it's exam time again, RMS voltage of chopped sine and power factor - Page 1
Forum section: electronics / beginners

author: Simon
author_rank: Unknown
timestamp: October 20, 2021, 06:47:00 pm
content: So the question is:A single phase full-wave AC voltage has a resistive load of R = 10Ω and the input voltage Vs = 100 V (rms), 50 Hz. The delay angle of thyristors T1 and T2 are equal and π/3. Determine:i. RMS output voltage V0ii. Input power factorMy answers (now submitted) were 96.5V and 0.93. Working out attached.

author: The Electrician
author_rank: Frequent Contributor
timestamp: October 21, 2021, 01:39:25 am
content: Quote from: Simon on October 20, 2021, 06:47:00 pmYou haven't shown the circuit or waveforms, so I can only guess.If the grid waveform is red, the blue waveform is what we get after the thyristors do their thing, with a delay angle of Pi/3. I have shown the waveform frequency with omega = 1 rather than 50 Hz, but that doesn't matter in this case.Is this what your problem is using for the waveform after the thyristors?

author: Andy Watson
author_rank: Super Contributor
timestamp: October 21, 2021, 02:17:26 am
content: I hope you answered with more detail and explanation than you have posted here !I have travelled via the ONC, HNC and HND route (and further), here is something that I wish had been explained to me right at the start [actually, it probably was explained - I just didn't perceive the message] You are not being tested on your ability to remember equations; the exams are an assessment of your understanding and ability to apply your acquired knowledge.What this means is: Writing down the correct numerical answer gets you possibly one mark, two marks if the examiner is feeling generous or looking for straws to grasp to get you up to the pass-mark.How many marks were allocated to this question?How many of those marks did you accumulate?I see The Electrician has beaten me to an answer. I do not know the exam rubric, however, there are probably more marks to be awarded to the Electrician for demonstrating an understanding of the question than the verbaitim regurgitation of an equation, ... an equation which is wrong, ... consider the outcome of setting \$\alpha\$ to zero (it should resolve to the standard sinewave R.M.S. - but it doesn't!)"I have travelled via the ONC, HNC and HND route ..." and yes, it is 'kin frustrating!Edit. I think the equation should have been \$ V_s \sqrt{\frac{1}{\pi} \left( \pi - \alpha + \frac{\sin 2 \alpha}{2} \right)}\$ which would return a Vrms of 89.7V.

author: The Electrician
author_rank: Frequent Contributor
timestamp: October 21, 2021, 04:02:56 am
content: Andy, you're right about the formula. I did the same thing you did, setting a to zero and saw that the result was wrong. I used a math software and calculated the formula from the fundamental expression for RMS:

author: Simon
author_rank: Unknown
timestamp: October 21, 2021, 07:15:11 am
content: Quote from: Andy Watson on October 21, 2021, 02:17:26 amOh Andy, if only, this university basically does multiple tick box answers except they don't show you specific answers to tick off. It's awful. So according to the feedback my calculation of the voltage is correct, but the calculation of the power factor is wrong. As you can see from my answer I literally copied out the equation as given, this would not be the first equation that this tutor has got wrong. I have been told "not to worry, the next assignment is easy and if you do well on that it will average out and you will still get a distinction" Yep that is university education these days. I know this in "only" distance learning but I should be getting the same quality of materials as a full time student. Granted I am proof reading this stuff but I would have though that is for english and spelling not technical detail. But then I have had to make numerous notes on the materials often suggesting alternative rewrites of entire paragraphs of gibberish.If you want to see more of the absurdity of this university see here:https://www.eevblog.com/forum/beginners/time-for-a-laugh-at-the-exam-question/new/?topicseen#newIt looks like I am going to have to go to war with this tutor, most of the other modules are like this, the style is distinctive so I know it's the same person and they are the ones that have always marked the worse written modules. I have had no answer to my queries about my errors as the feedback is brief and inadequate so it will be a formal complaint soon.

author: Simon
author_rank: Unknown
timestamp: October 21, 2021, 07:16:57 am
content: Quote from: Andy Watson on October 21, 2021, 02:17:26 amSo how is that different from what I have. The square brackets I took to be just to make sure everything was under the root.

author: The Electrician
author_rank: Frequent Contributor
timestamp: October 21, 2021, 07:32:50 am
content: Quote from: Simon on October 21, 2021, 07:16:57 amYou have sin 2a/a, but it should be sin 2a/2

author: Simon
author_rank: Unknown
timestamp: October 21, 2021, 12:23:56 pm
content: Ah, yes that is the formula from the module. There were square brackets on the formula in the book because it was raised to the one half power so I put it in the square root instead. But that is the formula!So is the power factor correct?

