Motors and W versus VA - Page 1
Forum section: electronics / beginners

author: Simon
author_rank: Unknown
timestamp: January 15, 2019, 01:33:21 pm
content: So assignment time again and oh boy what a mess:7. A factory has an average demand of 520 000 units per week. Themaximum demand is 25 MVA at 0.8 power factor and the minimumpower factor of 0.6 occurs when the demand is 11 MVA.The factory is charged at 2.5 pence per unit with a surcharge of 0.2 penceper unit for each 500 kW by which the maximum demand exceeds18 MW and a further surcharge of 3% (of charge, plus surcharge) forevery increment of 0.05 by which the minimum power factor fallsbelow 0.8.There is a large drive which operates continuously and is powered by aninduction motor with draws 2 MW at a power factor of 0.8 lagging.This motor is replaced by a synchronous motor which draws the samepower but runs at a power factor of 0.8 leading.Assuming the maximum demand penalty does not change(i) Show that the maximum demand power is 20 MW.(ii) Show that the total weekly charge for the factory is £19 219.20.(iii) Calculate the new power factor and reactive penalty charge when demand is 11 MVA.(iv) Show that the new total weekly cost is £17 644.50.(v) If the synchronous motor costs £250 000, calculate the time requiredto recover the cost of the motor.(i), I don't get it, if the consumption is 25MVA @ 0.8 PF that IS 20MW (real power) before you try to do any PF correction(ii) but I don't know for how long the peak power is drawn and if this is a 5 day 8 hour shift or if a 24/7 setup, I make it £13'000 on the average consumption so that is quite a lot of exceeding limits(iii) So i have do do the complex numbers(iv) same as (ii)(v) I still don't know the periods for which it has any benefit so can't compute.

author: nsrmagazin
author_rank: Regular Contributor
timestamp: January 15, 2019, 02:10:52 pm
content: Watts == real power, the losses are substracted.VA == effective power, the losses are not substracted.

author: mvs
author_rank: Frequent Contributor
timestamp: January 15, 2019, 03:10:58 pm
content: Quote from: Simon on January 15, 2019, 01:33:21 pmYes, before any correction 20MW at 0.8PF is 25MVA. If you correct PF to say 0.9, then you will get only 22.22MVA apparent power at the same real power.Quote from: nsrmagazin on January 15, 2019, 02:10:52 pmVA is Watt, just used to indicate apparent power. There are no direct losses in both, apparent power just has beside real also imaginary part of power in it.

author: rstofer
author_rank: Super Contributor
timestamp: January 15, 2019, 03:49:30 pm
content: I really dislike "units per week". What is a "unit".Since we know there is a continuous load of 2 MW we can calculate that this accounts for 336 MWhr/week so it might be reasonable to assume a "unit" is aMWhrkWhr.That gives us a baseline that might be useful(i) 25 MVA * 0.8 pf => 20 MW as you said.(ii) You pay forMWhrkWhrs (known value) plus penalties for demand plus penalties for power factor.So, the first part is 520,000 * $0.025 (I don't know how to get a pound sign) or $13,000 just for energyNow add in the demand of $0.002 * (20 MW - 18 MW) *2 (for 500 kW) * 520,000 or $$4,160Subtotal so far is $17,160.(0.8 - 0.6) -> 0.2 pf penalty / 0.05 -> 4 units of pf penalty. 4 * 0.03 * 17,160 => $2,059.20Total $19,219.20 <== The right answer! Amazing!That was exhausting - I haven't even had breakfast!More to follow if someone doesn't jump in first. The dogs want out and I need coffee!

author: Simon
author_rank: Unknown
timestamp: January 15, 2019, 04:19:25 pm
content: Quote from: rstofer on January 15, 2019, 03:49:30 pmYou missed something, the power factor is 0.6 when 11MW are used. there are two mutually exclusive conditions than can occur in unspecified quantities, high demand with a per unit surcharge but good power factor OR low consumption not incurring excess usage cost but a power factor penalty. you cannot have both at once. BAM I saw that in the first reading of the stupid question. As always these clowns never fail. Yes you have hit on a correct combination but it was not obtained by the rules stated.

