A watt's a what?

[brkwind2]

As usual.  I'm trying to figure out how much is to be gained by replacing the lamp we have on most of the time with a 12vdc fluorescent light.  Currently is a 120 vac compact fl.  My Kill a watt meter claims that the bulb uses 10 or 11 watts at 120vac coming out of my black & decker inverter.  The HF meter on the 12vdc side feeding the inverter says it is taking 1.5 amps.  1.5 amps time 12 volts =18 watts.

Not what I expected.  I thought if I was using 10 watts on the 120v side, I'd see 100 watts on the 12v side.  Are my figures wrong?  Is my inverter majic?  Are my meters junk?  Will I ever get my black belt in electrickery?

[Ungrounded Lightning Rod]

If you were seeing 10 AMPS on the output you'd be seeing 100 AMPS (plus inverter losses) on the input.

You should see roughly (slightly more due to inverter losses) the same number of watts on the input as you pull from the output.  If the inverter were 100% efficient they'd match.

Watts are power - energy per unit time.  Average power in equals average power out plus losses as a consequence of the conservation of energy combined with the negligible energy storage capacity of the inverter.

Amps are charge carrier flow per unit time and volts are "electrical pressure", pushing the charge carriers around.  Power is flow times pressure drop, i.e. watts = volts times amps.  Energy is the integration of watts over time, i.e. watthours = watts times hours of run time (or kilowatt hours equals kilowatts times hours).

[brkwind2]

Ok, so I'm trying to treat watts like amps.  So I should see roughly the same watts in & out taking into account for conversion losses.  I'm using 1.5 amps on the 12vdc side which would be .15 on the 120vac side & .15 times 120=18 watts.  Close to what I had on the dc side which means my inverter is extremely efficient or my meters are off a little.  In any case it looks like replacing the 120vac compact flourescent with a 12vdc flourescent isn't going to be a super big difference.  I need to look more at LED's.  Still, couldn't you assume at least a 10% increase in efficiency going to the 12vdc fl.instead of converting for the 120 vac fl. if your getting the same amount of light per watt?

Thanks for the reply.

[brkwind2]

No, make that I'm using 10 or 11 watts according to the kill a watt but I'm feeding it 18 watts so I'm losing 8 watts in the conversion.  So much for my majic inverter.  Maybe there would be a noticeable difference with a 12v light & no conversion.  Anyway, thanks for the info.

[bob g]

the biggest problem i see, is the quality of instumentation

and even with high quality meters, you can get erroneous readings

your B&D inverter is likely around 80% efficient

so if we are to believe your killawatt, which i have found to be pretty darned

accurate the lamp burns 10-11watts

lets call it 11 watts, which is probably pretty close

now 11 watts / 80% = 13.75

so i would expect your dc supply side to take about 14 watts

14 watts / 12 volts = 1.17 amps

so i would expect that your harbor freight amp meter is off, but why

well maybe cause its cheap? but just as likely

and in my experience...

the inverter places some wierd AC crap on the DC buss which skews the reading, or...

a combination of cheap meter and crap on the DC buss.

believe me i am yanking out what few hairs i have on my head right now fighting to get accurate amp reading because of crap overlaid on my DC buss making it impossible

to get anywhere near accurate readings.

yours skew to the inefficient side of things, mine skew to so far into the overunity camp that i should be selling tickets to witness the event!

anytime the reading exceeds what i know to be a reasonable efficiency i automatically become suspect of every facet of my measurement and instrumentation.

cheer up,, it only gets worse!

bob g

[Ungrounded Lightning Rod]

One thing to consider is that your meters may not read amps very well for a CFL.  These devices start with a rectifier and a filter capacitor.  If the capacitor is sized large enough to keep the lamp from flickering it will retain an appreciable charge from one half-cycle to the start of the next.

The result is similar to the cutin speed phenomenon on a windcharger:  On the upswing the voltage will rise but no current will flow until it exceeds the residual charge voltage of the capacitor and on the downswing the current will drop to zero shortly after the peak of the waveform.  So there will be a high current near the peak of the AC waveform and no current the rest of the time, rather than a current waveform matching the voltage waveform as you'd see with a resistive load.

