Im doing a degree in electrical engineering and renewable energy and one module we cover is sustainabilitty and business management, its alot of power factor correction, this demonstration seems to make perfect sense and is doing just what it should. sweeeeet !!
The extra amperage associated with reactive power isn't a big issue for small motors at home. But you do pay for the extra line losses in your household wiring which becomes heat in your wires, and is a relatively small loss. You as a residential customer aren't charged for anything but true power (watts), which includes anything that becomes heat. But this video is a demonstration of a principle which becomes important to large industrial motors where the utility company penalizes the customer for low power factors, where extra amps are bouncing back and forth between the grid and the industrial load. The reason being that the excess current causes I^2R losses in the power grid system, which is heat, which is true power or watts, and it's taking place on the power company's side of the watt hour meter. The amperage in the wires causes the wires to heat, and this loss goes up with the square of the amperage. The wires are heated by the amperes and it doesn't matter if the current is in phase with the volts or not. P=I^2R and as you can see, volts aren't even part of this equation. As for your circuit breaker, the same thing applies, it most definitely will react to the amps and it doesn't care what the phase shift of the current is. At home, if you can reduce your 1.5 hp. bench grinder from 17 amps down to 14 amps (by the use of a capacitor near the motor) on your 20 amp circuit, it will leave more head room for other things like lighting that is on the same circuit. The motor still will use 17 amps, but reactive amps are now bouncing back and forth between the capacitor and the motor, and your line and circuit breaker will only see the 14 amps. Your circuit breaker looks at amps, and will trip just the same whether the amps are in phase or 90 degrees out of phase with the voltage. - Retired electrician.
I don't agree with your last statement. The reactive power bounces between the capacitor and inductive motor. If the capacitor is near the motor then the reactive power will not be seen at the circuit breaker. And this was shown in the video where 2A is flowing before the splitter but 4.6A between the capacitor and motor. 2.6A of reactive power never gets to the circuit breaker. - Electronics Technician
@@dartplayer170 - I think we're saying the same thing. I agree with everything you just said. The numbers in my example are different, but the same concept is at play. 14 amps at the circuit breaker, but 17 amps between the capacitor and the motor.
It isn't phoney, it could be better but he has a good point. At home you are charged for kilowatt-hours but a factory has an extra charge to pay if its power-factor is poor, because that means that the power company has to supply unnecessary current. A factory using PF-correction thus saves money, hopefully more than the cost of the PF correcting equipment. The factory still pays for kWh but avoids the penalty for drawing unnecessary current because the losses in the power distribution system are reduced.
A good question Chris - we have encouraged some manufacturers to at least offer this as an option. As you can imagine, introducing a new component requires lengthy UL listing tests, etc.
As per some other videos on TH-cam, the real power should normally be fairly constant. The variation of about 10W would indicate that some electricity would be saved using the capacitor. Unless my computations are way wrong, I evaluated the capacitor to be in the 57uF range. Am I calculating it right or am I way off ? What would be the power factor values, before and after the addition of the capacitor ? 22.2% and 46.25% ?
Thanks for responding, LEDesigngroup!! I was pleasantly surprised that you answered, considering that this video was posted some years ago. Another question for you: Why did the power drop by only 10% when the current was reduced by 60%?
The total motor current isn't actually decreasing. The 10W decrease in power is presumably due to the resistance of the extension cord being avoided. Most people never consider how much energy is wasted when electricity flows though wires and connectors. The length of power cords is critical. Many power tools, for example, will have labels that warn against the use of extension cords. Anyway, I'm an electrical engineer, and I thought this was a great video. He explained things VERY clearly, which makes me wonder why there are so many nay-sayers. How many times does he have to say "distribution system" before you understand that he's talking about LOCAL currents? The fact that he accidentally said "120kW" instead of "120W" gives you an idea as to the level of power they usually deal with. Btw, I'll once again make that same comment that I've made a thousand times before - if the "Tesla" stuff was anything other than the nonsensical ramblings of a mentally ill, washed-up inventor, I'd gladly adopt them for myself. We electrical engineers do *NOT* like paying the power companies any more than you do.
With all respect to your careful effort in trying to save us money I have an electrical engineering degree and have scoured the internet on whether the consumer is actually being charged for apparent power. Overwhelmingly every source that I've consulted indicates that we are charged only for resistive loads - not the inductive component. If what your saying is true then we might as well not use cfl bulbs since they also cause substantial phase losses..
