Learning about power losses in the system

[Steve]

I never realized just how much power is lost when using an inverter.

I thought for a while my 12v 240ah deep cycle golf cart batteries might not be delivering their rated amperage.
I can burn a couple 120v 15w CFL's outside on a back yard lamp post overnight (About 9hrs) through my inverter and it takes two sunny days for my panels delivering 6.8ah to replace what the inverter + the CFL's consume.

Using a killa-watt meter, the two lamps on the post consume only 0.42ah @120v.

To test the batteries I connected a single 50W 12v incandescent bulb directly on my batteries. The lamp drew about 4ah tested on my meter.

I started with a full battery @ 12.76V.
I left the single bulb on for 9hrs
The battery tested 12.27v afterward.
(The battery tested unloaded after "resting" unloaded for several hours)

Hmm. This seems to me what I'd expect from my batteries, and is about what I was getting using the two CFL's and inverter.

So why do the CFL's drain about the same as wasteful incandescent?

One reason I figure is inverter loss.

For example I calculate my inverter loss to be 12.6% under the CFL load.
My CFL's draw 0.42ah but the ah draw between my inverter and battery is 3.35ah. Wow, what a difference between inverter input and output. It's close to the incandescent draw!

I am considering buying some 12v CFL's to avoid the loss. Two bulbs would consume around 2.2ah which is about 1ah less than the 120v ones through the inverter.

It may be a good consideration.
When my project is all done I hope to end up with two yard posts containing 3 bulbs each, and maybe two more single bulbs one on the shed and another on the remote car port.
That would be about 18ah consumption using 12v or about 27ah with 120v CFL's per hour.

In other words,
162ah consumption using 12v for 9hrs of night lighting
243ah consumption using 120v for the same duty.

Choosing between 162ah vs 243ah should be a simple choice, but I know that DC and wire impedance may be an issue with a 12v system.

The first yard post is about 150 feet from the batteries, second post on the string about 200 feet from the batteries, the single shed CFL is around 250 feet from the batteries and the final light under the port is around 300 feet.

For anyone familiar with my project, you'll know (and I know) my current solar project won't support that kind of consumption. But I hope to eventually expand to 8 panels delivering about 60ah for maybe 240-300ah of power on a good day. I also plan to expand my battery capacity to a total of 720ah, for a practical usage capacity of 360ah.
I realize the lights may not come on every night.

I hope I didn't make this post too confusing. That is one thing I'm very good at. If this is a dumb plan I know you will tell me.

[Sunking]

You are missing something very important which no DIY knows about unless they have an electrical background. You applying DC principle to an AC circuit. It will not work.

You are not even reading your Kill-A-Watt meter correctly. You are looking at watts which is DC. You need to be looking at the Volt-Amps and power factor, the wattage is meaningless without the Power Factor.

So let's take a example your meter said 50.4 watts right? But what is the power factor? For inexpensive CFL my guess is about .6 PF. So with a .6 PF the Volt-Amps = 50.4 / .6 = 84 VA.

Now to calculate the load on the batteries we need to know the inverter efficiency. Lets say at this power level it is 90%. so the load on the batteries will be 50.4 / .6 / .9 = 94 watts. So when you were thinking all I had was 30 watts of CFL, you actually have a 94 watt load. The system uses 3 times more power than you thought. Surprise.

FWIW the wattage listed on the CFL is the tube wattage and does not include the ballast wattage. That is why your meter read 50 watts instead of 30 watts. However even with the ballast power a CFL is far more efficient than incandescent lamps.

[Mike90250]

Sunking beat me to it with the power factor.

Also, with a non-sine inverter, the ballast will burn more, and the inverter has roughly 20% loss too. A modern, efficient inverter like the Morningstar suresine, may give better results than a Kragen Auto $28 inverter.

Straight 12V , you will have a lot of wire loss.

[bigsmile]

Is there a case against CLF when the meter is not net metering?

I mean if the CLF has a power factor of .5, and for simplicity assume that all other things have power factor of 1, and the meter is the newer kind that won't spin backward. Will the CLF be charged double the power it actually uses by the utility company?

[russ]

The power company has to supply more power than it can bill for I believe.