author: rstofer
author_rank: Super Contributor
timestamp: January 15, 2019, 04:21:38 pm
content: Quote from: mvs on January 15, 2019, 03:10:58 pmThere are losses in that the utility has to provide MVA and MVAhr and can only bill for MW and MWhr. MVA is always >= MW.In delivering MVAhr, the utility has to provide the excess Amps and this causes additional I2R losses in their lines for which they can't charge. Those I2R losses are quite real and expensive. So they add a penalty for truly poor power factor (less than 0.to recover those losses.

author: Kasper
author_rank: Frequent Contributor
timestamp: January 15, 2019, 04:21:42 pm
content: i)VA * PF = Wii)It says something runs continuously so I would assume the factory runs 24/7 and I would write that assumption in my answer. If the numbers don't add up then revisit the assumption.The average weekly demand should help you figure out the duty cycles of the machines.

author: Simon
author_rank: Unknown
timestamp: January 15, 2019, 04:26:32 pm
content: Average consumption is 9.3MWh, it also states that the minimum PF is at 11MW so the PF is variable. I am sorry but the question is bullshit as it's not a case of having to make assumptions, more like make shit up as there are 3 unknown variables.

author: Kasper
author_rank: Frequent Contributor
timestamp: January 15, 2019, 04:29:14 pm
content: To make matters worse, they don't say what the minimum load is they just say what the min PF is.Min load is atleast the load of the large drive which operates continuously, maybe more.

author: rstofer
author_rank: Super Contributor
timestamp: January 15, 2019, 04:32:28 pm
content: Quote from: Simon on January 15, 2019, 04:19:25 pmAnd that 11 MW 0.6 pf occurs for 2 minutes very early on Thursday morning whereas the peak demand occurs for a few minutes on Tuesday afternoon (heat wave?). The point is, peak values set the limits but don't always account for very much consumption. In fact, that's why there is a demand penalty. The utility has to provide for your peak demand requirements but they only get to bill for the average.I spent MANY years working on utility bills as an electrical type engineer. All I did throughout my career was power, lighting and very big power. I only used FFT once in 30 years! Transistors never came up!I think I followed the rules explicitly. They just provided too many facts to answer this question but they may be needed for the next part. I haven't slugged through that.Hint: You don't want your plant to have a leading power factor. That's why the next question is looking at the situation where the plant demand is 11 MW.I'm not saying your'e wrong but I think I'll stick by my analysis. There are two many calculations to accidentally come up with the right answer. To the penny (or pence, I suppose).

author: Kasper
author_rank: Frequent Contributor
timestamp: January 15, 2019, 04:33:27 pm
content: I too just woke up and without writting anything down, I am guessing you may get it if you can work the maxs and mins into your equations.Otherwise you might need to answer with a graph with a shaded region showing all possible answers.

author: Kasper
author_rank: Frequent Contributor
timestamp: January 15, 2019, 04:37:26 pm
content: rstofer sounds a bit more qualified than I. I took one power systems class years ago.

author: Simon
author_rank: Unknown
timestamp: January 15, 2019, 04:38:12 pm
content: Quote from: rstofer on January 15, 2019, 04:32:28 pmI submit that they have made a mistake themselves and missed their own question in calculating the answer.

author: rstofer
author_rank: Super Contributor
timestamp: January 15, 2019, 04:38:23 pm
content: Quote from: Simon on January 15, 2019, 04:26:32 pmIt's not a BS question! It comes up all the time in plants around the world. It is usually covered in depth in a course called Engineering Analysis. Really! They actually teach this stuff. A bit of engineering and a boatload of finance.Engineers design and build stuff. But they first have to get the money approved and to do that, they have to do a little Engineering Analysis. What does the project do? Why do we care? How much will it cost? How much will we save?How long until we break even?I haven't done the second part but it's going to be fairly straightforward and imaginary numbers aren't going to be necessary. In effect, they're buried in the power factor. There may be a little vector algebra, I haven't gotten to that part.