Hello, I have 2 grid tie inverters for solar panels on the same circuit and have a power factor of 0.57 using a Kill-a-Watt plug-in wall unit. Would using a KVAR unit plugged in parallel with my grid tie inverters correct increase the power factor or would I just need to run a larger size wire to my main breaker panel?
so the motor still takes the same current but the strain on the distribution system as a well as the power losses it causes is reduced considerably. good to know thank you sir.
Fascinating. I have a single phase 3,500w pure sine wave inverter (for my home solar). How would I hook up one of those capacitor banks to decrease the current (power) draw on the inverter (which would be considered as my "distribution center"). The house has many fans-motors which is powered by the inverter. Cost? Thanks.
In this scenario shown in the video, simply adapt the diagram to include your inverter in leu of the wall receptacle. I'm thinking you would add a capacitor bank at the inverter-fed electrical panel and land the capacitor conductors in the panel via a breaker. But honestly I've never done this before and you might want to verify what your actual PF is before investing time and money in PF correction equipment.
In AC, Apparent power is voltage times current. Correcting power factor reduces power factor, but doesn't change Active power, which is what homes are charged for. Industries are PENALIZED for having a bad power factor, they are not charged for the CURRENT they draw. Measuring current is not measuring your house expenses.
good day guys, i was wondering if i can install such capacitors to household appliances with motors like refrigerators or ACUs, thanks guys! more power!
whats cool about this demo is at the point of split you can measure the current drop. electricity goes into the caps, what happens when the caps are full and electron lock happens. do you have an explain.
The answer lies in the efficiency differential between the two systems. A HP is a HP and a KW is a KW, however the 5000 hp motor at a 1.0 PF will draw less current than the .8 pf motor the 5000 kw/50000 kva generator is simply more efficient at a 1.0 pf than the .8 pf unit, however their output capacity is the same.
@Lionelmanalaysay - it is possible although not common. Any electric motor can have a capacitor attached in parallel to capture the reactive current of the windings. Caution should be used however to insure that the capacitor is sized properly for the load it is expected to carry. thanks!
yeap, without a doubt a great demonstration, however the motor is clearly pulling 4.4-4.5 amps on the 1 socket twin conductor lead and 4.6-4.7 on the twin socket lead before the capacitors are installed and i am wondering why ? Would have been nice if old mate had have put his clip meter on the capacitor bank to !
There is no load on the motor- so of course the pf is low. Most of those little motors will get pretty close to unity if well loaded. Why is the "wasted"amperage an issue? I have never understood that. They are out of phase with voltage so there is no heating power to any of the equipment, including the breaker.
The "wasted" amperage isn't a big issue for small motors at home. But you do pay for the extra line losses in your household wiring which becomes heat in your wires, and is a relatively small loss. You as a residential customer aren't charged for anything but true power (watts), which includes anything that becomes heat. But this video is a demonstration of a principle which becomes important to large industrial motors where the utility company penalizes the customer for low power factors, where extra amps are bouncing back and forth between the grid and the industrial load. The reason being that the excess current causes I^2R losses in the power grid system, which is heat, which is true power or watts, and it's taking place on the power company's side of the watt hour meter. The amperage in the wires causes the wires to heat, and this loss goes up with the square of the amperage. The wires are heated by the amperes and it doesn't matter if the current is in phase with the volts or not. P=I^2R and as you can see, volts aren't even part of this equation. As for your circuit breaker, the same thing applies, it most definitely will react to the amps and it doesn't care what the phase shift of the current is. At home, if you can reduce your 1.5 hp. bench grinder from 17 amps down to 14 amps on your 20 amp circuit, it will leave more head room for other things like lighting that is on the same circuit. The motor still will use 17 amps, but reactive amps are now bouncing back and forth between the capacitor and the motor, and your line and circuit breaker won't see them. Your circuit breaker looks at total amps, and will trip just the same whether the amps are in phase or 90 degrees out of phase with the voltage. - Retired electrician.