author: The Electrician
author_rank: Frequent Contributor
timestamp: October 21, 2021, 03:31:10 pm
content: Quote from: Simon on October 21, 2021, 12:23:56 pmCalculating the power factor is complicated; see:https://en.wikipedia.org/wiki/Power_factorYou will have to deal with the blue waveform in reply #1. You will need to calculate the distortion power factor as discussed in the wikipedia article, and combine that with the displacement power factor (also discussed in that article).

author: Simon
author_rank: Unknown
timestamp: October 21, 2021, 05:33:38 pm
content: I don't think they expected that much depth. They made it resistive for a reason. At the most basic level what is the calculation?

author: TimFox
author_rank: Super Contributor
timestamp: October 21, 2021, 05:44:44 pm
content: At the University of Chicago Physics Department, there is a very famous "Candidacy Examination" required of students before they can pursue a PhD program. There is at least one published collection of problems from the exams that is a popular study aid for physics students:https://press.uchicago.edu/ucp/books/book/chicago/U/bo3624023.htmlAfter passing the exam, I was chatting with a professor who had been twice on the committee that poses the questions for the exam. He recounted that the first time, he naïvely posed his question in the form: "Prove thatA=B". Each question is graded by the professor who had submitted the question, so he got back dozens of exams where the answer always ended: "Therefore,A=B, Q.E.D." He then had to read all of the lines above the final answer to see if the student had correctly solved the problem.Learning from his experience, when on the committee for the second time, he asked "Find an expression forA."

author: Simon
author_rank: Unknown
timestamp: October 21, 2021, 06:12:14 pm
content: I am afraid there is nothing clever about this tutor. I am supposed to be proof reading their new module, I am actually rewriting a fair bit of it as what they have put down as their version of hard facts ranges from the illiterate to the factually incorrect. Such gems as "a motor controller produces torque for the motor"......

author: bdunham7
author_rank: Super Contributor
timestamp: October 21, 2021, 07:05:09 pm
content: Quote from: The Electrician on October 21, 2021, 03:31:10 pmFirst of all, I'm confused by the discussion of power factor when there is a resistive load. The very reason for calculating an RMS voltage for a non-sinusoid is to determine the power into a resistive load compared to DC or a perfect sinusoid. So I think the power factor is one here no matter what the input voltage looks like.Now if you come back and say that I'm an idiot and the 'power factor' is clearly meant to apply relative to the 100V sinusoid input, well, OK. But then the voltage in the PF calculation is 100 volts and the current isthe integral ofthe blue waveform divided by R. The RMS voltage of the blue waveform is not directly relevant to the PF computation, but it does allow you to easily calculate the RMS current from which you can then determine the PF. Irms= Vrms/10, then W (power) = I2R and PF = W/(100 * I).As for displacement and distortion components, it really isn't necessary to calculate them separately either from mathematical functions or from actual measurements. In either case if you can define (by mathematical expression) or measure (by sampling) V and I, then you can calculate W (instantaneous power) from those for each point. Then you can integrate W, V2and I2over one period (or a longer time) and PF = Wavg/(Vrms*Irms).Edit: I also get 89.7 VRMS as the answer to the first part. I didn't use a formula, just integrating sin2(x) from 0 to 𝝅, then 𝝅/3 to 𝝅, dividing the former by the latter, briefly pondering why my result didn't match, then taking the square rootof the result.

author: Simon
author_rank: Unknown
timestamp: October 21, 2021, 07:42:54 pm
content: Quote from: bdunham7 on October 21, 2021, 07:05:09 pmremember this tutor falls into the category of "those who can't, teach!" I got the voltage formula wrong but apparently my voltage calculation is correct and my power factor wrong. I think by power factor they mean the amount of power that had to be generated compared to that used.

author: Simon
author_rank: Unknown
timestamp: October 21, 2021, 07:55:24 pm
content: Gems such as:The power supplied by the batteries needs to be converted into torque for the motor according to the motorcharacteristics. The controller allows control of motor drive performance. The controller should beconfigured by an experienced technician as it controls reduced power at times of low SOC or hightemperature. Also some of its features overlap with those of the BMS so care must be taken to avoidconflicting messages. Once set, there is no need for the driver to change anything beyond some simplepreferences around the strength of the regenerative braking or the charging times.The controller ‘s features and outputs include

author: ejeffrey
author_rank: Super Contributor
timestamp: October 21, 2021, 09:34:35 pm
content: Quote from: bdunham7 on October 21, 2021, 07:05:09 pmThe question as posted asks for "input power factor" which would presumably be at the input to the TRIAC and take into account the switching waveform.

