Just for Fun:

[rattus]

Have it your way:

Have it your way:

One Volt times one Amp at the same instant is one Watt.
It doesn't matter what happened before or happens afterwards.
It is still one Watt.
Describing as anything else is nonsensical.

Makes sense to me because the vi product is, in general, the algebraic sum of real and reactive powers. If you can speak of instantaneous vars, you can speak of instantaneous vas.

And, I know of no hard and fast rule on this matter; it seems to merely be a metter of personal preference. So to each his own.

 

[rattus]

Try again:

Try again:

OK, try to open the attachment for some really good numbers.

 

[rattus]

Ok one more thing again,

If you are saying the above red line is true than how about this see below


P = W / t = F * V (force * velocity) .

F = kg * m/ (s square) ---1

v = m/s ------2


F * V = kg (meter square) / ( second * sqaure)



1W = 1Joul / second = 1kg * (meter sqaure) / (second cube)


It means there is only watt not VA

Ham, indeed the basic unit is the Watt, but some choose to use VA to indicate that the load may contain reactive elements. Matter of choice--no big deal.

 

[Smart $]

OK, try to open the attachment for some really good numbers.

Having numbers, we can solve for (separate) real and reactive powers even easier than the [partial] graphical solution I posted earlier. We simply take advantage of the relationship P = avg(p(t)). In other words the average of real power is equal to the average of apparent power. From there, if we want correlating instantaneous values, simply use the equation:

p_real(t) = P ? (1 – cos(2ωt)), where data begins at v = 0

p_real(t) = P ? (1 + cos(2ωt)), where data begins at v_max

If data begins at another point in a voltage cycle, a Δφ adjustment needs to be included in the equation. Though not readily apparent, the above equations are adjusted by Δφ = ?90?.​

For reactive power, just subtract real from apparent.

Also a reminder about taking an average in Excel. When data is divided equally into one cycle (i.e. includes 0? and 360? divided by "n" samples), the average must be taken with "n-1" consecutive samples. Regarding rattus' numbers, they have been rounded, so a little bit of error creeps into the result... but I estimated it is less than 0.00001% error... so nothing to worry about.

 

[Besoeker]

Makes sense to me because the vi product is, in general, the algebraic sum of real and reactive powers.

The instantaneous vi product is Watts. Nothing more. Nothing less.

 

[Smart $]

The instantaneous vi product is Watts. Nothing more. Nothing less.

If...

1 Watt = 1 Volt-Ampere​

...what is the problem with using VA as a unit of measure?

 

[mivey]

...what is the problem with using VA as a unit of measure?

Because he is constraining it to be one instant in time. You are discussing the collection of instantaneous values over a space of time. Besoeker knows there is a difference between a value at one instant and a collection of instantaneous values. Maybe it is just a matter of terminology preference.

 

[rattus]

Because he is constraining it to be one instant in time. You are discussing the collection of instantaneous values over a space of time. Besoeker knows there is a difference between a value at one instant and a collection of instantaneous values. Maybe it is just a matter of terminology preference.

Mivey, I don't see the rationale. True, a single point tells us little--all the more reason to use vas because we don't know the nature of the circuit. If we can have "instantaneous vars" which I have seen described, then we can have "instantaneous vas", and I have heard of no law of physics that makes this practice wrong.

 

[winnie]

The watt is an SI derived unit, and could equally be called joules per second or volt amperes, or in terms of SI base units, square meter kilograms per second cubed. All are valid expressions of the unit.

As commonly used, volt amperes is the unit used for the product of RMS voltage and RMS current, which is explicitly not something that happens at one instant in time. Because of this common usage, 'instantaneous volt amperes' would be quite confusing, in the same way that 'apparent instantaneous power' is confusing...it is quite _apparent_ that instantaneous volts times instantaneous amps gives instantaneous power, but in common usage 'apparent power' is the product of RMS voltage and RMS current.

-Jon

 

[mivey]

Mivey, I don't see the rationale. True, a single point tells us little--all the more reason to use vas because we don't know the nature of the circuit. If we can have "instantaneous vars" which I have seen described, then we can have "instantaneous vas", and I have heard of no law of physics that makes this practice wrong.

I've don't know of a real law violation either. I don't know why Bes does not want to acknowledge "instantaneous reactive power". But, I would probably choke on "instantaneous volt amps" as I would rather just say "instantaneous power"

The watt is an SI derived unit, and could equally be called joules per second or volt amperes, or in terms of SI base units, square meter kilograms per second cubed. All are valid expressions of the unit.

As commonly used, volt amperes is the unit used for the product of RMS voltage and RMS current, which is explicitly not something that happens at one instant in time. Because of this common usage, 'instantaneous volt amperes' would be quite confusing, in the same way that 'apparent instantaneous power' is confusing...it is quite _apparent_ that instantaneous volts times instantaneous amps gives instantaneous power, but in common usage 'apparent power' is the product of RMS voltage and RMS current.

