Power Factor verses Efficiency

[Besoeker]

kwirred:
RMS value apply to AC voltage and current.

Not only. It applies to all.
Take a hexaphase rectifier. Each of the six coils delivers all of the current in one direction for one sixth of the period. Each leg has to be rated at Idc/sqrt(6).
But I'm sure you knew that....

 

[kwired]

kwirred:
RMS value apply to AC voltage and current. Their product ie power for a purely resistive load is not a RMS quantity but an average quantity. You may find it by taking products of instantaneous values of AC voltage and current for a resistive load and finding the average of them. The result will be the power through the resistive load.

I understand that, I was trying to say the effective voltage - RMS, which is kind of the DC equivalent, multiplied by current is what POCO is using to determine power, then factor in time and you get the KWHR they are billing you for. They don't bill you for any average, they bill you for actual KWHR used.

Any averages they come up with are related to demand charges, but are in KW not KWHR.

Any power factor charges they come up also are in KVA - maybe for an average demand for a specific time, but certainly not as a unit of energy consumed.

 

[gar]

130416-1350 EDT

kwirwd:

A normal spinning disk kWh meter and probably most of the new electronic meters take the instantaneous product of the instantaneous voltage and current, average this instantaneous product, and then integrate the average value over time. In the electronic meters there are a large number of samples (the instantaneous values) per cycle.

In the Cirrus CS 5461 A chip the rate of providing instantaneous power is adjustable, and also the time period for summation to obtain an output average power. The TED 1000 system provides average power every 1 second. Independent of the instantaneous power measurement the chip also measures RMS voltage and current and from this Irms*Vrms. Thus, there is available information to calculate average power factor over the time interval of the average power measurement.

A spinning disk meter does not measure any RMS values.

.

 

[weressl]

Lazlo:
It is not correct to say there is no time element in KW. Because KW is energy consumed per unit time.

Nope, that would be kWhr, kWmin, kWsec, etc. Stop wasting everybody's time and go back(?) to school...:roll:

 

[weressl]

But you might expect people to learn.

In some cases I am going with low expectations. I mean how long is one expected to bang his head against the brick wall......

 

[GoldDigger]

I mean how long is one expected to bang his head against the brick wall......

Only until it breaks. And I will let you figure out which "it" is.

PS:

I understand that, I was trying to say the effective voltage - RMS, which is kind of the DC equivalent, multiplied by current is what POCO is using to determine power,

NOOOOOO!
I am sure that was not what you were intending say, and the words just came out wrong.

 

[Besoeker]

In some cases I am going with low expectations. ...

Understood.

 

[kwired]

130416-1350 EDT

kwirwd:

A normal spinning disk kWh meter and probably most of the new electronic meters take the instantaneous product of the instantaneous voltage and current, average this instantaneous product, and then integrate the average value over time. In the electronic meters there are a large number of samples (the instantaneous values) per cycle.

In the Cirrus CS 5461 A chip the rate of providing instantaneous power is adjustable, and also the time period for summation to obtain an output average power. The TED 1000 system provides average power every 1 second. Independent of the instantaneous power measurement the chip also measures RMS voltage and current and from this Irms*Vrms. Thus, there is available information to calculate average power factor over the time interval of the average power measurement.

A spinning disk meter does not measure any RMS values.

.

You lost me some here, how can you have an average of an instantaneous value? Don't you need a sample from a set interval and take averages of those samples. I can see those samples being as often as milliseconds or even less apart. Otherwise the average of a single instantaneous value is that value divided by 1 = same value.

 

[GoldDigger]

You lost me some here, how can you have an average of an instantaneous value? Don't you need a sample from a set interval and take averages of those samples. I can see those samples being as often as milliseconds or even less apart. Otherwise the average of a single instantaneous value is that value divided by 1 = same value.

