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3 phase instantaneous power 1

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esconfra

Mechanical
Dec 5, 2003
5
IT
in a power measurement application I found this formula for the instantaneous power

Pin = Ia Vac + Ib Vba + Ic Vcb
Ia, Ib, Ic = line currents
Vac, ... = line to line voltages

can someone verify the correctness of this?
many thanks
 
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It doesn't compute. Whenever you have a three-phase system and want to study what happens instantaneously or in steady state, you usually look at the equivalent single phase system.

And then you get phase-neutral instantaneous voltage times line instantaneous current. If you don't have a star system, then you have to transform from delta to star, which involves sqrt(3). Anyhow, I would feel more confident using line-neutral instantaneous volts and currents and then sum the three together. As done in picture below:

image_tw24ft.png




Gunnar Englund
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
 
In addition to Gunnar's post, (I like to resolve the loads to single phase also, when possible)
There is the added factor of power factor. Both displacement power factor and distortion power factor.
With AC you can not work with Volts and Amps alone. Even when the power factor is unity, calculations which do not show that unity power factor is considered may be suspect.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
It is about instantaneous power, Bill. So PF or any other complication doesn't apply. It is microsecond (or any chosen instant) power that is the question. That's why I showed just that.

Gunnar Englund
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
 
I believe that the OP formula works for a delta or three wire system if you assume that all quantities are vectors (i.e. complete with phase angles or expressed in complex notation)

One thing that you have to be careful about here is ensuring that you are consistent with the use of RMS or 'actual' magnitudes.
 
tinfoil:

I agree with Skogs here that the factors you bring up here are not useful for instantaneous power calculations. (They are useful for average power calculations over one or more cycles for idealized sinusoidal signals.)

To the original question, I don't think it makes any sense to multiply the instantaneous current through a phase by the voltage across two phases to get the instantaneous power in that single phase.
 
I agree with Skogs. You should just do phase to Vn(t)*Ia(t) for each phase. I don't know what you are getting from this though.

------------------------------------------------------------------------------------------
If you can't explain it to a six year old, you don't understand it yourself.
 
Below is an Illustration of Instantaneous power for AC system with sinusoidal driving sources on the time domain
Hope this help

Instantaneous_Power_o87f41.jpg
 
I think that this is the right moment to ask about power and energy measurements in general. I do that in a separate thread and would appreciate comments on the extremely high readings that plagued around 300,000 customers (in Sweden) when they "dimmed" electric heating to save money.

Instead, they got electric bills that were exorbitantly high. Se thread started in the Electric power & transmission & distribution Forum a few minutes from now.

Gunnar Englund
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
 
many thanks for your answers
i will propose to modify the software substituting the line to line voltages (Vac...) with Va=(Vac-Vba)/3
this should work in a 3 wire system
 
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