author: The Electrician
author_rank: Frequent Contributor
timestamp: October 21, 2021, 10:12:41 pm
content: Quote from: Simon on October 21, 2021, 05:33:38 pmI didn't notice that the load is a pure resistance. Since PF is just (real power)/(apparent power), you can show your calculation for those two quantities which is easy for a resistive load. Then the PF is the proper ratio of them.

author: The Electrician
author_rank: Frequent Contributor
timestamp: October 21, 2021, 10:55:58 pm
content: Quote from: Simon on October 21, 2021, 07:42:54 pmHow could your voltage calculation give a correct result if you got the result by using a wrong formula? The value you got for the RMS voltage is wrong, and the derived power factor (PF) is wrong.Looking at:https://en.wikipedia.org/wiki/Power_factorThe very first thing it says is: "In electrical engineering, the power factor of an AC power system is defined as the ratio of the real power absorbed by the load to the apparent power flowing in the circuit", which is what I said in the previous post, and as bdunham7 also said.If we define Vg = grid voltage (100 volts for this problem), Vr = RMS voltage applied to the 10Ω resistor, Ir = RMS current. You calculated PF as Vr2/Vg2. You must use the correct value for Vr in whatever formula you use, of course.This is not correct because PF = (real power)/(apparent power) = (Vr*Ir)/(Vg*Ir) = Vr/Vg (for a resistive load)Your result is wrong for two reasons--you used the wrong Vr and you squared the ratio Vr/Vg.

author: bdunham7
author_rank: Super Contributor
timestamp: October 21, 2021, 11:45:08 pm
content: Quote from: Simon on October 21, 2021, 12:23:56 pmIf the current is 8.97A and the input voltage is 100V, then the apparent power is 897 VA but the real power is I2R = 804.6 W. PF = W/VA = 0.897.

author: The Electrician
author_rank: Frequent Contributor
timestamp: October 21, 2021, 11:58:55 pm
content: Quote from: bdunham7 on October 21, 2021, 11:45:08 pmOr even easier, if the grid voltage is 100, and the RMS voltage applied to the resistor is 89.7, the PF = 89.7/100 = .897 as I showed in reply #17.

author: Simon
author_rank: Unknown
timestamp: October 23, 2021, 12:37:09 am
content: Quote from: The Electrician on October 21, 2021, 10:55:58 pmThe tutor said the voltage calculation was right but the power factor was wrong. I asked for an explanation but never got one. I reasoned that it was more likely I got the voltage calculation wrong and so this thread.

author: The Electrician
author_rank: Frequent Contributor
timestamp: October 23, 2021, 05:16:59 am
content: Quote from: Simon on October 23, 2021, 12:37:09 amIn the first post of this thread you gave a formula for Vrms that says it came from topic 3 lesson 5. That formula includes a term "sin 2a/a" which is wrong. If your tutor believes (incorrectly) that the formula is correct, he might (mistakenly) say that your voltage calculation is right even though it is not. Is this incorrect formula (that includes the term "sin 2a/a") really given exactly like that in topic 3 lesson 5, or did you make a typo when you copied it out?Several people, including me, have calculated that formula from the basic definition of RMS, and we all got "sin 2a/2" as the correct term in the formula.You might check with the tutor and see if he believes that the formula with "sin 2a/a" is correct. If he does, you might suggest gently that he check into it.You did get the power factor wrong by squaring the ratio (real power)/(apparent power); it should not be squared.

author: Simon
author_rank: Unknown
timestamp: October 23, 2021, 09:11:49 am
content: Yes I made a typo in the voltage formula. It's the power factor I thought I had right. To calculate the power factor I divided the used power by the generated power, the terms where squared because this is the two power calculations over each other V^2/R. The tutor has made their fair of mistakes in writing the module, one which someone else had to point out to me. I'm not great with the maths on this sort of thing. I triod various ways of calculating that, resolving the complete V^2/R first and then dividing or simplifying out the resistances that are the same and I got the same results. But the fundamental question is, is dividing the used power by the supplied power the correct way to calculate "this" power factor.

