Just for Fun:

[Cold Fusion]

If one had a series of v(t)i(t) product values, as many as needed, but no additional information, how would one go about separating the real and reactive powers from the instantaneous apparent power?

Are we to assume sinusoidal AC? ...and linear load?

Of course.

Hummmm ... I'm going to have to ignore "instantaneous apparent power" and translate that you wish to reconstruct the two single frequency sinusiods making up V and I and determine the phase angle between them.

Since there are no real and reactive components to instantaneous measurements, there is no determinent calculation available.

So, this is just a curve fitting issue. All curve fitting problems require assumptions. Depending on the assumptions, one point could be enough.

cf

 

[Smart $]

To avoid confusion, don't call it "complex". There are no complex numbers in p(t).

I'm not calling it complex. Just made a comment that it could literally be acceptable... nothing more, nothing less. I agree omitting the adjective "complex" would make the discussion less confusing, more coherent.

And, we are describing the vi product as sinusoids and an offset, but still no imaginary values. We would have to use rotating phasors to do that and we don't want to get into that quagmire just now.

With any sinusoidal waveform (cartesian) there are imaginary numbers (trigonometry by its very nature involves three numbers and three angles for every single relationship). One chooses to ignore that fact entirely at their discretion (or the behest of their mentors ).

But I'll shut up to avoid the "quagmire just now"...

...carry on

 

[Smart $]

I only see one waveform. Didn't your original question express two values?

And, what's with the DC offset?

The original question stated, "If one had a series of v(t)i(t) product values..." It did not say we had v(t), i(t), and v(t)i(t).

rattus will have to confirm.


As for the DC offset, that is an inherent property of the v(t)i(t) product.

 

[rattus]

The original question stated, "If one had a series of v(t)i(t) product values..." It did not say we had v(t), i(t), and v(t)i(t).

rattus will have to confirm.


As for the DC offset, that is an inherent property of the v(t)i(t) product.

''

Confirmed! Furthermore, there is in general a phase lead or lag between v and i. Also, the reactive components are in the load. Without reactance we would have no lead or lag.

And, FWIW, with digital measurements, we express the delay in units of time. With AC power measurements, we express the delay in radians or degrees.

 

[Smart $]

I'll bite...

...

Do I win a cigar??

I doubt there's any materialistic prize being awarded, especially in a thread titled "Just for fun". Though I'm not "the judge", I don't believe you've won a "just for fun" cigar

BTW, I know how... just waiting to see if someone else comes up with the answer.

 

[rattus]

I doubt there's any materialistic prize being awarded, especially in a thread titled "Just for fun". Though I'm not "the judge", I don't believe you've won a "just for fun" cigar

BTW, I know how... just waiting to see if someone else comes up with the answer.

Give Steve a match and tell him to keep going.

 

[Smart $]

Give Steve a match and tell him to keep going.

You just did, why should I? (reduntant)

 

[rattus]

I doubt there's any materialistic prize being awarded, especially in a thread titled "Just for fun". Though I'm not "the judge", I don't believe you've won a "just for fun" cigar

BTW, I know how... just waiting to see if someone else comes up with the answer.

Go ahead Smart, looks like no one else will.

 

[Hameedulla-Ekhlas]

To avoid confusion, don't call it "complex". There are no complex numbers in p(t).

And, we are describing the vi product as sinusoids and an offset, but still no imaginary values. We would have to use rotating phasors to do that and we don't want to get into that quagmire just now.

If one had a series of v(t)i(t) product values, as many as needed, but no additional information, how would one go about separating the real and reactive powers from the instantaneous apparent power

ok thanks for information but one more thing.

Since you can not call it a complex than in your first post is mentioned of instantaneous apparent power. Is it right?

 

[rattus]

ok thanks for information but one more thing.

Since you can not call it a complex than in your first post is mentioned of instantaneous apparent power. Is it right?

Yes, "instantaneous apparent power" is descriptive of the vi product which would carry the units of volt-amperes.

 

[Hameedulla-Ekhlas]

Yes, "instantaneous apparent power" is descriptive of the vi product which would carry the units of volt-amperes.

thanks rattus and can you give me any book reference or souce reference to mention the instantaneous apparent power

 

[rattus]

thanks rattus and can you give me any book reference or souce reference to mention the instantaneous apparent power

Not exactly, but,

"instantaneous reactive power, instantaneous quadrature power, instantaneous reactive volt-ampers, etc," are defined in,

[Kerchner & Corcoran, Alternating-Current Circuits, John Wiley, 1951].

I have taken the liberty of including "Instantaneous Apparent Power" in the "et cetera". It fits the situation perfectly.

 

[rattus]

Here!

Here!

OK, I tried to upload a spreadsheet with no success, but here are some numbers you can cut and past into Excel or other spreadsheet program. The copy didn't work very well, so look out for glitches in the waves.

