Ask Me Help Desk - What's the necessity of power factor improvement?

What's the necessity of power factor improvement?
Is it just because it will improve the power output also?
Is there any other reason?

A low power factor value results in higher amperage being drawn by a certain load. If the PF could be improved, this may result in smaller wire sizes, and possibly lower amperage rated switchgear ratings, etc.

Improving the PF will also help with reducing heat created at the distribution equipment due to lower amperage load.

In your own words, what is power factor?

PF=ratio of resistance to impedance.

What is amperage?current?

Are you asking what is amperage and current?

Please explain you question.

I am not going to start from the beginning of Electrical Theory.

Where do you find the definition of PF=ratio of resistance to impedance?

"PF=ratio of resistance to impedence" - This is what we were taught in the class.
I know what is current.I was just unfamiliar with the word amperage.I underatand it's the eletrical energy flowing through a circuit.Am I right?

Seek out a better course on PF. While I understand what they mean, it can be misleading. PF is a ratio between true power in watts, and apparent power in volt amps.

Their point is that pure resistance will create a PF of 1, and the power will be rated in watts. And impedance, while assuming an inductive or capacitive load in an AC circuit, will result in a PF of less than 1 for an inductive load, and greater than 1 for a capacitive load.

Since we are normally dealing with AC, even a circuit that is pure resistance load, such as an incandescent lamp, the circuit still is to be measured in impedance, as there is the capacitive element of the wiring alone.

If you know current, the original term is amperage, from the name of André-Marie Ampère, a pioneer in electrical theory.

If your learning electricity theory, and want to understand it's roots, you should be studying about Ampere, along with several other pioneers in the electrical field, Ohm, Volta, Joules, and of course, Nikola Telsa.

If I have an AC circuit with pure resistance, such as an incandescent lamp, the current will be in phase with the voltage, and the PF will be 1, and the watts will be calculated:

Power (watts)= E (electromotive force or volts) x I (Intensity or amps)

Assume the voltage is 240 volts and a current draw of 10 amps.
P=EI
P=240 x 10
P=2400 watts

Once you have an inductive load, such as a motor, or any coil, there are inefficiencies in the circuit due to the characteristics of a coil, small currents are induced in the coil that flow in reverse to the current being drawn by the coil. These currents act as a resistance to the primary current flow, and cause the current to lag behind the voltage, thus resulting in a PF of less than 1.

This now causes the "resistance" of the circuit to be called " impedance, as this will include the combination of any pure resistance of the circuit along with any inductance and capacitance.

While there are other calcs to determine impedance, let's just assume the motor we have has a PF of .85, or 85%.

AC Power (Volt-Amps)= E x I x PF

Assume the motor is at 240 volts and the current measures is 10 amps, and the PF is .85.

P=E x I x PF
P= 240 x 10 x .85
P=2040

Note there is now an additional 40 watts of power in the circuit.

The inductive characteristic of the motor, added to the impedance of the circuit, and added 40 more watts of power to be consumed, which does no work, and is dissipated as heat.

This is very simplistic, but enough to show the relationship of PF, and why it can be considered as measurement of inefficiency.

In a facility that has many motors, transformers, ballasts, Induction Furnaces, etc. the PF even thou only at .85, when the normal load is great, the lost power can be great.

Lagging PF caused by inductive loads can be corrected by adding capacitors, or capacitance,which will cause the current to Lead the voltage, or attempt to. This will help improve the PF, raise the PF of .85 up to closer of 1. This will help reduce the primary current that is being drawn to do work, and reduce heat and stress on an electrical system, maybe even allow smaller wire to be used.

PF is also calculated by using trigonometry, illustrated by a Power Triangle, using Vectors, by calculating the angle of the hypotenuse of a right triangle, shown here:http://www1.eere.energy.gov/industry...fs/mc60405.pdf

Much of what I have provided will explain in greater detail what I have provided.

No where in what I have provided will show that is a ratio between resistance and impedance, but as you can see from this power triangle, is what they mean, and another point of view to explain the relationship.

If you are learning electricity, I recommend you start at the basics of Direct Current, then work up to Alternating Current.

I hope this has helped more than confuse.

