Just for Fun:

[Smart $]

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

OK... the 'modulo' function [MOD(#, divisor) yields remainder in Excel]. You're using this to take full cycles of t out of a set of incremental t's exceeding one cycle, and work with repititions 0≤t≤(n-1) where n is the t equivalent of 360?. Not the way I'd set it up, but if it works for you...

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

It appears for Apparent Power you are taking the absolute values of the v-i product (VA). It don't work is all I can say... at least I don't see how. Inverting (negating) the negative values is not the same as offsetting all values such that the minimum value is at zero. When you average after inverting, the result will be wrong. I had to think about this twice earlier because a few respondents questioned my method of an 'absolute average' vs. average of absolutes. The first thought was I know I'm right, and the second thought was confirming in Excel.

 

[Smart $]

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

Did you get my embedded Excel file open?

 

[mivey]

OK... the 'modulo' function [MOD(#, divisor) yields remainder in Excel]. You're using this to take full cycles of t out of a set of incremental t's exceeding one cycle, and work with repititions 0≤t≤(n-1) where n is the t equivalent of 360?. Not the way I'd set it up, but if it works for you...

As usual, I started from a file I already had and modified it accordingly. Probably not what I would do from a fresh start either.

It appears for Apparent Power you are taking the absolute values of the v-i product (VA). It don't work is all I can say... at least I don't see how. Inverting (negating) the negative values is not the same as offsetting all values such that the minimum value is at zero. When you average after inverting, the result will be wrong. I had to think about this twice earlier because a few respondents questioned my method of an 'absolute average' vs. average of absolutes. The first thought was I know I'm right, and the second thought was confirming in Excel.

Yeah forget what I had before. I was mixing average and instantaneous formulas. It would never have worked.

Did you get my embedded Excel file open?

Yes, and it makes it easier to solve for the intermediate voltages and/or currents without trying to compound the instantaneous formulas. So I re-worked the plots using the following (I used a series R-L circuit):
Φv = MOD(360?t?60,360)
V = |Vpk|?COS(Φv)
Φi = Φv - 45?
I = |Ipk|?COS(Φi)
W = I^2?R
VR = I?R
VL = V - VR
var = VL?I
Apparent = |V|?|I|
VA = V?I

I now have positive watts, which makes me feel better about the results (I at least left out a DC term before) and my vars are now centered. So I have (if the update worked):

 

[Smart $]

.. So I have (if the update worked):

Hmmm... your [Real Power] Watts appears off, for one. It has negative values. Its minimum value should be zero (0).

Second, your [Reactive Power] vars is off. It should be centered about the x-axis (a scale average of 0).

Third, your still using absolute values of VA for Apparent Power.

Anyway, it's been some time since I charted a series RL circuit, so to make sure I was doing it right, I cheated and used QUCS circuit simulator and plug in some R and L values (so the parameters don't match my earlier example)...

The simulation yields the following results...

The simulator doesn't do power, so I exported data to Excel... View attachment 4620

 

[Hameedulla-Ekhlas]

Hmmm... your [Real Power] Watts appears off, for one. It has negative values. Its minimum value should be zero (0).

Discussing and proving and doing farther works, I really enjoy it guys to have such a great and good conversation and discussing.

Ok, anyway, Smart$: very nice graph in excel but let me ask you one question in graph:
I have just post one instant of time as an example here from your excel sheet.

Here at time equal to 0.020828 you have got the following data.

vX(t)*i(t) ( Reactive Power ) = 0.99670 Var
vR(t)*i(t) ( Real Power ) = 1.11807 Watt
v(t)*i(t) ( Apparent power which is called by you) = 2.11478 VA.

Lets takes consideration of one instant of time not farther and see

S ^ 2 = P^2 + Q^2

By puting the value of P and Q at the same instant of time, will we get the exact value of your excel VA? Lets see

S^2 = (1.11807 )^2 + (0.99670)^2

S = 1.4978289 VA



I am wondering why this formula does not match your excel VA (2.11478) and Now can you tell me the difference between these two.


Ok. If I am wrong please correct me at any instant of time.

 

[Hameedulla-Ekhlas]

Hmmm... your [Real Power] Watts appears off, for one. It has negative values. Its minimum value should be zero (0).

See one more concept in the excel sheet example

Here at time equal to 0.020828 you have got the following data.

vX(t)*i(t) ( Reactive Power ) = 0.99670 Var
vR(t)*i(t) ( Real Power ) = Watt

Now see according to your excel sheet cell v(t)*i(t) which equals to

v(t)*i(t) = vX(t)*i(t) ( Reactive Power ) + vR(t)*i(t) ( Real Power ) = 0.99670 + 1.11807 =2.1147. I got your V.A number by just direct adding the real and reactive power.

Now see what will you call this power

Complex power ?
Apparent power ?

Since we know that complex power is S = P + jQ which we can not add directly
Apparent power equals to S ^ 2 = P^2 + Q^2

 

[mivey]

Hmmm... your [Real Power] Watts appears off, for one. It has negative values. Its minimum value should be zero (0).

Second, your [Reactive Power] vars is off. It should be centered about the x-axis (a scale average of 0).

