Watts vs Volt-Amps - huh ??

[inetdog]

Now if you are running a generator it is important with fuel burn.

Could you elaborate a bit on why a higher VA (lower power factor) load will require more fuel energy from a generator but will not require more battery energy from an inverter?

I would expect the two cases to be fairly similar.

[Sunking]

Not a bad thing to do, but keep in mind that the reactive component of the current is not necessarily sucking energy from the batteries. It is causing resistive losses in your interconnecting wiring, and definitely causing resistive losses inside the inverter.

Ditto but it is not enough to worry about.

All one needs to know and do is not exceed the inverters ratings. If the inverter is rated 1 Kva @ .8 PF then do not exceed those ratings. All you need to know is the inverters worse case efficiency which is already taken care of in my previous post. Otherwise you are just spinning your wheels.

You can prove it to yourself with a very easy controlled experiment I have done many times. Here is an example. 12 volt battery > 1 Kva Inverter > 500 Kva load at .7 PF. All I need to know is at the input of the inverter I see 12.4 volts @ 31.4 amps or 390 watts inverter input. That means my inverter met its 90% efficiency. It takes 390 watts of battery power to power a 350 watt true power load. I don't give a crap about 500 KVA as it is not real so quit chasing that ghost.

Now if you are running a generator it is important with fuel burn.

[Sunking]

Gotcha. Thanks. I'm accelerating up the learning curve now. I wanted confirmation that what I saw was real and needed some nudges in the right direction with regards to where I should be heading. That link helps too. I seem to remember some of it from ECE many, many moons ago.

The quest is quite worthwhile. At current conditions and setup, my greenhouse is consuming an effective drain on my battery bank of 3.6 kwh per day. In a perfect world (PF=1), that drain would only be 2.1 kwh. That is some serious improvement potentially waiting for me to exploit.

Thanks again to all the responses.

Robert take this advice. You are looking in the wrong spot. Don't worry about VAR's. What is important is watt hours and efficiency of the chargers, battery charging efficiency and inverter efficiency.

Here is how electrical work is done.

• Measure everything with a micrometer.
• Mark all measurements with Chalk
• Cut with an axe
• Collect check your done

Quit sweating details and just use proven design concepts. It has enough over kill to CYA built in. If you were my employee I would fire you for wasting so much time. Get the job done and collect the fee. Determine daily watt hours, forget VARs' it is an imaginary issue and chasing ghost. Multiply the daily WH by 1.5 and your done as that is how much energy you must generate at the panels. That accounts for all the losses. Now relax and take my advice. I have been doing this professionally for over 30 years, I know what I am talking about.

[inetdog]

Me thinks perhaps my lights have an electronic ballast, and it's the capacitive value of that reactive power that is correcting my pump power factor (and the inductive power of my pump electromagnets that are correcting the light ballast). If true, one pump w/ two lights and/or two pumps w/ one light running should move the measured power factor around.

Great explaination.... http://openbookproject.net/electricC.../AC/AC_11.html

and great lead inetdog!! I can probably decrease my waste and optimize my battery drain by balancing this greenhouse leg of my circuit with a power correction capacitor or inductor. Since the pumps run constantly, I'll need to bring it in and out of circuit as the light come on and off. Neat stuff.

Not a bad thing to do, but keep in mind that the reactive component of the current is not necessarily sucking energy from the batteries. It is causing resistive losses in your interconnecting wiring, and definitely causing resistive losses inside the inverter. Depending on the exact design of the inverter, it may be causing little or no difference in the current drawn from the battery or a large difference in the current drawn from the battery and a corresponding waste of power inside the inverter.
So if you go from 200VA to 150VA by correcting the PF in the circuit you will generally not be saving 50W in battery drain.

[Robert1234]

Gotcha. Thanks. I'm accelerating up the learning curve now. I wanted confirmation that what I saw was real and needed some nudges in the right direction with regards to where I should be heading. That link helps too. I seem to remember some of it from ECE many, many moons ago.

The quest is quite worthwhile. At current conditions and setup, my greenhouse is consuming an effective drain on my battery bank of 3.6 kwh per day. In a perfect world (PF=1), that drain would only be 2.1 kwh. That is some serious improvement potentially waiting for me to exploit.