In residential settings the problem with power factor is on the utility side as you are billed for kWh only.

[bigsmile]

But Kwh is still read from the meter. So it's all boil down to how the meter reads the power consumption.

Also, my understanding is that the utility don't really need to generate the extra volt-amperes caused by power factor being less than 1. It just need to have capacitors that turn them back into phase.

[Steve]

Thank you both for the responses.

You provided a most valuable calculation formula and I did not understand PF.

If I plug my K-W meter into a normal 120v outlet then one single 15w CFL into it my PF reads 0.63. The VA reads 24. If I understand this calculation correctly assuming 90% inverter efficiency:

It's watts divided by PF divided by inverter efficiency equals battery load.
Is that the correct fomula?

The single 15w CFL bulb in this calculatoin would be 15(W)/.63(PF)/.9 (Inverter loss)= 26.5ah battery load.

If that is correct, running that single CFL on the inverter for 9hrs would consume 238.5ah. Wow! My current battery is rated 240ah. No wonder the SOC is only about half after one 9 hour night on my post lamp - and my panels take two good sunny days to recover it.

Can anyone guestimate how long a 12v power line can be before prohibitive losses occur?

I'm glad I'm asking these questions and people are willing to be helpful.
Thank you again.

[bigsmile]

I think how much energy CFL waste in the form of reactive power depends on how the inverter handle it. If the inverter can store the reactive power and feed it back to the CFL, it may not waste that much energy.

[russ]

Sunking or Mike should be along before long. They are the guys who know what they are talking about.

[Sunking]

Is there a case against CLF when the meter is not net metering?

No not really. Utilities cannot charge customers for VAR,s only watts. It is commercial and industrial customers who have to pay for VAR's + Watts. Residential PF does not add up to a hill of beans.

[Sunking]

Also, my understanding is that the utility don't really need to generate the extra volt-amperes caused by power factor being less than 1. It just need to have capacitors that turn them back into phase.

No no no. The generator still has to generate the VAR's. VAR's are reflected back onto the wiring and burnt off as heat. Large industrial customers spends 10's of thousands of dollars installing capacitor PF correcting networks to correct the phase angle which lowers their bill significantly.

[Sunking]

The single 15w CFL bulb in this calculatoin would be 15(W)/.63(PF)/.9 (Inverter loss)= 26.5ah battery load.

The single 15w CFL bulb in this calculatoin would be 15(W)/.63(PF)/.9 (Inverter loss)= 26.5ah battery load.

No forget AH, it does not enter the equation. Batt load watts = 15 W / .63 / .9 = 46.2 Watts. However you are still mis-reading the Kil-A Watt meter. Read the meter watts and PF. You have not included the ballast power A 15 watt CFL should burn about 25 watts as you noted in the OP. Now you get the real data. well almost real because we are assuming your inverter is 90% efficient at a low power level. I can promise you it is not that efficient.

Anyway to calculate the load on the battery in watts = 25 w / .63 / .9 = 44 watts per bulb. 44 wats at 12 volts = 3.6 amps. So if the bulb burns for 2 hours you use 3.6 amps x 2 hour = 7.2 AH

[Sunking]

I think how much energy CFL waste in the form of reactive power depends on how the inverter handle it. If the inverter can store the reactive power and feed it back to the CFL, it may not waste that much energy.

No it do it does not work th eway you think. Power just doesn't go out there are circulate. It is burned off as heat on the wiring between the inverter and light. Heat = power = watts

[bigsmile]

In order for the energy to be burnt out as heat on the wiring, the current need to be really strong. I don't think it's anywhere near that strong.

According to OP, it's 0.42 Amp (I'm assuming that by ah, the OP meant amp). That won't cause any significant heat loss from the wiring.

I think in order for there to be significant heat loss, the inverter itself has to have significant resistance, which will mean that the inverter has low efficiency.

No it do it does not work th eway you think. Power just doesn't go out there are circulate. It is burned off as heat on the wiring between the inverter and light. Heat = power = watts

[Steve]

Reference

Ok, I believe I've got it now. I calculate the load and inverter to determine the watts, then divide the watts with 12(v) to get the ah load to the battery.

This is most useful information. Thank you again.