-Jon

I agree. Mixing in the units can be confusing. Apparent power is a positive value that results from multiplying the magnitude of two positive or negative instantaneous values. I would prefer:
"instantaneous reactive power"
"instantaneous real power"
"instantaneous power"
"instantaneous power in the resistor"
"instantaneous power in the inductor"
"instantaneous power in the capacitor"

not
"instantaneous Xunits power"
"instantaneous Xunits"

 

[Smart $]

The watt is an SI derived unit, and could equally be called joules per second or volt amperes, or in terms of SI base units, square meter kilograms per second cubed. All are valid expressions of the unit.

Agreed? and thank you.

...what is the problem with using VA as a unit of measure?

Because he is constraining it to be one instant in time. You are discussing the collection of instantaneous values over a space of time. Besoeker knows there is a difference between a value at one instant and a collection of instantaneous values. Maybe it is just a matter of terminology preference.

There is no difference in the units of measurement, whether it be a collection of instantaneous values or just one out of the lot.

Mivey, I don't see the rationale. True, a single point tells us little--all the more reason to use vas because we don't know the nature of the circuit. If we can have "instantaneous vars" which I have seen described, then we can have "instantaneous vas", and I have heard of no law of physics that makes this practice wrong.

I've don't know of a real law violation either. I don't know why Bes does not want to acknowledge "instantaneous reactive power". But, I would probably choke on "instantaneous volt amps" as I would rather just say "instantaneous power"

As commonly used, volt amperes is the unit used for the product of RMS voltage and RMS current, which is explicitly not something that happens at one instant in time. Because of this common usage, 'instantaneous volt amperes' would be quite confusing, in the same way that 'apparent instantaneous power' is confusing...it is quite _apparent_ that instantaneous volts times instantaneous amps gives instantaneous power, but in common usage 'apparent power' is the product of RMS voltage and RMS current.

I agree. Mixing in the units can be confusing.

I don?t believe we have been attempting to use (or coin) the phrase 'instantaneous volt amperes', though it seems to me to be no different than ?instantaneous watts? or ?instantaneous power?. The objective in using any such term is merely to convey the type of measure, or measures, under consideration. The objection seems to be in using the term ?instantaneous apparent power?. Whether one chooses to accept this nomenclature as a viable term is their preference, but given steady-state values representing Volts (an RMS value), Amperes (another RMS value), Apparent Power (an ?absolute? average value), Real Power (an average), and Reactive Power (another ?absolute? value), they each can and do have a corresponding waveform comprised of instantaneous values. So why is it so difficult to accept putting the adjective ?instantaneous? in front of those terms to indicate we are discussing the waveforms, collection of instantaneous values, continuous function, etc. and not the DC-equivalent, steady-state values? I have yet to fathom it taking such a major leap to understand.

Apparent power is a positive value that results from multiplying the magnitude of two positive or negative instantaneous values. ?

What about when current leads or lags the voltage. There are instances when one is positive and the other is negative, yielding a negative product???? These instances are included in apparent power. You will find the DC-equivalent, steady-state value used for Apparent Power is one-half the range of these instantaneous values, not the actual instantaneous-scale average. In fact, the instantaneous-scale average of both real power and apparent power are equal.

 

[Cold Fusion]

... In fact, the instantaneous-scale average of both real power and apparent power are equal.

What is an "instantaneous-scale average"? I'm not familiar with that term.

cf

 

[mivey]

There is no difference in the units of measurement, whether it be a collection of instantaneous values or just one out of the lot.

Then I'm not sure what Bes' rub is

I don?t believe we have been attempting to use (or coin) the phrase 'instantaneous volt amperes', though it seems to me to be no different than ?instantaneous watts? or ?instantaneous power?. The objective in using any such term is merely to convey the type of measure, or measures, under consideration. The objection seems to be in using the term ?instantaneous apparent power?. Whether one chooses to accept this nomenclature as a viable term is their preference, but given steady-state values representing Volts (an RMS value), Amperes (another RMS value), Apparent Power (an ?absolute? average value), Real Power (an average), and Reactive Power (another ?absolute? value), they each can and do have a corresponding waveform comprised of instantaneous values. So why is it so difficult to accept putting the adjective ?instantaneous? in front of those terms to indicate we are discussing the waveforms, collection of instantaneous values, continuous function, etc. and not the DC-equivalent, steady-state values? I have yet to fathom it taking such a major leap to understand.

Apparent power is not a continuous waveform like real and reactive power. For sinusoidal voltages and currents, there is a Theorem of Conservation of Complex Power so it is handy to track real and reactive power separately to compare power delivered vs power received. We can separately plot the real and reactive power and get sinusoidal waveforms. But, the apparent power plot would be a mutation of the p(t) = v(t)*i(t) curve because it is always positive or S = √(P? + Q?).

