Power Factor verses Efficiency

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kwired

Electron manager
Location
NE Nebraska
Thanks mivey for your effort.
Both KVA and KW are powers. Do you agree or not?
Yes, but motor output power is KW and not KVA, so in the equation for efficiency there is no place for KVA.

KVA can have an impact on efficiency, like in a long circuit where there are losses from voltage drop. Those lossses are given up as heat in the conductor. The loss here is not lost by the motor but is a loss to the whole system.
 

Sahib

Senior Member
Location
India
mivey:
Assuming that you agree, I want to state that some utility supplies power in KVA (Maximum demand) to its consumer and the extent it is used by the consumer is given by the power factor of the consumer. If the power factor is 75%, the KVA power used is 0.75*KVA. If the power factor is 50%, the KVA power used is only 0.5*KVA. So from the utility viewpoint, power factor of a consumer is measure of efficiency of utilization of KVA power supplied to the consumer.
 

kwired

Electron manager
Location
NE Nebraska
mivey:
Assuming that you agree, I want to state that some utility supplies power in KVA (Maximum demand) to its consumer and the extent it is used by the consumer is given by the power factor of the consumer. If the power factor is 75%, the KVA power used is 0.75*KVA. If the power factor is 50%, the KVA power used is only 0.5*KVA. So from the utility viewpoint, power factor of a consumer is measure of efficiency of utilization of KVA power supplied to the consumer.
Base energy charges are still in KW. Demand charges are additional charges.
 

mivey

Senior Member
mivey:
Assuming that you agree, I want to state that some utility supplies power in KVA (Maximum demand) to its consumer and the extent it is used by the consumer is given by the power factor of the consumer. If the power factor is 75%, the KVA power used is 0.75*KVA. If the power factor is 50%, the KVA power used is only 0.5*KVA. So from the utility viewpoint, power factor of a consumer is measure of efficiency of utilization of KVA power supplied to the consumer.
Some do indeed bill for kVA. Others charge for excess kvar load. Some do not directly bill for kvar costs.

Are you trying to relate efficiency to use of capacity or percent utilization of resources?
 

mivey

Senior Member
You mean their nature is same but their effect changes with respect to load connected. Correct?
kW is a direct measure of the rate of energy consumption. Efficiency tells us how much of that consumed energy was lost in transport.

kVA is related to the rate of energy exchange, permanently consumed and/or temporarily stored then later returned.
 

Besoeker

Senior Member
Location
UK
That's about as plain as it gets.
Thank you for that.

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.
Indeed it is. But you might have expected someone who purports to be an engineer to understand the difference.
 

Besoeker

Senior Member
Location
UK
Yes. Same fundamental units, but not measurements of the same thing.
Don't agree. The fundamental unit for power is the Watt (the Scots were into everything ;)) but kVA has no such fundamental unit. It is not a measure of power. It may be called apparent power but power it is not.
 

Besoeker

Senior Member
Location
UK
I do not know why confusion should arise.
Nor do I. But, as you have demonstrated, it clearly has.

It is no nonsense. For that to exist, you need to show that it contradicts some known fact. Can you, please?
You want me to prove that kWin/kWout is not the same as kW/kVA?? Surely that is self evident? Or ought to be.

But here are a couple of things for you to consider.

Take a synchronous motor. You can control its excitation such that it takes lagging, unity, or leading power factor. Let's take the unity case.
At rated load, the efficiency might be 95%. So not such a huge disparity at that point. Run it at no load. You can still have unity power factor but the efficiency is 0%. Huge disparity.

Take a PFC bank, and for present purposes, ignore losses. They are usually pretty small anyway. So you have a PF of zero. Efficiency is indeterminate since there is power neither in nor out.

Perhaps consideration of those two situations will disabuse you of the notion that you can relate efficiency and power factor as you have done but I'm not sanguine about the prospects of that.
 
Some do indeed bill for kVA. Others charge for excess kvar load. Some do not directly bill for kvar costs.

Are you trying to relate efficiency to use of capacity or percent utilization of resources?

Billing for kVA is a surcharge that is allowed to be incurred because it reflects the conversion equipment size that is necessary to transmit the energy. It is NOT an amount of power as it lacks the time element of consumed power. It is measured as an averaged maximum demand fro 15/30/60 minutes in a billing period.
 

mivey

Senior Member
Don't agree. The fundamental unit for power is the Watt (the Scots were into everything ;)) but kVA has no such fundamental unit. It is not a measure of power. It may be called apparent power but power it is not.
I'll disagree with that unless you provide better information. Allow me to explain where I'm coming from.

The unit, symbol, and dimension for the three power units are:

volt ampere, VA, ML2T-3
volt ampere reactive, var, ML2T-3
watt, W, ML2T-3

thus they all break down to the fundamental units of kg*m2*s-3 given that you like metric and all.
 

mivey

Senior Member
Billing for kVA is a surcharge that is allowed to be incurred because it reflects the conversion equipment size that is necessary to transmit the energy. It is NOT an amount of power as it lacks the time element of consumed power. It is measured as an averaged maximum demand fro 15/30/60 minutes in a billing period.
For the best accuracy (especially where harmonics are involved), it is measured as apparent energy. The product of Vrms and Irms are integrated over a brief time interval. These results are then integrated over a longer period to obtain apparent energy (VAh).

This apparent energy is then used to calculate the VA demand. Watthours and watts are obtained in the traditional manner. With the VA and W measurements resolved, we then calculate the var and distortion power using the relationship:
var2 + D2 = VA2 - W2
 
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Sahib

Senior Member
Location
India
But you might have expected someone who purports to be an engineer to understand the difference.
Sure, I am holding an Engineering post in a Government organisation and am designated as an Engineer. But you are welcome to doubt and challenge my Engineering ability. I well know there can be differences between a poet and a Professor of poetry, for example.
 

Sahib

Senior Member
Location
India
It is NOT an amount of power as it lacks the time element of consumed power.
No. It persists at least for the maximum demand period and so has the corresponding time element and so the utility has to 'rate' the efficiency of its usage by the consumer with the power factor.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
First please come to terms with KVA, KW and KVAR as all powers.

Since the only difference between VA and Watts is the presence of the cos(theta), which is a dimensionless number, I will accept the argument that they have the dimensional units of power, are derived by a method which can, in a special case, yield power, and are referred to as power. Other than that,....
In the same sense, r2, where r is the radius of a circle, is an area. It is just not the area of the circle. :)
To get the area of the circle you multiply by the dimensionless number, pi.
 

Sahib

Senior Member
Location
India
Since the only difference between VA and Watts is the presence of the cos(theta), which is a dimensionless number, I will accept the argument that they have the dimensional units of power, are derived by a method which can, in a special case, yield power, and are referred to as power. Other than that,....
In the same sense, r2, where r is the radius of a circle, is an area. It is just not the area of the circle. :)
To get the area of the circle you multiply by the dimensionless number, pi.

Power is the capacity to do work. In this regard, not only KW can do work but also KVA and even KVAR can do work under appropriate conditions. By KVA, KW and KVAR are all powers, I just mean that.
 
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