If your ammeter is an averaging type (mechanical movement) multiplying a voltage and a current reading will estimate your power as lower than it is, because most of the current happens when the voltage is high, resulting in more watts compared to an average current flowing over the whole cycle.  An electronic ammeter may have any of a number of oddball responses to an oddball waveform, depending on how that particular meter works.  The only way to get the true power with goofy waveforms is to use a true integrating wattmeter, which averages the instantaneous product of the current and voltage.

So don't trust the output metering.

Yes, using a 12V fluorescent will eliminate the inefficiency of the inverter and should save a significant amount of power.  (Especially if the inverter is large compared to the load.  A significant fraction of the inverter's losses are "idling power", which goes up with the size of the inverter and doesn't depend on the actual load.)  You should save at least 10% and probably more by going to 12V CFLs.

LED arrays are just getting near the efficiency of compact fluorescents.  But they potentially can exceed fluorescents considerably and are improving rapidly.  They're more pricey now but the price is dropping and the color output balance improving with time as well.  I'd wait a couple years before switching to them.  (By which time the CFLs you're buying now will be about due for replacement.  B-)  )

[Ungrounded Lightning Rod]

Yep that sounds about right.

[brkwind2]

I never thought about that.  Having ac on the dc side from the inverter.  I'll put the meter on ac & see what I get just for curiosity.  Good info.  thanks.  Good luck with your problem.  

[Flux]

What really matters is the light you get out for a given input power. Unless you have suitable photometers to measure this you are just guessing.

Fluorescents all work from high voltage, any attempt to run from a battery needs an inverter. A big inverter intended to run many lights will have a significant standing loss that may be out of proportion with only one light. It may reach over 90% efficiency with more load.

The low voltage fluorescents also have an inverter in them and this is sized to the lamp so its standing loss may be lower. I have generally not found these low voltage lamps with integral inverter to be as good as their high voltage brothers as far as lumens/watt go so you may have less of a wattage drain from a low voltage unit but if you check with a photometer you may find it's not as bright, but if you can't notice the difference then you will be happy with the lower current drain.

It is possible to build high efficiency individual inverters for low voltage lamps but they are complicated and to keep cost low for a not very common commercial product the manufacturers use fairly crude circuits.

Measuring power with inverters and nasty waveforms is also a problem and I doubt that any of your measurements are particularly accurate.

Don't use a large inverter to drive one small lamp. Your B & D inverter will do much better running a string of lamps and the efficiency will be higher.

LEDs offer a major improvement if you want focussed local lighting. For general room lighting they are not so good but are improving all the time and I have a feeling that the days of the compact fluorescent are numbered. Again LEDs have to be driven properly to manage good efficiency, cheap fittings with current limiting resistors are not going to get you the highest efficiency.

Flux

[bob golding]

hi, i have both 11 watt 240 volt cfl's and a led batten with 65 leds. using a doc wattson meter and a 24 volt inverter i get the following readings. cfl 18 watts led 4 watts. the light output is about the same using the "does it light up the space well enough" method.  the main disadvantage is the light is a little too blue,and the leds are still expensive. 25 UK pounds about 40 dollars. you can buy 12 volt led battens but have yet to try those. the leds are much more rugged and last for 50000 hours or so.

bob golding

[cardamon]

One other consideration is the cost of the 12 vdc cfl's.  They are maybe 15 bucks compared to  a few bucks for the high voltage ac ones.  I did a cost benefit analysis on running 120v vs 12 v to my shop for lighting, and found it more economical (not even considering the larger wire size and cost for 12v) to deal with less efficiency  from the inverter and pocket the $90 saved on the bulbs, or in theory apply it to more solar capacity which would more than make up for the losses

[Photowhit]

(not even considering the larger wire size and cost for 12v)

I think this is an important point, more so since you don't appear to have a good grasp of the basics yet(it'll come...)

Running lower voltage (AC or DC) means you'll lose more current in resistance(heat) in the same size wire.

In learning about this you might search for "wire loss" chart or calculator. Be sure you check out the differences for different voltages.

If your right under your panels this might not matter much, but once your "transporting" current wire loss is an important componet.

.

[Madscientist267]

Tried throwing some stiffening caps before and after the shunt to suppress the transients?