THE PHYSICS OF... POWER

[Besoeker]

Here's a hint... it ain't the average of the 3 individual phase PF's!

Phil

So, how would you do it? And how would instantaneous values feature?
Which was my point.

 

[FionaZuppa]

Leaving starting out of consideration and assuming a 100% efficient motor, so that the mechanical output power equals the electrical input power:

The engine (prime mover of the generator) needs to be able to produce 1kW continuously if you need the motor to work at full rated load. The one answer you ask for is (1).
The generator itself needs to source 1kW but also needs to be rated for 1.25kVA output (continuous).

100% efficient motor? so a motor with no R value, purely inductive? thus the ideal motor is an ideal inductor w/ fully loaded output (sounds like a xfrmer), and when unloaded the poco screams that their wires are melting. just cant win when it comes to power generation vs power consumption.

 

[GoldDigger]

100% efficient motor? so a motor with no R value, purely inductive? thus the ideal motor is an ideal inductor w/ fully loaded output (sounds like a xfrmer), and when unloaded the poco screams that their wires are melting. just cant win when it comes to power generation vs power consumption.

If you do not specify an efficiency for your motor the question cannot be answered, since you list only the motor output.
I could have picked any efficiency and given the corresponding answer. I chose 100%.

If the efficiency of the motor is given by "e" and expressed in percent, then the power needed for the prime mover is 100/e kW and the generator must be able to source (100/e)(1.25) kVA.
Does that make you happier? The question still can not be answered by any of your four choices.

And, no a perfectly efficient motor would have no lumped constant R component but would still present a mostly resistive load to the generator because of what I will call the dynamic resistance, R_d, which converts input electrical energy into mechanical work. That effective R_d varies with the motor load.
An inductor with "a fully loaded output" is a very unfortunate way to try to describe it. It is in a sense a transformer though, in that it transforms electrical energy into mechanical energy, the input current depends on the output loading, and there is a purely reactive idling current when there is no load.

 

[FionaZuppa]

If you do not specify an efficiency for your motor the question cannot be answered, since you list only the motor output.
I could have picked any efficiency and given the corresponding answer. I chose 100%.

If the efficiency of the motor is given by "e" and expressed in percent, then the power needed for the prime mover is 100/e kW and the generator must be able to source (100/e)(1.25) kVA.
Does that make you happier? The question still can not be answered by any of your four choices.

And, no a perfectly efficient motor would have no lumped constant R component but would still present a mostly resistive load to the generator because of what I will call the dynamic resistance, R_d, which converts input electrical energy into mechanical work. That effective R_d varies with the motor load.
An inductor with "a fully loaded output" is a very unfortunate way to try to describe it. It is in a sense a transformer though, in that it transforms electrical energy into mechanical energy, the input current depends on the output loading, and there is a purely reactive idling current when there is no load.

hold up, isnt efficiency tied directly to kVAR and R ? if my motor is just a big pure R then the e = 0 and kVAR = 0 (no mag field). in both cases each can have same amps thus looks like same load to the generator. if i have giant kVAR and zero R then its pure kVAR but there is no work done when there is no load on the motor (R_d=0). if i have just kVAR with R=0 then i am not wasting any watts, all the input in converted to output via mag coupling at a function R_d (R_d=0 then e=0 = no work done anywhere).

V_rms÷(2∙π∙f∙L) is free wheel motor amps when R=0, R_d=0, and there are no mechanical losses.


how does kVAR affect input power to gen, like this
it takes a little more work to push charge into inductor, but pure reactance will aid during the other half cycle which requires just a tad less work by the input engine to the gen. over time the tad more and tad less avg's out to a "rms" power #. this input jitter can be seen on input shafts using special monitoring devices, the more kVAR there is the bigger the amplitude of the jitter.

 

[Phil Corso]

The engine (prime mover of the generator) needs to be able to produce 1kW continuously if you need the motor to work at full rated load. The one answer you ask for is (1).
The generator itself needs to source 1kW but also needs to be rated for 1.25kVA output (continuous).

Assuming the motor, generator, and engine are perfect machines (meaning zero-losses) the generator providing 1kW @ 0.8pf to the motor, must also be rated 1kW @ 0.9pf, which equals 1.25kVA! However, the engine driving the generator must be rated 1kW, corresponding to 0.746Hp!

Phil

 

[GoldDigger]

hold up, isnt efficiency tied directly to kVAR and R ? if my motor is just a big pure R then the e = 0 and kVAR = 0 (no mag field). in both cases each can have same amps thus looks like same load to the generator. if i have giant kVAR and zero R then its pure kVAR but there is no work done when there is no load on the motor (R_d=0). if i have just kVAR with R=0 then i am not wasting any watts, all the input in converted to output via mag coupling at a function R_d (R_d=0 then e=0 = no work done anywhere).

