Hello everyone, Thanks in advance for any wisdom you can offer regarding this matter, I just joined physicsforums.com and am now wondering why I haven't been a member all along... My crudentials: I'm an electrician that went part way through college for electrical engineering, however Its been a while and I would like someone who can confirm my logic regarding power factor correction. I have a 1765 rpm, 40 horse power, 3 phase induction motor running at 480vac. full load amps are 46. it has a power factor of 87.5 I'm trying to get the power factor near unity, say 95. I have done calculations, however the end number gave me a single unit of run capacitor micro farads. Sense this is a 3 phase motor do I divide the correction value by 3 and use that divided value in parallel per phase? Or, do I need to divide it by 1.73 to get it to the single phase values? Its odd how hard it is to find correct information re guarding this, it seems like it would be a super common practice given the amount of asynchronous induction motors in the world... Thanks again smart people =)
Based on your numbers above: You need to think of the power triangle. 46*480*1.73=38.2 KVA Hypotenuse of triangle Multiply KVA*PF 38.2*.875=33.4 KW (KWatts or Horsepower delivered at shaft) X direction of triangle Now you need to just use pythagream theroem or trignometry rules to find third side which is VARS VARS are Y direction of triangle. 33.4^2 + b^2 = 38.2^2 b = 18.5 KVars. This is the entire reactive power made by the motor as it sits at .875 PF or uncorrected. To go to a .95 factor, simply divide your watts (horsepower at shaft) by .95 to get your new KVA. New KVA is 35.2 KVA. Now redo your power triangle again. 33.4^2 + b (New Vars) = 35.2^2 New vars = 11.1 KVARS You will then need the difference between 18.5 KVARS and 11.1 KVARS. Which is a total 7.4 KVARS worth of capacitor bank. I believe three phase cap banks are rated as a whole and are often adjustable in ranges. So perhaps a 5 to 10 KVAR cap bank would work for this motor. So to really answer your question, I would not divide the entire 7.4 KVARS by 1.73....but simply by 3 to get 2.46 KVAR of each capacitor for each of the three lines. But again, they are ussually ordered in a bank. Now, let the greater minds prevail and we will get to the bottom of this:)
Incidentally, if I go backwards from your numbers. 40HP X 746 watts = 29.8 KWATTS Now if I take your full load amps of 46 I get 38.2 KVA as shown above. 29.8KWATTS divided by 38.2 KVA and I get a power factor of .78. See what I mean? Something is not adding up. Was your full load amps taken with amp meter? Was power factor given by a meter or figured by hand?
Thanks, that was incredibly helpful. That was the amprage and pf number on the motor name plate, perhaps I should go mesure the actual amperage and use that for the calculations?
Actually, there are two different horsepower ratings here. 29.8 KW and 33.4 KW. 29.8 KW at the shaft.......33.4 KW right before shaft so to speak. The difference between the two is lost in heat and other losses I believe. The 38.2 KVA takes in account the reactive power and the loss of heat to make 29.8KW at shaft...or 40 HP. Therefore, I would conclude that what in the post #2 of the thread to be correct. Post #3 just gives us something to think about.
ah, thats true... I forgot about the heat loss. an odd thing that I just looked at, was that I have a 60 horsepower motor that draws 46 amps at 480 as well... that must be a much better motor? based on the horsepower conversion number it would be 51.2 horsepower. It seems like everything is somewhat of an estimation
My recent motor horsepower chart shows a 60 HP motor pulling 77 Full load amps at 480 volts. Heck at 600 volts it shows a 60 HP motor at 62 full load amps. Is that 46 amps on the nameplate? Incidentally, a 60 HP motor running at a perfect power factor would require at least 54 Full load amps. Something is definitely fishey about the 46 amps number.....it just can't be if there is really 60 HP at shaft.
yea, all of these numbers are just nameplate numbers. Ill bring my amp meter on monday and see what they really draw. I haven't really cared to conferm that until now. Its one thing to be off by 5 amps or so due to voltage variations and whatnot, but thats a serious difference... hey, thanks again for your time... thats super awesome of you
one quick question while I'm thinking about it, I have a few charts that try to represent this math. they seem to range from somewhat close to pretty dang off from the actuall math, which is why i'm here in the first place... there is one commonality between them though. as the motor rpm decreses from 3600 to 1750 or lower the kvars required scale up inversely... I'm wondering if there is another element that I should be considering when doing the math??