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Posted: 7/26/2007 11:58:23 AM EDT
I'm going to try to make this question as clear as possible so here it goes.

I'm trying to figure out how many watts of power a battery will put out before it needs to be charged. This would be for a battery bank in the home.

Lets use a 20 watt bulb as an example.

These are the specs for an Optima deep cycle marine battery:

BCI Groups: D34
Terminal: Dual Post
Voltage: 13.2V
Cold Cranking Amps: 750
Reserve Capacity: 120 Min.
Capacity (C/20 Rate): 55 Ah
Size: 10" L x 6.8" W x 7.8" H
Weight: 48.3 lbs

How do I figure out how much power this will put out before it needs to be recharged?

TIA.

D_N

Link Posted: 7/26/2007 12:23:54 PM EDT
[#1]
you have to take into consideration how effeciant your inverter is unless your gonna be running 12v light bulbs. you cant just run a light bulb off a 12v battery.
Link Posted: 7/26/2007 12:24:47 PM EDT
[#2]
get LED bulbs if you want this to work well
Link Posted: 7/26/2007 12:33:15 PM EDT
[#3]
I think this is right...



If you are using 12Volt lamps:


Load = 20Watts@12Volts =

Watts / Voltage = Amperes

  20/12 = 1.6666 amps per Hour

Capacity / Load = Hours

 55 / 1.6666 = 33.001 Hours



Vulture.

Link Posted: 7/26/2007 12:35:13 PM EDT
[#4]
Link Posted: 7/26/2007 1:05:15 PM EDT
[#5]

Quoted:
I think this is right...



If you are using 12Volt lamps:


Load = 20Watts@12Volts =

Watts / Voltage = Amperes

  20/12 = 1.6666 amps per Hour

Capacity / Load = Hours

 55 / 1.6666 = 33.001 Hours



Vulture.



This is how long it COULD last, but, for practical purposes, you would not want to discharge your battery more than 50%. So my estimate would be 1/2 that. (~15 hours)

Others have posted that a 20watt bulb (incandescent) would be highly inefficient,  and a 9watt CF or a 3 watt LED would be even better. In those cases, your usable capacity before 50% drain would be over 4 1/2 days with the LED.

Just something to think about.
Link Posted: 7/26/2007 4:09:33 PM EDT
[#6]
Good question, I am doing a study using FR. bulbs.
Link Posted: 7/26/2007 4:16:55 PM EDT
[#7]
A fast way to do it....

its a 12v battery at 55Ah

So..

12v * 55Ah = 660Wh

660Wh / 20w = 33h

Ive metered a 300w inverter to use 6w or so by itself so

660Wh / (20w + 6w) = 25h

The calculation has actually come pretty close in real world usage so I continue to use it. Inverters usually turn off when the battery voltage drops under 10.5v. Usually you dont want to run the battery under 11 if you can help it.

-Foxxz
Link Posted: 7/26/2007 4:29:28 PM EDT
[#8]

Quoted:
you have to take into consideration how effeciant your inverter is unless your gonna be running 12v light bulbs. you cant just run a light bulb off a 12v battery.


 You can get 12 volt bulbs,  then you dodge the loss going through the inverter.  RV's have been using them for years.
Link Posted: 7/26/2007 4:37:05 PM EDT
[#9]

Quoted:
you have to take into consideration how effeciant your inverter is unless your gonna be running 12v light bulbs. you cant just run a light bulb off a 12v battery.


the most efficient way to use the battery's stored power would be to skip the inverter completely and simply use 12vdc bulbs (from an automotive or marine store) or LEDs designed for automotive use.  the other advantage is that the bulbs/LEDs will run on any input voltage, whereas the inverter will drop out once the battery terminal voltage sags below around 11vdc or so.

ar-jedi
Link Posted: 7/26/2007 4:41:56 PM EDT
[#10]

Quoted:

Quoted:
you have to take into consideration how effeciant your inverter is unless your gonna be running 12v light bulbs. you cant just run a light bulb off a 12v battery.


 You can get 12 volt bulbs,  then you dodge the loss going through the inverter.  RV's have been using them for years.