author: Simon
author_rank: Unknown
timestamp: January 15, 2019, 04:39:53 pm
content: Quote from: Kasper on January 15, 2019, 04:37:26 pmYou don't need engineering, just go through the logic and you see that it can't work unless you make massive and totally unfounded assumptions. The answer is calculated as demonstrated by applying two conditions that do not occur together.

author: rstofer
author_rank: Super Contributor
timestamp: January 15, 2019, 05:10:02 pm
content: Quote from: Simon on January 15, 2019, 04:39:53 pmLet's try to go through it piece by piece. AFAIK, I didn't make any assumptions, I took what was given.1) I guess we can agree that "units" is reallyMWhrskWhrs but it doesn't matter for this problem because everything is measured in "units".2) "The factory is charged $0.025 per unit" for energy and uses 520,000 units ->$13,000(I really dislike the wording of "A factory has an average demand of 520,000 units" when "units" are energy and demand is really power. It should say "A factory has a consumption of 520,000MWhrskWhrs" - but that is just semantics. Poor wording but ok...3) There is a demand penalty above 18 MW and the factory has a maximum demand of 20 MW which we get from the 25 MVA at 0.8 pf. Remember, these are the peak numbers at the end of the week, not average values. Demand has been up and down, pf has been all over the map but, at the end of the week, the plant looks like 25 MVA at 0.8 or 20 MW. The excess demand is 20 MW - 18 MW or 2 MW which is 4 blocks of 500 kW demand.Each block is charged at $0.002 so the penalty is $0.008 * 520,000 ->$4160.Thesubtotalis $13,000 + $4,160 =>$17,160Every number, up to this point is straight out of the problem statement, no assumptions of any kind.4) The power factor penalty is added to the subtotal (stated in the problem) and is based on the minimum power factor of 0.6 as it is the minimum below 0.8. So, we have 0.2 units of pf penalty (0.8 - 0.6) which is 4 increments of 0.05. Multiply the 4 increments times 0.03 (penalty percentage) * $17,160 (subtotal so far) =>$2,059.205) Now add up the subtotal and the pf penalty $17,160 + $2,059.20 =>$19,219.20I believe the book has the right answer, I'm pretty confident in my number crunching and this method of calculating power bills is fairly common. You buy the energy, you pay a charge for maximum demand and another charge for power factor below 0.8. Around here, the pf penalty is based on the average, not some minimum or maximum but the problems says minimum, so be it.I don't see where I made any assumptions, every number is straight out of the problem statement and it seems to work out. I'll stand by it until somebody knocks me over.

author: rstofer
author_rank: Super Contributor
timestamp: January 15, 2019, 05:30:54 pm
content: I stand corrected, "units" is kWhrs, not MWhrs. It doesn't change the math because the math was always in terms of "units".That 2 MW motor accounts for 336,000 kWh or about 65% of the consumption.

author: Simon
author_rank: Unknown
timestamp: January 15, 2019, 06:12:27 pm
content: The problem with your solution is that you assume a constant use of 20MWh which over a week (24/7) amounts to 3360 MW not 520 MW, you also charge the penalty for poor power factor which only occurs at 11MW on a usage case of 20MW.Yes this is clearly how the answer has been worked out but it is wrong!If the factory runs 24/7 the average usage at 520MWh per week is 3.095 MWh