Quote:

Originally Posted by tkrussell

Seek out a better course on PF. While I understand what they mean, it can be misleading.(1) PF is a ratio between true power in watts, and apparent power in volt amps.

Their point is that pure resistance will create a PF of 1, and the power will be rated in watts. And impedance, while assuming an inductive or capacitive load in an AC circuit, will result in a PF of less than 1 for an inductive load, and(2) greater than 1 for a capacitive load.

If I have an AC circuit with pure resistance, such as an incandescent lamp, the current will be in phase with the voltage, and the PF will be 1, and the watts will be calculated:

Power (watts)= E (electromotive force or volts) x I (Intensity or amps)

Assume the voltage is 240 volts and a current draw of 10 amps.
P=EI
P=240 x 10
P=2400 watts(apparent power)

AC Power (Volt-Amps)= E x I x PF

Assume the motor is at 240 volts and the current measures is 10 amps, and the PF is .85.

P=E x I x PF
P= 240 x 10 x .85
P=2040(active power)

Note there is now an(3) additional 40 watts of power in the circuit.

(1) Considering the power triangle PF is the ratio of active power to the avg. power.But if you consider the impedence triangle??.It is the ratio of resistance to the impedence.

(2) When does PF becomes greater than 1??Always 0<=PF<=1.If you think it will become greater than one please site an example.

(3)where did the additional 40 watt come from??active power is 2040 and apparent pwer is 2400.The difference is 360.

(1) I'll have to think about this one.
(2) Looks like a boo boo.
(3) PF can actually be between +1 and -1. P = V*I*Cos(theta). PF = invcos(theta); where theta is the phase difference between voltage and current assuming voltage and current are both sine waves. PF =1 when cos(theta) = 0. PF can be -1 when they are 180 deg out of phase. Power factor is usually less than one because typical loads are motors which are inductive. Power factor correction is employed at some industrial sites which add capacitance to compensate. You can see it from this equation: Z = SQR (R^2 + (XL - XC)^2)).

(3) I'm kind of right. Look here: Power factor - Wikipedia, the free encyclopedia

Power Factor is defines as the Absolute value of (cos theta) and the sign is indicated by leading or lagging.

(1) The equations governing the impedance triangle are:

ABS(Z) = SQRT( R^2 + X^2). Theta = arctan(X/R)
where Z = ABS(Z)<theta = R + jX
the "<" is read as "at" which is a typical notation
j is the sqrt(-1) because I is used customarily for current

PF is related to the phase angle of voltage and current for sinusoidal sources. It really is the relationship of two powers, resistive and apparent and dimensionally a ratio of a linear term (Z/R) and a squared term from P=I^R or P=(V^2)/R, PF cannot be the ratio of the impedance to resistance.

Quote:

Originally Posted by KeepItSimpleStupid

PF is related to the phase angle of voltage and current for sinusoidal sources. It really is the relationship of two powers, resistive and aparent and dimensionally a ratio of a linear term (Z/R) and a squared term from P=I^R or P=(V^2)/R, PF cannot be the ratio of the impedence to resistance.

Then what is the ratio of impedance to resistance from an impedance triangle called?

I don't think it has a name except "ratio of impedance to resistance"

Similar things are:

SWR

Standing wave ratio - Wikipedia, the free encyclopedia

And Damping factor

Damping factor - Wikipedia, the free encyclopedia

Ok

What about the problem.there is no additional 40W

Calculator malfunction: either human or computer

[QUOTE=KeepItSimpleStupid]I don't think it has a name except "ratio of impedance to resistance"

PF is the cosine of the angle between the current and voltage in an ac circuit.
Now from the power triangle it is the ratio of active to apparent power.But if you consider the impedance triangle? Cosine of the angle between the current and voltage is the ratio of resistance to impedance.

Define the impedance triangle? Are you talking Xc, Xl and the absolute value of (Z)?

Yes.I am talking about Xc,Xl and Z

Xc is the magnitude along one axis, Xl is the magnatude along the other axis, Z is the result. The length is typically called |Z|. Z is the length at a specific angle or the polar representation of the vector.

Ratio of what? |Z|/Xc or |Z|/Xl or something else? When Xc = Xl, you have a pure resistance

Let the phase angle be theta.Consider the impedance triangle of an RL circuit.Here R is along the x axis,Xl along the yaxis and Z the resultant.Let theta be the angle between R and Z.

