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Power factor correction capacitors are often used in industrial applications to correct a lagging power factor caused by inductive loads as utilities often charge a premium for service to a plant with a lagging power factor.
The power factor of a load is not unity if the voltage and current are out of phase. Inductive loads cause the current to lag the voltage, i.e., a lagging power factor. Capacitive loads result in an opposite effect. As a result, a properly sized capacitor bank can correct a lagging power factor eliminating the premium.
This is not an issue with residential service.
Dean
Power factor correction capacitors are often used in industrial applications to correct a lagging power factor caused by inductive loads as utilities often charge a premium for service to a plant with a lagging power factor.
The power factor of a load is not unity if the voltage and current are out of phase. Inductive loads cause the current to lag the voltage, i.e., a lagging power factor. Capacitive loads result in an opposite effect. As a result, a properly sized capacitor bank can correct a lagging power factor eliminating the premium.
This is not an issue with residential service.
Dean
No on the house use. I have plastic injection molding machines. I have two meters one reads the motor spikes that a regular meter does not/can not read caused by higher hydraulic pressure needed at different times in the cycle in molding parts. I get an extra charge from the second meter.
I installed two capacitors to reduce this added cost. It smooths out the peeks.
I installed two capacitors to reduce this added cost. It smooths out the peeks.
buickanddeere
Well-known Member
The main idea with the PF correction was to crutch along an undersized generator.
MarkB_MI
Well-known Member
- Location
- Motown USA
What a power factor correction capacitor does is to reduce the phase angle between current and voltage when you have an inductive load. Power factor is defined as the cosine of the phase angle between current and voltage; when the phase angle is zero the power factor is one (unity). Power factor is also the ratio of power (in watts) to volt-amps. So when the current and voltage waveforms are in phase, power equals volt-amps.
Now generators are rated in "watts" but really they're rated in volt-amps. So if your load is 5000 watts with a .8 power factor, a so-called "5000 watt" generator ain't gonna cut it because you need 6250 volt-amps and you only have a 5000 volt-amp generator.
Measuring power factor at a generator isn't easy with the instruments most people have at home, so calculating the required correction factor probably isn't possible. However, for an individual 120 volt appliance, the "Kill-a-Watt" meter will tell you watts, volt-amps and power factor. So you could conceivably calculate the inductive reactance of all your loads and add them up, but it's really not worth the aggravation. Your home has enough resistive load that the overall power factor probably isn't too bad except when a refrigerator or air conditioner starts up.
Power factor correction is very common in large single-phase welders. If your welder had PF correction as an option, just order the PF correction capacitor for your welder and you're good to go.
I can walk through the math to calculate the value of a power correction capacitor for a given volt-amp load at a given power factor, but not tonight.
Now generators are rated in "watts" but really they're rated in volt-amps. So if your load is 5000 watts with a .8 power factor, a so-called "5000 watt" generator ain't gonna cut it because you need 6250 volt-amps and you only have a 5000 volt-amp generator.
Measuring power factor at a generator isn't easy with the instruments most people have at home, so calculating the required correction factor probably isn't possible. However, for an individual 120 volt appliance, the "Kill-a-Watt" meter will tell you watts, volt-amps and power factor. So you could conceivably calculate the inductive reactance of all your loads and add them up, but it's really not worth the aggravation. Your home has enough resistive load that the overall power factor probably isn't too bad except when a refrigerator or air conditioner starts up.
Power factor correction is very common in large single-phase welders. If your welder had PF correction as an option, just order the PF correction capacitor for your welder and you're good to go.
I can walk through the math to calculate the value of a power correction capacitor for a given volt-amp load at a given power factor, but not tonight.
Anonymous-0
Well-known Member
I think the higher end AC/DC welders have power factor correction capacitors as an option. I don't think they're available on the low end AC buzz boxes. Dave
An inductor stores energy in the form of a magnetic field. In an AC system, this energy is a function of the current through the inductor, and the voltage across the inductor is a function of rate at which the current through the inductor is changing. The storage and release of energy in the magnetic field creates a voltage which adds to the incoming voltage, and causes the voltage waveform across the inductor to shift in time with relationship to the current waveform.
