Power factor and VA vs Watts
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ggunn
PE (Electrical), NABCEP certified
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- Austin, TX, USA
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- Consulting Electrical Engineer - Photovoltaic Systems
zog
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Are you asking about how they relate to each other in a power system or specific to power factor testing? Similar yet different explanations.
I am trying to figure out the correlation between inductance and motors, and power factor and why VA is higher than Watts. Also what is actually hapening to cause the amperage to lag the voltage on time vs magnitude graphs. Is it caused by the magnetic field that builds due to the inductance? So many questions!
james_mcquade
Senior Member
In a perfect world, you would have volts, amps, and resistance.
When viewing the voltage and current waveforms on an oscilliscope, the two curves would be on top of each other, in unison hence the name unity power factor.
But when dealing with ac circuits you must also include a term called reactance (ac resistance) which can be composed of inductive loads and capactive loads.
When viewing the voltage and current waveforms on an oscilliscope when reactance is involved, the two curves will be viewed as separate graphs. Either the voltage curve will lead the current curve (inductive type of loads), or the current curve leads the voltage curve(capactive loads).
It has been too long since i had to explain this, but the phase angle between the voltage and current waveforms determines the power factor.
The more inductive loads you have, the more inductive reactance you have, which gives you a lagging power factor. You have a leading power factor with mostly capactive loads.
Almost all power companies require inductries to have a .95 lagging power factor. if you have to much inductive load, you can add capacitors to your power system to help get your power factor closer to unity or "1" meaning the voltage and current waveforms are together.
the opposite is true for mostly capacitive loads.
power (watts) = volts * amps * cosine(theta)
this cosine(theta) is what is known as the power factor.
I hope this information helps.
James.
When viewing the voltage and current waveforms on an oscilliscope, the two curves would be on top of each other, in unison hence the name unity power factor.
But when dealing with ac circuits you must also include a term called reactance (ac resistance) which can be composed of inductive loads and capactive loads.
When viewing the voltage and current waveforms on an oscilliscope when reactance is involved, the two curves will be viewed as separate graphs. Either the voltage curve will lead the current curve (inductive type of loads), or the current curve leads the voltage curve(capactive loads).
It has been too long since i had to explain this, but the phase angle between the voltage and current waveforms determines the power factor.
The more inductive loads you have, the more inductive reactance you have, which gives you a lagging power factor. You have a leading power factor with mostly capactive loads.
Almost all power companies require inductries to have a .95 lagging power factor. if you have to much inductive load, you can add capacitors to your power system to help get your power factor closer to unity or "1" meaning the voltage and current waveforms are together.
the opposite is true for mostly capacitive loads.
power (watts) = volts * amps * cosine(theta)
this cosine(theta) is what is known as the power factor.
I hope this information helps.
James.
zog
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- Charlotte, NC
PF is the cosine of the impedance angle between Real and Apparant power. So (PF x VA)= Watts.
Basically yes, an inductor resists a change in current flow, the voltage induced opposes the force that created it. Inductive reactance or a lagging PF will cause curent to lag voltage. The opposite is true in a capacitive circuit, or a leading power factor. (ELI the ICE man)
And so many answers, I am trying to explain it as basic as I can. My advice is to read up on the other threads and come back with specific questions.
Hameedulla-Ekhlas
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by formula,
P = sqrt3 V*I*cos(tetha)
S = sqrt3 V*I
if P = 400 watt, S = 500 VA
pf = P/S = 0.8
kingpb
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When reviewing the previous archived post I would like to caution the reader as, I think the beer and foam analogy is adequate to pictorially convey the general concept on how to explain KW and KVAR to the non-technical person, however, I must interject that this example is overly simplified, in that KW and KVAR are vectors, and therefore cannot be combined by regular addition, of which the analogy does not actually state, but certainly eludes too.
Also, it conjures up images that the foam (reactive power) has no use, when in fact, the interconnection of motors and other non-resistive loads on a power system do rely on reactive power (VAR) support from the electrical system.
To get a good discussion going; in theory, given a 15kV, 25,000Hp motor starts across the line. How many VAR's would be needed at the bus, where the motor is connected, at T0+ starting? Assume a 0.17 motor starting power factor. Also, what size generator is needed, assuming a 0.9pf rated machine, and no transformer in between? Ignore cable and what not.