Is it safe? Suppose you unplug the power cord of the capacitor unit when the voltage is maximum. Thereafter somebody touches one of the prongs on the power cord. The voltage can be 120 X sqrt(2) volts. Since the capacitance is large, there is lots of energy. Lots of current at high voltage can flow through the human body.
jbarnes A motor does not produces but consumes electrical power. To answer your question both the motor and generator consume/produce the same amount of power- 5000hp. You want to have a high PF because it will bring down the line currents which will reduce voltage drops and IR lossesn the line
in an industrial application, if you have enough of a voltage drop while running vfd's and inverters you could destroy them. this is a small example and doesnt reflect the true benefits of power factor correction. this is one 1hp motor running on 110 volts.... he is telling you that when your running a plant with 100 motors ... this can and will come into play. when i worked for basf we had several capactior banks setup, so does every single platform in the gulf.
PLZ correct me iam worng .active power is the power deliverd to a motor which runs the motor.reactive power is the power produced by megnatic field of motor which oppose the orginal power and cause lagging of current form voltage.iam worng or right PLZ reply and correct the active and reactve power
Improving Power Factor reduces reactive power thus decreasing Apparent Power & does not do anything to Active Power. The monthly bill is based on Active Power hence your bill won't reduce.
@gottagobideo You have to remember ohms law, to calculate Watts you take the amperage draw "current" multiplied by the voltage and that is your wattage or what you are billed on. If you reduce the amperage or "load" your wattage will be reduced by your multiplication factors being lower.
For all you noobs out there: 120*4.5 = 540 VA = Apparent Power. If you multiply 540 VA by the power factor (which seems to be around 0.222 very very low) that should give you your 120 W of real power.
I’m an average guy that isn’t in the field but heard the term “power factor correction unit”. I still don’t understand it after watching this video. Lol can anyone just summarise without maths in a few layman’s sentences of what it is/does? And how it differs to a Power Filter?
Can anyone help me answer 2 questions? 1. Which is producing more power a. A motor drawing 5000 hp at .8 power factor, or b. A motor drawing 5000 hp at 1.00 power factor. 2. Which is producing more power? a. A generator producing 5000 KW, 5000 KVA at unity power factor, or b. A generator producing 5000 KW, 6250 KVA at .8 power factor. Thanks
Something is wrong. We apply Capacitor bank to reduce the Reactive power but why are you explaining that Real Power was reduce by 9 Watts?? It should be the Reactive Power to be Reduce or Apparent Power & Same Real Power since no change in the Load. If your Apparent Power was reduce then your Total Current will Reduce also same goes with your electric billing. I believe you are wrong in proving but I got what are you trying to say here .
So if I connect a solenoid to the mains, the power company will get angry at me because my average power is zero and I'm sending them back during the negative half cycle what they gave me during the positive half cycle? Are you sure? ;)
Well, I don't think they have to pay you since its almost impossible to make the power factor to be exactly zero, so that means you will have always some power consumption that you have to pay for and then there are also copper losses and the losses you have in your electronics and solenoids. So they won't be paying you anything, but you will be paying much much much less t them ;)
Ryugamine Mikado I actually went as far as testing this idea. I connected a coil of wire with a ferrite core to increase the inductance in series with an electric heater rated at 1200 watts. The average power calculator on net was telling me that the average power consumed by my system will be 0.003 watts, but when I connected everything, my power meter (bought from China) was showing about 1200 watts. I did not have time to play with it much but I think I try to see if I can get it to give better results tomorrow ;)
A watt is amps times voltage. A kilowatt is 1000 watts . A kilowatt hour is 1000 watts per hour A reduction in amps means less watts. Fewer watts over time less kWh fewer kWh less money
Power Factor correction is not a new idea. Industry with large reactive loads have been doing this for years. I run a capacitor bank in my fusebox and save on average $35 a month on electricity.
No load on the motor so this is extreme example. Typical "Black Box" technology. This puts true Power Factor Correction companies in a bad light. Power Factor Correction Capacitors, in almost all cases, should only be implemented to avoid Power Factor Penalties from a utility, not for KWH savings. There are other benefits for installing capacitors, however KWH savings is not the driver.
Another explanation would be, the white shoe boys like things the way they are, because if they really wanted to make electricity cheaper for the consumer they would have embraced Teslas over unity devices and no loss transformers for electrical distribution to residential and commercial buildings.
Yet another UNLOADED motor and bullshed show & tell from a person seeking to profit from selling to suckers that don't have an education in how electricity works.
from Fluke.com - "The most common cause of imaginary power is motor inductance, and is greater when motors are not loaded to their recommended capacity."
you are insane. your motor has no load. you are building reactive power in the motor winding. try this: let the motor run for 24 hours. Call me when the smoke alarm goes off.
good day guys, i was wondering if i can install such capacitors to household appliances with motors like refrigerators or ACUs, thanks guys! more power!
