Electronics Math and Power Factor Not Adding Up - Page 1
Forum section: electronics / beginners

author: tommygdawg
author_rank: Contributor
timestamp: January 09, 2016, 06:00:43 pm
content: Hey all,I've got an HMI light (one of those ones that runs off a ballast) and I'm trying to do some basic math but it's not coming out right. I've plugged the device into a Kill-a-Watt to measure the amperage and wattage. Here's my issue:The light is rated for 1200 watts. According to the Kill-a-watt, I'm actually drawing 1040 at approximately 12.6 amps. I took my measured amperage times the voltage and got a wattage of 1,512 watts (12.6*120). Why the discrepancy? I then wanted to understand the power factor of the unit, but the math for that doesn't work out at the 1,040 watts measured. To get my power factor I have to take the rated wattage and divide it by the actual draw (1200/1040). This doesn't work out, it gives me a power factor of 1.15 which is impossible. Now, if I divide by the wattage the math gives me it works (1200/1512 = .79). I should also note that if divide measured wattage by voltage (1,040/120) I get around 8.67 amps. So which is it? Am I running at 8.67 amps or 12.6?Do these question make sense? I suppose the TL;DR version is: why does my meter read 1,040 watts at 120 volts and 12.5 amps, when the math tells me that Ishouldbe reading 1,500 watts at 120 volts and 12.5 amps?Thank you!

author: helius
author_rank: Super Contributor
timestamp: January 09, 2016, 06:20:35 pm
content: You constantly refer to 120 volts but you never said you measured that. Voltage at the outlet can vary +/- 10% or sometimes even more, and exactly 120VAC is unusual.When a device is "rated" for a certain power output that usually means it is the maximum safe output that the design anticipated. So if there is a dependency of power on the available voltage (could be interpreted as a "resistive" characteristic, or simply as no line regulation), this figure only applies at the highest anticipated supply voltage, which could be ~140VAC.

author: Yansi
author_rank: Super Contributor
timestamp: January 09, 2016, 06:22:44 pm
content: Power factor is the ratio ofactivereal power andimaginaryapparent power.imaginaryapparent power S is your Volts times Amps you measure with (trueRMS) multimeter. The active power P is what you measure with a special device called wattmeter.cos phi = P / SNote: It also does apply that S^2 = P^2 + Q^2When the power factor "cos phi" reaches 1, it means there is no reactive power Q (Q=0) and S = P.In reality, the situation is a little more complicated, because there is also a deformation power D, which causes distortion.For your 1040 watt measured by wattmeter and 1512 voltamperes (beware!! imaginary power is in VA not W!) gives a power factor of 0.69 (1040/1512)EDIT: Corrected terminology.

author: retrolefty
author_rank: Super Contributor
timestamp: January 09, 2016, 06:47:39 pm
content: Power = V x I is true only if the attached load is a pure resistance. This is one case where the Wikipedia explaines it pretty well. Welcome to the world of AC where there is resistance, inductive reactance, and capacitance reactance to account for.https://en.wikipedia.org/wiki/Power_factor

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 09, 2016, 09:29:53 pm
content: Quote from: Yansi on January 09, 2016, 06:22:44 pmReal to apparent*.The equation is right though

author: Yansi
author_rank: Super Contributor
timestamp: January 09, 2016, 09:47:24 pm
content: Sorry, I have some gaps in the exact english ee terminology, sometimes.I've fixed that. Thanx for correction.

author: tommygdawg
author_rank: Contributor
timestamp: January 10, 2016, 07:58:55 am
content: Hey all, I'm still confused but I'll be a bit more specific (and let's just drop power factor for the moment).I did measure my voltage at the wall and it's right around 120 plus/minus literally 1-3 volts. I don't remember the exact number but it's right around 120. So, for the purposes here, let's use that number.My main issue is I don't understand where the discrepancy in the numbers is coming from. I'm measuring a pull of 12.5 amps (at the wall) at a voltage of 120 volts. I'm also measuring (at the wall) a wattage of 1040 watts.According to the math, 120*12.5 = 1500 watts. If I did it in reverse then the math gives me incorrect amperage (1040/120 = 8.67 amps). So my question is, why am I measuring a wattage of 1040 when the math tells me I should be measuring 1500 (or conversely measuring an amperage of 12.5 when the math tells me I should read 8.67)?Just for the record, I happen to know that these lamps draw quite a high current and there's no way it's *only* drawing 8.67 amps, so it's much more likely that the 12.5 amps figure is more accurate.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 10, 2016, 09:08:10 am
content: If you could look at the current and voltage waveforms, you would see them either out of phase, or current looking spikey. Power != V*I because math.Tim

