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)
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.
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?
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.
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.
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.
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.
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.
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.
compensation is necessary for low electricity payment. i simply measure the inductance of all running machines than i add negative value of capacitance to make imaginary axis to nearby zero and remaining one is real resistor axis. your problem writing style is not common i cant understand the written values.
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.
my thoughts exactly regarding the p.f 56%, thought I was loosing my head lol
Solid video. I have an exam on this on Friday.
"Drop like it's hot." You've given a giggle to a student desperately studying for finals.
I hope it helped !
Cos inverse one(.56) = 55.94 degree.
Why it is 38°? Please
What would be a good cap circuit if any of the motors can each be switched on or off???
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.
wow that's heaps you're gona help my exam tmrw!!
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
Peter Alexand thanks. You're right. I need to redo that video with an update. Did the annotations show up in the video?
it sure did. threw me off track initially though :P
I asked my mother and she said she doesn't give a rat's ass.. great step by step explanation. cheers
Great videos dude, hope you can start getting more views and recognition
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)
very well explained . good work
So the needed capacitance is 397 microfarad?
My calculations came out to 400 microFarad, so I would say, disregarding rounding errors, our answers agree.
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.
hey dude,its impressive..........keep it up.
what ever you did a great job
Thanks ,but cos-1(0.56)=56
sir Invcos(0.56)=55.94 deg. correct me if I am wrong......
+MrMalangii You're definitely right! I don't know how I let that get by me. I will make a revision. Thanks for bringing it to my attention!
+MrMalangii I added an annotation until I get the revised film done. Thanks again.
anyway this video is very helpful, thanks and salute to the owner.
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.
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?
boshbosh mag
my collection for the 3rd moter is 29.58 kw , 36°if moter kvar is 21
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.
+worldsgreatestride yeah. You don't divide it by 3. You have to use root 3. I definitely will be doing a video on three phase.
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.
what will I do if the loads are in series?
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.
Excellent video! You should of took it one step further and showed how to convert the Kvar to microfarads.
+worldsgreatestride good call. I'll work on that.
Can you explain what I line is and how it is different from normal I
one final thing. Voltage squared divided by Xc equals power 34.5 KVAR. Solve Xc and then find C.
So would our corrected capacitor be 542.2uF? Using formula C= 1000 × Q(kVAR) / (2πf×V^2)
my calc came to 401uf
I got 400 microF@@jaysysumakaRatMaster3
how is 21 kvar have 21 kw when the phase angle is 45°. while we know that kW is greater than kvar??????? if there is a phase shift
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.
how could 56% p.f becomes 38 degree? in my calculation using cos-1. (.56) is = 55.94
+Leandro Tolentino I made a mistake :) I added an annotation to the video. Good catch though.
♥️
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.
Cos-1 0.56=55.9 at the first triangle! The 38degree is wrong!
Most triangles are wrongly constructed. 1st triangle input on base, 2nd triangle output on base, 3rd triangle completely wrong.
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.
that 38degrees is cos-1(.79)
jaden telebrico error cos-1(56%)
56degre
You forgot about the square root of 3.
"What you say bout my mama"!!!
I think you make a mistake in the 3rd moter
✌✌✌
746 the number to remember
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.
You dont explain this at all for people my god
This wasn't helpful. Some of my instructors expect us to know and remember everything. You're the same thing.
You obviously need to spend some more time with your textbook.
compensation is necessary for low electricity payment. i simply measure the inductance of all running machines than i add negative value of capacitance to make imaginary axis to nearby zero and remaining one is real resistor axis. your problem writing style is not common i cant understand the written values.