Improving Power Factor reduces reactive power thus decreasing Apparent Power & does not do anything to Active Power. The monthly bill is based on Active Power hence your bill won't reduce.
Im doing a degree in electrical engineering and renewable energy and one module we cover is sustainabilitty and business management, its alot of power factor correction, this demonstration seems to make perfect sense and is doing just what it should. sweeeeet !!
The extra amperage associated with reactive power isn't a big issue for small motors at home. But you do pay for the extra line losses in your household wiring which becomes heat in your wires, and is a relatively small loss. You as a residential customer aren't charged for anything but true power (watts), which includes anything that becomes heat. But this video is a demonstration of a principle which becomes important to large industrial motors where the utility company penalizes the customer for low power factors, where extra amps are bouncing back and forth between the grid and the industrial load. The reason being that the excess current causes I^2R losses in the power grid system, which is heat, which is true power or watts, and it's taking place on the power company's side of the watt hour meter. The amperage in the wires causes the wires to heat, and this loss goes up with the square of the amperage. The wires are heated by the amperes and it doesn't matter if the current is in phase with the volts or not. P=I^2R and as you can see, volts aren't even part of this equation. As for your circuit breaker, the same thing applies, it most definitely will react to the amps and it doesn't care what the phase shift of the current is. At home, if you can reduce your 1.5 hp. bench grinder from 17 amps down to 14 amps (by the use of a capacitor near the motor) on your 20 amp circuit, it will leave more head room for other things like lighting that is on the same circuit. The motor still will use 17 amps, but reactive amps are now bouncing back and forth between the capacitor and the motor, and your line and circuit breaker will only see the 14 amps. Your circuit breaker looks at amps, and will trip just the same whether the amps are in phase or 90 degrees out of phase with the voltage. - Retired electrician.
I don't agree with your last statement. The reactive power bounces between the capacitor and inductive motor. If the capacitor is near the motor then the reactive power will not be seen at the circuit breaker. And this was shown in the video where 2A is flowing before the splitter but 4.6A between the capacitor and motor. 2.6A of reactive power never gets to the circuit breaker. - Electronics Technician
@@dartplayer170 - I think we're saying the same thing. I agree with everything you just said. The numbers in my example are different, but the same concept is at play. 14 amps at the circuit breaker, but 17 amps between the capacitor and the motor.
I'm curious whether other electrical devices such as computer, light bulb, oven will benefit from power factor correction. Do they?
Many newer computer power supplies have power factor correction built in to make them more efficient - electrical devices probably not
It isn't phoney, it could be better but he has a good point. At home you are charged for kilowatt-hours but a factory has an extra charge to pay if its power-factor is poor, because that means that the power company has to supply unnecessary current. A factory using PF-correction thus saves money, hopefully more than the cost of the PF correcting equipment. The factory still pays for kWh but avoids the penalty for drawing unnecessary current because the losses in the power distribution system are reduced.
You're welcome Glenn- thank you for the kind feedback!
A good question Chris - we have encouraged some manufacturers to at least offer this as an option. As you can imagine, introducing a new component requires lengthy UL listing tests, etc.
A video is worth a thousand words. excellent this really helped
THANKS FOR THE EDUCATIONAL VIDEO!!
What is the capacitance, in uF (microFarads) of that capacitor?
As per some other videos on TH-cam, the real power should normally be fairly constant. The variation of about 10W would indicate that some electricity would be saved using the capacitor. Unless my computations are way wrong, I evaluated the capacitor to be in the 57uF range. Am I calculating it right or am I way off ? What would be the power factor values, before and after the addition of the capacitor ? 22.2% and 46.25% ?
Nice explanation.....The current drawn from supply reduces to 2.6 Amps from 4.5 Amps after the addition of capacitor...
But the motor itself still take 4.5
Thanks for responding, LEDesigngroup!! I was pleasantly surprised that you answered, considering that this video was posted some years ago.
Another question for you: Why did the power drop by only 10% when the current was reduced by 60%?
