Measure apparent power, watt and powerfactor (3 phase)

[Maverickmax]

how to calculate power factor

Hi

I am taking samples at least 1ms sample of 50Hz sine wave on 3 phase and I manage to get Vrms and Irms from 3 phase. But I am not sure how to calculate apparent power and watt.

3Ph Vrms =(V1rms + V2rms + V3rms)/3
3Ph Irms =(I1rms + I2rms + Irms)/3


3Ph Apparent power = sqrt(3)* 3Ph Vrms * 3Ph Irms (Is that correct?)
3Ph Watt = 3Ph Vrms * 3Ph Irms (Is that correct?)

pf=kW/kVA (Is that correct?)

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Added after 1 hours 19 minutes:

It appears that apparent power is correct. I am still unsure about kW beacause the power factor is unknown.

kW = (1.73*Irms*Vrms)/1000*power factor.

Since I don't know what load will be. Assuming that load is resistive hence power factor would be unity (1).

So how earth can I calculate the kWatt then calculate powerfactor by dividing kVA?

Please help me asap

MM

 

[FvM]

three phase apparent power

if Vnrms are the Y-voltages, apparent power would be 3*3Ph Vrms * 3Ph Irms.

Real power expression is wrong, it has to be averaged from the product of momentary currents and voltages instead.

 

[Maverickmax]

power factor calculation 3 phase

if Vnrms are the Y-voltages, apparent power would be 3*3Ph Vrms * 3Ph Irms.

Real power expression is wrong, it has to be averaged from the product of momentary currents and voltages instead.

Are you referring to product of sum (instantanous voltage and instantaneous current)?

For example,

kW= ( (Vi*Ii) + (Vi*Ii) + (Vi*Ii) + (Vi*Ii) + (Vi*Ii) ....... up to 100) /(50*1000)

MM

 

[FvM]

calculating apparent power

Are you referring to product of sum (instantanous voltage and instantaneous current?

I would rather call i at sum of products. A suitable number of samples is a integer number of full waves.

With a known resistive load, you don't need to measure real power explicitely. If the current waveform is distorted, a 1 ms sampling interval possibly gives inaccurate results, cause it doesn't reproduce the peak current.

 

[Maverickmax]

apparent power 3 phase

So mathematick term.....


P(kW)=Sum( (Vi*Ii) + ....... Up to 100 )/(100*1000) ??

P(kVA)=Sum( (Sqrt(3)*(Vi*Ii)) + ............... up to 100)/(100*1000) ???

For example,

Vrms =400V
Irms = 250V

kW = 100kW
kVA = 173kVA

Hence power factor would be 0.577??? I am soooooooo confused

Are they correct?

 

[FvM]

3 phase power watts

V1 usually designates a Y-voltage, and V12 a delta-voltage. It seems, that you are measuring delta voltage acording to the value of 400. So the apparent power calculation would be correct with the sqrt(3) factor.

For the real power measurement, you must use Y-voltages, cause current and voltages have to be in-phase. Or you have to perform a delta-Y transform before multiply. I don't know, what you have actually been summing in your example for the real power measurement, but it don't work with delta voltages.