Here what is cos(theta)? Ratio of resistance to impedance or PF

Quote:

Originally Posted by Cho

PF=ratio of resistance to impedence.

Yes mr cho I agree with you.. but if you are going to improve the pf of a certain industrial plant you have to use the power triangle formula... to determine the size of your capacitor bank to be connected to your system.

:)

Quote:

Originally Posted by tkrussell

Seek out a better course on PF. While I understand what they mean, it can be misleading. PF is a ratio between true power in watts, and apparent power in volt amps.

Their point is that pure resistance will create a PF of 1, and the power will be rated in watts. And impedance, while assuming an inductive or capacitive load in an AC circuit, will result in a PF of less than 1 for an inductive load, and greater than 1 for a capacitive load.

Since we are normally dealing with AC, even a circuit that is pure resistance load, such as an incandescent lamp, the circuit still is to be measured in impedance, as there is the capacitive element of the wiring alone.

If you know current, the original term is amperage, from the name of André-Marie Ampère, a pioneer in electrical theory.

If your learning electricity theory, and want to understand it's roots, you should be studying about Ampere, along with several other pioneers in the electrical field, Ohm, Volta, Joules, and of course, Nikola Telsa.

If I have an AC circuit with pure resistance, such as an incandescent lamp, the current will be in phase with the voltage, and the PF will be 1, and the watts will be calculated:

Power (watts)= E (electromotive force or volts) x I (Intensity or amps)

Assume the voltage is 240 volts and a current draw of 10 amps.
P=EI
P=240 x 10
P=2400 watts

Once you have an inductive load, such as a motor, or any coil, there are inefficiencies in the circuit due to the characteristics of a coil, small currents are induced in the coil that flow in reverse to the current being drawn by the coil. These currents act as a resistance to the primary current flow, and cause the current to lag behind the voltage, thus resulting in a PF of less than 1.

This now causes the "resistance" of the circuit to be called " impedance, as this will include the combination of any pure resistance of the circuit along with any inductance and capacitance.

While there are other calcs to determine impedance, let's just assume the motor we have has a PF of .85, or 85%.

AC Power (Volt-Amps)= E x I x PF

Assume the motor is at 240 volts and the current measures is 10 amps, and the PF is .85.

P=E x I x PF
P= 240 x 10 x .85
P=2040

Note there is now an additional 40 watts of power in the circuit.

The inductive characteristic of the motor, added to the impedance of the circuit, and added 40 more watts of power to be consumed, which does no work, and is dissipated as heat.

This is very simplistic, but enough to show the relationship of PF, and why it can be considered as measurement of inefficiency.

In a facility that has many motors, transformers, ballasts, Induction Furnaces, etc. the PF even thou only at .85, when the normal load is great, the lost power can be great.

Lagging PF caused by inductive loads can be corrected by adding capacitors, or capacitance,which will cause the current to Lead the voltage, or attempt to. This will help improve the PF, raise the PF of .85 up to closer of 1. This will help reduce the primary current that is being drawn to do work, and reduce heat and stress on an electrical system, maybe even allow smaller wire to be used.

PF is also calculated by using trigonometry, illustrated by a Power Triangle, using Vectors, by calculating the angle of the hypotenuse of a right triangle, shown here:http://www1.eere.energy.gov/industry...fs/mc60405.pdf

Much of what I have provided will explain in greater detail what I have provided.

No where in what I have provided will show that is a ratio between resistance and impedance, but as you can see from this power triangle, is what they mean, and another point of view to explain the relationship.

If you are learning electricity, I recommend you start at the basics of Direct Current, then work up to Alternating Current.

I hope this has helped more than confuse.

Quote:

Originally Posted by tkrussell

Their point is that pure resistance will create a PF of 1, and the power will be rated in watts. And impedance, while assuming an inductive or capacitive load in an AC circuit, will result in a PF of less than 1 for an inductive load, and greater than 1 for a capacitive load.

PF cannot be greater than 1, based on the formula P = VIcos(theta); cos(theta) which is the power factor.

Cos(theta) cannot be bigger than 1.

You can use the terms leading and lagging.