THE PHYSICS OF... POWER

[wwhitney]

Q is reactive power, and I don't know that it makes sense to consider it instantaneously.

Sorry, that's not correct, we can look at the difference between the total power waveform and the I²R contribution to it. But it's late here, so I'll expand on that tomorrow if I have a chance.

Cheers, Wayne

 

[Besoeker]

were you mocking use of "sin"?

Yes.

 

[GoldDigger]

Ok, dumb it down for me. What is dynamic torque?

It was an analogy to the distinction between "dead" or static load on a roof or floor versus the "live" or dynamic load from people or things moving, jumping, etc. on that same roof or floor. I was trying to indicate the fact that the resistive portion is a steady, or at least unidirectional, torque on the shaft, similar to a static load while the reactive portion alternates in direction with an amplitude that might be comparable to the average torque, hence dynamic.

 

[mbrooke]

It was an analogy to the distinction between "dead" or static load on a roof or floor versus the "live" or dynamic load from people or things moving, jumping, etc. on that same roof or floor. I was trying to indicate the fact that the resistive portion is a steady, or at least unidirectional, torque on the shaft, similar to a static load while the reactive portion alternates in direction with an amplitude that might be comparable to the average torque, hence dynamic.

Makes more sense. So PF basically causes the generator to oscillate back and forth.

 

[FionaZuppa]

so power follows I, as in in-phase with I ?? yes?

you know VA is just a special case of I²Z, yes? in better terms, VA(t)=I(t)²Z

you can call Z complex if you like, but with no X then Z=R, just another special case of Z :thumbsup:

so you agree, power(t) (be it real or imaginary) happens as a function of I(t) ??

 

[GoldDigger]

so power follows I, as in in-phase with I ?? yes?

you know VA is just a special case of I²Z, yes? in better terms, VA(t)=I(t)²Z

you can call Z complex if you like, but with no X then Z=R, just another special case of Z :thumbsup:

so you agree, power(t) (be it real or imaginary) happens as a function of I(t) ??

Not really.
Unless you decide that the sign of the power transfer does not matter and you ignore the fact that the rate of energy transfer into magnetic field is proportional to dI/dt and not the instantaneous I(t).
It is, indirectly, a function of I(t), but not a function only of the instantaneous magnitude of I at that moment in time.
As a very simple counterexample, resistive power transfer is a function of I and R and the same function works equally well for constant DC as for AC.
But for constant DC I'(t) = dI(t)/dt =0. Which does not mean that there is no energy stored in the inductor but rather that the stored energy is not changing and so there is no power at all at time t.

If you want to get argumentative about it, you can try saying the dI/dt for a sine wave is directly proportional to the amplitude and the frequency of I itself, just phase shifted. I do not see that as adding anything to the understanding of the situation though.

 

[Besoeker]

Makes more sense. So PF basically causes the generator to oscillate back and forth.

You know that ain't so.

 

[iwire]

So PF basically causes the generator to oscillate back and forth.

You know that ain't so.

I don't know about that, I wired up a genset just like this one and watched the entire unit oscillate for a couple couple of hours.

Of course it may have more to do with the concert lighting rig it was running than power factor. Still it was fun to watch it bounce around in time with the music.

 

[Besoeker]

I don't know about that, I wired up a genset just like this one and watched the entire unit oscillate for a couple couple of hours.

Of course it may have more to do with the concert lighting rig it was running than power factor. Still it was fun to watch it bounce around in time with the music.

But not at 60Hz.

 

[iwire]

But not at 60Hz.

Very true.

 

[mike_kilroy]

The extra torque is generally not a problem. It averages out to zero, BUT if most of the flywheel mass is on the prime mover side rather than the generator it would increase the dynamic torque on the coupling between them. That could be a problem even for a short term overcurrent, whether resistive or reactive. (e.g.stopping an engine driven generator by shorting the generator output is not a good idea. :happysad

Working with torque, couplings, response times, every day, I will guarantee you without having to do the math (T=jw/t) there is no significant increased torque load, dynamic or otherwise, on that coupling.

Inertia J is way big, the speed change w could only be really really really small (+/-.1rpm [.01rad/sec] guess) and time of 1/4 of 6msec is way to small to effect a change.

 

[FionaZuppa]

Not really.
Unless you decide that the sign of the power transfer does not matter and you ignore the fact that the rate of energy transfer into magnetic field is proportional to dI/dt and not the instantaneous I(t).
It is, indirectly, a function of I(t), but not a function only of the instantaneous magnitude of I at that moment in time.
As a very simple counterexample, resistive power transfer is a function of I and R and the same function works equally well for constant DC as for AC.
But for constant DC I'(t) = dI(t)/dt =0. Which does not mean that there is no energy stored in the inductor but rather that the stored energy is not changing and so there is no power at all at time t.

If you want to get argumentative about it, you can try saying the dI/dt for a sine wave is directly proportional to the amplitude and the frequency of I itself, just phase shifted. I do not see that as adding anything to the understanding of the situation though.

no power in DC? what about 0 to I_max
for AC max mag field is at I_max of any period between zero crossings, yes, you agree? regardless of differential, the mag field energy follows the I curve, yes, you agree?

my argument is made, power # follows amps, power is tied to I² , this is how you find your P & Q #'s, then S via vector sum of P&Q, yes?

