Power factor

[mbrooke]

understand it conceptually, not mathematically
basically electric (caps leading) and magnetic (induct lagging) fields rising/collapsing in an oscillating manner between source/sink
basically required to 'charge' the system, perform no real work, but still must be supplied

Well I view it like this... correct me if wrong... when the cap or reactor is "charging" it draws energy in terms of KW slowing down the generator, however it is then sent back speeding up the generator as what energy was drawn out is now being put back. Kind of like pumped hydro, but instead of drawing and returning energy over hours it takes place over a cycle. Would this be a correct analogy?

 

[mbrooke]

the rotor speed is constant, synched to the frequency or the prime mover
the rotor and the field relationship varies based on load/pf

But load slows down the rotor, so when a capacitor charges in its half cycle, the rotor must experience some drag. When the polarity reverses that energy is now sent back speeding up to rotor.

 

[winnie]

But load slows down the rotor, so when a capacitor charges in its half cycle, the rotor must experience some drag. When the polarity reverses that energy is now sent back speeding up to rotor.

If this were a single phase generator then you _might_ see some sort of pulsing torque as energy is stored and returned. You are correct that energy is shuttling between the capacitor and the generator; what is not clear is if that energy is stored _mechanically_ in the rotor, or stored in the magnetic field of the generator.

-Jon

 

[mbrooke]

If this were a single phase generator then you _might_ see some sort of pulsing torque as energy is stored and returned.

Why would a 3 phase generator be exempt? If a cap is only connected across one phase and neutral I could see the same happening.

You are correct that energy is shuttling between the capacitor and the generator; what is not clear is if that energy is stored _mechanically_ in the rotor, or stored in the magnetic field of the generator.

-Jon

Million dollar question. Is it known?

 

[Ingenieur]

If this were a single phase generator then you _might_ see some sort of pulsing torque as energy is stored and returned. You are correct that energy is shuttling between the capacitor and the generator; what is not clear is if that energy is stored _mechanically_ in the rotor, or stored in the magnetic field of the generator.

-Jon

This^^^

pf is 0 so no real work is done
no load on rotor, hence no force or change in speed
you do not increase fuel or throttle due to VAr
maybe abit due to IR losses but moot in the scope of things

only when real work like heating or mechanical conversion is done do you need to throw another log on the fire

 

[mbrooke]

This^^^

pf is 0 so no real work is done
no load on rotor, hence no force or change in speed
you do not increase fuel or throttle due to VAr
maybe abit due to IR losses but moot in the scope of things

only when real work like heating or mechanical conversion is done do you need to throw another log on the fire

But for that brief moment during the first half cycle you are indeed useing KW. What is I then take the capacitor at this exact instant and use it to light an LED? Do I get free energy?

 

[winnie]

mbrooke,

There clearly are high frequency components in generator torque; the torque output of something like a diesel engine is not constant, the magnetic torque that the generator sees is not constant, etc. The flywheel is supposed to damp these things out; but if these torques hit a mechanical resonance in the system.....

It is very likely that putting a pure capacitive load on a generator will change the torque spectrum.

A google search for 'generator torque spectrum' or 'generator air gap flux fourier analysis' pulls up too many possibly interesting papers for me to quickly summarize, and they quickly jump beyond my level of experience.

-Jon

 

[mbrooke]

mbrooke,

There clearly are high frequency components in generator torque; the torque output of something like a diesel engine is not constant, the magnetic torque that the generator sees is not constant, etc. The flywheel is supposed to damp these things out; but if these torques hit a mechanical resonance in the system.....

It is very likely that putting a pure capacitive load on a generator will change the torque spectrum.

A google search for 'generator torque spectrum' or 'generator air gap flux fourier analysis' pulls up too many possibly interesting papers for me to quickly summarize, and they quickly jump beyond my level of experience.

-Jon

I will check it out. But, can't to much reactive or capacitive load damage a generator for this exact reason?

