presenter recommends) our analysis has /also/ shown us that these example systems are fubar. Reaction?
Please upload more videos related to power system and electrical machine
Sir, for the mismatched system base, how come the Vbase of zone 3 is still 20kV? I'm kinda confused...
You have to follow the ratio of the transformers to get your voltage base in each zone. The 220kV/22kV transformer ratio is still 10:1, so 200 kV in the middle zone must still give you 20 kV in the load zone. We just need to be careful to adjust the impedance of the transformer so it matches the system base voltages because the 220/22 kV is on a different base.
oh i see... thank you sooo much@@VisualElectric_
1000V, 1 ohm impedance should give a current of 1000 A. But your calculations from pu is giving 990.1 A. Why the difference?
Can I have your email please?