author: flynwill
author_rank: Regular Contributor
timestamp: January 10, 2016, 04:37:18 pm
content: If your kill-a-watt is like mine it will happily report Volts, Amps, Watts, VA, Hz and Power factor. (As well as tally kWh if you leave it plugged in).When you measure Volts and Amps separately and multiply them together you get "VA" (for Volt-Amps) what Yansi referred to as "apparent power".If the load were purely resistive (eg traditional incandescent light bulb) the Voltage and Current waveforms would be exactly in phase and the VA and Watts readings would match.However with a reactive load, like your ballasted lamp, the waveforms are out of phase. Which means during the cycle of the AC waveform there are times when power is flowing towards the load and other times when the power is flowing back into wall (having been momentarily stored in the magnetic fields of the ballast). So to measure power in this situation you must measure the instantaneous current and voltage at many points through the AC cycle, do the Watts=Volts*Amps calculation at each point (keep track of whether the power is positive or negative) and add it all up. That's exactly what your Kill-a-watt is doing which is what makes it such a nice little meter for $40.If you select "VA" on the kill-a-watt it should display the same 1500 you get when you multiply the Volts and Amps. Select power factor (PF) it should report Watts/VA.The VA number is important. Those 8.67 Amps are real amps and are heating up the wiring just like 8.67 amps of purely resistive load would. That's why power transformers are usually rated by "VA" and not by "Watts".

author: coppice
author_rank: Super Contributor
timestamp: January 10, 2016, 04:44:07 pm
content: Quote from: flynwill on January 10, 2016, 04:37:18 pmThat's not quite true. A water heater is a resistive load. An incandescent light bulb isn't. The temperature, and therefore resistance, of a lamp filament varies so much through the mains cycles that most lamps produce about 10% to 20% THD in their current waveform, even when the voltage waveform is a pure sine wave. The harmonics in the current affect the RMS current measurement, and hence the apparent power measurement. They do not affect the active power measurement, because this is only sensitive to things that correlate between the voltage and current waveforms. The result is that active and apparent power readings do not match, even though the fundamentals of the voltage and current waveforms may be exactly in phase.

author: flynwill
author_rank: Regular Contributor
timestamp: January 10, 2016, 05:17:55 pm
content: Quote from: coppice on January 10, 2016, 04:44:07 pmHmm well there may well be incandescent bulbs out there where the filament fluctuates in temperature as you say, but I would assert that most don't (or the effect is very small). Since I had the bits easily to hand I plugged a 150 W incandescent bulb into my kill-a-watt, and the PF display bounced back and forth between 0.99 and 1.00. (Of course I haven't tried to do the math, perhaps you could have 10-20% THD in the current waveform andstill have a power factor 0.99).But yes there are plenty of loads (like just about any DC power supply) where the current waveform will be in phase with the voltage but very distorted leading to differences between VA and Watts just as you say.In any case this is way beyond the OP's concerns.

author: coppice
author_rank: Super Contributor
timestamp: January 11, 2016, 01:51:38 am
content: Quote from: flynwill on January 10, 2016, 05:17:55 pmI assume this is a faith based belief, because if you try actual current waveform THD measurement on normal commercial incandescent lamps for room lighting you will find considerable THD. I don't know how the kill-a-watt measures power factor. Do you? There are several approaches. The influence of ADC noise on small current measurements means most meters derive power factor from active and reactive measurements. However, reactive measurement is also problematic when harmonics are present, unless you have the cash needed to build a full scale power quality analyser.