Power is a function of the instantaneous product of voltage and current. When the voltage and current waveforms line up (i.e., are "in phase"), power is simply the product of voltage and current. When voltage and current waveforms do not line up, the instantaneous product of the voltage and current waveforms is something less than the product of voltage and current in phase. The ratio of real to apparent power is called the "power factor" Consequently, for a constant power and constant voltage (set by the source), more current is required to provide a given power with an inductive load.
A capacitor stores energy in the form of an electric field between the plates in the capacitor. In a capacitor, the energy stored is proportional to the voltage, and current through the capacitor is a function of the rate of change of the voltage. In this case, the release of energy is a current which adds to the load current through the capacitor, causing the current waveform to shift in time. In this case, the time shift between voltage and current is in the opposite direction than the shift through an inductor.
Again, if the current and voltage waveforms are not synchronized in time, more current is required for a given power.
The energy stored in a reactive (inductive or capacitive) element is referred to as "reactive power", and does no work. Real power is the product of the amplitude of the voltage and current that are in phase with each other. Apparent power is the product of voltage and current, ignoring phase. Resistive loss through wiring is a function of the actual current and is independent of the phase of the current. If we think of the actual current as the vector sum of reactive (+/- 90 degrees) and real (0 degrees) current, then it can be seen that a reactive load requires more net current to produce a given amount of real power, and hence results in higher line losses. This is why the power company worries about large reactive loads.
Since most large loads are inductive (motors, transformers, welders, etc.), capacitors are used to shift the waveforms back to being snychronized in time, i.e, peak voltage occurs at the same time as peak current.
For residential applications, the effect is small enough to be safely ignored. By the time you have a generator with enough surge capacity to handle things like refrigerator motors starting, you'll have plenty of capacity to accommodate the small amount of phase shift.
If you're trying to run a large reactive load close to the capacity of a generator, such as a buzz box welder without power factor correction, then the generator will likely trip on overcurrent before reaching anywhere near the actual generator load capacity.
Hope this helps a little. Understanding reactive loads and their effects is not a trivial undertaking unless you have a bunch of electrical & electronics training.
Keith
Power is a function of the instantaneous product of voltage and current. When the voltage and current waveforms line up (i.e., are "in phase"), power is simply the product of voltage and current. When voltage and current waveforms do not line up, the instantaneous product of the voltage and current waveforms is something less than the product of voltage and current in phase. The ratio of real to apparent power is called the "power factor" Consequently, for a constant power and constant voltage (set by the source), more current is required to provide a given power with an inductive load.
A capacitor stores energy in the form of an electric field between the plates in the capacitor. In a capacitor, the energy stored is proportional to the voltage, and current through the capacitor is a function of the rate of change of the voltage. In this case, the release of energy is a current which adds to the load current through the capacitor, causing the current waveform to shift in time. In this case, the time shift between voltage and current is in the opposite direction than the shift through an inductor.
Again, if the current and voltage waveforms are not synchronized in time, more current is required for a given power.
The energy stored in a reactive (inductive or capacitive) element is referred to as "reactive power", and does no work. Real power is the product of the amplitude of the voltage and current that are in phase with each other. Apparent power is the product of voltage and current, ignoring phase. Resistive loss through wiring is a function of the actual current and is independent of the phase of the current. If we think of the actual current as the vector sum of reactive (+/- 90 degrees) and real (0 degrees) current, then it can be seen that a reactive load requires more net current to produce a given amount of real power, and hence results in higher line losses. This is why the power company worries about large reactive loads.
Since most large loads are inductive (motors, transformers, welders, etc.), capacitors are used to shift the waveforms back to being snychronized in time, i.e, peak voltage occurs at the same time as peak current.
For residential applications, the effect is small enough to be safely ignored. By the time you have a generator with enough surge capacity to handle things like refrigerator motors starting, you'll have plenty of capacity to accommodate the small amount of phase shift.
If you're trying to run a large reactive load close to the capacity of a generator, such as a buzz box welder without power factor correction, then the generator will likely trip on overcurrent before reaching anywhere near the actual generator load capacity.
Hope this helps a little. Understanding reactive loads and their effects is not a trivial undertaking unless you have a bunch of electrical & electronics training.