wt vi
0 0.866025404
0.043633194 0.953181073
0.087266389 1.039673436
0.130899583 1.124844235
0.174532778 1.20804527
0.218165972 1.288643332
0.261799167 1.366025021
0.305432361 1.439601417
0.349065556 1.508812562
0.39269875 1.573131716
0.436331944 1.632069373
0.479965139 1.685176983
0.523598333 1.732050365
0.567231528 1.772332786
0.610864722 1.805717672
0.654497917 1.831950944
0.698131111 1.850832952
0.741764306 1.862219993
0.7853975 1.866025404
0.829030694 1.862220224
0.872663889 1.850833413
0.916297083 1.831951631
0.959930278 1.805718579
1.003563472 1.772333907
1.047196667 1.732051692
1.090829861 1.685178505
1.134463056 1.632071079
1.17809625 1.573133592
1.221729444 1.508814595
1.265362639 1.439603591
1.308995833 1.366027319
1.352629028 1.288645736
1.396262222 1.208047764
1.439895417 1.124846798
1.483528611 1.03967605
1.527161806 0.953183717
1.570795 0.866028057
1.614428194 0.778872378
1.658061389 0.692379985
1.701694583 0.607209135
1.745327778 0.524008031
1.788960972 0.443409881
1.832594167 0.366028085
1.876227361 0.292451564
1.919860556 0.223240279
1.96349375 0.158920968
2.007126944 0.09998314
2.050760139 0.046875347
2.094393333 0.0
2.138026528 -0.040280857
2.181659722 -0.073665956
2.225292917 -0.09989945
2.268926111 -0.118781684
2.312559306 -0.130168954
2.3561925 -0.133974596
2.399825694 -0.130169648
2.443458889 -0.118783066
2.487092083 -0.09990151
2.530725278 -0.073668679
2.574358472 -0.040284221
2.617991667 0.0
2.661624861 0.04687078
2.705258056 0.099978023
2.74889125 0.158915339
2.792524444 0.22323418
2.836157639 0.292445043
2.879790833 0.366021191
2.923424028 0.443402666
2.967057222 0.52400055
3.010690417 0.607201446
3.054323611 0.692372145
3.097956806 0.778864447
3.14159 0.866020097
3.185223194 0.953175786
3.228856389 1.03966821
3.272489583 1.124839109
3.316122778 1.208040283
3.359755972 1.288638522
3.403389167 1.366020425
3.447022361 1.43959707
3.490655556 1.508808496
3.53428875 1.573127963
3.577921944 1.632065962
3.621555139 1.685173939
3.665188333 1.732047712
3.708821528 1.772330543
3.752454722 1.805715856
3.796087917 1.83194957
3.839721111 1.85083203
3.883354306 1.86221953
3.9269875 1.866025404
3.970620694 1.862220686
4.014253889 1.850834334
4.057887083 1.831953004
4.101520278 1.805720394
4.145153472 1.77233615
4.188786667 1.732054346
4.232419861 1.685181549
4.276053056 1.63207449
4.31968625 1.573137345
4.363319444 1.50881866
4.406952639 1.439607939
4.450585833 1.366031915
4.494219028 1.288650546
4.537852222 1.208052751
4.581485417 1.124851925
4.625118611 1.039681276
4.668751806 0.953189004
4.712385 0.866033365
4.756018194 0.778877665
4.799651389 0.692385211
4.843284583 0.607214262
4.886917778 0.524013018
4.930550972 0.443414691
4.974184167 0.366032681
5.017817361 0.292455911
5.061450556 0.223244344
5.10508375 0.158924721
5.148716944 0.099986552
5.192350139 0.046878391
5.2359833334 0.0
5.279616528 -0.040278614
5.323249722 -0.073664141
5.366882917 -0.099898076
5.410516111 -0.118780762
5.454149306 -0.130168491
5.4977825 -0.133974596
5.541415694 -0.13017011
5.585048889 -0.118783988
5.628682083 -0.099902883
5.672315278 -0.073670494
5.715948472 -0.040286464
5.759581667 0.0
5.803214861 0.046867736
5.846848056 0.099974612
5.89048125 0.158911586
5.934114444 0.223230115
5.977747639 0.292440695
6.021380833 0.366016595
6.065014028 0.443397856
6.108647222 0.523995563
6.152280417 0.60719632
6.195913611 0.692366918
6.239546806 0.778859161
6.28318 0.866014789
6.326813194 0.953170499

 

[Besoeker]

Yes, "instantaneous apparent power" is descriptive of the vi product which would carry the units of volt-amperes.

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.

 

[Hameedulla-Ekhlas]

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.

I am agree with you completely.

 

[Besoeker]

OK, I tried to upload a spreadsheet with no success, but here are some numbers you can cut and past into Excel or other spreadsheet program. The copy didn't work very well, so look out for glitches in the waves.

You can copy it into a host like photobucket as an image and post it from there.

 

[Smart $]

OK, I tried to upload a spreadsheet with no success, ...

Copy in Excel, paste in Word (as an Excel object), save Word file, then upload Word file.

 

[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.

 

[Hameedulla-Ekhlas]

Yes, "instantaneous apparent power" is descriptive of the vi product which would carry the units of volt-amperes.

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

 

[Smart $]

Go ahead Smart, looks like no one else will.

Well here's a depiction of the first part, separating the real power portion.

From our data we can easily distinguish four points in time:

1. p_max
2. p_min
3. zero crossing 1
4. zero crossing 2

The time between p_max and p_min is pi/2 radians (90?).

The zero crossings occur when v = 0 and i = 0. The time between these occurences is equal to θ in time units. We cannot distinguish which one comes first if it is not known whether this data represents a net capacitive or a net inductive circuit. Depicted above is inductive. It could just as easily been capacitive, in which case the timing measures and p_real pk would be to the right rather than left as in the depiction.

Knowing sinusoidal waveforms are symmetrical, the time between zero crossing 1 and p_min, and p_min to zero crossing 2 is θ/2.

Continuing, we know v_peak occurs pi/2 radians before and after the v=0 zero crossing.

We also know p_real max (aka p_real pk) occurs at both v_max and v_min, while p_min occurs at v = 0...

So who is going to continue the analysis and show how the reactive power is separated?