THE PHYSICS OF... POWER

[wwhitney]

we dont ignore R in wires

Sure, lots of times you have to account for the resistance of wires, e.g. to see if they need upsizing for voltage drop.

But if you are trying to understand some other concept, the resistance of the wires may be a confounder that you should temporarily ignore. Then once you've understood the new concept, if the resistance of the wires does really matter to the issue at hand, you can add that effect back in.

Cheers, Wayne

 

[wwhitney]

The old mechanical meters literally integrated the instantaneous values, since the force driving the disk was proportional directly to the instantaneous product of the current and the voltage, and the motion was the integral of that force.

Speaking of such meters, suppose we had a perfect mechanical meter (massless, frictionless, large enough diameter to see very small changes in angle). Then if the only load connected to the meter was a perfect capacitor, the meter would wiggle back and forth at 120 Hz. Right?

Cheers, Wayne

 

[GoldDigger]

Speaking of such meters, suppose we had a perfect mechanical meter (massless, frictionless, large enough diameter to see very small changes in angle). Then if the only load connected to the meter was a perfect capacitor, the meter would wiggle back and forth at 120 Hz. Right?

Cheers, Wayne

Yup, it would.

Glad you observed the 120 Hz instead of 60Hz.

 

[4hunter4]

Si hoc legere scis nimium eruditionis habes

Si hoc legere scis nimium eruditionis habes

Power is power.
Apparent power isn't power.
It's as simple as that.

Very Cute = Si hoc legere scis nimium eruditionis habes = If you can read this, you're overeducated

 

[FionaZuppa]

from post #1, if purely reactive load means the generator has to do no work, then why is it accounted for? and when a plate says 120v 10A is that meant to mean apparent load?

 

[GoldDigger]

from post #1, if purely reactive load means the generator has to do no work, then why is it accounted for? and when a plate says 120v 10A is that meant to mean apparent load?

It is accounted for because the generator windings and the interconnecting wires must handle the very real current associated with the reactive load.

And the nameplate current is always the actual current, regardless of phase angle. As such it does not directly tell you what the load power actually is. You would need to know at least the PF too.

For switches, fuses, breakers, etc. their rating is for first of all the current they can carry (for more than just a few seconds) and second for the current they can interrupt and the current surge on make for a particular load.
The first part does not depend on the characteristics of the load.
The second part will be different for resistive loads than for inductive or motor loads. The inrush may also make for a different rating for tungsten lamp filament loads.

In the case of a generator the nameplate lists current regardless of power factor and also lists maximum real power by listing a wattage. You may also see it listed by real wattage and worst case allowed power factor.

 

[Phil Corso]

if you had a load where voltage lagged current by 180 degrees would you have any power?

Fiona (pls excuse familiarity)...

The answer is no! Not mathematically! But it can be realized physically if both the v and i wave-shapes were derived from PT and CT secondaries connected to a Wattmeter... with one pair of leads reversed!

If an analog meter the needle would be driven down-scale!

Regards, Phil Corso

 

[Phil Corso]

It is accounted for because the generator windings and the interconnecting wires must handle the very real current associated with the reactive load.

In USA, generator nameplate lists kVA! In country using SI units, then nameplate lists kW!

Phil

 

[winnie]

if you had a load where voltage lagged current by 180 degrees would you have any power?

If the current is out of phase with the voltage by 180 degrees, then you have real power with unity power factor, but the direction of power flow is in the reverse direction.

You might see this at a residential meter with installed solar and 'net metering'. When the home is generating to the grid the current will be in the vicinity of 180 degrees out of phase with the voltage.

-Jon

 

[Carultch]

In USA, generator nameplate lists kVA! In country using SI units, then nameplate lists kW!

Phil

KVA are part of the SI units. And algebraically, they appear to be the same thing as kilowatts. The reason why we distinguish the units, is to understand that the concept is just a product of Volts and Amps, that is only part of the picture in calculating the power transmitted by the circuit.

A similar example is the SI unit of torque is Newton-meters, which at first glance, seems to be Joules, the SI unit of energy. But torque is not energy. Not until it rotates the target object through a distance. A Newton-meter-radian, is equal to a Joule of work.

 

[Besoeker]

In USA, generator nameplate lists kVA! In country using SI units, then nameplate lists kW!

Phil

Listed as kVA in the UK.
UK uses SI.

 

[Phil Corso]

Listed as kVA in the UK.
UK uses SI.

