Inductive loads cause current to be higher than it needs to be. With the use of a power factor correction capacitor, current is able to be decreased without decreasing the output of the loads.
Even with the minor arithmetic error, this was very helpful. Strangely, sometimes when you stumble, the skinned knee causes enough of a wound to heal stronger. I paused the video right at the beginning and did the question on my own. Only thing is, even motors rated for continuous will use usually have an intermittent duty cycle, with variance of PF by rpm, so the numbers jump around. Brilliant teaching.
this wrong angel of 38 degree is the inverse cos of the 79% that you can read above the 56%. So he said the right percentage but calculated accidentally with the wrong number.
great vid , just shame about the error's , your first powertriangle , 4,7KW with a powerfactor of 56% give's an angle of 55,94° as stated , this give's us Tang = 1,479.. , so Kvar is not 3,6 Kvar but 6,95 Kvar. Logic , with an angle of 56° opposite is greater then the adjacent side
Is this a three-phase circuit? I would assume it is, based on the source's voltage level, and you were also making reference to LINE voltage and current. If that's the case, the KVA (apparent) power should be equal to: Sqrt(3) * V(Line) * I(Line)
Other than you made a small mistake on the angle of the first Triangle, you did not calculate the capacitor you need to increase to 0.95 power factor. I liked everything else you did to explain the process.
Hello! It`s very bad thing a reactive power in DC/AC converters. The converter consums a power from the battery (in case of 50Hz transformer or in case of lowpass filter capacitor in PWM pure sine converters) - but this power is not working any work - it`s only recharge a capacitor or an inductor.
could it be said that since the voltage is 480V and the capacitor is 34.5Kvar that, the capacitor would need to be j6.678ohm (480V)^2 / 34500Var ? My prof wants that answer in ohms.
Let me see if I got this for 3 phase. Calculate it the same as you illustrated. Whatever the cap value turns out to be divide it by 3 and connect the caps up in a delta configuration. Sound right? Perhaps you need to do a 3 phase video just to set the record straight.
Great! Please let me know when the video comes out. Be sure to add the cap values in microfarads this time. And how to figure the working voltage would be a nice addition too.
so actually what you are trying to say is that the reacted power will be reduced due to adding capacitor NOT that we need the difference in KVAr right?
Because in a right triangle with a Theta angle of 45, the legs of the triangle will be of equal length, so VARs and Watts will be the same. The phase shift always results in the VA or Apparent Power being greater than the True Power and the Reactive Power, because the legs of of Right Triangle are always shorter than the hypotenuse. 45 degrees is where the legs are equal in length. Greater than 45 degrees results in KVARs being greater than True Power. Less than 45 degrees results in True Power being greater than KVARs. Just picture it in your head.
there's confusion in what you are doing because at times you treat power factor as theta and vise versa. power factor = cos theta, theta =arcos(pf) but you mix the two throughout your presentation.
Then you just add all your loads together and treat it as one load. However it's very unlikely that you would ever have such a case. Remember these pieces of equipment are rated at a standard voltage (480V is the popular LV). You won't connect such loads in series, since they'll each have individual voltage drops. Also, if one piece of equipment experienced a fault or cable damage all the loads would be affected. For those reasons, these industrial loads are placed parallel to each other.
Hi, I am a follower of your channel, I like your channel, I am wondering if you can help me please. I need your help in explaining LV Capacitor Bank step by step but the most important is; 1- the calculation of reactors in base of the harmonics in an industry using VFD controlled motors for example or any wave distorting source, with an example using power factor meter that gives the harmonics in the facility and how to deal with those harmonics. And which are the cases that we don't need to use reactors in series with the capacitors, that if we are using Capacitor duty Contactors to limit the Inrush current. 2- discharging resistors issue, how to calculate them and how to connect them exactly? Is the connection made via timer relay? With a diagram, if possible. I found diagrams on the net showing the resistors connected directly on the Capacitors terminals without any timing relays and that did not convince me, the proper time permissible to discharge in case of on off loads like motors used to cut marble or variable mechanical load that changes the pf in the way that we need to cut the Capacitors out for some moments.Thank you in advance. My best regards and respect.
