Originally posted by Sunking
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Watts vs VoltAmps  huh ??
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SunnyBoy 3000 US, 18 BP Solar 175B panels. 
So it's "real"!! You can make more effective use of the power by how you arrange your components  Cool
Originally posted by inetdog View PostIf 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.
I'm a bit visual in my thinking, so it seems DC can be thought of as a 3D 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 XY 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?Last edited by Robert1234; 07112013, 08:57 AM. Reason: Need to more clearly state my final questionComment

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.Comment

Originally posted by Robert1234 View PostI 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?
In an AC circuit resistance to current is called Impedance. It is a [U]complex ratio[/U] 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 [U]without respect[/U] 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 1.41MSEE, PEComment

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.Comment

Originally posted by Robert1234 View PostMe 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 [I]decrease my waste and optimize my battery drain[/I] 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.
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.SunnyBoy 3000 US, 18 BP Solar 175B panels.Comment

Originally posted by Robert1234 View PostGotcha. 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.
Here is how electrical work is done.
[LIST][*]Measure everything with a micrometer. [*]Mark all measurements with Chalk [*]Cut with an axe [*]Collect check your done [/LIST]
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.MSEE, PEComment

Originally posted by inetdog View PostNot 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.
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 > [B]500 Kva[/B] 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 [B][COLOR=#ff0000]390 watts[/COLOR][/B] inverter input. That means my inverter met its 90% efficiency. It takes [B]390 watts[/B] of battery power to power a [B]350 watt[/B] 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.MSEE, PEComment

Originally posted by Sunking View PostNow if you are running a generator it is important with fuel burn.
I would expect the two cases to be fairly similar.SunnyBoy 3000 US, 18 BP Solar 175B panels.Comment

Originally posted by inetdog View PostCould 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.
Every work with PWM motor speed controllers in a DC series wound motor? Why is it you can have a 600 amp controller pushing 600 amps into the motor, while only taking 100 amps from the battery? I will give you a clue. Back EMF.MSEE, PEComment

Good thing I don't do this for $$ then.
Forgive my ignorance, but quite honestly I'm "sweating the small stuff" simply because I'm curious. If understand what you are saying correctly, it is that the inverter capacitors (behind where I measure) that should correct for the reactive power I measured. Somewhat curious as to how that works as the required capacitive value moves around even in my simple examples.
Referring back to the data I posted, I did exactly as you suggest  measured DC Watts and compared to AC Watts.
For pumps, lights, pumps & lights respectively:
AC Watts  34.8, 102.0, 134.0
DC Watts (at inverter)  35.3, 158.5, 183.9
% Efficiency  99%, 64%, 73%
This is a Magnum 4024 inverter. Published efficiency is 87% min I believe. Measured AC/DC efficiency is well below that in 2 of the 3 examples.Comment

Originally posted by Robert1234 View PostGood thing I don't do this for $$ then.
Forgive my ignorance, but quite honestly I'm "sweating the small stuff" simply because I'm curious. If understand what you are saying correctly, it is that the inverter capacitors (behind where I measure) that should correct for the reactive power I measured. Somewhat curious as to how that works as the required capacitive value moves around even in my simple examples.
Referring back to the data I posted, I did exactly as you suggest  measured DC Watts and compared to AC Watts.
For pumps, lights, pumps & lights respectively:
AC Watts  34.8, 102.0, 134.0
DC Watts (at inverter)  35.3, 158.5, 183.9
% Efficiency  99%, 64%, 73%
This is a Magnum 4024 inverter. Published efficiency is 87% min I believe. Measured AC/DC efficiency is well below that in 2 of the 3 examples.
In a generator, the same sort of exchange of energy into and out of the generator will cause a varying mechanical load placed on the engine by the generator, and the flywheel effect of the motor/generator system will be able to level this out reasonably well, minimizing the extra fuel needed. [B]But[/B] the full current is actually flowing through the generator windings, and although the extra power consumption may not be an issue, the heating of the generator windings will be. For that reason, generators too can be specified in terms of both the maximum real power (capacity of the prime mover engine) and maximum VA at a PF above a specified limit (usually somewhere between .6 and .8, so check the actual generator specs.)
I am not sure why your inverter is so much more efficient in supplying a very inductive motor load than a resistive load (possibly some of the efficiency factors that Dereck mentioned), but that makes it a particularly good inverter for that use. And there might be a problem with your AC power measurements.SunnyBoy 3000 US, 18 BP Solar 175B panels.Comment

Originally posted by Robert1234 View PostFor pumps, lights, pumps & lights respectively:
AC Watts  34.8, 102.0, 134.0
DC Watts (at inverter)  35.3, 158.5, 183.9
% Efficiency  99%, 64%, 73%MSEE, PEComment

Have to agree on the numbers looking weird.
To directly answer your query, DC Power is measured by using a multimeter at the inverter input lugs (volts) and using a clamp meter on the 30 positive cable at the inverter. Precision of the measurement was pretty good (low standard deviation) but I cannot account for the accuracy other than saying this is pretty decent test equipment (Fluke). I have multiple volt meters and they all read the same. If there is a significant error, it is probably in the amps measurement. I do have another amp clamp that I can get another reading with (if I can dig it up). I also have a 150 amp shunt system but elected not to install it on these lines figuring I'd just clamp them whenever I wanted. Unless I was going to set up some method to integrate the shunt reading, I really couldn't see much of a purpose for it other than having a fancy digital display to look at.
Thanks once again for the discussion.
P.S. As this is a hobby, I data log all the daily charge / discharge readings. Have done so with resistive loads prior to this for months on end and have good records of coulombic efficiency for charge / discharge and can make extremely accurate predictions for when the inverter will go off line when the drain has high power factors on the AC side. The readings as of late with the new loads APPEAR to suggest the system is draining at the voltamp rate rather than the watt rate. I'll know better tomorrow am. We've had about two weeks of really overcast skies and rain here. Unless we get some really good sun today (to the tune of 120 amphr into the batteries minimum), the inverter will choke at 7 am tomorrow due to low volts when the lights kick in at the aquaponic units (assumin the drain is indeed following the voltamp relationship). If it's following AC watts, my batteries should only be 40% of the way to shut down and all will be well with the world.Comment

Originally posted by Robert1234 View PostHave to agree on the numbers looking weird.
To directly answer your query, DC Power is measured by using a multimeter at the inverter input lugs (volts) and using a clamp meter on the 30 positive cable at the inverter. Precision of the measurement was pretty good (low standard deviation) but I cannot account for the accuracy other than saying this is pretty decent test equipment (Fluke). I have multiple volt meters and they all read the same. If there is a significant error, it is probably in the amps measurement. I do have another amp clamp that I can get another reading with (if I can dig it up). I also have a 150 amp shunt system but elected not to install it on these lines figuring I'd just clamp them whenever I wanted. Unless I was going to set up some method to integrate the shunt reading, I really couldn't see much of a purpose for it other than having a fancy digital display to look at.
Ohm's Lab makes really good ones and there are dozens of other less expensive ones out there. Just make sure not to buy one any larger than you anticipate needing. Best accuracy is at or near full scale.MSEE, PEComment
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