Electrical engineers needed

stevemcol

OK, I've been playing with Excel for the last half hour but I still get apparent power greater than true power. Unless you're talking about instantaneous samples where true power is a positive value and apparent power is zero.

DVardysShadow

So what is the calc you are doing?

stevemcol

DVardysShadow wrote: »

So what is the calc you are doing?

339volt (peak) sine wave colunm A
Current in column B. In phase. zero value if voltage <240V. Assuming 50ohm load.

Instantaneous apparent power in column C (V x I) approx 2300VA peak.

So current & power look like square waves with sine wave tops.

RMS (true) power in column D (Peak V x Peak I x 0.707) 1624 Watts

DVardysShadow

stevemcol wrote: »

339volt (peak) sine wave colunm A
Current in column B. In phase. zero value if voltage <240V. Assuming 50ohm load.

Instantaneous apparent power in column C (V x I) approx 2300VA peak.

So current & power look like square waves with sine wave tops.

RMS (true) power in column D (Peak V x Peak I x 0.707) 1624 Watts

Current should be a rectangular wave with a flat top

true power is instantaneous power summed over one cycle

- the way you are doing it is time averaging V and I before doing the power calc. and the peak values 0.707 makes an implicit assumption that you are working with sine waves - the peaks and 0.707 are effectively rules of thumb derived from working out sine waves from first principles. as you are not working with sine waves, you are 'no longer in Kansas' and you must work it out from first principles:

true power = instantaneous power summed over one cycle. You will probably need integral calculus and your head might hurt.

stevemcol

DVardysShadow wrote: »

Current should be a rectangular wave with a flat top

true power is instantaneous power summed over one cycle

- the way you are doing it is time averaging V and I before doing the power calc. and the peak values 0.707 makes an implicit assumption that you are working with sine waves - the peaks and 0.707 are effectively rules of thumb derived from working out sine waves from first principles. as you are not working with sine waves, you are 'no longer in Kansas' and you must work it out from first principles:

true power = instantaneous power summed over one cycle. You will probably need integral calculus and your head might hurt.

I agree, I shouldn't be using root 2 for RMS and that I need to integrate. I'm interested so I'll keep going with excel but integrating current is only going to further reduce true power.
Smoke and mirrors going on here I fear. I'd like to see a worked example rather than try and reconstruct it from hints.