The total motor current isn't actually decreasing. The 10W decrease in power is presumably due to the resistance of the extension cord being avoided. Most people never consider how much energy is wasted when electricity flows though wires and connectors. The length of power cords is critical. Many power tools, for example, will have labels that warn against the use of extension cords. Anyway, I'm an electrical engineer, and I thought this was a great video. He explained things VERY clearly, which makes me wonder why there are so many nay-sayers. How many times does he have to say "distribution system" before you understand that he's talking about LOCAL currents? The fact that he accidentally said "120kW" instead of "120W" gives you an idea as to the level of power they usually deal with. Btw, I'll once again make that same comment that I've made a thousand times before - if the "Tesla" stuff was anything other than the nonsensical ramblings of a mentally ill, washed-up inventor, I'd gladly adopt them for myself. We electrical engineers do *NOT* like paying the power companies any more than you do.
With all respect to your careful effort in trying to save us money I have an electrical engineering degree and have scoured the internet on whether the consumer is actually being charged for apparent power. Overwhelmingly every source that I've consulted indicates that we are charged only for resistive loads - not the inductive component. If what your saying is true then we might as well not use cfl bulbs since they also cause substantial phase losses..
Hello, I have 2 grid tie inverters for solar panels on the same circuit and have a power factor of 0.57 using a Kill-a-Watt plug-in wall unit. Would using a KVAR unit plugged in parallel with my grid tie inverters correct increase the power factor or would I just need to run a larger size wire to my main breaker panel?
U r Great. High Standard explanation
well explained sir, thank you...........
@erdnallewen glad we could assist - thank you for checking out the video!
sounds like an interesting program - best of luck and thank you for your input!
so the motor still takes the same current but the strain on the distribution system as a well as the power losses it causes is reduced considerably. good to know thank you sir.
Just a curious hobbiest here . So basically I observed that you effectively reduced the amount of amperage?
Great video. Nice work.
Fascinating. I have a single phase 3,500w pure sine wave inverter (for my home solar). How would I hook up one of those capacitor banks to decrease the current (power) draw on the inverter (which would be considered as my "distribution center"). The house has many fans-motors which is powered by the inverter. Cost? Thanks.
In this scenario shown in the video, simply adapt the diagram to include your inverter in leu of the wall receptacle. I'm thinking you would add a capacitor bank at the inverter-fed electrical panel and land the capacitor conductors in the panel via a breaker. But honestly I've never done this before and you might want to verify what your actual PF is before investing time and money in PF correction equipment.
In AC, Apparent power is voltage times current. Correcting power factor reduces power factor, but doesn't change Active power, which is what homes are charged for. Industries are PENALIZED for having a bad power factor, they are not charged for the CURRENT they draw. Measuring current is not measuring your house expenses.
good day guys, i was wondering if i can install such capacitors to household appliances with motors like refrigerators or ACUs, thanks guys! more power!
I said it usually doesn't lower the watts , sometimes there is a small reduction.
VERY
CLEAR!!!!!!!!
whats cool about this demo is at the point of split you can measure the current drop. electricity goes into the caps, what happens when the caps are full and electron lock happens. do you have an explain.
The answer lies in the efficiency differential between the two systems. A HP is a HP and a KW is a KW, however the 5000 hp motor at a 1.0 PF will draw less current than the .8 pf motor
the 5000 kw/50000 kva generator is simply more efficient at a 1.0 pf than the .8 pf unit, however their output capacity is the same.
@Lionelmanalaysay - it is possible although not common. Any electric motor can have a capacitor attached in parallel to capture the reactive current of the windings. Caution should be used however to insure that the capacitor is sized properly for the load it is expected to carry. thanks!
What are the characteristics of the capacitor? Is it a current, resistance, formula to get the capacitor value in Farads? Thanks!
Can you advise for a refrigeration facility
yeap, without a doubt a great demonstration, however the motor is clearly pulling 4.4-4.5 amps on the 1 socket twin conductor lead and 4.6-4.7 on the twin socket lead before the capacitors are installed and i am wondering why ? Would have been nice if old mate had have put his clip meter on the capacitor bank to !
What happens when the circuitbreaker goes into action; is there still power on the terminated grid because of the capacitor? So: is it dangerous?
The cap is in parallel with the motor windings so the cap would dissipate when power is removed.
Nice vedio , and it's a good idea but i want know what can i calculate the value and the voltage of capacitor that i want to use
There is no load on the motor- so of course the pf is low. Most of those little motors will get pretty close to unity if well loaded. Why is the "wasted"amperage an issue? I have never understood that. They are out of phase with voltage so there is no heating power to any of the equipment, including the breaker.