author: rstofer
author_rank: Super Contributor
timestamp: January 15, 2019, 06:31:42 pm
content: Quote from: Simon on January 15, 2019, 06:12:27 pmThey gave us the 20 MW demand (power) not 20 MWhr (which is energy). That is a maximum instantaneous demand and not any part of an 'average'. At some point during the week, they lit off an electric furnace and increased the demand for a short period. The integral of the demand over the short period is energy and applies to the 520,000 but it doesn't run for the entire week and may not contribute all that much to total consumption. You do not take 20 MW demand and multiply by 7 * 24 (and * 1000) to get kWh.QuoteYes, the AVERAGE demand is about 3 units and 2 of those units comes from that continuous motor. All of the rest comes from various loads that cycle on and off over the week. At some point, the loads come on together to create some maximum demand.I think you are confusing energy and power, consumption (kWhr) and demand (kW). They are not at all interchangeable because the loads are varying over time (except that one motor).You simply can't do much of anything knowing the average demand other than making a baseline on a demand chart to show average. But it will be an average, there will be times when the demand is less than average and times when it is more than average. Demand tends to be low on the graveyard shift and high on the day shift. More people, more lights, more computers, more coffee pots, more everything!I had exactly this situation at the last place I worked. We had a plant maximum demand of around 2000 kW until we fired off a huge heater that added 20 MW. It didn't run for very many hours per month (like maybe 5) but once it fired off in a billing period, we were billed for that maximum demand, even if it ran for just 30 minutes (the demand billing interval). It didn't cost much for kWhrs but it broke the bank of kW. I was able to install a $3M high voltage substation (115kV->12kV) to get a better rate schedule for demand and reduced our annual electric bill by a couple of million $. I don't remember the actual numbers but in the time I was there, I reduced an $8 million annual bill down to a million and change. It's the little things...Think about the Engineering Analysis:Build a substation for $3MSave $2M per yearPayback 1.5 years!It took a couple of paragraphs on one sheet of paper and about 30 minutes to get approval and that included walking back and forth across the campus.

author: Simon
author_rank: Unknown
timestamp: January 15, 2019, 07:23:44 pm
content: I see, some assholes they have at this university I am studying with! The module for this question is about motors, not how power is billed and how on earth am I to know how power companies think when it is totally irrelevant to the subject I have been taught.So you say that despite the fact that they meter the power instantly rather than simply clock up a total use (or how else do they know about the peaks) they bill the whole period at a price that assumes both opposing worse case scenarios simultaneously for the whole consumption?If that is the case then it makes sense and I'd hate to be the customer and the question is rather irrelevant to the subject material. Last I checked i could not find a textbook on the inner workings of power companies and the course only talks about the bare bones of motors.

author: rstofer
author_rank: Super Contributor
timestamp: January 15, 2019, 07:26:54 pm
content: Let's try the last half - I'm a lot less certain of this arithmetic.First, the plant : 11 MVA at 0.6 pf => 6.6 MW and 8.8 MVARThe existing motor: 2.0 MW at 0.8 pf => 2.5 MVA and 1.5 MVARThe new motor, instead of lagging by 1.5 MVAR, leads by 1.5 MVAR for a net gain of 3 MVAR to the plant.The new plant: 6.6 MW and 5.8 MVAR => 0.75 pf, 6.6 MW (no change expected) and 8.8 MVANow we only have 1 increment of power factor penalty ((0.8 - 0.75) / 0.05) instead so 0.03 * $17,160 or $514.80.We save $1544.40 per week.$250,000/$1544.40 = 162 weeks