In an analog device (like an analog multiplier in this case) the speed of the disk is proportional to the integral of the instantaneous values over one cycle, or using a less precise term, the average of those values.
More precisely, the motion of the disk over a time interval is proportional to the integral of the voltage-current product over that time interval. Which when divided by the time interval gives the speed. The use of the word average is not limited to discrete digital samples.
If I time your car traveling one mile in 60 seconds, I do not have to take any individual measurements of your speed to do that, I just record the time at the start and the finish, giving me the average speed.
Sorry, got carried away lecturing.

 

[gar]

130416-2350 EDT

kwired:

First, consider the electronic type meter. Suppose we make an instantaneous measurement of both current and voltage each 0.0001 seconds (0.1 millisecond) long. This implies an average over that time. Actually these are some sort of average over that time, but for most purposes at 60 Hz there is little change during that time so we define the result as the value at that time.

Since one full cycle is 16.7 milliseconds this means there are 167 samples per cycle. Multiple this "instantaneous" voltage and current, and the result is the average power for the particular 1/167 of a cycle. Do this for each of these sample periods and add all the sample products for one full cycle and divide by 167 and you have the average power over that cycle.

Next multiply the value of average power for each sample by 0.0001 and you have the average energy for each sample time. Do this for each sample period for as long a duration over which you desire to measure energy and add all these together. The result is your energy over that time. Note: this did not use long term RMS values.

Second, consider the spinning disk meter. By design it electrically measures average power over some short time. It never measures voltage or current separately. Rather it produces a torque on the disk proportional to the instantaneous product of current and voltage. The inertia of the disk and the reflected inertia of the gear train performs the averaging of the instantaneous torque. The gear box and dials are the summing device that performs integration over time.

Still lost, then I will try to further clarify.

.

 

[GoldDigger]

Second, consider the spinning disk meter. By design it electrically measures average power over some short time. It never measures voltage or current separately. Rather it produces a torque on the disk proportional to the instantaneous product of current and voltage. The inertia of the disk and the reflected inertia of the gear train performs the averaging of the instantaneous torque. The gear box and dials are the summing device that performs integration over time.

Very well (and compactly) stated. My only quibble is that for the torque to integrate the way you describe it, there has to be another component to the meter design, namely a decelerating force (counter torque) on the disk which is proportional to its speed, and therefore makes the speed proportional to the applied torque.One way of doing this is with a damping permanent magnet at some point around the disk.

 

[Sahib]

Nope, that would be kWhr, kWmin, kWsec, etc. Stop wasting everybody's time and go back(?) to school...:roll:

I wonder if it is you who should go to school, for you do not yet accept that power is energy consumed per unit time.

 

[Besoeker]

I wonder if it is you who should go to school, for you do not yet accept that power is energy consumed per unit time.

The watt is an INSTANTANEOUS measure of power.
Please take the time to review post #99.
I had hoped it would aid your understanding.......

 

[GoldDigger]

I wonder if it is you who should go to school, for you do not yet accept that power is energy consumed per unit time.

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. But the time interval does appear directly in the name of the derived unit we commonly use for energy, the kW-hr. The basic dimensionality of energy is kg-m²/sec²; (kg-m²/sec³)-hr is a truly mixed unit. It just happens to be convenient in the real world.

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

 

[Sahib]

goldDigger:
:thumbsup:

 

[Sahib]

The watt is an INSTANTANEOUS measure of power.

How would you measure average i.e active power using INSTANTANEOUS measure of power i.e KW?

 

[Sahib]

Further, how do you distinguish between 50KW of active (average) power and 50 KW of NSTANTANEOUS power?

 

[Sahib]

If you ponder over my last questions, you may realize that KW can, not only, be a measure of INSTANTANEOUS power but also a measure of average (active) power.

 

[Besoeker]

How would you measure average i.e active power using INSTANTANEOUS measure of power i.e KW?

Ah. So you want AVERAGE power now rather than power? Different thing of course.
An analogy.
You make a journey of 40 miles in one hour. You can calculate average speed. But that tells you nothing at all about the actual speed at any one point. Speed, like power, is an instantaneous measure.

 

[ggunn]

I think you guys are pushing up against the uncertainty principle here. There really is no such thing as an instantaneous measurement.