Keith
MarkB_MI
Well-known Member
- Location
- Motown USA
OK, now that I've had a little sleep I'll take a crack at calculating the value of a power factor correction capacitor. What you are trying to do is to create a capacitive reactance that is the same as the inductive reactance in the load. When the two are the same, the net reactance is zero.
OK, for our example, let's say you have a 2000 volt-amp, 240 volt load with a power factor of 0.8. How you determined that is not my business; it could be from the manufacturer's specs, phase-angle voltmeter, oscilloscope or whatever.
The phase angle (theta) is the arccosine of the power factor:
theta = arccos(0.8) = 37 degrees
The total impedance is the voltage divided by current.
The current is 2000 volt-amps / 240 volts = 8.3 amps.
So the total impedance is 240 volts / 8.3 amps = 29 ohms.
(It turns out we didn't need to know the voltage; it falls out of the equation.)
The inductive reactive impedance (XL) is the total impedance times the sine of the phase angle:
XL = 29 * sine(theta) = 29 * sine(37 degrees) = 29 * 0.60 = 17 ohms.
Now we know the capacitive impedance must equal the inductive impedance. The impedance of a capacitor equals 1 divided by (2 * pi * frequency * capacitance). So: capacitance = (1 / (2 * pi * frequency * impedance))
Assuming we're in North America, the line frequency is 60 Hz, so:
capacitance = 1 / (6.28 * 60 * 17) = 1 / 6408 = .000156 farads = 156 microfarads. I think.
Hopefully someone will check my work and point out the various errors. Also, I think this wouldn't be exactly correct with a motor, since the back emf affects the volt-amps; it's been a LONG time since I did this.
OK, for our example, let's say you have a 2000 volt-amp, 240 volt load with a power factor of 0.8. How you determined that is not my business; it could be from the manufacturer's specs, phase-angle voltmeter, oscilloscope or whatever.
The phase angle (theta) is the arccosine of the power factor:
theta = arccos(0.8) = 37 degrees
The total impedance is the voltage divided by current.
The current is 2000 volt-amps / 240 volts = 8.3 amps.
So the total impedance is 240 volts / 8.3 amps = 29 ohms.
(It turns out we didn't need to know the voltage; it falls out of the equation.)
The inductive reactive impedance (XL) is the total impedance times the sine of the phase angle:
XL = 29 * sine(theta) = 29 * sine(37 degrees) = 29 * 0.60 = 17 ohms.
Now we know the capacitive impedance must equal the inductive impedance. The impedance of a capacitor equals 1 divided by (2 * pi * frequency * capacitance). So: capacitance = (1 / (2 * pi * frequency * impedance))
Assuming we're in North America, the line frequency is 60 Hz, so:
capacitance = 1 / (6.28 * 60 * 17) = 1 / 6408 = .000156 farads = 156 microfarads. I think.
Hopefully someone will check my work and point out the various errors. Also, I think this wouldn't be exactly correct with a motor, since the back emf affects the volt-amps; it's been a LONG time since I did this.
woodbutcher
Well-known Member
Excellent discussion. I've had the same questions in the past, and now they are answered. Thanks to all.
Butch
Butch
That's a lot to absorb. OK, I've got a 7 1/2 HP single phase electric motor on a compressor having two running capacitors. Do the capacitors serve two purposes? 1. Storing up for the peak loads. 2. A sink to avoid consuming electricity to counter the out of phase electricity produced by the spinning motor (slow down motor).
Will doubling the capacitors stop my lights from flickering?
Will doubling the capacitors stop my lights from flickering?
Anonymous-0
Well-known Member
If the generator is way too small PF capacitors might not make any difference. An inverter welder would use a lot less power and also give you DC current. Inverters are also a lot smaller and lighter than a conventional welder which makes them a lot more portable. Dave
buickanddeere
Well-known Member
Sorry, no, no and half of some. You are thinking DC instead of AC.
A larger running capacitor which would reduce or eliminate reactive power. Voltage drop from carrying unused reactive power would be reduced.
A larger running capacitor which would reduce or eliminate reactive power. Voltage drop from carrying unused reactive power would be reduced.