Beesoaker...

I now promote you to A.F.C. ... The Adumbrative-Fact-Checker!

Phil

 

[Besoeker]

Beesoaker...

I now promote you to A.F.C. ... The Adumbrative-Fact-Checker!

Phil

And you shouldn't assert things that are simply not true.

 

[dionysius]

reactive power IS dissipated AND absorbed by reactive loads
it performs no net work though

I have to disagree with this. 100% reactive loads do NOT dissipate power. The phasor representation is the most direct way to show this. Only resistive loads dissipate power.

I am open to learning if I am incorrect. Lets not try to lose this in semantics by redefining the widely accepted notions of 'dissipation' and/or 'power'.

 

[GoldDigger]

I have to disagree with this. 100% reactive loads do NOT dissipate power. The phasor representation is the most direct way to show this. Only resistive loads dissipate power.

I am open to learning if I am incorrect. Lets not try to lose this in semantics by redefining the widely accepted notions of 'dissipation' and/or 'power'.

We have already found several different definitions of power and even transfer, so I agree with your hope that we can at least keep the plain English meaning of dissipate to mean convert to thermal energy or disperse is a non-reversible way. (With no moral judgement implied, since that is a different definition. )
Dissipating power is a perilous semantic concept given the lack of unanimity about "power", even though dissipating energy is fairly clear.

 

[Carultch]

I have to disagree with this. 100% reactive loads do NOT dissipate power. The phasor representation is the most direct way to show this. Only resistive loads dissipate power.

I am open to learning if I am incorrect. Lets not try to lose this in semantics by redefining the widely accepted notions of 'dissipation' and/or 'power'.

Think about it this way:
Real power exits the circuit. It could be resistive dissipation, or it could be the mechanical work done by a motor. Or countless other applications.
Reactive power stays in the circuit, and amounts to net-zero energy exchange over the course of an AC cycle. Its effects are either a delay of the current from the voltage (inductive / positive), or the opposite (capacitative / negative).
Negative real power means that power enters the circuit from a source.

 

[FionaZuppa]

from post #1 (and the few that followed) it shows how reactive components (vectors) of "power" are accounted for. the sum of reactive components + pure resistance (vector summation) = impedance (what the load "looks like" to the source), the consumption of power load-side is from the pure DC resistance of the elements, the remainder of that power is absorbed (consumed) elsewhere.

but here's a interesting question. if the reactive components load-side are not responsible for power consumption then the generator does not need to do work, but if no work is done by the generator for those components then where exactly is that reactive power being "absorbed" (aka consumed)? i have a good idea where that power is wasted, but wasting power means the generator is doing work for the lost power associated with the reactive components of the load. would a poco just hate a load that was a giant inductor super-cooled to zero ohms DC ??

 

[Besoeker]

Beesoaker...

I now promote you to A.F.C. ... The Adumbrative-Fact-Checker!

Phil

And I've already told you that I don't soak bees. I prefer them battered and fried.

 

[GoldDigger]

from post #1 (and the few that followed) it shows how reactive components (vectors) of "power" are accounted for. the sum of reactive components + pure resistance (vector summation) = impedance (what the load "looks like" to the source), the consumption of power load-side is from the pure DC resistance of the elements, the remainder of that power is absorbed (consumed) elsewhere.

but here's a interesting question. if the reactive components load-side are not responsible for power consumption then the generator does not need to do work, but if no work is done by the generator for those components then where exactly is that reactive power being "absorbed" (aka consumed)? i have a good idea where that power is wasted, but wasting power means the generator is doing work for the lost power associated with the reactive components of the load. would a poco just hate a load that was a giant inductor super-cooled to zero ohms DC ??

Reactive power is not consumed and does not require any change in the average power produced by the prime mover at the generator. There will, however, be I²R losses associated with the reactive current in all of the wiring and transformers along the way. That will require real power to be generated. And the ampacity of the generator and delivery wiring will have to be higher than if the generator served only resistive loads. For that reason, POCO will be unhappy about your inductor.

 

[K8MHZ]

Reactive power is not consumed and does not require any change in the average power produced by the prime mover at the generator. There will, however, be I²R losses associated with the reactive current in all of the wiring and transformers along the way. That will require real power to be generated. And the ampacity of the generator and delivery wiring will have to be higher than if the generator served only resistive loads. For that reason, POCO will be unhappy about your inductor.

That is a very good explanation. Thanks.