THE PHYSICS OF... POWER

[FionaZuppa]

Depends on the time period, and on the phase offset between V and I.

In other words, the power delivered is (V^2 / Z) * cos theta. That is also the rate of work required.

Cheers, Wayne

0<PF<1, take whatever period you need to express it in terms of energy per coloumb since both watts and amps have common denominator.

as for the work required, you can verify cos(Φ)V²/Z when you have the math for J/coloumb

i believe physics says you need more work to increases amps when volts remain constant. this does not agree with cos(Φ)V²/Z.

 

[Carultch]

0<PF<1, take whatever period you need to express it in terms of energy per coloumb since both watts and amps have common denominator.

as for the work required, you can verify cos(Φ)V²/Z when you have the math for J/coloumb

i believe physics says you need more work to increases amps when volts remain constant. this does not agree with cos(Φ)V²/Z.

If you are going to increase Amps when Volts remain constant, that means impedance Z has to decrease. V = I*Z.

Z can either decrease in its real component, in which case you will need more work to increase Amps. Or Z can decrease in its imaginary component, in which case you are making better use out of the work you already have, in order to increase Amps.

If Z's imaginary component decreases, then that means that phi becomes closer to 0, and cos(phi) becomes closer to 1. And this is true whether the magnitude of Z remains constant (a trade of reactance for resistance), or whether the real resistance remains constant.

 

[GoldDigger]

0<PF<1, take whatever period you need to express it in terms of energy per coloumb since both watts and amps have common denominator.

as for the work required, you can verify cos(Φ)V²/Z when you have the math for J/coloumb

i believe physics says you need more work to increases amps when volts remain constant. this does not agree with cos(Φ)V²/Z.

Physics tells us that for a pure inductance the voltage drop is directly proportional to dI/dt and so you cannot arbitrarily change one or the other.
The current at any given moment is the time integral of dI/dt from a time at which the current was zero.
So the current at time t depends not on the voltage at time t but on the history of the voltage over the previous interval.
We have to assume that Z total and its real and imaginary components remain constant for the integral to be easy to calculate.

 

[Besoeker]

why would i integrate to find max energy from mag field, it happens at only two points for AC sin, 90 and 270, not the summation of instantaneous values over a period.

Hmmm = Ws anybody?

 

[wwhitney]

0<PF<1, take whatever period you need to express it in terms of energy per coloumb since both watts and amps have common denominator.

Joules/coulomb is just volts. So the energy delivered (= work required) in joules/coulomb is just cos(Φ)V, where I've switched from theta to phi to follow the physics convention.

as for the work required, you can verify cos(Φ)V²/Z when you have the math for J/coloumb

Since I've already derived cos(Φ)V²/Z, I'll just go the other way, take that expression and divide by I = V/Z to get the above.

Cheers, Wayne

 

[FionaZuppa]

Joules/coulomb is just volts. So the energy delivered (= work required) in joules/coulomb is just cos(Φ)V, where I've switched from theta to phi to follow the physics convention.

Cheers, Wayne

cos(Φ)VC ??

so if PF=0 the work required to push V/Z amps is zero?

 

[Carultch]

cos(Φ)VC ??

so if PF=0 the work required to push V/Z amps is zero?

On average, yes.

Instantaneously, in the case of a pure inductive load, it requires positive work to accelerate charges from rest, and negative work (work done by the energy storage elements onto the prime mover) to slow them down back to rest. Speed up, slow down, speed up in the other direction, slow down.

In the case of a pure capacitative load, it requires positive work to accumulate the charge, and negative work to manage the discharge. On net, this is zero work done by the prime mover.

 

[wwhitney]

cos(Φ)VC ??

If C is coulombs, that would be the work. You asked for work/coulomb, so that's cos(Φ)V.

Cheers, Wayne

 

[Carultch]

If C is coulombs, that would be the work. You asked for work/coulomb, so that's cos(Φ)V.

Cheers, Wayne

Typically it is a q that is used for representing charge, and a C for the unit of Coulombs.

 

[wwhitney]

Instantaneously, in the case of a pure inductive load, it requires positive work to accelerate charges from rest, and negative work (work done by the energy storage elements onto the prime mover) to slow them down back to rest. Speed up, slow down, speed up in the other direction, slow down.

So suppose we take a small single phase 60 Hz portable generator, and connect it to just an inductor, so the power factor is close to zero, and carefully measure the engine shaft rotational speed. Will the 120 Hz power oscillation show up as a 120 Hz oscillation in the shaft rotational speed? Or in some other observable mechanical quantity of the generator?

Cheers, Wayne

 

[GoldDigger]

So suppose we take a small single phase 60 Hz portable generator, and connect it to just an inductor, so the power factor is close to zero, and carefully measure the engine shaft rotational speed. Will the 120 Hz power oscillation show up as a 120 Hz oscillation in the shaft rotational speed? Or in some other observable mechanical quantity of the generator?

Cheers, Wayne

Yes.

 

[Besoeker]

Yes.

How so?

 

[wwhitney]

How so?

