VA vs Watts motor load from genny/inverter

icarusicarus Solar Expert Posts: 5,436 ✭✭✭✭
Displaying my own ignorance here, but I have a question.

Situation is this:  1/2 hp motor load  draws ~7.2 amps running (understand starting current is MUCH higher).  Using a Kill-a-watt, it draws the 7 amps, shows ~250 watts running, shows 8-900 VA running and a PF of .6.  I have been running this on a Honda Eu1000 genny with success.  So the question is, if I run this on an inverter how much power am I going to draw from the inverter (not counting system loses).  Am I going to draw ~7.2*120v= 864 watts from the battery, or will I be drawing the ~250 watts shown on the KoW meter?

I am thinking about changing the motor out for a 1/4 hp motor that draws about 5.5 amps.  This is a simple old fashioned wringer washer that was gasoline powered, but I have installed the motor on, so the load on the motor is pretty small.

Thanks for any advice.

Tony

Comments

  • BB.BB. Super Moderators, Administrators Posts: 32,230 admin
    Tony,

    The Kill-a-Watt meters are probably not very accurate when you have a non-linear (non-sine wave) waveform (small inverter genset with relatively large inductive load).

    The general equation for power/power factor/etc. is:
    • Power (Watts) = Volts (RMS) * Current (RMS) * Cosine angle between voltage and current = Volts * Current * Power factor (0.0 to 1.0 range)
    So, if you are measuring 850 VA (volt*amps) and a PF of 0.6, the Watts would be:
    • ~850 VA (from kaw) * 0.6 (from kaw) = 510 Watts (what the "math says")
    • 7.2 Amps * 120 Volts = 864 VA (very close to 850 VA from kaw)
    • 250 Watts reading from KaW meter (does not make sense based on above)
    From what little I have played with using a Kill-a-Watt meter, the math usually would work out like:
    • 250 Watts / 864 VA = 0.29 PF
    I would suspect a relatively loaded motor (generally, the heavier the load on a motor--closer to nameplate rating), the "better" the PF (higher the number) should be.

    If we assume that the motor is closer to 0.6 PF (a reasonable number, 0.29 is not for a loaded induction motor), I would suspect that 510 Watts is closer to reality... Anyway, you can confirm by measuring the current on your DC bus to the AC inverter input... The power drawn from the DC Bus will be Watts (Vrms *Irms) and you can quickly confirm which it is the more accurate number):
    • 510 Watts (AC RMS) * 1/0.85 AC inverter eff * 1/12 volt nominal DC bus = 50 Amps RMS @ 12 VDC bus
    • 250 Watts (AC RMS) * 1/0.85 AC inverter eff * 1/12 volt nominal DC bus = 24.5 Amps RMS @ 12 VDC bus
    I used 12 volts in the above... If you have a 24 vdc battery bus, use 24 volts--Or more accurately use the measured Vbus (like 28.0 volts during day when charging the battery bank). AC inverters are "constant power" DC loads... Power=Volts*Amps ... If volts fall, amps must increase to keep constant power (250/510 watts / etc.).
    Notice that I am using RMS (root mean square) which basically converts from your arbitrary AC wave form to the DC (or RMS AC) amount.

    DMMs come in both "direct reading" and RMS reading types. The direct reading assumes a sine wave and simply measures the "peak voltage" and multiplies by 1/(sqrt 2):
    • 170 VAC sine wave peak / 1.414 (sqrt of 2) = 120 VAC RMS
    And 120 VAC RMS "contains the same amount of power" as 120 VDC.

    Different waveforms have different conversion factors (sine wave area under the square of the square of the curve is the sqrt 2 factor).

    "Cheap meters" read the peak wave from voltage (or current) and use the sqrt of 2 as a conversion factor (assumes sine waves). The Kill-a-Watt is a "cheap meter"--I would assume. And they read once or several times a seconds the voltage/current/etc.

    The more expensive meters are True RMS reading, and they take something like 50,000 reading a second and "do the math" (sum of the squares, then square root of the average) to get the "True RMS" value.

    So, the long way around is that the meters you have probably are not reading a "clean sine wave" from the inductive motor+small inverter-genset and are not giving you accurate numbers.

    And with a simple 2 lead meter, even a True RMS can only give you Vrms and Irms and you can calculate VArms (volts*amps).

    To measure real Wattage for arbitrary waveforms, you need to measure both Vrms and Irms (and phase between the two) to calculate true Watts. The KAW can measure both--But is just not always that accurate.

    When you measure the DC current for the inverter--You have another issue here... The DC current to the AC inverter is actually a "sine squared" waveform with a 120 Hz frequency... Remember your AC inverter is outputting a 60 HZ 120 VAC waveform, and when the volts crosses "zero volts", there is no power transfer--So the input current to the inverter is that 120 Hz "pulse train". And to accurately measure that current pulse train, you need a True RMS reading current meter... Here is an example:

    https://www.amazon.com/gp/product/B019CY4FB4 (True RMS reading AC/DC Current Clamp DMM--Medium priced)

    Sorry for the long answer... AC and pulsed waveforms are a pretty complex subject--And adding Power/PF/etc. measurements into the equation--It is very difficult to get "accurate" power measurements with typical equipment around the house. Even with "lab grade" equipment, two different brands will give slightly differing results (even though both are calibrated meters). And I would classify "accurate" power measurements on solar with home equipment as being "accurate" if within 10% of true value. And as you can see the conversion between VA and Watts (Power Factor) can easily range from 0.5 to 0.8 (0.95 is "near" perfect power factor correction). So just the PF SWAG is almost a 50% error by itself... Let alone the other factors (like 1.414 to 1 peak sine wave to RMS conversion).

    Here is a website that does a very good job of putting most all of the AC issues into one (very long/complex) page:

    http://www.screenlightandgrip.com/html/emailnewsletter_generators.html

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • icarusicarus Solar Expert Posts: 5,436 ✭✭✭✭
    Thanks Bill,

    so what it looks like is that it should (as currently configured) draw ~550 watts running, which sounds about right.  That explains why it seems to run fine on the little Eu1000.  Doesn’t always start when it is cold, probably due to thick oil in the gear box, but a couple of false starts and it runs.  I guess I will let sleeping dogs lie

    thanks as always,

    tony
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