Hint: It's a lot of foam!
Also, it conjures up images that the foam (reactive power) has no use, when in fact, the interconnection of motors and other non-resistive loads on a power system do rely on reactive power (VAR) support from the electrical system.
To get a good discussion going; in theory, given a 15kV, 25,000Hp motor starts across the line. How many VAR's would be needed at the bus, where the motor is connected, at T0+ starting? Assume a 0.17 motor starting power factor. Also, what size generator is needed, assuming a 0.9pf rated machine, and no transformer in between? Ignore cable and what not.
Hint: It's a lot of foam!
ggunn
PE (Electrical), NABCEP certified
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- Austin, TX, USA
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- Consulting Electrical Engineer - Photovoltaic Systems
At the root of it are two basic facts: You cannot instantaneously change the current going through an inductor and you cannot instantaneously change the voltage across the terminals of a capacitor.
Real power is expressed in Watts, and it is the power that is sent to the load, and transfered to some other form - like heat or work needed to move the equipment attached to a motor. It leaves the electrical system, and is usually gone forever (i.e. Power disipated through a heater is lost as heat transfered to the room.) Resistors disipate real power in the form of heat.
Imaginary power (imaginary is just a name) is power that is transfered to the load (say a motor), and stored in the load during one part of a cycle. However, imaginary power is transfered back to the source during another part of the cycle. So this power just gets transfered back and forth. Its measured in Volt-Amps-Reactive (VAR's). Inductors and capacitors store and return power in the form of VAR's.
Most circuits have a combination of real power and imaginary power. They add together vectorally, and the combination of these two types of power is called Volt-Amps (VA's.)
Imaginary power (imaginary is just a name) is power that is transfered to the load (say a motor), and stored in the load during one part of a cycle. However, imaginary power is transfered back to the source during another part of the cycle. So this power just gets transfered back and forth. Its measured in Volt-Amps-Reactive (VAR's). Inductors and capacitors store and return power in the form of VAR's.
Most circuits have a combination of real power and imaginary power. They add together vectorally, and the combination of these two types of power is called Volt-Amps (VA's.)
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PowerQualityDoctor
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- Israel
I think that the most intuitive explanation is with a horse and a track:
To understand power factor, visualize a horse pulling a railroad car down a railroad track. Because
the railroad ties are uneven, the horse must pull the car from the side of the track. The horse is
pulling the railroad car at an angle to the direction of the car?s travel. The power required to move the
car down the track is the working (real) power. The effort of the horse is the total (apparent) power.
Because of the angle of the horse?s pull, not all of the horse?s effort is used to move the car down the
track. The car will not move sideways; therefore, the sideways pull of the horse is wasted effort or
nonworking (reactive) power.
The angle of the horse?s pull is related to power factor, which is defined as the ratio of real (working)
power to apparent (total) power. If the horse is led closer to the center of the track, the angle of side
pull decreases and the real power approaches the value of the apparent power. Therefore, the ratio
of real power to apparent power (the power factor) approaches 1. As the power factor approaches 1,
the reactive (nonworking) power approaches 0.
To understand power factor, visualize a horse pulling a railroad car down a railroad track. Because
the railroad ties are uneven, the horse must pull the car from the side of the track. The horse is
pulling the railroad car at an angle to the direction of the car?s travel. The power required to move the
car down the track is the working (real) power. The effort of the horse is the total (apparent) power.
Because of the angle of the horse?s pull, not all of the horse?s effort is used to move the car down the
track. The car will not move sideways; therefore, the sideways pull of the horse is wasted effort or
nonworking (reactive) power.
The angle of the horse?s pull is related to power factor, which is defined as the ratio of real (working)
power to apparent (total) power. If the horse is led closer to the center of the track, the angle of side
pull decreases and the real power approaches the value of the apparent power. Therefore, the ratio
of real power to apparent power (the power factor) approaches 1. As the power factor approaches 1,
the reactive (nonworking) power approaches 0.
Hameedulla-Ekhlas
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- AFG
Now, explaining electrical equation through dynamics example. Alot of other factors come in consideration like friction force. How you relate this to ( real power ), ( reactive power ) and (apparant power) during pulling.
Isnt it better to explain it through real and imaginary axis.
Real power is real axis which is horizontal.
Reactive power is imaginary axis whis vertical.
