# Power Factor verses Efficiency

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#### Besoeker

##### Senior Member
See if this might help refresh your memory:
It confirms what I have been saying.

?Power is the time rate of doing work?
?Power is the time rate of transferring or transforming energy?
?Power is the measure of the rate at which work is being done.?
?The rate at which work is done is a measure of power.?
?The rate at which this work is provided is power.?

Rate is an instantaneous value. Just as in my analogy with speed and distance.

#### mivey

##### Senior Member
Rate is an instantaneous value. Just as in my analogy with speed and distance.
So you have been wanting to confirm instantaneous power is an instantaneous value? OK, but no one is disagreeing with that.

However, instantaneous power was not what Sahib was talking about before. Instantaneous power is not the power we are talking about with utility bills.

You want to focus on instantaneous power. Do you have a pratical use for calculating the power at a particular instant in time? If so, state it and we can discuss the merits. Otherwise, get happy with the fact that we are discussing the active power when discussing utility billing.

Also get happy with the fact that the power the OP was talking about was true power (or average power, active power, real power, wattage, rate of consumed energy, etc) NOT instantaneous power. Power factor is real power divided by apparent power. Instantaneous power is not what we are talking about. AC power oscillates.

Nothing wrong with discussing instantaneous power and its relationship to the rest of the topics, but continuously stating that power is an instantaneous value when we are obviously talking about something other than non-instantaneous power is not productive.

Now you want to introduce that all rates are instantaneous? In what world? How about interest rates? You think interest rates mean continuous compounding?

#### mivey

##### Senior Member
... when we are obviously talking about something other than non-instantaneous power is not productive.
when we are obviously talking about non-instantaneous power...

#### Besoeker

##### Senior Member
However, instantaneous power was not what Sahib was talking about before.

What he said was :
It is not correct to say there is no time element in KW
It's wrong.

#### Sahib

##### Senior Member
It's wrong.

No, because

It is somewhat simplistic to say that "time does not appear in kW",

since the dimensionality of power in the SI system is kg-m2/sec3.

You can look at the list of common derived units in the SI system here.

“Power is the time rate of doing work”
“Power is the time rate of transferring or transforming energy”
“Power is the measure of the rate at which work is being done.”
“The rate at which work is done is a measure of power.”
“The rate at which this work is provided is power.”

Rate is an instantaneous value.
Please note that instantaneous and average power become same when v and i are constant. So 'rate' could also represent average.

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#### mivey

##### Senior Member
What he said was :

It's wrong.
Agreed. I went back and looked and he was trying to say S (V_rms*I_rms) was a function of time and it is not. Instantaneous power, VA(t) = V(t)*A(t), is a time function.

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.
Note the time element weressl was talking about was for real 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.
It appears he might be saying the kVA calculated during the peak demand period. Nothing wrong with that as we do that all the time. Still does not make S a time function because there is no integral across time. It uses peak values.

#### mivey

##### Senior Member
Please note that instantaneous and average power become same when v and i are constant. So 'rate' could also represent average.
Instantaneous power is a time function, S is not.

#### Sahib

##### Senior Member
mivey:

Power=Energy/time

If the ratio 'Energy/time' is constant (for active power), it does not mean that the ratio 'Energy/time' i.e active power is not function of time because it does not contradict the definition of a function: for each value of input time, there is a corresponding ( constant) value of power.

#### Besoeker

##### Senior Member
:

Power=Energy/time
That would give you average power over the time period considered. It tells you absolutely nothing about how that power was delivered.

From a kW profile over a period you can determine kWh.
From kWh and period you can't determine the kW profile.

#### Sahib

##### Senior Member
From kWh and period you can't determine the kW profile.
Why not, please, if initial condition given?

#### Besoeker

##### Senior Member
Why not, please, if initial condition given?
What if the load varies throughout the period which what happens in real life?

#### Besoeker

##### Senior Member
Agreed. I went back and looked and he was trying to say S (V_rms*I_rms) was a function of time and it is not. Instantaneous power, VA(t) = V(t)*A(t), is a time function.
Appreciated, thanks. I would have said a time varying function but I know what you mean.

#### Sahib

##### Senior Member
That would give you average power over the time period considered. It tells you absolutely nothing about how that power was delivered.

Wrong. As I stated earlier, instantaneous power becomes equal to average power, when v and i are constant.

From kWh and period you can't determine the kW profile.

From kWh and period you can determine the kW profile exactly by differentiating the energy expression as a function of time.

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#### Sahib

##### Senior Member
What if the load varies throughout the period which what happens in real life?
Please see my last sentence of my last post above.

#### mivey

##### Senior Member
As I stated earlier, instantaneous power becomes equal to average power, when v and i are constant.
I thought you were discussing utility billing determinants for AC systems.
From kWh and period you can determine the kW profile exactly by differentiating the energy expression as a function of time.
All roads lead to Rome but seeing someone in Rome does not tell you how they got there. You are not considering path dependency.

#### Sahib

##### Senior Member
I thought you were discussing utility billing determinants for AC systems.

I was. But B did not recognize average power as much as instantaneous power. For example

Nonsense.
kW is power, not energy. It is an instantaneous value.

You are not considering path dependency.
Path dependency is for integration and not differentiation of energy to obtain power.

#### Besoeker

##### Senior Member
Wrong. As I stated earlier, instantaneous power becomes equal to average power, when v and i are constant.
Moving from the general to a unique set of circumastances, eh? What if they vary? Which what happens in real life. That invalidates your argument at a stroke.

From kWh and period you can determine the kW profile exactly by differentiating the energy expression as a function of time.
And where do you think you can get an expression for energy as a function of time if not from the kW profile in the first place?

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#### mivey

##### Senior Member
Path dependency is for integration and not differentiation of energy to obtain power.
That will give you average kW, not the kW profile you spoke of. Bes said:
From kWh and period you can't determine the kW profile.
Which means you have and energy value and a time value. You have no path information, just information about the endpoints.

#### ggunn

##### PE (Electrical), NABCEP certified
Since this is a code forum, I suppose we have a responsibility to be clear.
As clear as the Code is, anyway...

#### GeorgeB

##### ElectroHydraulics engineer (retired)
Wrong. As I stated earlier, instantaneous power becomes equal to average power, when v and i are constant.
I think we are both under-thinking and over-thinking this. One eye opening experience in the usefulness of the calculus was when I was shown that a 120 degree 3 phase source and load, whether with a PF of 1 or 0 or anything in between is that instantaneous power (sum of the 3 phases) is constant, thus the lack or torque ripple in an (ideal) 3 phase system.

If I (uniformly) load to 1kW for 1 hour, then 2kW for 1 hour, then 6kW for the third hour, I have consumed 9kWh for an average load over that time of 3kW. My instantaneous power at all times during the 1st hour was 1kW (then 2 then 6 for subsequent hours. At no time was my instantaneous power the same as my ___3 HOUR 3 kW___ average power, but each hour had instantaneous power equal to that hour's average.

Note that a volt meter's and amp meter's readings would not allow this calculation unless it was either DC or unity PF AC ... or I had an independent method of determining PF.

(edited to add) Any PF other than 1 requires the AVERAGE current to be higher (at constant AC voltage) for the same power. At least some losses are current dependent, I^2*R usually primarily (cannot cite a source), and magnetic secondarily (again, no source). The additional losses impact efficiency.

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