author: The Electrician
author_rank: Frequent Contributor
timestamp: October 23, 2021, 10:00:02 am
content: I'll assume that by "supplied power" you mean the (real) power that comes directly from the grid before it passes through the thyristors, and by "used power" you mean the power that heats up the resistor. These two powers are the same; they have to be with a negligible difference due to the small power loss in the thyristors. If the real power delivered to the resistor (used power) were substantially less than the real power coming in from the grid (supplied power), something would have to be getting hot, and thyristors don't get very hot when controlling around 10 amps.However, if by "supplied power" you mean what you get when you multiply the voltage from the grid by the current supplied by the grid, that would not be a "real" power, but rather it is the "apparent" power. If that's what you mean, you should call it "apparent power" because that's what it is in this case.Power engineers don't use the words "supplied power" and "used power". It's better to use the standard words used in the power engineering business. Those words are "real power", "apparent power", "reactive power" and "complex power", and they are explained in detail on the wikipedia page:https://en.wikipedia.org/wiki/AC_power.Ordinarily a person can calculate power by just multiplying voltage and current, but with AC it gets complicated.For your problem, if you multiply the voltage and current coming from the grid (before it passes through the thyristors) you will get the apparent power. If you multiply the voltage and current applied to the resistor, you will get the real power. The power factor is the ratio of those two (in the right order, or course).

author: Simon
author_rank: Unknown
timestamp: October 23, 2021, 01:42:13 pm
content: so the power factor is indeed the real power that goes through the resistor over the apparent power which is what the grid was ready to supply but did not get used. using the voltage I got albeit wrong the power factor is therefor correct. But apparently the voltage is right and the power factor wrong.

author: bdunham7
author_rank: Super Contributor
timestamp: October 23, 2021, 04:25:57 pm
content: Quote from: Simon on October 23, 2021, 01:42:13 pmSeveral of us appear to be trying to tell you the same thing although perhaps I've been too brief. I don't know about this learning module or what the teacher is trying to get across with a problem like this, but frankly learning a 'formula' for calculating the PF of a phase-controlled resistive load is not something I would take the time to do--or pay for. I'm pretty sure the answers are 1) 89.7V and 2) 0.897 and althoughThe Electrician'ssimplification also gets you there, I much prefer to let that 0.897 'fall out' of the full calculation.The problem with trying to reduce PF to concepts that you can intuitively understand, like "what the grid was ready to supply but did not get used", is that they're just not quite right. The most common basic understanding of PF starts with an ideal AC voltage source that can source or sink any amount of current, thethe PF is a characteristic of a load to which this voltage is applied. Saying that reactive (or 'recirculating') current did not get used is not really all that helpful and could be very misleading.PF as a concept--and the math--can be applied to any system including DC, but it was devised as a way of understanding AC power systems, so it isn't typically referred to as 'power factor' outside of that context. The basic idea can be useful in other contexts--like an audio amplifier driving a complex multi-driver speaker system. The math is a lot harder there and I've never seen an explicit PF calculation by audio engineers.Anyway, for a source voltage V and load current I, Vrms= (∫V2dt)1/2and Irms= (∫I2dt)1/2and both are always positive numbers--so apparent power is always positive. Real power Wavgis simply ∫ VI dt and can be a positive or negative number. In the case of purely resistive load, I=V/R, so W simply becomes Vrms2/R or Irms2R. This is where that 0.897 'falls out'.Apologies for my poor math notation skills and any silly errors I may have made.PF = Wavg/(Vrms* Irms).If all three values are provided or obtained by measurement, the calculation is straightforward. If one of the values is missing and cannot be obtained by using some constraint in the problem (like a resistive load) then you know that you can't calculate PF. If the math is hopelessly complex, then at least you still understand conceptually what the PF is even if you can't calculate it. For an example, take a half-wave circuit consisting of a heating element driven by an AC source through an ideal rectifier. Even if you don't have a formula and only have basic algebra skills, you should be able to figure it out.

author: Simon
author_rank: Unknown
timestamp: October 23, 2021, 09:47:31 pm
content: I'm trying to simplify it to the level that the tutor is to make a comparison and figure out who is right about what. Welcome to modern UK university studying! The tutor is frankly incompetent! they seem to have some knowledge somewhere but also seem to be often writing abut topics they are not familiar with, have just read up about and are now regurgitating the information incorrectly whilst producing near illiterate words to fill a page. Honestly a copy writer could do a better job!