Power Factor verses Efficiency

Status
Not open for further replies.

ToolHound

Senior Member
If, in very casual terms, power factor is the ratio of the
brigtness out of the light bulb over the volt amps into the light bulb circuit, then how is efficiency different from power factor.


Power Factor = ( watts ) / (volt amps )

Power Factor = True power / Apparent power


Efficiency = Output / Input



Somehow, efficiency seems to be same as power factor.

What am I missing ?

How is Power Factor different from efficiency ?

Thanks.
 

Besoeker

Senior Member
Location
UK
If, in very casual terms, power factor is the ratio of the
brigtness out of the light bulb over the volt amps into the light bulb circuit, then how is efficiency different from power factor.


Power Factor = ( watts ) / (volt amps )

Power Factor = True power / Apparent power


Efficiency = Output / Input



Somehow, efficiency seems to be same as power factor.

What am I missing ?

How is Power Factor different from efficiency ?

Thanks.

Power factor of a load is kW/kVA
Efficiency is kW output/kW input

They are not the same.

For example a typical medium sized cage induction motor might have an efficiency of 93% and a power factor of 0.85.
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
In addition to what Bes said, your light analogy is incorrect.

The ratio of the brightness (luminous flux) of a light source to the power used by the light source is 'Luminous Efficacy' (often expressed as lumens per watt), not 'Power Factor.'
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
130402-1255 EDT

ToolHound:

Consider a DC permanent magnet motor. Efficiency = power output from the shaft divided by the input electrical power to the motor. This is easy to see here.

In an AC motor you can write exactly the same equation. Power input to an AC motor is not volts*amps. It is power measured in watts or equivalent. VA input to an AC motor under any conditions will be more than power input. Thus, power factor would always be less than efficiency if you used motor power output for the power component of the power factor equation.

PF = Power/V*A

But usually power factor is defined as the motor input power divided by the RMS V*A measurement.

A typical example from a textbook. A 1/4 HP capacitor run motor has the following values:
Load Effic PF
.00 0.0 0.46
.25 0.4 0.65
.50 0.6 0.75
.75 0.7 0.83
1.0 0.75 0.85
1.1 0.77 0.86
1.5 0.70 0.90

Efficiency and Power Factor are not the same.

.
 

Sahib

Senior Member
Location
India
The utility supplies power to its consumers measured in KVA. The power factor shows how efficiently this power is used by the consumers. In this sense, the power factor is a measure of efficiency of usage of KVA power by the consumers.
 

Besoeker

Senior Member
Location
UK
The utility supplies power to its consumers measured in KVA.
My electricity supply is measured (and billed) in kWh. I'd be surprised if yours isn't too.

The power factor shows how efficiently this power is used by the consumers. In this sense, the power factor is a measure of efficiency of usage of KVA power by the consumers.

That is, at best, very misleading.
Efficiency is simply Power out/Power in.
If you have a motor with a low power factor you can improve the power factor as seen by the supply by fitting a power factor correction at the motor. It will make no difference to the motor efficiency. If it was 93% before it will be 93% after.
 

Sahib

Senior Member
Location
India
My electricity supply is measured (and billed) in kWh. I'd be surprised if yours isn't too.

For homely purposes, you are correct. But in case of a factory, where two part electricity tariff is availed, it is not completely true: here power supplied (Maximum demand) is measured in KVA and energy in KWH.

Efficiency is simply Power out/Power in.
If you have a motor with a low power factor you can improve the power factor as seen by the supply by fitting a power factor correction at the motor. It will make no difference to the motor efficiency. If it was 93% before it will be 93% after.

I meant the efficiency of usage of power (supplied by the utility) by the consumer and not about the efficiency of individual appliances.

That is, at best, very misleading.
I want to submit it is a case of misunderstanding and not misleading.
 

Besoeker

Senior Member
Location
UK
I want to submit it is a case of misunderstanding and not misleading.
I want to submit that it is misleading.
Comparing power factor and efficiency is the very point that the that the OP was having trouble understanding.
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Consulting Electrical Engineer - Photovoltaic Systems
If, in very casual terms, power factor is the ratio of the
brigtness out of the light bulb over the volt amps into the light bulb circuit, then how is efficiency different from power factor.


Power Factor = ( watts ) / (volt amps )

Power Factor = True power / Apparent power


Efficiency = Output / Input



Somehow, efficiency seems to be same as power factor.

What am I missing ?

How is Power Factor different from efficiency ?

Thanks.

No. An incandescent light bulb is a purely resistive load. That means it has a PF of 1.0, which would make it 100% efficient by the above logic, and the most efficient light source possible, which is, of course, not the case.
 
The utility supplies power to its consumers measured in KVA. The power factor shows how efficiently this power is used by the consumers. In this sense, the power factor is a measure of efficiency of usage of KVA power by the consumers.