That was an issue with Photobucket not updating. The file appeared correct on my Photobucket page but there must be some delay in a cache update on their website. Anyway, it appears correct this morning as I logged in on a different machine. Maybe they update their cache at night.

Third, your still using absolute values of VA for Apparent Power.

Because that is what it is by definition. It is the product of the ammeter reading and the voltmeter reading.
Apparent Power = |S| = |E?I*| = |E|?|I|

It is not the same as V?I because the V?I considers the phase angles.

 

[Smart $]

Discussing and proving and doing farther works, I really enjoy it guys to have such a great and good conversation and discussing.
Ok, anyway, Smart$: very nice graph in excel but let me ask you one question in graph:
I have just post one instant of time as an example here from your excel sheet.
Here at time equal to 0.020828 you have got the following data.
vX(t)*i(t) ( Reactive Power ) = 0.99670 Var
vR(t)*i(t) ( Real Power ) = 1.11807 Watt
v(t)*i(t) ( Apparent power which is called by you) = 2.11478 VA.
Lets takes consideration of one instant of time not farther and see
S ^ 2 = P^2 + Q^2
By puting the value of P and Q at the same instant of time, will we get the exact value of your excel VA? Lets see
S^2 = (1.11807 )^2 + (0.99670)^2
S = 1.4978289 VA
I am wondering why this formula does not match your excel VA (2.11478) and Now can you tell me the difference between these two.
Ok. If I am wrong please correct me at any instant of time.

See one more concept in the excel sheet example
Here at time equal to 0.020828 you have got the following data.
vX(t)*i(t) ( Reactive Power ) = 0.99670 Var
vR(t)*i(t) ( Real Power ) = Watt
Now see according to your excel sheet cell v(t)*i(t) which equals to
v(t)*i(t) = vX(t)*i(t) ( Reactive Power ) + vR(t)*i(t) ( Real Power ) = 0.99670 + 1.11807 =2.1147. I got your V.A number by just direct adding the real and reactive power.
Now see what will you call this power
Complex power ?
Apparent power ?
Since we know that complex power is S = P + jQ which we can not add directly
Apparent power equals to S ^ 2 = P^2 + Q^2

The answer to your query is simply that S, P, and Q are [absolute] averages, not a set of instantaneous values at one instant. Consider (oh, boy, I can hear it already) each has a counterpart set of continuous instantaneous values and we?ll call them, s(t), p(t), and q(t). These values can be added directly.

For S being Apparent Power, s(t) is the continuous instantaneous apparent power having an absolute average S.
For P being Real Power, p(t) is the continuous instantaneous real power having an absolute average P.
For Q being Reactive Power, q(t) is the continuous instantaneous reactive power having an absolute average Q.
Note this p(t) is not the same as p(t) = v(t)i(t). Rather it equals v_resistive(t)i(t) for series RCL circuits and v(t)i_resistive(t) for parallel RCL circuits.

There are a few ways to perceive the ?absolute average? referred to above. One is to take the maximum amplitude minus the minimum amplitude and divide the result by two (I used this in my Excel file near the top). This is actually the same as one-half the range of values. Another way would be to offset all the values by the negated amount of the minimum amplitude value, then take the average of all offset values. Of course when we take an average it is using the values comprising one or multiple periods of time equal to that of a voltage cycle, or in the case of power as it be, one-half of a voltage cycle.

 

[mivey]

The simulation yields the following results...

The simulator doesn't do power, so I exported data to Excel...

I adjusted my resistor & inductor values to match yours and put in the 20.828 ms delay. Although we did not plot all of the same curves, my data appears to agree with yours:

 

[Smart $]

That was an issue with Photobucket not updating. The file appeared correct on my Photobucket page but there must be some delay in a cache update on their website. Anyway, it appears correct this morning as I logged in on a different machine. Maybe they update their cache at night.

Yeah, I've had problems trying to update the same name file. The only solution I've found for immediate solution is to change the filename or upload it to a different album. On this end I'm still seeing the same graph (I think... looks the same anyway).

Because that is what it is by definition. It is the product of the ammeter reading and the voltmeter reading.
Apparent Power = |S| = |E?I*| = |E|?|I|

It is not the same as V?I because the V?I considers the phase angles.

Well I can tell you that your average, S, will be in error if you do it that way. Let's work a backward example...

S = 5
P = 4
Q = 3

S? = P? + Q?
5? = 4? + 3?
25 = 16 + 9

View attachment 4621

 

[Smart $]

I adjusted...

That looks much, much better

 

[Hameedulla-Ekhlas]

The answer to your query is simply that S, P, and Q are [absolute] averages, not a set of instantaneous values at one instant. Consider (oh, boy, I can hear it already) each has a counterpart set of continuous instantaneous values and we?ll call them, s(t), p(t), and q(t). These values can be added directly.

Ok I am asking you too much question but it is just get the result. If I am wrong and get comment to correct myself, I will be happy.