Apparant power is their magnitude
power factor is the ratio between the real and apparant as well as other relations.
ggunn
PE (Electrical), NABCEP certified
- Location
- Austin, TX, USA
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- Consulting Electrical Engineer - Photovoltaic Systems
The problem with analogies is at their heart they are, well... analogies. You can get across a basic understanding of something using an analogy, but the details get messy and diversionary. You can't look at it too closely.
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Vectors can also be assigned to the pulling analogy, too. The difference is electrical power is a circular product by its nature (i.e. by the way it is generated, regarding POCO supplied AC), whereas pulling power is linear (in the example anyway).
Real power axis is inline with the railroad tracks.
Wasted power (aka effort) axis is perpindicular to the railroad tracks and intersects with the load's center of mass (it moves with the load).
Apparent power is inline with the tow line.
The shallower the angle of pull, the less wasted effort.
PS: Friction isn't a force. In regards to motors, they have to overcome friction also... and the power to do so is included in the electrical measurements.
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gar
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The basic definition of power factor is
PF = POWER/VA
where VA is based on RMS measurement of V and A. Although I have never seen a qualification that the voltage be a sine wave this may be a necessary useful assumption. There may be some logic to a requirement that the voltage be a sine wave. In most practical applications this may be approximately correct. On the other hand currents can be quite distorted.
The above definition says nothing about any angle. Only when you require that both voltage and current be sine waves can one assign any meaning to an angle being uniquely associated with power factor.
What is the phase angle of the current to to a bridge rectifier with a load capacitor in parallel with a resistance if we use the waveform peaks as a means of relating to an angle?
Is the power factor of this rectifier, capacitor, and resistor unity?
There is no useful angle relating to the waveforms here to define or calculate power factor.
.
The basic definition of power factor is
PF = POWER/VA
where VA is based on RMS measurement of V and A. Although I have never seen a qualification that the voltage be a sine wave this may be a necessary useful assumption. There may be some logic to a requirement that the voltage be a sine wave. In most practical applications this may be approximately correct. On the other hand currents can be quite distorted.
The above definition says nothing about any angle. Only when you require that both voltage and current be sine waves can one assign any meaning to an angle being uniquely associated with power factor.
What is the phase angle of the current to to a bridge rectifier with a load capacitor in parallel with a resistance if we use the waveform peaks as a means of relating to an angle?
Is the power factor of this rectifier, capacitor, and resistor unity?
There is no useful angle relating to the waveforms here to define or calculate power factor.
.
zog
Senior Member
- Location
- Charlotte, NC
Back to the topic, I found this nice little site a few years back, click on the flash movie for a good ecplanation of the voltage current relationships in a graphical form.
http://www.intmath.com/Complex-numbers/9_Impedance-phase-angle.php
zog
Senior Member
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- Charlotte, NC
The angle being refered to has nothing to do with the relationship between voltage and current, that is a differtent topic and is called the phase angle. The angle I was refering to is called the impedance angle and is shown on a power triangle which show the relationship between true, reactive, and apparant power. (Watts being the x axis, vars being the y axis and VA being the hypotonuse).
The angle of this ?power triangle? graphically indicates the ratio between the amount of dissipated (or consumed) power and the amount of absorbed/returned power. It also happens to be the same angle as that of the circuit's impedance in polar form. When expressed as a fraction, this ratio between true power and apparent power is called the power factor for this circuit. Because true power and apparent power form the adjacent and hypotenuse sides of a right triangle, respectively, the power factor ratio is also equal to the cosine of that phase angle.
gar
Senior Member
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- Ann Arbor, Michigan
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- EE
100406-2236 EST
zog:
You can not define an impedance angle for a non-linear either.
The original question was about power factor and what I am trying to point out is the actual definition of power factor. So only when you have a linear load is it meaningful to associate a unique angle with power factor.
.
zog:
You can not define an impedance angle for a non-linear either.
The original question was about power factor and what I am trying to point out is the actual definition of power factor. So only when you have a linear load is it meaningful to associate a unique angle with power factor.
.
:-? It sounds like you are saying that the phase angle between the current and voltage doesn't depend on the impedance angle.
The phase angle between the voltage and current is determined by the impedance angle by ohms law: I=V/Z.
So for example, if the impedance angle is 45 degrees, we can see there will be a 45 degree shift between the voltage and current.
Steve
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