THE PHYSICS OF... POWER

[Sahib]

See Post #334

In your example, the concepts are different, with different names though having same units. So Your example is not so relevant when applied to my last post.

 

[Carultch]

units are J/s
but let me make it very clear, J/s by itself does not mean heat or work. its already been stated a few times, lossless X_C coupled to a lossless X_L
you can pre-charge the cap or run a magnet over the inductor to get things moving, after that if you observe any part of the system you will observe J/s of energy moving back and forth. adding charge to cap or moving magnet over the inductor requires some input work, but after that the observation is J/s at any given time, yet there is no power at any given time. add 10^{-100000000000000000000000000} of R and then and only then will your input work slowly dissipate into heat.

the use of the term "power" should be dropped, talk about everything in eV, then you'll see the clarity.


all of the reactive power is wasted because there exists R in real world. the only way reactive "power" is not wasted is if the system is lossless gen on one side, and a lossless X connected to gen, you would need work to push charge into the X, that work is then returned, net zero over time. we see this with super conductors where the load side is pure X. it requires work to push charge across R, no R then no work. real world is LRC. it requires zero avg work to push 10^{100000000000000} Amps_continuous through a wire that has zero R.

As an example, let's go back to your 1kW motor with an 80% power factor. Suppose it is supplied through a pure resistive distribution circuit that has a resistance of 1 ohm, and operates with 120V across the motor terminals.


For the 80% power factor, 10.4A are needed to operate the motor. The I^2*R losses in the distribution line are 108 Watts. The efficiency of the distribution circuit is 92.5%.


If you replace the motor with a 1 kW resistive heater, the current is 8.3A. The I^2*R loss in the distribution line is 67.6 Watts. The efficiency of the distribution circuit is 96.1%. Adding a capacitor in parallel with the motor will help the motor behave closer to a 1kW resistive load, by cancelling out the reactive power.


Compare 92.5% efficiency with 96.1% efficiency. This is the consequence of non-unity power factor in the load. It requires more current from the source, which means more resistive losses in the distribution. And therefore more power generated at the source to compensate.

 

[Sahib]

Incorrect.

Why?

 

[FionaZuppa]

Reactive power is not wasted.


Reactive power causes a need for more real power generated, than real power delivered to the load. What you really want to look at, is (real power delivered to load)/(real power supplied at source).


And if you want to determine power lost because of the reactive nature of the load, what you want to compare is (real power generated) - (real power delivered) for the load as is, compared to what that difference would be, if you had an resistive load of equivalent power in its place.

so,,,,,,,,,,,,,, VA_gen vs kW_load, yep, there's a line in the sand, the meter. meter charges resi cust for kW, poco eats the diff.

 

[FionaZuppa]

Compare 92.5% efficiency with 96.1% efficiency. This is the consequence of non-unity power factor in the load. It requires more current from the source, which means more resistive losses in the distribution. And therefore more power generated at the source to compensate.

sounds all right to me. now tell me, how do you push charge across R. i think there is some work involved. :thumbsup:

 

[Besoeker]

Why?

VAr is not a measure of watts. It is not power.

 

[FionaZuppa]

VAr is not a measure of watts. It is not power.

well, it kinda is right.
with Φ=0 VA=W holds very true
with Φ !=0 VA=W still holds true, but, the term "power" loses its real meaning because jX is energy in units of Joules that are not consumed by the load, yet that energy comes and goes described by watts (aka J/s) :thumbsup:

but then again, VAr is not power if using real definition of power when applied at the load, but then again, the amps associated to VAr has to traverse R of the poco, hmmm, back to power again.

it boils down to P_in=P_used+P_lost, or, for the power folks VA_gen=kW_customer+kVAr_lost
so as you can see, R of poco cannot easily be lowered, but kVAr can, simply add some X_C to the transmission line endpoint that is highly inductive.

 

[Sahib]

VAr is not a measure of watts. It is not power.

VAr is a unit. Do you agree? If you agree, then tell for which it is a unit.

 

[Carultch]

sounds all right to me. now tell me, how do you push charge across R. i think there is some work involved. :thumbsup:

A voltage difference pushes charge across R. And this precisely is work, as time accumulates the real power loss in the resistance of the transmission line.


The nominal voltage at the source would have to compensate for this, in order to maintain 120V at the load. If it isn't compensated for at the source, then the voltage available at the load will be less.

 

[Carultch]

VAr is a unit. Do you agree? If you agree, then tell for which it is a unit.

VAr is a unit...yes.
VA is a unit...yes.
W is a unit...yes.


VAr is a unit for a mathematical construct involved in the transmission of power, that is given the term reactive power.
VA is a unit for a mathematical construct involved in the transmission of power, that is given the term apparent power.
W is a unit for the actual power that is generated and consumed.