Thanks again to all the responses.

[Sunking]

I assume Z is the VA reading, X is the resistive load, and thus Y must be the reactive load. How does the reactive load exhibit itself?

Robert you are on the right track but mixing up unlike terms.

In an AC circuit resistance to current is called Impedance. It is a complex ratio of Voltage to Current phase relationship in an AC circuit. The Impedance is the Resistance plus Reactance and expressed mathematically as:



Where

Z = Impedance
R = Resistance
X = Reactance
j = +/- lead or lag

In the DC world of things there is no Reactance or Impedance. Life is simple and Ohm's Law is the domain of DC.

In a reactive circuit, current and voltage are not in phase. In an Inductive circuit like a motor Voltage Leads Current (ELI), and in a capacitive circuit Current Leads Voltage (ILE). Inductance is opposite of Capacitance and they cancel each other out. So what you end up with Is True Power + Reactive Power = Volt Amps aka VAR's. True power is the heat value, the part that does the real work. In a DC circuit Power = Voltage x Current. In an AC circuit voltage and current are not IN PHASE with each other. So you end up with two power products. The True Power will be the Product of Voltage x Current at a moment in time with respect to the phase relationship. Appearant Power is the Product of Voltage x Current without respect to the phase relationship and will always be higher than True Power. This is where Power Factor comes into play. PF = a number of 1 or less. If current and voltage are in phase the PF = 1 which means it is a purely resistive circuit just like DC. If the voltage and current are out of phase PF is less than 1.

So if I were to say to you you are using 10 KVa with a PF = .8 you know immediately you are using 8 Kw of True Power, and 2 Kw are being reflected back to the source heating up the wiring and and windings in your generator. You are also burning 2 KW of excess fuel as waste heat. This is why commercial and industrial users get billed for Apparent power. Residential customer do not.

That is far as I am going to take you right now. If you are a scientist, then you know high school level Trigonometry and can find out more on your own. Here is a good simple place to start your journey. The trouble you are having is you are trying to just add things together and you cannot do that. It is a complex formula and also depends if the circuits are in Series or Parallel. You canno tsum 1 + 1 = 2 as it is really 1.41

[Robert1234]

Me thinks perhaps my lights have an electronic ballast, and it's the capacitive value of that reactive power that is correcting my pump power factor (and the inductive power of my pump electromagnets that are correcting the light ballast). If true, one pump w/ two lights and/or two pumps w/ one light running should move the measured power factor around.

Great explaination.... http://openbookproject.net/electricC.../AC/AC_11.html

and great lead inetdog!! I can probably decrease my waste and optimize my battery drain by balancing this greenhouse leg of my circuit with a power correction capacitor or inductor. Since the pumps run constantly, I'll need to bring it in and out of circuit as the light come on and off. Neat stuff.

[Robert1234]

So it's "real"!! You can make more effective use of the power by how you arrange your components - Cool

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.

Maybe not over my head, but certainly up to my eyeballs though I'm a ChE by training; make my living as a research scientist; and have dabbled in nuclear, civil, mechanical, and electrical engineering during my life. Solar & farming are a couple of my more intense hobbies for now.

I'm a bit visual in my thinking, so it seems DC can be thought of as a 3-D world (if we take into account time) while AC adds a 4th dimension. This gives us new things to consider, and I've learned that in ALL the sciences I've been involved in that one axiom always remains true, "there ain't no free lunch". I get the vector math with it's imaginary components, and would like to more fully understand and clarify it's components though as they portray themselves in the "real" world I live in. Let's stick with the X-Y coordinate examples (X^2+Y^2=Z^2) as that may be easiest to talk from since that is where we started.

I assume Z is the VA reading, X is the resistive load, and thus Y must be the reactive load. How does the reactive load exhibit itself? Does it come out as magnetic flux? Why has this component changed by my combination of components?

[inetdog]

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

That you could not get the PF of the parallel combination by dividing the sum of the two wattages by the sum of the two VA numbers.

[Sunking]

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

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

[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.

[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?

[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.

[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]

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.