Watts vs Volt-Amps - huh ??

[Robert1234]

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.

[SunEagle]

[ATTACH=CONFIG]2966[/ATTACH]

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.

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.

[Robert1234]

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

[Robert1234]

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)

[inetdog]

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.

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.

[SunEagle]

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.

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.

[inetdog]

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.

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.

[SunEagle]

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!

Got it. Thanks for the lesson.

[inetdog]

Got it. Thanks for the lesson.

You are welcome. I am just very surprised that nobody pointed it out earlier. I guess you are in good company.

[SunEagle]

You are welcome. I am just very surprised that nobody pointed it out earlier. I guess you are in good company.

I made an assumption that since both loads were on the same electrical phase the amps would be additive. Forgot about vectors.

[inetdog]

I made an assumption that since both loads were on the same electrical phase the amps would be additive. Forgot about vectors.

Vectors have this nasty habit of coming up whenever PFs are less than 1.0.

[SunEagle]

Vectors have this nasty habit of coming up whenever PFs are less than 1.0.

Yep. Add in harmonic distortion and you then have more issues correcting the PF. That light fixture probably does have the switched power supply like a computer which will generate harmonics.

[Naptown]

Would you two like to get a room?
Sorry no smilies on the mobile version or sarcastic font for that matter.
That went over my head. Think it was understood by the OP?

[inetdog]

That went over my head. Think it was understood by the OP?

Which OP? Mike90250 back in 2012 or Robert1234?

If it went over Robert's head, we can aim a little lower if he asks. Or be happy to discuss it interactively when he comes back.
What surprises me most is that Dereck let that point slip by.

A very simple example just to set the tone is this:

Q: When you have a motor with low PF and you want to add a PF correction capacitor to it, how can that work? The PF of the capacitor is close to zero, so how it can it improve the PF of the motor/capacitor combination?

Answer: Vectors.

[Sunking]

What surprises me most is that Dereck let that point slip by. .

I guess I missed something. What point are you referring too?