Trouble with calculating 3 Phase Power

[roro36]

Hi,

I think I am correct, but just wanted to check. We have some furnaces and I want to work out what power they run at. I have a scope with a current clamp and voltage probe. I have set the scope to read out power and power factor. That is where my problem lies, I think. I should have set it to voltage and pf and read the current off an aditional meter? In any event. It has worked out the power for me. What I want to know, is what is this power value referring to? I attached a drawing that I have used to try and understand it in my head.

Because I have connected the voltage probe to read Line-Line, ie ~400V, the power that it is reading is effectively P=VI*pf.. But it is a 3 phase system. So because I set the voltage probe acoss Line-Line, I need to multiply this Power value by root3 to get the power of the machine. If i had set the voltage probe to read Line-Neutral, I would multiply the power reading by 3? I know I have assume equal currents and that the meter is working out single phase Power I think, the setting just says Power and says Voltage Probe A, Current Probe B.

Please let me know if this correct.

 

[uart]

Without knowing the specifics of you scope/meter I think that you may get an anomalous power factor reading due to the voltage measurement being line to line. That is, (unless the scope/meter somehow knows this and is correcting for it), the line to line voltage measurement will throw off the phase angle by 30 degrees.

If you have L1,L2,L3 labeled correctly in that phase sequence, then your voltage measurement will lead by 30 degrees, giving a reported power factor of 0.866 (current lagging) even if the actual power factor was unity!

I think you'd need to first correct for this error in the reported pf and then multiply by sqrt(3) to get the total power, assuming a balanced load of course.

 

[uart]

So to be specific, with you're example measurements of 400V 120A and pf=0.8. I would calculate the true phase angle as $\cos^{-1}(0.8) - 30 = 6.9$ degrees, and hence find the true power factor of about 0.99.

Using these value I get a total 3 phase power of approx 82KW as opposed to the 66kW you calculate.

 

[roro36]

Thanks, I understand what you are saying. It would then be best to redo the test making sure the probe was referenced from Neutral. Would it make any difference if it were referenced to earth?

 

[jim hardy]

If the phases are balanced, ie equal currents in all 3,
then
kva 3 phase = Vline X Iline X sqrt(3)
and kw = kva X pf
and at unity pf there'll be 30 degrees between line-to-line volts and line amps

an oven i would expect to be resistive hence unity pf

do your measurements look like a balanced resistive load?