The "wasted" amperage isn't a big issue for small motors at home. But you do pay for the extra line losses in your household wiring which becomes heat in your wires, and is a relatively small loss. You as a residential customer aren't charged for anything but true power (watts), which includes anything that becomes heat. But this video is a demonstration of a principle which becomes important to large industrial motors where the utility company penalizes the customer for low power factors, where extra amps are bouncing back and forth between the grid and the industrial load. The reason being that the excess current causes I^2R losses in the power grid system, which is heat, which is true power or watts, and it's taking place on the power company's side of the watt hour meter. The amperage in the wires causes the wires to heat, and this loss goes up with the square of the amperage. The wires are heated by the amperes and it doesn't matter if the current is in phase with the volts or not. P=I^2R and as you can see, volts aren't even part of this equation. As for your circuit breaker, the same thing applies, it most definitely will react to the amps and it doesn't care what the phase shift of the current is. At home, if you can reduce your 1.5 hp. bench grinder from 17 amps down to 14 amps on your 20 amp circuit, it will leave more head room for other things like lighting that is on the same circuit. The motor still will use 17 amps, but reactive amps are now bouncing back and forth between the capacitor and the motor, and your line and circuit breaker won't see them. Your circuit breaker looks at total amps, and will trip just the same whether the amps are in phase or 90 degrees out of phase with the voltage. - Retired electrician.
If you'll notice from the demonstration the wattage dropped from 120 watts to 110 watts so the wattage required to run the motor was reduced
What's the difference between a quart of beer and a quart of vodka? One is much more efficient at getting you wasted.
Is it safe? Suppose you unplug the power cord of the capacitor unit when the voltage is maximum. Thereafter somebody touches one of the prongs on the power cord.
The voltage can be 120 X sqrt(2) volts. Since the capacitance is large, there is lots of energy. Lots of current at high voltage can flow through the human body.
You can find something like this on the Avasva page. Full step-by-step instructions right on your desk.
But dont all the motors come with a capacitor while purchasing ?
jbarnes
A motor does not produces but consumes electrical power. To answer your question both the motor and generator consume/produce the same amount of power- 5000hp. You want to have a high PF because it will bring down the line currents which will reduce voltage drops and IR lossesn the line
mainly only for inductive loads eg. motors, discharge lighting etc.
in an industrial application, if you have enough of a voltage drop while running vfd's and inverters you could destroy them. this is a small example and doesnt reflect the true benefits of power factor correction. this is one 1hp motor running on 110 volts.... he is telling you that when your running a plant with 100 motors ... this can and will come into play. when i worked for basf we had several capactior banks setup, so does every single platform in the gulf.
PLZ correct me iam worng .active power is the power deliverd to a motor which runs the motor.reactive power is the power produced by megnatic field of motor which oppose the orginal power and cause lagging of current form voltage.iam worng or right PLZ reply and correct the active and reactve power
thank you sir you are really amazing
excellent video
Where do you get these capacitors? I can't locate them, will ans electric warehouse have them?
Improving Power Factor reduces reactive power thus decreasing Apparent Power & does not do anything to Active Power. The monthly bill is based on Active Power hence your bill won't reduce.
it does so
@gottagobideo You have to remember ohms law, to calculate Watts you take the amperage draw "current" multiplied by the voltage and that is your wattage or what you are billed on. If you reduce the amperage or "load" your wattage will be reduced by your multiplication factors being lower.
measure the voltage without the capacitor ,and measure again the voltage with the capacitor....
if it so simple to improve the motor performance, I often wondered, why don't the motor manufacturer just add in a capacitor to make a better motor?
Cause they just want to sell motors
For all you noobs out there:
120*4.5 = 540 VA = Apparent Power. If you multiply 540 VA by the power factor (which seems to be around 0.222 very very low) that should give you your 120 W of real power.
so so so so gooodddddddddddddddd
nice jobbbbbbb
Y goes to negative
wow this guy is good
I’m an average guy that isn’t in the field but heard the term “power factor correction unit”.
I still don’t understand it after watching this video.
Lol can anyone just summarise without maths in a few layman’s sentences of what it is/does?
And how it differs to a Power Filter?
Can anyone help me answer 2 questions?