In a word, NO.

The efficiency of a motor is by definition the ratio of output mechanical power to input electrical power.
The reactive portion of the input current does not correspond to input power. Period.
If there is no input and no output, then the efficiency is undefined and cannot be measured. If there is input an no useful output, then the efficiency is zero.

The efficiency losses resulting from bearing friction, air friction, and inductive losses in the rotor and stator material do not correspond to a measurable static R in the stator windings, but they are still there. And they can reduce the efficiency without reducing the power factor of the motor, since they still represent a purely resistive load.

BTW, R_d would be considered a parallel not a series resistance, so Rd = 0 is not a good thing. The efficiency would indeed be zero until the breaker opened, then it would once again be undefined. :angel:

 

[FionaZuppa]

In a word, NO.

The efficiency of a motor is by definition the ratio of output mechanical power to input electrical power.
The reactive portion of the input current does not correspond to input power. Period.
If there is no input and no output, then the efficiency is undefined and cannot be measured. If there is input an no useful output, then the efficiency is zero.

The efficiency losses resulting from bearing friction, air friction, and inductive losses in the rotor and stator material do not correspond to a measurable static R in the stator windings, but they are still there. And they can reduce the efficiency without reducing the power factor of the motor, since they still represent a purely resistive load.

BTW, R_d would be considered a parallel not a series resistance, so Rd = 0 is not a good thing. The efficiency would indeed be zero until the breaker opened, then it would once again be undefined. :angel:

the only way to generate a mag field is via inductive kVAR. electrical amps to mag field to output power. so how is efficiency not tied to kVAR ?? please explain.

Assuming the motor, generator, and engine are perfect machines (meaning zero-losses) the generator providing 1kW @ 0.8pf to the motor, must also be rated 1kW @ 0.9pf, which equals 1.25kVA! However, the engine driving the generator must be rated 1kW, corresponding to 0.746Hp!

Phil

what? math please.

 

[GoldDigger]

the only way to generate a mag field is via inductive kVAR. electrical amps to mag field to output power. so how is efficiency not tied to kVAR ?? please explain.

There may not be a motor with zero kVAR, but that has nothing to do with the efficiency of the motor.
An ideal (100% efficient) motor can have almost any kVAR and any PF you choose to design it for. The magnetic field is still there whether it is generating output power or not.
Also, if you use a permanent magnet rotor, the inductive part of the input current can be very small, even though a magnetic field is generated by the stator coils.

 

[FionaZuppa]

There may not be a motor with zero kVAR, but that has nothing to do with the efficiency of the motor.

no inductive kVAR then you cannot have a motor (as we typically know it), period.

if there is no inductive component and the generator supplies 240Vac_rms @ 100A then what do you have, nothing but a heater and a "motor" with zero efficiency if we use your definition of Power_out(motor)/Power_in(motor)


a good efficient motor has no R and lots of inductive kVAR, and when you load the motor the kVAR turns into watts. if we want to use the basic nomenclature, potential to kinetic energy, etc.

 

[Phil Corso]

what? math please.

EE-0001 when I went to school! Am I wrong in assuming you are an EE?

Phil

 

[Phil Corso]

Assuming the motor, generator, and engine are perfect machines (meaning zero-losses) the generator providing 1kW @ 0.8pf to the motor, must also be rated 1kW @ 0.9pf, which equals 1.25kVA! However, the engine driving the generator must be rated 1kW, corresponding to 0.746Hp! Phil

Sorry Fiona...

My oops. (Most feared word in an operating room!) PF should have been 0.8!!! Beesoaker must have been away from his post!

Phil

 

[FionaZuppa]

EE-0001 when I went to school! Am I wrong in assuming you are an EE?

Phil

you are wrong phil, me a nuclear engineer and my early days of study were EE and CE, but those days have come and gone for me :thumbsup:
but my question is still valid, pls show the math from your EE-0001 class.

edit: i see from your last post that you corrected your statement.

and as a side note, kVAR is like a short to a generator, take a pure inductor and lower the frequency, amps will approach infinity.

 

[FionaZuppa]

for some this will look familiar and how it relates to the vector components of electrical power

 

[dionysius]

for some this will look familiar and how it relates to the vector components of electrical power

Fiona......please elaborate what the point is here.......

