Power factor

[mbrooke]

It is common industry and educational terminology to refer to the sinusoidal component as "AC" and the constant (or decaying or rising) component as "DC".

Don't have time at the moment to read the thread in detail but from scanning the latest it seems for the others to deny that common terminology is just them being argumentative. It is a very, very long-standing use of terminology.

Wouldn't falling be a better word than decaying? I think of decaying as something discharging like a capacitor, not a generator switching direction in polarity. I am not trying to be picky, just a bit confused at the terminology, or at least how to correctly apply it.

 

[Smart $]

Brain fart. What is RMS vs the total power in a wave forum?

The RMS value is significant for AC (sinusoidal) voltage and current because it represents the equivalent DC voltage and current. When we analyze the AC power waveform, it is the mean value that represents the DC steady-state equivalent... not the RMS value.

Let's use an example:

AC sinusoidal circuit
±1.4142 volt peaks
±1.4142 ampere peaks
Current in phase with voltage (i.e. pf 1)

We know the RMS value to be the peak divided by the square root of 2... so 1V_RMS, 1A_RMS. With a power factor of 1, the circuit consumes 1 watt.

Now when considering the power waveform, P=E*I*pf, the low values are zero at the voltage and current zero crossings, and the positive peaks are 1.4142V*1.4142A which is 2W_{INSTANTANEOUS}. Doing an RMS calculation on the power waveform, the result is 1.5 (as calculated in Excel). Not a significant value is it?

 

[Ingenieur]

Wouldn't falling be a better word than decaying? I think of decaying as something discharging like a capacitor, not a generator switching direction in polarity. I am not trying to be picky, just a bit confused at the terminology, or at least how to correctly apply it.

He's talking about the DC component, not the ac, but it could apply to both since the sum is the total signal

Decaying is the accepted terminology
the rate is e^kt of some form

 

[Ingenieur]

So RMS I times RMS V = RMS watts (actually work exerted)? I say this because most EE books have actual work as a complex equation, at least for 3 phase power.

Watt is power, not work
power = work/time

S = P + j Q
total power S = complex(real power P + reactive power Q)

 

[mbrooke]

He's talking about the DC component, not the ac, but it could apply to both since the sum is the total signal

Decaying is the accepted terminology
the rate is e^kt of some form

This is me being picky, but Id never use the word decaying for a reversal in current flow depicted across a graph.

 

[mbrooke]

Watt is power, not work
power = work/time

S = P + j Q
total power S = complex(real power P + reactive power Q)

I know, but but isn't a watt a unit of measurement of power expressed over time? Unless you mean that brief period of time when work is being used to push reactive power into say a capacitor?

EDIT! :slaphead: I see I used the term work to express watts. My mistake. I sometimes will use terms interchangeably when they can have a very different meaning.

 

[Smart $]

I know, but but isn't a watt a unit of measurement of power expressed over time?

No. The measure of power over time is energy. The electrical industry uses the kilowatt-hour unit.

 

[mbrooke]

No. The measure of power over time is energy. The electrical industry uses the kilowatt-hour unit.

Power (work) over a deified period of time (1 second) is watts, right?

watt

wät/

noun
noun: watt; plural noun: watts; symbol: W

• the SI unit of power, equivalent to one joule per second, corresponding to the power in an electric circuit in which the potential difference is one volt and the current one ampere.

 

[mbrooke]

Joule:

joule


The dimension of power is energy divided by time. The SI unit of power is the watt (W), which is equal to one joule per second. Other units of power include ergs per second (erg/s), horsepower (hp), metric horsepower (Pferdestärke (PS) or cheval vapeur (CV)), and foot-pounds per minute.

 

[Smart $]

Power (work) over a deified period of time (1 second) is watts, right?

Joule:

Please visit:
http://www.diffen.com/difference/Energy_vs_Power

 

[wwhitney]

So RMS I times RMS V = RMS watts (actually work exerted)?

No, if I and V are in phase and of the same shape, then RMS I * RMS V = average power. There's no RMS on the power, the point of the RMS on the I and the V is so that you can get the true average power.

Cheers, Wayne

 

[mbrooke]

No, if I and V are in phase and of the same shape, then RMS I * RMS V = average power. There's no RMS on the power, the point of the RMS on the I and the V is so that you can get the true average power.

Cheers, Wayne

Sinking in I guess whats holding me back is that my terminology has always been incorrect.

 

[mbrooke]

Please visit:
http://www.diffen.com/difference/Energy_vs_Power

Thanks But are the terms I posted at least correct?

 

[Sahib]

So RMS I times RMS V = RMS watts (actually work exerted)? I say this because most EE books have actual work as a complex equation, at least for 3 phase power.

It is not that 'RMS I times RMS V = RMS watts'. It is simply 'RMS I times RMS V = watts'. Because watts is the average of product of instantaneous values of voltage and current taken over a period. For a purely resistive load, it is equal to

watts=RMS I times RMS V.

For a purely reactive load

watts=0.

 

[mbrooke]

It is not that 'RMS I times RMS V = RMS watts'. It is simply 'RMS I times RMS V = watts'. Because watts is the average of product of instantaneous values of voltage and current taken over a period. For a purely resistive load, it is equal to

watts=RMS I times RMS V.

For a purely reactive load

watts=0.

Very well explained. Thank you This helps me understand much better now.

 

[mbrooke]

pretty much
let's look at it in series, math is easier
R = 1 ohm
Xl = 1 ohm
Xc = 1 ohm

X = reactance
Xl = j 2 Pi f L or j Xl
Xc = 1 / ( j 2 Pi f C) or -j Xc
j = sqrt (-1)
f = system frequency
Pi = 3.14159..., not apple

let Zeq be total ckt impedance
if R: Zeq = R
if R and Xl: Zeq = R + jXl
if R, Xl and Xc: Zeq = R + jXl +(-jXc) or R + j(Xl -Xc)
so if Xc = Xl the reactive portion cancels

PF = cos (arctan ((Xl-Xc)/R))
so if (Xl - Xc) = 0 PF=1
as the ratio varies so does the PF
it can be leading/lagging or positive/negative depending on convention
Xl > Xc is usually considered lagging/negative, absorbs reactive power
Xc > Xl is usually considered leading/positive, supplies reactive power
although it actually is an exchange

the net power factor is the sum of the R, L and C values
so a lot of L, a little C with mostly R will drop it below 1 (lagging)

Not to stray off, but for this series circuit what would the watts be at 120 volts? Ditto for the impedance?

 

[Sahib]

Not to stray off, but for this series circuit what would the watts be at 120 volts? Ditto for the impedance?

If R is fixed and power factor is varied and so is the impedance, the watts is also fixed.

 

[mbrooke]

If R is fixed and power factor is varied and so is the impedance, the watts is also fixed.

I agree here. But lets say the capacitive or reactive impedance is increased, wont watts go down since voltage will begin to dip at the resistor? Think of a lighting ballast...

 

[mbrooke]

This is what I have in mind:


http://www.vintage-radio.com/repair-restore-information/valve_dropper-calcs.html

As used in LED lamps:


http://www.bigclive.com/ledlmp.htm

 

[Besoeker]

Because watts is the average of product of instantaneous values of voltage and current taken over a period.

A watt is an instantaneous value. See post 55.