Mark, Well, I think JLN measures the input power the right way. That is - by integration of the instantaneous products of current and voltage, using his dual trace oscilloscope. See the following picture on his web site. http://jlnlabs.imars.com/mahg/tests/mahgjln4.jpg
However, he seems to contradict himself by stating that his analog and digital ammeters and voltmeters agree with his oscilloscope within 2%. http://jlnlabs.imars.com/mahg/tests/index.htm Surely, the power calculated by multiplying the 2-channels of an oscilloscope, and the power calculated by multiplying the Average or the RMS current & voltage shown by analog/digital meters, SHOULD BE DIFFERENT for 5-10% duty cycle pulses!!! Mark, since you have your calcs handy, could you estimate the typical power calculation error when using data from TrueRMS ammeter and voltmeter for 5% rectangular pulses, like you just did for the Averaging ammeter and voltmeter ? Regards, Horace At 00:12 2005-10-28, Mark Bilk wrote: >I think there's a big question of how Naudin measures the input >power, given that his input is pulsed. He doesn't explicitly >say how he measures it. > >http://www.gifnet.ch/lab/mahg/mahg2d.htm > >Suppose you have a pulsed DC source connected to a resistor. > >Vp = peak voltage >Va = average voltage >R = resistance >Ip = peak current >Ia = average current >Pp = peak power >Pa = average power >Po = thermal power output >d = duty cycle, less than or equal to 1 > >Then > >Pp = (Vp^2) / R ( ^ is exponentiation) > >Pa = d * (Vp^2) / R ( * is multiplication) > >Po = Pa > >Va = d * Vp > >Ip = Vp / R > >Ia = d * (Vp / R) > >Now suppose someone thinks that you can measure the average power >by multiplying the average voltage times the average current. >Let's call this Pf (f for fake or false). > >Pf = Va * Ia = d^2 * (Vp^2) / R > >Pf = d * Pa > >Pf = d * Po > >If the duty cycle is the 5% that Naudin uses, the fake average >input power is only 5% of the actual average input power. And if >the increase in water heat measured by calorimetry is due solely >to resistive heating, i.e., Po = Pa -- the output power equals the >true average input power -- then the output power divided by the >fake average input power would equal 1/d. > >In other words, if COP means output power divided by input power >(I can't find an actual definition of COP on the Web), then > >COPf = fake COP = Po / Pf = 1/d > >For Naudin's 5% duty cycle, the fake COP would be 20. > >He measures about 21.
