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Amps are a vector! You can add two different vector amperages and get a shorter sum vector than either of the two vectors you added. Look at the sides of a triangle!
Since you can end up with less amps than either of the two individual contribution values, it is not at all surprising that you ended up with less than their numeric sum.Leave a comment:
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I knew there was a possibility of the lights having an electronic ballast which could be the reason for the question about the power factor and the reduced measured load.
What I don't get is that the "kill a watt" measured each item's amps and then again together yet the total measured amperage was less then the two individual readings added together. While the total wattage can change due to the power factor the amps should not.Leave a comment:
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SPOILER: You cannot just add VA numbers.
There is a fundamental problem with the OP's approach. VA is a scalar number. It, and PF, do not have a direction.
The motors have a PF of .35 and are highly inductive.
The fluorescents have a PF of .61, and if they are using an electronic rather than magnetic ballast will have a high distortion PF and the phase PF will be capacitive.
When you add the two wattages, the sum is meaningful and is the total true power wattage.
But instead of adding the two VA figures, you have to add the two currents as vectors. That result will be less than the arithmetic sum of their magnitudes, and so the overall power factor can be higher than for the individual loads rather than being somewhere in between.
A perfect inductor alone has a PF of .00. A perfect capacitor alone has a PF of .00. Put them in parallel (choosing your values correctly based on the frequency) and you end up with an infinite impedance and a power factor of 0/0. For a real circuit, you will be left with the stray resistances of the two components and a PF of 1.0
If you put them in series, you end up with a different resonant circuit with near zero impedance and a PF of infinity/infinity. When you take the stray resistances into consideration, you are left with a very low resistance and a PF near 1.0.Leave a comment:
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Does it make a diff that I am only utilizing one leg of 120 from the Magnum inverter? (The other leg isn't even hooked up yet)Leave a comment:
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I've never seen such as this before untill I set up these pumps and these light fixtures. That is ALL that is on the circuit. Nothing else. I can tell you that the lights appear to be the same intensity in both measurements when they were on, and that the pumps were measured to be putting out the same gph whether the lights were on or not.
Very strange, isn't it
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While I can't dispute what you saw I can say that the readings don't make sense unless the loads changed between each readings.
I don't know how the pumps could be drawing 0.8 amps and the lights pulling 1.37 amp and yet together they are only pulling 1.58 amps. The total amp draw should be the sum of the two individual amp readings ( 2.17) unless during the "total" reading something was not "on".
There is a possibility that both pumps were not running at the same intensity as they were when you measured them separately. Or the lighting fixture is using a switched power supply and the load is not consistent. It only powers up for a brief time but stay lit until the next full power spike.
Can you use something else to test like a fan or incandescent lamp. Both are continuous type of loads.Leave a comment:
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PF.gif
Interesting....
First three columns are "raw" measurements (except where noted).
4th column is column 1 & 2 added together.
5th Column are values assuming the Kil-A-Watt accurately measures the wattage.
6th Column is Column 2 & 5 added together.
Comments Welcome & Encouraged!!!
P.S. Those Lights are 4 x 32 watt (published power) T8 bulbs as noted earlier. Guess my memory isn't THAT bad after all.Leave a comment:
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AC measurements come from the Kill-A-Watt meter - both Watts and VA. The pumps & lights are on a common line that I can feed through the meter. That way I can measure them individually and/or combined simply by turning them on/off. I can use the clamp meter to confirm amps back at the AC legs leaving the inverter (and have in the past), but have not as of yet on this paticular setup. Previously, it has lined up very well with Kill-A-Watt measurements (within 5% or so) but I'll note those measurements were on resistive loads (high PF).
DC Measurements are from a multimeter. The DC amps specifically were measure on the 3-0 positive lead to the inverter using the clamp meter. I elected not put put a shunt in this system.
Yes, loads were measured individually and the VA total didn't add up. That's what I am struggling with. Additionally, the Kill-A-Watt meter power factor is significantly higher on the combined load than either of the individual loads. Very strange and certainly unexpected (at least by me).
I'll get new measurements today or tomorrow and we'll go from there.Leave a comment:
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They are usually pretty accurate. Have you measured each of your loads individually as well as the total load on your inverter?
I'm trying to get an understanding of how you get both your wattage and VA measurements. If you are using multiple meters to measure (a watt meter, and amp meter and a volt meter) there may be some inaccuracies between them.
You mentioned calculating 183 watts DC based on a 6.9 amp and 26.5 volt measurements. How are you getting your AC wattage, amps and volts?Leave a comment:
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Yep... 4x32 Watt T8 bulbs - 108 Watts measured... It measured lower than expected. Like I say, memory isn't what it used to be. Maybe it was 118, but it is in my head that it was lower than the 4x32 value. I'll get back with the rechecked values, but if I read your responses correctly, I think I should conclude that VA should have been additive as I originally expected?
Are those little Kill-A-Watt meters accurate with regards to VA and Watts? (Cause that's what I'm using for AC Watts and VA)
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In section B you have 4 x 32 watt lamps that you only measure 108 watts total. I think it would be closer to 4 x 32 = 128 watts plus ballast losses of ~ 10% so add another 12 watts for 140 watts total. with 177 VA that is closer to 0.79 PF. Considering the majority of your load (lighting) has the higher pf it is possible your overall pf is around 0.75.
I would check all of your measurements again to understand you true Wattage load and then measure the voltage and amps to get your VA.Leave a comment:
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Ok dudes. I thought I was starting to understand power factor with regards to magnetic flux, etc, but now I'm not so sure. Does this make sense to you guys?
Individually Measured...
(a) I have two small pumps running on my aquaponics.
34.3 Watts Total
97.5 VA (0.35 PF)
(b) I have flourescent grow lights (two ballasts I believe)
4x32 Watt T8 bulbs
108 Watts measured
177 VA (0.61 PF)
Put them together on the same circuit
(c) Combined (all units on) I measured
136 Watts
181 VA (0.75 PF)
WTF???? I might be off on (b) a little bit because I'm recording it here from memory, but I'm pretty sure I'm close. Measurements (a) and (c) have been repeated a couple times now on the AC side. It just hit me last night when I was keying the data in my spreadsheet that the power factor was higher on (c) than I expected. I pulled out my DC clamp meter this morning before work to see if the inverter power was balancing and on the DC side I got at the inverter input: 6.9 amp @ 26.5v (183 DC Watts) - pretty much what I'm measuring at (c). I'm gonna remeasure everything again tonight to make sure I'm right on what I'm remembering.
Does this make sense? Is it real? Can PF of the combined items be better than the PF of the individual components?Leave a comment:
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