Learning about power losses in the system

[Sunking]

Thank you Sunking.

I've done some reading about battery characteristics and have come across exactly what you posted. I was unsure about how much the higher loads would actually decrease ah performance. Thanks for the links.

Peukert is most notable in Lead Acid batteries, not so much in other chemistry like Lithium and and NiMh

[Steve]

Thank you Sunking.

I've done some reading about battery characteristics and have come across the theory you posted. I was unsure about how much the higher loads would actually decrease ah performance. Thanks for the links.

[Sunking]

Steve not trying to confuse you any more, but there is also another factor you might are overlooking called the Peukert Law. Simple definition is: As the rate increases, the battery's available capacity decreases.

When you buy a deep cycle battery the amp hour rating is specified at a 20 hour discharge rate.Example for a 200 AH battery, the 20 hour discharge rate = C/20 = 200 / 20 = 10 amps. So in theory if you took a fully charged 200 AH battery, put a 10 amp load on it, the battery should last 20 hours before being completely discharged. With me so far?

But what happens is say I put a 100 amp load on the battery? How long will it supply 100 amps? The formula is H = AH/A = 200/100 = 2 hours right? WRONG, dead wrong because Mr Peukert just robbed you. Your 200 AH battery went from 200 AH down to 100 AH. so the real answer is 100 ah / 100 a = 1 hour, not 2.

Look at this specification for a Surrette EIW 250 It is a popular golf cart battery that can be used in RE applications. It is specified a 6 volt 250 AH battery at 20 hours. Now look and see what the various AH rating at say 100 hours, 10 hours, down to 1 hour. The range is 333 AH down to 90 AH.

The reason I bring this up is because you tried to figure out the time large loads will last like a fridge, AC, and heater. Your calculations would be way off.

[Steve]

I did some fun calculations.
Note that I am not going to actually use these heavy loads, but listed them anyway. I wouldn't drain my battery to 100% DOD either. These are just some 240ah battery running times based on a 90% efficient inverter:

CFL 15W/.63PF/.9 = 26W/12v = 2.2ah = 110 hrs
TV 20" 29W/.57PF/.9= 57W/12v = 4.75ah = 50 hrs
TV Rear Projection 110W/.65PF/.9 = 188W /12v = 15.7ah = 15 hrs
Fridge, kitchen 178W/.99PF/.9= 200W /12v = 17ah = 14 hrs
Air Conditioner 725W/1.0PF/.9= 805W/12v = 67ah = 3 hrs
Space Heater 1266W/.99PF/.9 = 1421W/12v = 118ah = 2 hrs

Note: The Fridge and AC was measured while the compressor was on. It didn't measure starting or fan (only) readings.

I've read that very low current draws may deliver longer AH times.
High current draws will last much shorter than the rating.

[bigsmile]

The watt is still calculated by the electricity meter. As I understand, you can't really measure watt directly, it has to be calculated from some sort of average of current multiply voltage.

As your other post suggests, the digital electricity meter considers energy flows both way to be electricity used. I can't think of another way of realizing this than using the integral over time of the absolute value of the product of the instantaneous current and voltage. If it's done this way, certainly when the CFL starts to push current back into the grid, that will be counted as electricity used. That means the utility company will be billing more than the CFL actually used. Actually, not only it's charged for the extra energy it doesn't use, it's also charged again when it returns this energy back to the grid.

On the other hand, the old style electricity meter takes the integral of the product of instantaneous current and voltage, not the absolute value, so the old meter will be charging the real power used.

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.

[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.

[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

[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

[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]

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]

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.

[russ]

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

[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.

[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]

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.