author: rstofer
author_rank: Super Contributor
timestamp: January 15, 2019, 07:40:54 pm
content: Quote from: Simon on January 15, 2019, 07:23:44 pmIn the real world, the revenue meter accumulates both kWh and kVARh - two separate indicators. At the end of the month these values are used to compute average power factor for the billing month.Independent of all that, the meter also accumulates kWh and kVARh in 30 minute buckets called demand intervals. The billing demand is the bucket with the highest accumulated kWh. Thebillingdemand interval may change across industries. Maybe some company has a really huge demand for just 15 minutes. The utility may decide that a 15 minute demand interval is more appropriate to capture the peak and change to a different rate schedule. One can imagine some benefit in turning on a large short duration load 1/2 way before the end of one interval and running it half way into another interval to cut the apparent demand. The utility won't give you a signal that shows the beginning and end of the interval for that very reason.We always got billed for the monthly power factor, not some instantaneous version like stated in this problem. Nevertheless, the idea is the same. You pay a penalty for poor power factor to help the utility recover from I2R losses.Not knowing anything about your program other than the interesting problems you present, I can't really say where this fits. We dealt with power factor in AC circuits and the final question is directly on point. The motor isn't changing in MW, just MVAR so the plant isn't changing in consumption or demand but there is less of a penalty for poor power factor. Since that motor runs 24/7, it makes sense that it just contributes MVARs which reduces the plant's MVARs and this results in a change in the minimum power factor. Which saves money and a 3 year payback isn't too bad!Let's see, I loan the company $250,000 and they pay me back $80,000 per year forever. Hell yes!I paid the money, I keep the savings - that's the way it works - I wish!

author: rstofer
author_rank: Super Contributor
timestamp: January 15, 2019, 07:54:08 pm
content: I have always been of the opinion that facilities projects should be allowed a longer payback than the conventional 2 years. I am wrong but it has always been my opinion.The reason is, there are more projects than money. The projects with the fastest payback get done first regardless of 'goodness' or other factors. This project takes 3 years - absolutely it should be done!I am also right! Facilities projects last essentially for the life of the plant. Other projects may pay back faster but they crash and burn as technology changes. I worked in high tech, everything changed. But facilities changed slowest and the tech part changed daily.Energy saving projects last forever. Variable speed drives, improved lighting, upgraded HVAC, rate schedule specific electrical systems. These things don't change very fast. Ten years seems about right!And I could whine and snivel all I wanted but if I didn't make the payoff, the project was DOA.

author: mvs
author_rank: Frequent Contributor
timestamp: January 15, 2019, 07:58:51 pm
content: Quote from: rstofer on January 15, 2019, 04:21:38 pmI have talked about physical definition, what apparent power is or consist of. Losses are not an integral part of it.In practical usage, high uncompensated reactance may lead to some economical losses (like due inefficient capacity usage of infrastructure) or some power losses in the grid (I2R in wires). But it is important to understand, that imaginary/reactive part is not equal to power loss. Power plant will not burn more coal to generate it. That is the reason why the utility can not charge for it directly.Quote from: rstofer on January 15, 2019, 04:21:38 pmPenalties may apply for uncompensated inductive reactance only. Capacitive reactance may compensate inductive reactance of the neighbors, so it is not generally billed by utilities. Hoseholds are also excluded in most countries.

author: Gregg
author_rank: Super Contributor
timestamp: January 15, 2019, 08:07:01 pm
content: @Simon: Maybe the university is trying to teach you about dealing with incompetence and the usual mix of those who talk a lot and don’t say anything useful.Wait until you try to sell real of energy savings to management. Instead management will spend more by hiring a consulting group with pretty power point presentations of pure guesswork in regard to power savings and claim to the stockholders that they are saving at least the advertised amount and get a huge bonus for nothing as they never follow up with real data.What rstofer is saying about facilities projects is absolutely spot on. The only projects that are implemented are those that will provide some manager(s) with bigger bonuses and/or promotion possibilities.

author: Simon
author_rank: Unknown
timestamp: January 15, 2019, 08:07:05 pm
content: Quote from: rstofer on January 15, 2019, 07:40:54 pmBut despite the average being calculated this problem counts any instantaneous occurrence as the averaged norm.

author: Simon
author_rank: Unknown
timestamp: January 15, 2019, 08:11:34 pm
content: Quote from: Gregg on January 15, 2019, 08:07:01 pmThe university is incompetent. At work I am one of 4 that are studying distance learning with this university and all say the same, the materials are rubbish, i am the furthest ahead as i was the first to start and I am further into my modules of choice and the further in you go the worse you get. The material has some strange and unorthodox ways of explaining stuff and questions like this one are totally irrelevant to the material supplied or so strangely worded that you have to study the convoluted material.