You could show the apparent power as S = |E|?|I|e^(jΦ) but that is not going to rotate like the voltage and current phasor.

What about when current leads or lags the voltage. There are instances when one is positive and the other is negative, yielding a negative product???? These instances are included in apparent power.

The magnitude of a negative number is a positive number. I did not mean the voltage and current had to have the same sign. The product of the two magnitudes is a positive number, regardless of which is positive or negative.

You will find the DC-equivalent, steady-state value used for Apparent Power is one-half the range of these instantaneous values, not the actual instantaneous-scale average. In fact, the instantaneous-scale average of both real power and apparent power are equal.

Yes, it is the product of the RMS values for voltage and current. Since we have a sinusoid, you can also use the instantaneous information to find the peak value then divide by sqrt(2). The time function for voltage and current (the instantaneous equations) can be written using the peak value or the RMS value. My statement was incomplete. Thanks for the correction.

 

[Smart $]

Apparent power is not a continuous waveform like real and reactive power.

Says who. I understand what you are saying, but I do not agree. Read on...

For sinusoidal voltages and currents, there is a Theorem of Conservation of Complex Power so it is handy to track real and reactive power separately to compare power delivered vs power received. We can separately plot the real and reactive power and get sinusoidal waveforms. But, the apparent power plot would be a mutation of the p(t) = v(t)*i(t) curve because it is always positive or S = √(P? + Q?).

You could show the apparent power as S = |E|?|I|e^(jΦ) but that is not going to rotate like the voltage and current phasor.
The magnitude of a negative number is a positive number. I did not mean the voltage and current had to have the same sign. The product of the two magnitudes is a positive number, regardless of which is positive or negative.Yes, it is the product of the RMS values for voltage and current. Since we have a sinusoid, you can also use the instantaneous information to find the peak value then divide by sqrt(2). The time function for voltage and current (the instantaneous equations) can be written using the peak value or the RMS value. My statement was incomplete. Thanks for the correction.

But all you are doing is using math to manipulate the data for convenience, to make all instantaneous values ≥ 0. The only difference between the v-i product curve and your 'mutated' apparent power curve is that you offset it. It is still the same curve. This is nothing more than what I referred to earlier as taking the 'absolute average", which is equal to one-half the range of the curve. The image depicts both what you are doing and what I just wrote. The v-i product curve is the green line wavy line, its range is indicated by dashed green lines, while its average is indicated by the straight, horizontal line. The purple lines indicate what you have done via math manipulation. Note the magnitude of the red and black arrowed lines are all equal. You can also plug any of these the magnitudes into your formula S? = P? + Q? as S and it works just fine.

If you need to confirm, work it out with this example:

|v| = 2V
|i_real| = 1A
|i_reactive| = 1A
θ = 45?​

 

[mivey]

If you need to confirm, work it out with this example:

|v| = 2V
|i_real| = 1A
|i_reactive| = 1A
θ = 45?
​

Here is what I got, unless I messed up the formulas:

 

[mivey]

Check my formulas

Check my formulas

Here is what I got, unless I messed up the formulas:

Here is what I used when plotting against time:
Φv = MOD(360?t?60,360)
V = |Vpk|?COS(Φv)
Φi = Φv - 45?
I = |Ipk|?COS(Φi)
W = V?I?COS(Φi - Φv)
var = V?I?SIN(Φi - Φv)
Apparent = √(W? + var?)
VA = V?I

 

[Smart $]

Here is what I got, unless I messed up the formulas:

Hmmm... looks a bit strange.

See View attachment 4619 (zipped Excel file embeddded).

 

[Smart $]

Here is what I used when plotting against time:
Φv = MOD(360?t?60,360)
V = |Vpk|?COS(Φv)
Φi = Φv - 45?
I = |Ipk|?COS(Φi)
W = V?I?COS(Φi - Φv)
var = V?I?SIN(Φi - Φv)
Apparent = √(W? + var?)
VA = V?I

What is MOD(360?t?60,360)?

The rest I understand.

For the example I gave:
|Vpk| = 2
Φv = 0?
|Ipk| = √2
Φi = -45?

θ = Φv - Φi = 45?, current lags voltage by 45?. You didn't even need this parameter in the original example values because i_real(t) and i_reactive(t) just happen to add up to i(t) peak at -45? respective of v(t). I just happen to know that's what it was and provided it without thinking. :roll:

 

[mivey]

What is MOD(360?t?60,360)?

Removes multiples of 360 from "360?t?60" and returns the remainder

I just happen to know that's what it was and provided it without thinking.

That's ok, I did the same. I actually entered 2 volts and set R & L to get |I| = 1. The angle worked out by itself, but I saw no need to get into R & L values (√2).

 

[mivey]

Hmmm... looks a bit strange.

I think I may have left out part of the equation.