 

[Ingenieur]

But for that brief moment during the first half cycle you are indeed useing KW. What is I then take the capacitor at this exact instant and use it to light an LED? Do I get free energy?

you are not, you are storing VAr (which can be discharged into a real load)

you are charging the cap
over 1 cycle net power is zero
when you put it across the led (or a resistor) you are discharging it

an ideal C or L does not consume real power
the phase angle is 90 deg (+/-, lead or lagging)
pf = cos (90 deg) = 0 or the inverse Q (VAr) = sin (90 deg) = 1
real power (watts) = pf x VA = 0
reactive power = sin 90 x VA = 1 x VA = VAr

 

[mbrooke]

you are not, you are storing VAr (which can be discharged into a real load)

you are charging the cap
over 1 cycle net power is zero

Over one cycle. Half cycle is consuming Watts, actual work.

when you put it across the led (or a resistor) you are discharging it

Correct, and actual watts lights that LED or heats the resistor.

an ideal C or L does not consume real power
the phase angle is 90 deg (+/-, lead or lagging)
pf = cos (90 deg) = 0 or the inverse Q (VAr) = sin (90 deg) = 1
real power (watts) = pf x VA = 0
reactive power = sin 90 x VA = 1 x VA = VAr

Correct, when the same power is sent back on the other half cycle.

 

[Ingenieur]

Over one cycle. Half cycle is consuming Watts, actual work.

Correct, and actual watts lights that LED or heats the resistor.

Correct, when the same power is sent back on the other half cycle.

it is absorbing (transferring) VA, not consuming watts
it is returned to the system next 1/2 cycle, not expended/consumed as work (power/time)

 

[mbrooke]

it is absorbing (transferring) VA, not consuming watts
it is returned to the system next 1/2 cycle, not expended/consumed as work (power/time)

Look at it like this. If I subjected the capacitor to only a 1/2 cycle (charge) and then discharged the cap through a light bulb to make it light, during that half cycle did I not have the generator do work?

 

[Ingenieur]

Look at it like this. If I subjected the capacitor to only a 1/2 cycle (charge) and then discharged the cap through a light bulb to make it light, during that half cycle did I not have the generator do work?

no

what would the voltage be across the cap at 1/2 cycle?

 

[winnie]

This starts to be a terminology debate.

The concept of VAr _requires_ that you only consider periods of an integral number of complete half cycles. It is meaningless to discuss VAr over part of a cycle. When discussing _time averaged_ power in this fashion, then you can differentiate between W and VAr. In this case you can use 'complex' numbers to do math on the values, if you consider W as 'real' and VAr as 'imaginary'. But in this case, the terms 'real' and 'imaginary' are describing the _numbers_, not the energy being accounted for.

Over a complete half cycle, 'real' power gets delivered to a load, whereas the 'imaginary' or 'reactive' power represents energy shuffling back and forth to the load, with no net delivered.

If you want to discuss power over periods smaller than a cycle, the 'VAr'/'real'/'reactive' no longer applies. You need to graph instantaneous power, meaning a graph of instantaneous volts times instantaneous amps.

The product of volts and amps could really do work. But this is not 'real' in the 'real versus imaginary' sense. If you look at a graph of instantaneous power, then for a resistive load you will see that the instantaneous power is always positive, and for a reactive load you will see that the instantaneous power is sometimes negative.

-Jon

 

[winnie]

mbrook,

You keep saying 1/2 cycle, but I think you really mean 1/4 cycle. Energy flows into the capacitor for 1/4 cycle, then out for 1/4 cycle, then in again.

-Jon

 

[Ingenieur]

mbrook,

You keep saying 1/2 cycle, but I think you really mean 1/4 cycle. Energy flows into the capacitor for 1/4 cycle, then out for 1/4 cycle, then in again.

-Jon

yep, power cycles at 2 x f
better to use degrees or radians
for a cap (or L)
90 deg 1/4 cycle V= peak, what is current?

 

[mbrooke]

no

what would the voltage be across the cap at 1/2 cycle?

The peak of the sine wave.

Edit. You got me It would not be the peak, it would be Im guessing half.

 

[mbrooke]

mbrook,

You keep saying 1/2 cycle, but I think you really mean 1/4 cycle. Energy flows into the capacitor for 1/4 cycle, then out for 1/4 cycle, then in again.

-Jon

My mistake, yes, 1/4 of the cycle. As the voltage drops the current starts to flow in the opposite direction back into the generator.

 

[Ingenieur]

The peak of the sine wave.

Edit. You got me It would not be the peak, it would be Im guessing half.

at 1/4 or 90 deg V is peak
but I is 90 deg out of phase = 0
power = V x I = peak x 0 = 0

 

[Sahib]

If you switch on a capacitor and then switch it off. the elapsed time may not be an integral multiple of 60 hz cycles. In that case some energy supplied by source is stored in capacitor with which you may do some useful work such as lighting a LED. But this does not work with an inductor. Why?