Anyway, I did not ask about the average absolute value. I asked you have at specific instant of time the following data

Here at time equal to 0.020828 you have got the following data. please tell me each one unit

vX(t)*i(t) ( Reactive Power ) = 0.99670 unit?
vR(t)*i(t) ( Real Power ) = 1.11807 unit?
v(t)*i(t) ( Apparent power which is called by you) = 2.11478 unit?


s(t), p(t), and q(t) are really new for me and I had not seen such a these thinks. I need others confirmation and explaination about this because I dont know about this

Well I can tell you that your average, S, will be in error if you do it that way. Let's work a backward example...

S = 5
P = 4
Q = 3

S? = P? + Q?
5? = 4? + 3?
25 = 16 + 9

As long as what mivey has posted and hinting, I am completely agree with him. There is no error in his graph and calculation. See the formula and calculation does not match your graph and calculation. That's why there is a difference.

 

[LarryFine]

That was an issue with Photobucket not updating. The file appeared correct on my Photobucket page but there must be some delay in a cache update on their website. Anyway, it appears correct this morning as I logged in on a different machine. Maybe they update their cache at night.

Yeah, I've had problems trying to update the same name file. The only solution I've found for immediate solution is to change the filename or upload it to a different album. On this end I'm still seeing the same graph (I think... looks the same anyway).

Guys, this may be due to a browser setting. I believe there is a setting for whether your computer always refreshed web images you've seen, or whether it uses the last version of that image you saw, for faster browsing.

I think you can manually refresh an image when you're viewing it, as well as change a browser setting so the newest version of the image is always downloaded when you re-visit the site and re-view the image. I think.

 

[david luchini]

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.

Besoeker, that's quite a statement, and completely wrong. Imagine a circuit with 1 volt and 1 amp where the current lags the voltage by 30 degrees (ie, a 0.866 power factor.) V=1<0 and I=1<-30.

Multiplying this 1V times the 1A, or (1<0) x (1<-30) yields 1<-30 VA (also written as 0.866-j0.5 VA.)

This is 1.0 VA, but it is only 0.866 WATTS (and 0.5 VARS.)

 

[Besoeker]

Besoeker, that's quite a statement, and completely wrong. Imagine a circuit with 1 volt and 1 amp where the current lags the voltage by 30 degrees

At the same instant in time there is no lag or lead.
Just instantaneous values.

 

[rattus]

Discussing and proving and doing farther works, I really enjoy it guys to have such a great and good conversation and discussing.

Ok, anyway, Smart$: very nice graph in excel but let me ask you one question in graph:
I have just post one instant of time as an example here from your excel sheet.

Here at time equal to 0.020828 you have got the following data.

vX(t)*i(t) ( Reactive Power ) = 0.99670 Var
vR(t)*i(t) ( Real Power ) = 1.11807 Watt
v(t)*i(t) ( Apparent power which is called by you) = 2.11478 VA.

Lets takes consideration of one instant of time not farther and see

S ^ 2 = P^2 + Q^2

By puting the value of P and Q at the same instant of time, will we get the exact value of your excel VA? Lets see

S^2 = (1.11807 )^2 + (0.99670)^2

S = 1.4978289 VA



I am wondering why this formula does not match your excel VA (2.11478) and Now can you tell me the difference between these two.


Ok. If I am wrong please correct me at any instant of time.

Ham, there is no power triangle with p(t). The powers add algebraically.

 

[rattus]

At the same instant in time there is no lag or lead.
Just instantaneous values.

Seems I have heard this unsupported claim before. In general, the lead or lag is present whatever you do even if you do nothing. It is obvious though that you cannot determine lead or lag from a single measurement. It is also obvious that the lead or lag affects the value of the function at any instant.

Instantaneous equations are continuous and take the form:

v(t) = Vm*sin(wt)-->Vm*sin(wt1)
i(t) = Im*sin(wt + phi)-->Im*sin(wt1 + phi)

Clearly, i(t) leads v(t) by the angle phi at time t1.

Bes, why don't you support your position with a reference. If you don"t we will think this is just your take on the matter.

 

[Hameedulla-Ekhlas]

Ham, there is no power triangle with p(t). The powers add algebraically.

Thanks rattus and what will be the unit of each real, reactive and apparent power?

 

[david luchini]

Seems I have heard this unsupported claim before. In general, the lead or lag is present whatever you do even if you do nothing. It is obvious though that you cannot determine lead or lag from a single measurement. It is also obvious that the lead or lag affects the value of the function at any instant.

Instantaneous equations are continuous and take the form:

v(t) = Vm*sin(wt)-->Vm*sin(wt1)
i(t) = Im*sin(wt + phi)-->Im*sin(wt1 + phi)

Clearly, i(t) leads v(t) by the angle phi at time t1.

I'm with rattus on this one. If Vm and Im are both 1 (ie, 1 Volt and 1 Amp) then at any instant in time (t1), v(t1)*i(t1) will provide a different value if angle phi is 0 (pf=1) or if angle phi is 30 (pf=0.866)

 

[rattus]

Thanks rattus and what will be the unit of each real, reactive and apparent power?

Ham, I prefer to use "watts, ivars, and ivas". The "i" indicates an instantaneous value. But, some do not agree with me or with some deceased engineering profs either.