author: rstofer
author_rank: Super Contributor
timestamp: January 15, 2019, 08:53:42 pm
content: Quote from: Simon on January 15, 2019, 08:11:34 pmBut I like this problem; I even delayed breakfast for the first part. It helps that I used to do this stuff for a living and have spent literally decades dealing with electric bills.I will be very interested to find out if I got the right answers to the second part. Don't forget to post back! It seems correct, the "Law of Neat Numbers" seems to apply to the pf correction, so I suspect it is right.Oh, wait! That's a made up law!

author: rstofer
author_rank: Super Contributor
timestamp: January 15, 2019, 09:47:41 pm
content: Quote from: Simon on January 15, 2019, 07:23:44 pmSince the subject is motors, maybe they are trying to show an application for a synchronous motor. If the load is essentially constant and the motor can provide leading VARs then it might make sense to replace an induction motor (always lagging pf) with the synchronous motor and reduce any penalty. Notice that they didn't mention a reward if you got above 0.8.Economically, it depends on having a relatively constant load over the billing interval. That's why the problem stated a continuous 2 MW motor. If the motor only ran one hour per week, it wouldn't produce enough VARs to be useful.At one company I worked for, aerospace as it turns out, there were two large air compressors serving the entire plant and they were both powered by 500 HP synchronous motors (I think... It's been 45 years). Aircraft manufacturing is a 3 shift operation and uses a LOT of compressed air.The stationary engineer that worked in the area would tweak the DC excitation as necessary for power factor correction.

author: Simon
author_rank: Unknown
timestamp: January 15, 2019, 10:06:30 pm
content: I find anything interesting but I judge questions that do not pertain to the material taught to be in bad taste when I am to be marked on them. I can learn, but I can't do clairvoyance.I would love to do the module on EMC next but I am literally shitting myself about how bad it may turn out to be and this is becoming an excise in getting the piece of paper with no expectation of learning much. Meanwhile in the real world I have plenty of learning to do of the sort you do not get qualifications for.

author: rstofer
author_rank: Super Contributor
timestamp: January 15, 2019, 11:25:17 pm
content: I have been pretty cavalier about the calculation of demand. It is not really instantaneous but rather the amount of energy consumed during the demand interval multiplied by some factor to account for units.Suppose the maximum consumption during any demand interval in the billing cycle is 100 kWh and the demand interval is 1 hour (by tariff). In that case, we divide 100 kWhr by 1 hr and we get 100 kW of demand.The length of the demand intervals may vary so it might be necessary to multiple by 2 if the demand interval is 1/2 hr.I'm not sure this helps but I did want to clear up the idea of instantaneous demand. No, we don't measure motor inrush. We measure consumption over some demand interval.

author: Simon
author_rank: Unknown
timestamp: January 16, 2019, 07:52:07 am
content: Yes I understand but the answer is rather odd as the interval is 1 week.

author: perieanuo
author_rank: Frequent Contributor
timestamp: January 16, 2019, 09:11:37 am
content: I worked in a megawatt consuming factory.when you have low cosphi you just buy a compensatory intelligent unit.do the calculations yourself or speak to dedicated tech, conclusions are the same.it's ok to buy one.Envoyé de mon iPad en utilisant Tapatalk

author: nsrmagazin
author_rank: Regular Contributor
timestamp: January 16, 2019, 10:37:17 am
content: And you have to pay for that imaginary power too!

author: rstofer
author_rank: Super Contributor
timestamp: January 16, 2019, 03:47:24 pm
content: Quote from: Simon on January 16, 2019, 07:52:07 amYes, that is odd. It makes the problem easier to approach, I think, but it would be an odd rate schedule that did something like that. Then too, billing intervals are seldom as short as one week yet the problem revolves around a weekly cost.Around here we have for kWh of energy, kW of demand (largest 30 minute window) and a penalty for power factor below 0.8. We do not pay directly for kVARh, it is buried in the power factor penalty which is a month long average.