 

[Besoeker]

VAr is a unit. Do you agree? If you agree, then tell for which it is a unit.

Volt amps reactive. Not power.

 

[FionaZuppa]

Volt amps reactive. Not power.

not 100% true.
how you deem to expand VA is a function of Φ
in fact, VA is power that has several terms in it.
VA when Φ=0 is a very simplistic view since most of the math collapses to simplified version, but this does not mean you cannot express it using the math in complex terms.

in order to push X_amps across ckt Z_ohms @ V_volts requires W_joules work. how much of that input work gets converted to output work at the endpoint depends on what the load looks like (R & X_C & X_L). The reactance looks like ± feedback to the gen, takes a tad more work to push charge into the reactance (as example, it takes work to create mag field in inductor or work to create electric field in cap), takes a little less input work while the reactance kicks back, but the overall VAavg_gen is simply VA.

 

[wwhitney]

As an example, let's go back to your 1kW motor with an 80% power factor. Suppose it is supplied through a pure resistive distribution circuit that has a resistance of 1 ohm, and operates with 120V across the motor terminals.

For the 80% power factor, 10.4A are needed to operate the motor. The I^2*R losses in the distribution line are 108 Watts. The efficiency of the distribution circuit is 92.5%.

If you replace the motor with a 1 kW resistive heater, the current is 8.3A. The I^2*R loss in the distribution line is 67.6 Watts. The efficiency of the distribution circuit is 96.1%. Adding a capacitor in parallel with the motor will help the motor behave closer to a 1kW resistive load, by cancelling out the reactive power.

Good example. Quick comment: Note that in the 80% power factor case, the extra resistive loss is 40 watts. While the VAr is 250 VA. So it is not correct to say that all the "reactive power" is wasted.

Cheers, Wayne

 

[Besoeker]

not 100% true.

Yes true. No ifs, not buts.

 

[FionaZuppa]

Yes true. No ifs, not buts.

hmmm, Φ=0, so if you size the gen in VA where load is 100Vac_rms @ 100amps with no reactance, what size gen in VA do i build? hmmm, VA suddenly looks like kW.

very simple, its the same type of Q as if i ask what size gen if my endpoint load is 100kW with .707PF

you can twist the terms around all you want. kW and PF is just a twisting to make it easier for power folks to understand how much of their VA is not paid for by customer and how much overhead they need to account for. 100kW in yields 80kW paid for by cust, then the $kW cust better be larger than the $kW gen, otherwise might as well go back to basket weaving.

but heck, if poco say my kWh price is 10c, and poco sees poor PF (inductive) on that feed, they install X_C to improve PF, will poco then lower my kWh costs ?? looks like the gasoline model. of course they wont, they'll claim they are just adding capacity to their system for future use :lol:, it just means less waste for poco (less input) that yields more net profit.

 

[wwhitney]

VAr is a unit for a mathematical construct involved in the transmission of power, that is given the term reactive power.
VA is a unit for a mathematical construct involved in the transmission of power, that is given the term apparent power.
W is a unit for the actual power that is generated and consumed.

Nicely put.

A further comment: consider two cases of a 1 VA load at 60Hz AC, with power factor either 0 or 1. For the case of power factor 1, the load is receiving 1W of average power. Every standard quarter cycle (1/240 seconds), the load receives 1/240 joules.

If instead the power factor is 0, the load receives 0W of average power. Each standard quarter cycle, some energy is moving either to or from the load, but that amount of energy is not 1/240 joules. It is 1/240 * (2/pi), see the math below.

My point is that VAr is not a real measure of power transfer anywhere in the circuit.

Cheers, Wayne

Math: in a previous post, we saw that the integral of sin(t)cos(t) over a quarter cycle was 1/2. In comparison the integral of sin(t)sin(t) over a quarter cycle is pi/4. The ratio of these numbers is 2/pi.

 

[FionaZuppa]

My point is that VAr is not a real measure of power transfer anywhere in the circuit.

right, VAr is tied to jX (the imaginary power component of VA, and only exists when Φ !=0), until you look elsewhere where it morphs into real.

 

[Ingenieur]

VAr is not a measure of watts. It is not power.

var or volt ampere (reactive)
v x a = joule/coulomb x coulomb/sec = joule / sec = watt

 

[Ingenieur]

What units are a measure of power?

Joule/sec
watt
va
va (reactive) or var
HP
PS
....

 

[Ingenieur]

To avoid confusion between active, reactive and apparent powers, they are measured in watt, var and va units. Each such unit is equivalent to joules/second. Hence the units measure the same thing: POWER.

BINGO!

all are joule/sec or va a power term
by convention
P Watt
Q var
S va