Truly awesome 'quick summary' which helped me tremendously with my understanding of PFC as indeed all your lectures are just great. I love the way you deliver the concepts in your own style for ease of understanding, so much so I believe I have watched them all and sometimes more than once! Thank you very much for sharing your knowledge and please do keep them coming! PS I calculated an additional 21.06uF for a total of 49.46uF for the capacitive requirement - I hope that's right!
Even after almost four years of the publication, this video helped me a lot today, so I could explain this concept to a dear friend of mine, thank you so much because knowledge is a gift
Thank you very much Dr Linares. I really understood much from this lecture and enjoyed the way you explained, it is amazing and I am sure there is a lot to learn from this tutorial time. by the way I found the answer of the last questions capacitor 49.46mF
I did the homework of the minute 27:50 of this great video in a step by step basis. Problem. Data Load S = 400KVA = 400 x10^3 VA Power factor(old)= 0.4 inductive Frequency=60 Hz Calculate Q delivered and C necessary to connect in parallel to the load in order to improve the power factor to 0.85 capacitive. P=ISI cos𝜃 = P=ISI pf(old)=400000(0.4)=1600W 𝜃old= +arcCos pf(old)=+arcCos(0.4)=+66.4218° The angle is positive because the old power factor is inductive. Using trigonometry I got Tan 𝜃old=Qold/P I solved for Qold Qold=P Tan 𝜃old=(160000) Tan 66.4218° Qold=366605.68VAR This is the reactive power without the capacitor. I calculated the new angle of the complex power S 𝜃new= -arcCos pf(new)=-arcCos(0.85)=-31.788° The angle is negative because the new power factor is capacitive. Using trigonometry I got Tan 𝜃new=Qnew/P I solved for Qnew Qnew=P Tan 𝜃new=(160000) Tan (-31.788)° Qnew=-99157.816 VAR This is the new reactive power after connecting the capacitors (load+capacitor) The reactive power that should be delivered to the load is: ΔQ=Qnew-Qold ΔQ=-99157.816-366605.68 ΔQ=-465763.49 VAR The negative quantity indicated that the power delivered should come from a capacitor. Calculation of the capacitance. I used the following formula: C=IΔQI/(2πf (Vs^2)) C=465763.49/(2π(60)(5000^2)) C=4.9419x10^-5 F C=49.419 μF My name is Carlos Vicente Dominguez. I recently graduated as a specialist in electric power systems from Central University of Venezuela in Caracas. Best regards from Venezuela.
Dear Ingeniero Dominguez, You honor me with the detail and care of your post. Especially coming from a graduated engineer from "La Casa que vence las Sombras". Most sincerely, L.L.
Thanks very much Dr. Linares, your fantastic ping pong analogy is wonderful, so many mathematical/power concepts explained in one efficient analogy. Thankyou
14:36 Through some months of thinking about this stuff just casually after taking intro to circuits (which I am happy I don't need to go further, because I am a Civil Engineering Undergrad Student) 14:38 makes perfect sense. No wonder i kept seeing 1/2 and in your older videos it's just 1. I remember I would go to the gym this past Summer and keeping trying to processes all the phasor relationships involving power while doing cardio... and always could not rationalize that difference in coefficient.
Thank you very much for those videos. Absolutely great. Is it in any case possible to do a separate video on reactive power and it‘s good use in transmission network in order to control and regulate voltage? It is such difficult to grasp the concept...would be very grateful. Thanks in advance!
Thanks for the feedback. I think that I made that video on the true nature of reactive power a long time ago. It is somewhere in this channel, but I don't remember where. If I come across it, I'll let you know.
Ahhh... Thanks for using the 'scope! I kinda got it before...now i really get it. Hmm... Is a switching type power supply considered resistive, inductive, capacitive, a combination or...? Just wondered, as i imagine linear transformers are inductive with heavy losses.. But switching ones are inductors switched by resistive semiconductors and also have tank capacitors...
+MyThundermuffin I'll let other viewers either confirm or challenge your result. Kudos for trying That is the best feedback I get for the usefulness of these video lessons. Cheers.
omega, the angular frequency in radians per second, is twice pi times the frequency of the source. In Cananda, the frequency is f = 60, so w = 2 pi 60 is approximately 377 rad/s.