MarkB_MI
Well-known Member
- Location
- Motown USA
The location of the power factor correction capacitor doesn't make much difference. Installing it inside the welder is the preferred location, because then it will reduce the current through the power cord. Placing it at the generator will still reduce the volt-amps that the generator has to produce, but the increased current in power cord will result in some voltage drop.
Actually, it doesn't hurt to add a little PF correction to any welder. It isn't normally done to cheap buzz-box welders because 1) it isn't needed (they don't draw much current), 2) it costs money for the extra part, and 3) capacitors aren't real reliable components and tend to fail a lot. If you decide to add a capacitor to your welder, place it downstream of the power switch.
Actually, it doesn't hurt to add a little PF correction to any welder. It isn't normally done to cheap buzz-box welders because 1) it isn't needed (they don't draw much current), 2) it costs money for the extra part, and 3) capacitors aren't real reliable components and tend to fail a lot. If you decide to add a capacitor to your welder, place it downstream of the power switch.
MarkB_MI
Well-known Member
- Location
- Motown USA
The capacitors on capacitor-start capacitor-run motors increase the starting and running torque of the motor by creating a rotating field. The capacitors shift the phase of the current through the start winding relative to the run winding, which results in a rotating field. The run capacitor does improve power factor, but that's incidental to its real purpose.
The capacitors do not store energy for peak loads, because they charge and discharge 120 times per second.
Don't mess with the values of the capacitors. Some engineer spent time calculating values that result in an acceptable compromise between torque and efficiency.
The capacitors do not store energy for peak loads, because they charge and discharge 120 times per second.
Don't mess with the values of the capacitors. Some engineer spent time calculating values that result in an acceptable compromise between torque and efficiency.
buickanddeere
Well-known Member
The start capacitor can usually be up sized 10-20% for a little more starting torque.
The run capacitor can be upsized until unity PF is achieved.
The run capacitor can be upsized until unity PF is achieved.
MarkB_MI
Well-known Member
- Location
- Motown USA
It's unclear to whom you are addressing your question, or even what your question is.
Capacitors are energy storage devices that store energy in the form of an electric field. (As opposed to inductors, which store energy in a [i:654c4848f0]magnetic[/i:654c4848f0] field.) Voltage through a capacitor cannot change instantaneously, but builds slowly as the capacitor charges up. (In an inductor, [i:654c4848f0]current[/i:654c4848f0] cannot change instaneously.)
When we place a capacitor in an ac circuit, it charges and discharges 120 times per second. The behavior of a capacitor can be observed as a ninety degree phase shift between voltage and current, with current leading voltage by ninety degrees. That is to say that the current reverses direction a quarter of a cycle (1/240th of a second) prior to the voltage changes direction. In the case of an inductor, current LAGS voltage by ninety degrees.
Now if you connect an inductor and a capacitor together in circuit, they tend to cancel each other out, with energy bouncing back and forth between the inductor and capacitor. The effect is similar to if you bounce a car: After you quit pushing on the car, it continues to bounce up and down as energy is transferred between the mass of the car and the springs, until the shocks use up all the stored energy.
Capacitors are energy storage devices that store energy in the form of an electric field. (As opposed to inductors, which store energy in a [i:654c4848f0]magnetic[/i:654c4848f0] field.) Voltage through a capacitor cannot change instantaneously, but builds slowly as the capacitor charges up. (In an inductor, [i:654c4848f0]current[/i:654c4848f0] cannot change instaneously.)
When we place a capacitor in an ac circuit, it charges and discharges 120 times per second. The behavior of a capacitor can be observed as a ninety degree phase shift between voltage and current, with current leading voltage by ninety degrees. That is to say that the current reverses direction a quarter of a cycle (1/240th of a second) prior to the voltage changes direction. In the case of an inductor, current LAGS voltage by ninety degrees.
Now if you connect an inductor and a capacitor together in circuit, they tend to cancel each other out, with energy bouncing back and forth between the inductor and capacitor. The effect is similar to if you bounce a car: After you quit pushing on the car, it continues to bounce up and down as energy is transferred between the mass of the car and the springs, until the shocks use up all the stored energy.
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