Reactive Power

[martinmistere]

HI i've got some problems while calculating reactive power... i post you what the IEEE 1459-2000 says.


I need to calculate the  Q that is  hooped of red.

I've post you also what the IEEE says about the Active power.
In that case i've converted the integration into a summation such as showed into the picture...
so i've done 1/(k*T) * summation of (v*i)

now how can i implement the reactive power?i'm in trouble with the double integral...

i've tried to do it as:

2*pigreco*f/(k*T)*summation of i * summation of v

but i don't think that it works because it is too much different from the reactive power calculated at F=50Hz such as Q=V*I*sin(fi) so there is something wrong...
in any case consider that v and i are synchronized.

Plz give me a  reply as soon as possible

[gwd]

Interesting question. If we believe that the equation 3.1.1 .3 is true in general

what it says is we can get the same result by integrating EITHER i*(dv/dt) OR v*(integral of i(t)). This only makes sense to me if I evaluate integral of i(t) as a function of time. So (with samples indexed by j) I think you want for the integral

sum over j of [v(j)*(sum of i(k) from k=0 to k=j)]

(where k is a dummy index)

Seems to me there ought to be limits on the integral of i, and for clarity this would normally be written with a dummy variable t'

Definitely try this out on test cases.

[martinmistere]

mmm i'ven't understood a lot...sorry
can you show me how to do it into a VI in labVIEW?because you say that k and j are different...but i'm syncronizating the signal and i'm sure that k=j so they have same number of elements.
tnx

Message Edited by martinmistere on 02-18-2007 04:05 PM

[gwd]

This is what I was thinking...but in test cases it is approximately but not exactly right.

[martinmistere]

i thank you for the VI but i see that you don't have multiply the value that exits from the sum for 2*pigreco*f/N
i've tried to do this but unfortunately it isn't correct because with my change i have had a reactive power of 20 while the reactive power of the fundamental (that i calculate as Vrms*Irms*sin(fi) as the IEEE says) is near to 0; with your VI i have had a reactive power of 8000 with a ractive power of the fundamental always near to 0.
This for me is strange because i have only 2 signals that are both sinusoidal one with amplitude 10V while the other 8V.
Can you give me some others advices?
Tnx a lot for all.

[carlo>]

Hi martinmis,

I think that the calculation of the double integral Q as made in 3.1.1.3 should not be considered in a numeric sense but into an analytical sense. In fact, if you consider the last term of the equation as you did, you can obviously see that the integration of i in dt over a period must give 0 as result. Try instead to consider the first term of the right member in which there is the integral of v in di: in so doing what you have to calculate is the integral of v multiplied for the derivative di.

Starting from that point, you know that Power factor (P) is 1  if voltage and current are in phase (teta [or fi as you called it]=0) while is 0 if the phase (teta) is 90 degrees. For the reactive power Q is the contrary. So the example you discuss on doesn't be good in this case because I expect to have Q=1 (without considering constant factor) because the sin and cos generated have a relative phase of 90 degrees.

So, what I suggest to do is to make a vi that multiply the current value of v for the derivative of i (difference between current value of i and previous one) and then make the summation of all the value v*(i2-i1) obtained in that manner. This should fix your problem.

carlo>   

[martinmistere]

mmm sorry probably i will be too demanding but i don't know how to use as good as possible the block of derivation.
Could you, if it is possible and if for you it is not  a problem, post a VI (or a photo) in which you show me how to implement the things that you have told me before?

In any case tnx a lot for your advice.

[carlo>]

Hi martinmis,

I post an example, I modify the VI posted by gvd. In this example I show you how to use shift register in order to do a derivative operation. If you take a look to the Block diagram, you will see that on the right side:

1. I leave unchanged gvd's calculation (for loop at the bottom).

2. I do calculation with i and v that are out of phase of 90 degrees i.e sin and cos (for loop in the middle) -> Q=1 as expected. You see that if you run the vi "sum" is equal to -pi=-3,14 that's should be right if you take into consideration the various normalization factors.

3. I do calculation with i and v that are in phase [teta=0] i.e sin and sin (uppermost for loop)-> Q=0 as expected. Note that if you increase the number of cicles that generate sin and cos functions in the left for loop of the vi Q-->0 faster.

Please, consider this example as a guideline, I let you to fix the various normalization constant.

I hope I've been of some help

carlo>

[martinmistere]

i have labview 8.0 can you convert it?or in any case can you make me a photo of the VI that i need to see?
tnx again

[carlo>]

Hi martinmis,

here it is the VI saved for LV 8.0. I also include an image of the block diagram.

I hope this help you.

carlo>