author: Simon
author_rank: Unknown
timestamp: January 16, 2019, 03:59:42 pm
content: It made the answer no easier as this calculation has nothing to do with the theory of operation of synchronous motors which this a module on.30 minute windows I think are common. My own home smart meter reports at 30 minute intervals.

author: rstofer
author_rank: Super Contributor
timestamp: January 16, 2019, 04:29:37 pm
content: Quote from: Simon on January 16, 2019, 03:59:42 pmWell, it kind of did apply in that the synchronous motor provided a leading 1.5 MVARs and replaced an induction motor with lagging 1.5 MVARs for a net change to the plant of 3.0 MVARs which improved the overall power factor and so on.I agree, it was the long way around to doing a simple pf calculation. Having to do all the cost calculation seemed a little outside the scope but, who knows, it might be important someday. Without the original calculation, it wouldn't be possible to compute the savings because we first needed the penalty. Yes, the problem could have been simplified.I do know this: I was never able to make pf correction pay. Whether I did it on the incoming 12 kV or at the 480V motor control centers, I could never pay it back for the $1k monthly penalty I was seeing. Remember the 2 years thing? Well, $24k doesn't buy a lot of electrical work! Now, if they would stretch the payback to 10 years, I could do some work. Funny thing is, the plant was still there 10 years later and still paying $1k/month penalty. Of course we should have done the project.

author: Simon
author_rank: Unknown
timestamp: January 16, 2019, 06:39:59 pm
content: The problem I have with this question apart from it's lack of information is that the module I am studying is the bare basics of electric machines. The concept of power factor correction with a synchronous motor has been expressed but this question is over thought as there is not much of a question here.

author: rstofer
author_rank: Super Contributor
timestamp: January 16, 2019, 07:20:24 pm
content: Quote from: Simon on January 16, 2019, 06:39:59 pmI had an entire semester course on Motors, Rotors and Dynamos called "Electric Machines". It was not my finest class and I don't remember anything about the material. I still have to book should I ever have a problem sleeping.

author: Simon
author_rank: Unknown
timestamp: January 17, 2019, 07:53:59 am
content: Quote from: rstofer on January 15, 2019, 07:26:54 pmAgain they have made a mess of this one. If the 2MW or 2.5MVA motor replaces an existing one then that 2.5MVA needs subtracting from the 11MVA before calculating the new power factor, so you have 8.5MVA with 0.6PF lagging and 2.5MVA with 0.8PF leading. This actually adds up to a new power factor of 0.8 lagging so there will be no penalties.

author: beesbees
author_rank: Newbie
timestamp: August 02, 2021, 05:42:40 am
content: Looks like they're still trotting out this question, I cannot for the life of me get the £17 644.50 in part iv. Working backwards, sideways and all the other ways.I'm pretty sure I have the new power factor which reduces the reactive penalty charge to 1*3% surcharge, £514.80. And the question states "assume the max demand penalty does not change"... These numbers just will not fit to the one provided in the questions.My method so far:take the existing factory, calculate the load without the induction motor by subtracting the real and reactive components,calculate a new power factor based on the new real and reactive values,calculate new MVAR and MW figures for an apparent power of 11 MVA,add in the synchronous motor to these new figures by subtracting 1.5 MVAR and adding 2MW to real,then calculating the new PF (0.71).But if we "assuming the max demand penalty does not change" then we need a reactive penalty of, erm, £484.50, to get the figure in IV...

author: Simon
author_rank: Unknown
timestamp: August 02, 2021, 05:51:30 am
content: They told me they were taking it out when I complained that it had nothing to do with the materials provided and was not something I could simply look up online. I was very fortunate to find someone on here that know how this works.I would suggest contacting them and explaining that the question is not relevant to the module unless they have updated the module which I doubt.