 

[Carultch]

you are wrong phil, me a nuclear engineer and my early days of study were EE and CE, but those days have come and gone for me :thumbsup:
but my question is still valid, pls show the math from your EE-0001 class.

edit: i see from your last post that you corrected your statement.

and as a side note, kVAR is like a short to a generator, take a pure inductor and lower the frequency, amps will approach infinity.

I made a spreadsheet that goes through the instantaneous current, voltage and power, for the situation in question. A 1kW motor on a 120V circuit, with an 0.8 power factor. Voltage/nominal voltage, current/nominal current, and power/nominal power. I can't seem to upload the original file, but you can email me and I'll send it. carultch@gmail.com.

Here's a graph of the results.


Observe that the power peaks at 2.25 kW, and has a minimum of -0.25kW. On average, it is 1kW.

This means that the prime mover has to supply up to 2.25 kW at a certain point in the cycle, but it also supplies considerably less throughout the rest of the cycle. In fact, for a brief period, the prime mover becomes a load of the circuit. If this were a rotating generator, it would temporarily behave as a motor during this part of the cycle. On net, it delivers the equivalent energy of a steady 1kW supply over a complete cycle.

This is the consequence of a reactive load on a power source. If there were no net reactance, the peak power would only be as high as 2 kW, and there would be no period when the prime mover becomes a circuit load. Due to the reactive load, the prime mover has to be able to shift its power contribution and the amplitude of the power variation is greater than what it would be for its pure resistive counterpart.

 

[FionaZuppa]

Fiona......please elaborate what the point is here.......

just a graph of what a generator may look like when kVAR becomes very large. no need to get any deeper than that, beyond the scope of this discussion.

the 1st few posts of this thread basically summarize "PF", "kVAR", "VA", "kW", but how varying these things affect generator and transmission is a topic for another day, etc.

think of kVAR like a tiny steam driven piston & wheel, except no steam and the piston is just a closed tube with sealed piston (no leaks), you turn the wheel by hand and in doing so it gets harder and harder as the piston compresses the air in the tube, but once you turn past the end of the stroke the energy that went into compression of air now is returned back to you. so instead of your hand turning the wheel you use a small fixed methane engine to turn the wheel, what happens? as long as there is no loss of energy in the wheel/piston you end up with cyclic pos and neg spikes of input power, but averaged over time the spikes seem to vanish, and, the net power consumed should sum to zero (hence, no loss of energy). want more kVAR from the wheel/piston example, then increase area and stroke of piston, etc. but, we live in real word of physics, not just math :thumbsup:

 

[FionaZuppa]

a capacitive motor ?? or the labels were switched?

 

[GoldDigger]

a good efficient motor has no R and lots of inductive kVAR, and when you load the motor the kVAR turns into watts. if we want to use the basic nomenclature, potential to kinetic energy, etc.

That is a very simplistic and very misleading statement.

First, it implies that the eventual wattage cannot be greater than the unloaded kVAR. (It does not say that, but I am afraid some will draw that conclusion.)
Second, it implies that the kVAR goes down as the motor is loaded. In fact the PF improves as the motor is loaded mainly because the resistive component goes up. The kVAR does not need to go down for the PF to improve (although it might).

And the analogy to the relation between potential and kinetic energy is just flat out wrong and therefore not particularly useful in understanding what is happening.

 

[FionaZuppa]

[/B]

That is a very simplistic and very misleading statement.

First, it implies that the eventual wattage cannot be greater than the unloaded kVAR. (It does not say that, but I am afraid some will draw that conclusion.)
Second, it implies that the kVAR goes down as the motor is loaded. In fact the PF improves as the motor is loaded mainly because the resistive component goes up. The kVAR does not need to go down for the PF to improve (although it might).

And the analogy to the relation between potential and kinetic energy is just flat out wrong and therefore not particularly useful in understanding what is happening.

take a 1:1 xfrmer that is pure inductive kVAR on both coils (no R), the device is 100 efficient. the secondary is open ckt (no load), jX_primary=1ohm, voltage = 100Vac_rms

its 100% kVAR, right? my generator is moving 100amps yet no power is dissipated anywhere, charge is just free-wheeling around the primary ckt at a rate of 100 coulombs/sec

now connect a 1ohm power resistor to the secondary, does the primary amps go up, down, or remain the same ?


if i have a fixed physical motor with Z_vector=R_constant+jX and its free wheeling, i then put load on output shaft, how exactly does that make R go up ??

 

[Phil Corso]

Fiona...

"Turning kVAr into kW!"

I agree with GoldDigger. Your interpretation is way-off! Suggest you refresh your knowledge of PF! BTW, I did offer to provide a simple explanation, in my original post, but subsequent posters showed zero interest.

If you would like a copy contact me off-forum!

Phil