1. Which is producing more power
a. A motor drawing 5000 hp at .8 power factor, or
b. A motor drawing 5000 hp at 1.00 power factor.
2. Which is producing more power?
a. A generator producing 5000 KW, 5000 KVA at unity power factor, or
b. A generator producing 5000 KW, 6250 KVA at .8 power factor.
Thanks
@ledesigngroup can you this method for ur fridge
They do, you just have to pay more for the motor and most people want to spend as little as possible.
It is atually 4.6A. We apologize if it is hard to read!
Something is wrong. We apply Capacitor bank to reduce the Reactive power but why are you explaining that Real Power was reduce by 9 Watts?? It should be the Reactive Power to be Reduce or Apparent Power & Same Real Power since no change in the Load. If your Apparent Power was reduce then your Total Current will Reduce also same goes with your electric billing. I believe you are wrong in proving but I got what are you trying to say here .
So if I connect a solenoid to the mains, the power company will get angry at me because my average power is zero and I'm sending them back during the negative half cycle what they gave me during the positive half cycle? Are you sure? ;)
wouldn't that mean they have to pay you?
Well, I don't think they have to pay you since its almost impossible to make the power factor to be exactly zero, so that means you will have always some power consumption that you have to pay for and then there are also copper losses and the losses you have in your electronics and solenoids. So they won't be paying you anything, but you will be paying much much much less t them ;)
yep, only if we reduce reactive power would the power factor approach zero, but still beats expensive bills
Ryugamine Mikado I actually went as far as testing this idea. I connected a coil of wire with a ferrite core to increase the inductance in series with an electric heater rated at 1200 watts. The average power calculator on net was telling me that the average power consumed by my system will be 0.003 watts, but when I connected everything, my power meter (bought from China) was showing about 1200 watts. I did not have time to play with it much but I think I try to see if I can get it to give better results tomorrow ;)
Life is illusion I see, please do and post the resukts :)
I've been doing this for years ,nobody had to tell me either just common sense to me
A watt is amps times voltage. A kilowatt is 1000 watts . A kilowatt hour is 1000 watts per hour A reduction in amps means less watts. Fewer watts over time less kWh fewer kWh less money
A kiloWatt hour (kWh) is kiloWatt times hour. kWh is energy not power! And energy is what you pay for.
Power Factor correction is not a new idea. Industry with large reactive loads have been doing this for years. I run a capacitor bank in my fusebox and save on average $35 a month on electricity.
Did anyone else see this is not in parallel it is in series ?
Sandra Hicks nope in parallel, but made you think lol
I was saying that you don't get charged for watts not amps. Correcting pf usually doesn't lower watts only amps. Go back to school.
*I would like to apologize for my earlier comment discounting the use of this device*.
You are using reactive power to model active power. Get a watt meter. Open your eyes.
Over size your capacitors?
really?
No load on the motor so this is extreme example. Typical "Black Box" technology. This puts true Power Factor Correction companies in a bad light. Power Factor Correction Capacitors, in almost all cases, should only be implemented to avoid Power Factor Penalties from a utility, not for KWH savings. There are other benefits for installing capacitors, however KWH savings is not the driver.
you get charged kilowatts/hour not amps. The watts stay the same so how does this save you money? It;s only saves industrial buildings money .
BONELESS CHICKEN RANCH
2:55
Dan
Another explanation would be, the white shoe boys like things the way they are, because if they really wanted to make electricity cheaper for the consumer they would have embraced Teslas over unity devices and no loss transformers for electrical distribution to residential and commercial buildings.
Yet another UNLOADED motor and bullshed show & tell from a person seeking to profit from selling to suckers that don't have an education in how electricity works.
from Fluke.com - "The most common cause of imaginary power is motor inductance, and is greater when motors are not loaded to their recommended capacity."
you are insane.
your motor has no load. you are building reactive power in the motor winding.
try this:
let the motor run for 24 hours. Call me when the smoke alarm goes off.
good day guys, i was wondering if i can install such capacitors to household appliances with motors like refrigerators or ACUs, thanks guys! more power!
Improving Power Factor reduces reactive power thus decreasing Apparent Power & does not do anything to Active Power. The monthly bill is based on Active Power hence your bill won't reduce.
Shiran Fernando totally agree
For your home use it will not but for big factories which are charged per KVA it does make sense.
Wrong. Big industrial plants get a power factor penalty and many get billed on kva, not kw