My theory was that while the engine has some rotational inertia, it is finite, and while the governor (is that the term?) will respond to fluctuations in power demand, it won't respond instantaneously, so the engine won't be able to precisely track the fluctuating power demand. So over a small fraction of a cycle, when the power demand slightly exceeds the power supplied, that would be reflected in a slight slowing of the shaft speed to the generator. Which should be observable if the shaft rotational speed is measured accurately enough.

But I'm not so confident I have the details correct in the above. And this assumes the engine is producing its power continuously, while in reality it is also running at a particular speed and producing its power in bursts.

Cheers, Wayne

 

[Besoeker]

My theory was that while the engine has some rotational inertia, it is finite, and while the governor (is that the term?) will respond to fluctuations in power demand, it won't respond instantaneously, so the engine won't be able to precisely track the fluctuating power demand. So over a small fraction of a cycle, when the power demand slightly exceeds the power supplied, that would be reflected in a slight slowing of the shaft speed to the generator. Which should be observable if the shaft rotational speed is measured accurately enough.

But I'm not so confident I have the details correct in the above. And this assumes the engine is producing its power continuously, while in reality it is also running at a particular speed and producing its power in bursts.

Cheers, Wayne

Do you mean bursts at power frequency i.e 60 or 50 Hz?

 

[wwhitney]

Do you mean bursts at power frequency i.e 60 or 50 Hz?

No, I was referring to the firing rate of the engine cylinders.

Cheers, Wayne

 

[GoldDigger]

My theory was that while the engine has some rotational inertia, it is finite, and while the governor (is that the term?) will respond to fluctuations in power demand, it won't respond instantaneously, so the engine won't be able to precisely track the fluctuating power demand. So over a small fraction of a cycle, when the power demand slightly exceeds the power supplied, that would be reflected in a slight slowing of the shaft speed to the generator. Which should be observable if the shaft rotational speed is measured accurately enough.

But I'm not so confident I have the details correct in the above. And this assumes the engine is producing its power continuously, while in reality it is also running at a particular speed and producing its power in bursts.

Cheers, Wayne

Correct.
The actual shaft speed would not be constant even with only a friction loading because of the uneven application of torque by the cylinders firing. The exact details of those minor speed fluctuations would depend on the number of revolutions per power stroke (two or four cycle), the number of cylinders, and their relative angle.
With a single phase generator the torque exerted by the windings on the rotating shaft will always be opposing the rotation and will peak at every peak in the current waveform. That would make the rotational torque a sinusoid with DC bias that touches zero every half cycle of the 60Hz waveform when the current crosses through zero and peaks at every half cycle when the current is at a maximum. That sinusoid will have a frequency of 120Hz (assuming US 60Hz power.)
Given that single phase generator, then the torque required to build up and decrease the current in the inductor would also present a sinusoidal torque which alternately opposes and aids the shaft rotation. The frequency of that extra torque would also be 120 Hz.

Besoeker's comment reminds me that if the generator is three phase with balanced loading on all phases the resistive component of the torque load will be constant over the full shaft rotation, unlike the single phase case.
The torque from the inductive load will also be at the second harmonic frequency, 120Hz, and will probably also cancel out across the three phases. Somebody should do that actual calculation to confirm that.

 

[Besoeker]

No, I was referring to the firing rate of the engine cylinders.

Cheers, Wayne

I'm pretty sure that the inertia of the prime mover, flywheel, and alternator make that insignificant otherwise you wouldn't get a clean sine wave.

 

[FionaZuppa]

How so?

I'm pretty sure that the inertia of the prime mover, flywheel, and alternator make that insignificant otherwise you wouldn't get a clean sine wave.

inductive kickback

gen pushes energy into reactive component, reactive component then pushes back. at 60Hz the push + pushback creates 120 "pulses" per sec.

gool for it, shaft sensors can see/monitor amplitude of the push and kickback. on avg the gen pushes (Push), to push into inductive load it does Push+more, when inductive load kicks back the gen does Push-more.

 

[Besoeker]

inductive kickback

gen pushes energy into reactive component, reactive component then pushes back. at 60Hz the push + pushback creates 120 "pulses" per sec.

gool for it, shaft sensors can see/monitor amplitude of the push and kickback. on avg the gen pushes (Push), to push into inductive load it does Push+more, when inductive load kicks back the gen does Push-more.

Have you measured the quality of the sinewave?
Carried out a harmonic analysis to determine its integrity?

 

[wwhitney]

With a single phase generator the torque exerted by the windings on the rotating shaft will always be opposing the rotation and will peak at every peak in the current waveform.

The torque peaks at a peak in the current waveform, not at a peak in the electrical power waveform? When the phase shift between current and voltage is non-zero, the two peaks will differ.

The torque from the inductive load will also be at the second harmonic frequency, 120Hz, and will probably also cancel out across the three phases. Somebody should do that actual calculation to confirm that.

As long as the non-constant portion of the torque from each phase is a sinewave, and the three sinewaves are 120 and 240 degrees out of phase, they will cancel. The frequency doubling isn't an issue.

Cheers, Wayne