Inertia J is way big, the speed change w could only be really really really small (+/-.1rpm [.01rad/sec] guess) and time of 1/4 of 6msec is way to small to effect a change.

isnt .1rpm = 0.00265rad/sec ??

 

[wwhitney]

so power follows I, as in in-phase with I ?? yes?

No. Here's a simple graph of sin(x) (e.g. V), sin(x+pi/2) (e.g. I), and 2sin(x)sin(x+pi/2) (e.g. 2*P). I scaled the product by 2 so that it stands out more. You can see that the power fluctuates at twice the frequency of the current or voltage, and that its peaks are offset from those of either current or voltage.

Cheers, Wayne

 

[wwhitney]

and the power dissipated by R is I(t)²R = real watts = power(t)

Agreed, when treating I(t) as real (not complex).

the Q is I(t)²X_L = not consumed watts = no power(t)

When I(t) is real, it is Q = I(t)I(t+pi/2ω)X. See below for a long answer.

Cheers, Wayne


For the moment, let's keep everything real and avoid any complex numbers.

Any single frequency AC waveform will be of the form Acos(ωt-φ) for some amplitude A (but RMS amplitude A/sqrt(2)), some cyclic frequency ω, and some phase angle φ. If we are working with a fixed frequency, then the waveform is determined by A and φ, two real numbers (subject to some limitations).

For a circuit with just a voltage source and a resistance R, the current waveform will be in phase with V, and I(t) = V(t)/R.

For a circuit with just a voltage source and an inductive reactance X, the current waveform will lag V by φ = pi/2 radians (90 degrees), and I(t) = V(t-φ/ω)/X.

For a circuit with a voltage source and a series resistance R and inductive reactance X, we need to define the impedance magnitude Z = sqrt(R^2 + X^2) and the impedance phase shift φ = arctan(X/R). Then the current waveform will be I(t) = V(t-φ/ω)/Z.

Now for this case we can ask what the voltage drop and power dissipation are across the resistance R. When the current I(t) flows through the resistance R it creates a voltage drop V_R(t)=I(t)R. The power dissipation is P_R(t) = I(t)V_R(t) = I²(t) R. This term is strictly nonnegative, and it peaks when I(t) peaks.

How about the inductive reactance X? We know the voltage drop across it leads the current by pi/2 radians, so V_X = I(t+pi/2ω)X. The power flow across the inductive reactance is P_X(t) = I(t)V_X(t) = I(t)I(t+pi/2ω)X. This term has average value 0, and it peaks pi/4 radians before I(t) does.

 

[mike_kilroy]

isnt .1rpm = 0.00265rad/sec ??

Not last time I looked. .1rev/min * (2*pi radians/rev) * (min/60 sec) = .01 rad/sec (I just round the 1/9.55 conversion factor since the .1 was a swag anyway).

 

[FionaZuppa]

Not last time I looked. .1rev/min * (2*pi radians/rev) * (min/60 sec) = .01 rad/sec (I just round the 1/9.55 conversion factor since the .1 was a swag anyway).

1 rev / min = 2pi rad / min

drop the time, really not needed now

1 rev = 2pi rad

.1 rev = .2pi rad

.1 rev / min * 1min/60sec = .2pi rad / min * 1min/60sec

.1rev/60s = .2pi rad / 60sec

0.00166rev/sec = 0.00333 pi rad/sec

drop the sec

0.00166rev = 0.00333 pi rad

0.00166rev/(0.00333 pi) = rad = 0.15867

did my math go south somewhere ?

 

[mike_kilroy]

1 rev / min = 2pi rad / min

drop the time, really not needed now

1 rev = 2pi rad

.1 rev = .2pi rad

.1 rev / min * 1min/60sec = .2pi rad / min * 1min/60sec

.1rev/60s = .2pi rad / 60sec

0.00166rev/sec = 0.00333 pi rad/sec

drop the sec

0.00166rev = 0.00333 pi rad

0.00166rev/(0.00333 pi) = rad = 0.15867

did my math go south somewhere ?

AH, YAH...

What are you trying to prove with your math? you seem to be trying to prove that 1 radian = .15867 rev?

why?

we all KNOW 1 rad= .15867 revs, or more commonly, 1 rev = 2pi radians

You can't just arbitrarily throw out time right and left! time sometimes (no pun intended) is important! Like in SPEED.

I SHOWED you the math to convert RPM to rad/sec. If you don't believe me, go to one of the 1,000,000 convert sites on line and enter the values yourself. I am not sure why you want to argue basic stuff so much! I only argue with your 'facts' you put up that are blatantly wrong.

 

[FionaZuppa]

Agreed, when treating I(t) as real (not complex).


When I(t) is real, it is Q = I(t)I(t+pi/2ω)X. See below for a long answer.

so two different amps at the same time? how is this shown in post 753? the amps waveform is shown as V(t)/Z for amplitude shifted by phi.

AH, YAH...

my bad on this one :thumbsup:

 

[GoldDigger]

so two different amps at the same time?

No. One of those terms represents the voltage across the load (which can be expressed as I times X where there is a phase offset from the actual current.) The other represents the current itself.

 

[wwhitney]

how is this shown in post 753?

See the last paragraph of that post.

Cheers, Wayne