377 rad/s? It is the frequency of the electric system of the country where the problem is solved. Here in Canada, f = 60 Hz, so w = 2 pi 60 approx 377 rad/s, in Europe, f = 50 Hz, and w = 2 pi 50 approx 314 rad/s.
Oh, it is because here in Canada, the electric power grid operates, same as in the U.S.A., at 60 Hz. Just replace 60 for 50 in the computations and you should be OK.
It is the frequency of the power system in your country. I'm in Canada, and the electric power grid operates here at 60 Hz, which makes omega approximately 377 rad/s.
+David Kamore You also confirmed the result that Mr. Bashir Mohamud got two months ago. At this point, three of you have got that answer, good work to the three of you.
No omega, do you mean Qc, the reactive power supplied by the capacitor? Yes, I'am doing the computations the way an engineer would, using absolute values for Q's and X's. In reality, the reactive power "absorbed" by the capacitor, Qc, is negative, so Qc = V^2/Xc gives us a negative reactance Xc, which is just fine because the reactance of a capacitor is negative Xc = -1/(wC) (where w is the frequency in rad/s, omega), and C turns out to be positive (as is omega, the frequency as well.)
+jessie hayden roculas From the power utility, it depends on the country, in Canada it's 60Hz, so two times pi times 60 is approximately 377 radians per second. In Europe it's 50Hz, etc. Wikipedia has a nice world map of electric generating frequencies.
In general, what I call Qold is the reactive power consumed by the load before the insertion of the capacitive reactor. It can be computed in several differnt ways depending on the data available. If we know the KVAs of the load and its power factor, pf, the Qold = KVA x sin(acos(pf)). Or, in proper notation, Q = S x sin(acos(pf)). Observe that Qold will be positive for inductive loads and negative for capacitive loads. I hope that this helps a little.
The frequency in electric power systems in Canada is 60 HZ, so the angular frequency, omega, is 2 x pi x f = 377 rad/s. It is the frequency of our electric power grid. (Same as in the USA).
GREAT VEDIO ,SIR I AM FROM INDIA ,ITS CLEAR MY DOUBT, SIR I HAVE A DOUBT IF S=P+IQ CAN WE USE MAGNITUDE OF S=SQUARE ROOT OF SUM OF SQUARE OF P AND Q. WE I USE THIS FORMULA I FIND MY ANSWER DIFFERENT AS BY USING Vrms*Irms plz help me
Yes, for a single phase load, the apparent power is the absolute value of the complex power that can be computed as you wrote above, using Pythagoras, it can also be determined by a simple real numbers multiplication of the RMS values of the voltage and the current. Both computations do concur, of course ... if done correctly.
In that case, the power triangle is upside down, and we need to add an inductive reactor to the load to bring closer to one the total power factor. The reactance of an inductor is wL (omega "ELL"), and it "absorbs" reactive power at a rate of Q = V^2/(wL). I hope that should give you an idea.
Q=Q1+Q2 where Q1=366.5kVAr and Q2=99.2kVAr. So Q=465.7kVAr. From this, we can get the reactance (Xc)=53.7ohms, then the C=49.4microFarad Did I pass? :D
Top Hat is a software tool to improve in-class student engagement. At the beginning of the term, each student acquires a Top Hat account which is his/her exclusive own. The teacher has two options: to prepare questions before the lecture and to upload them to the Top Hat site for his/her course, or to create a question "on the fly" during the lecture. In either case, the teacher logs in to the Top Hat site of the course (that has been set by the TH company on request) and issues the specific question. The question appears on the smartphones/tablets/laptops of the students in the class. They solve the problem and answer the question to their smartphones, etc. Top Hat keeps track of the answers and grades them. At the end of the term, TH presents the teacher with a spreadsheet with all the grades of each student in the class. Last July, I gave a workshop to professors of UoZ, HIT, NUST and CUT on teaching technology with technology and Top Hat was one of the topics.
The workshop was in Africa, in Zimbabwee, in Harare, sponsored by the North American IEEE (Institute of Electrical and Electronics Engineering) who invited me, and covered all travel and expenses of my stay in that beautiful land. Very good memories!
Improve? Only if the answers of the student are correct. The actual weight of the Top Hat average is not the grade of the course. It merely represents "in-class" participation and is usually less than 10% of the total course grade.
The other components are assignments (which are individual per student and fully graded), midterms (up to six), the laboratory, and the final exam. In short, the Top Hat grade is only a thin slice of the grading "cake". It does work as an incentive for engagement.