All utility billing is done in kWhr. The power factor is indicative of the leading or lagging of the current vector from the voltage vector due to the nature of the circuit, eg. capacitive or inductive load. The concurrent measurement of voltage and current over time results in the actual 'work' performed and mathematically the power factor, or cos(phi) will make that adjustment in the formula. It is called POWER factor and not EFFICIENCY factor because it expresses the difference between the appearent or phantom power and the actual power that produces useable work. Power factor is inherent nature of a circuit element, and 'correcting' it will not produce 'more' work, it simply makes more phantom power available as useable power. It also helps to reduce voltage drop and it is a favored 'magical' power of snake-oil enrgy booster gadget salesman.

Efficiency is the expression of conversion losses from one form of energy to another and inclusive of all losses that it takes to convert. The simplest example is electric motors, where the electrical energy input and the mechanical energy output on the shaft differs. [h=3]Primary and Secondary Resistance Losses[/h]The electrical power lost in the primary rotor and secondary stator winding resistance are also called copper losses. The copper loss varies with the load in proportion to the current squared - and can be expressed as
Pcl = R I2 (3)
where
Pcl = stator winding - copper loss (W)
R = resistance (Ω)
I = current (Amp)
[h=3]Iron Losses[/h]These losses are the result of magnetic energy dissipated when when the motors magnetic field is applied to the stator core.
[h=3]Stray Losses[/h]Stray losses are the losses that remains after primary copper and secondary losses, iron losses and mechanical losses. The largest contribution to the stray losses is harmonic energies generated when the motor operates under load. These energies are dissipated as currents in the copper windings, harmonic flux components in the iron parts, leakage in the laminate core.
[h=3]Mechanical Losses[/h]Mechanical losses includes friction in the motor bearings and the fan for air cooling.
 

Sahib

Senior Member
Location
India
I want to submit that it is misleading.
No.
True. Power factor and Efficiency are two different concepts. But still they could represent each other in some applications such as indicated in post#5

Comparing power factor and efficiency is the very point that the that the OP was having trouble understanding.
I am trying to bring out the similarities between the two to facilitate his understanding .........but under harsh opposition. :)
 

Sahib

Senior Member
Location
India
An incandescent light bulb is a purely resistive load. That means it has a PF of 1.0, which would make it 100% efficient by the above logic, and the most efficient light source possible, which is, of course, not the case.
But I have already stated as
I meant the efficiency of usage of power (supplied by the utility) to the consumer and not about the efficiency of individual appliances.
 

Besoeker

Senior Member
Location
UK
I am trying to bring out the similarities between the two to facilitate his understanding .....
By disseminating confusion?
Power factor and Efficiency are two different concepts. Trying to compare the two is nonsensical.
So please don't.
 

Sahib

Senior Member
Location
India
By disseminating confusion?
Power factor and Efficiency are two different concepts. Trying to compare the two is nonsensical.
So please don't.
I do not know why confusion should arise.
It is no nonsense. For that to exist, you need to show that it contradicts some known fact. Can you, please?
 

mivey

Senior Member
Power factor of a load is kW/kVA
Efficiency is kW output/kW input

They are not the same.

For example a typical medium sized cage induction motor might have an efficiency of 93% and a power factor of 0.85.
That's about as plain as it gets.

I want to submit that it is misleading.
Comparing power factor and efficiency is the very point that the that the OP was having trouble understanding.
Unfortunately, you are correct. This mix of fact and fiction is a common ploy used by sellers of devices that are supposed to save money, clean up "dirty" power, etc. This mix of terminology is weaved into sales brochures for the very purpose of misleading the distributors and consumers.

For example, here is a link showing the mixing of "efficiency" with "power factor": http://www.greenrevolutionems.com/wp-content/uploads/2013/03/PACS-in-Plain-English.pdf

Motors such as in air conditioners and pumps have low efficiency, typically in the range of 75% to 85% because of something called reactive power.
 

kwired

Electron manager
Location
NE Nebraska
I do not know why confusion should arise.
It is no nonsense. For that to exist, you need to show that it contradicts some known fact. Can you, please?
I thought he already show us, as well as a few other people, you are just not seeing it.
 

mivey

Senior Member
I do not know why confusion should arise.
It is no nonsense. For that to exist, you need to show that it contradicts some known fact. Can you, please?
As kwired noted, he already did that a while back:

Power factor = kW / kVA
Efficiency = kW output / kW input

Those are the facts.

Saying that kW / kVA = kW_out / kW_in contradicts a known fact.

Plain enough?
 

Sahib

Senior Member
Location
India
As kwired noted, he already did that a while back:

Power factor = kW / kVA
Efficiency = kW output / kW input

Those are the facts.

Saying that kW / kVA = kW_out / kW_in contradicts a known fact.

Plain enough?

Thanks mivey for your effort.
Both KVA and KW are powers. Do you agree or not?
 
Status
Not open for further replies.
Top