Yes, but you more than make up for any power "savings" by not being able to use compact florescent bulbs instead*. And, since 12 volt incandescent bulbs pull roughly ten times as much current as their 120 volt counterparts of the same wattage and brightness, you often end up with more losses in the 12 volt wiring than in an inverter.




*although there are a few 12 volt CF bulbs sold by specialty suppliers, mostly for the off-the-grid home market.
Link Posted: 7/26/2007 4:45:24 PM EDT
[#11]

Quoted:
Capacity (C/20 Rate): 55 Ah
Reserve Capacity: 120 Min.


one other thing to note...

the AH battery capacity above is specified at the "20 hr rate".  what this means is that if you discharge the battery at 55Ah/20h = 2.75A current, it will be dead in 20hrs.  

the reserve capacity above is specified at 25A load.  what this means is that if you pull 25A from the battery, it will be dead in 120mins.  

the above data is two points on the discharge curve.  note that i said curve.  at low current draws (like the OP's 20W bulb [1.66A] example), the battery will appear to have more than 55AH capacity.  conversely, at high discharge rates, the battery will appear to have less than 55AH capacity.  

ar-jedi

Link Posted: 7/26/2007 4:54:57 PM EDT
[#12]
I'm not sure what kind of system you are looking for.

In my case.  I found that the cost of heavy gauge wire to run 12Volt lights and accessories prohibitive.   So, I found the studs in several of my inside petitions.  Cut a hole, installed plastic fixture boxes.  Snaked 12/2 Romex up from the basement.  I installed Sconces that looked OK with 120v Compact Fluorescence's (G16 Decorative Globe).  

Under the basement stairs I put a used 8D battery (Free from Work), on a smart battery minder ($25). In a plastic Battery box ($45).  I have two converters.  One of the Walmart ($17) cheapies that does not have the juice wasting cooling fans.  I run ten-7WATT compact fluorescents off that.  7watts = 40watts conventional.

A three hundred watt  converter I scrounged off of a dead Jumper Pack powers a second line. Which just goes to a double outlet box under the kitchen counter.   I can keep my cable box, one of those eight inch screen TV's, computer modem, and lap top charged.

The longest I have used my system was a four day black out, after a storm.  I had light on demand, news, computer time.  For really little money. It sure beat sitting in the dark like a Viking!    
Link Posted: 7/26/2007 6:11:24 PM EDT
[#13]
OK, the math is simple! But you need to know the Amp Hour rating of the battery, not the cold cranking amps.

12V  (nominal)
40 AH (a typical number)

Your battery will provide (in theory, not reality) 40 amps for one hour. Or 20 amps for 2 hours, etc.

The reality is that the LESS you draw, the more power you can get out of your battery! Sometimes many times more.

Example: With a good 40AH battery, you could run a 12 watt bulb (draws 1 amp) for 40 hours. But you could run a 1 watt bulb for well over 1000 hours.

AND, if you choose to run a 200 watt bulb, it is likely that you will only get 1 or 2 hours.

Chris
Link Posted: 7/26/2007 6:15:17 PM EDT
[#14]
Buy some candles
Link Posted: 7/26/2007 8:06:35 PM EDT
[#15]

Quoted:

Quoted:
Capacity (C/20 Rate): 55 Ah
Reserve Capacity: 120 Min.


one other thing to note...

the AH battery capacity above is specified at the "20 hr rate".  what this means is that if you discharge the battery at 55Ah/20h = 2.75A current, it will be dead in 20hrs.  

the reserve capacity above is specified at 25A load.  what this means is that if you pull 25A from the battery, it will be dead in 120mins.  

the above data is two points on the discharge curve.  note that i said curve.  at low current draws (like the OP's 20W bulb [1.66A] example), the battery will appear to have more than 55AH capacity.  conversely, at high discharge rates, the battery will appear to have less than 55AH capacity.  

ar-jedi



This is correct. The rates are important. It's more than stored potential and simple unit analysis. The battery is a physical system with physical dependancies. You're dealing with chemical reactions that are dependant on everything from temperature to plate surface area to corrosion impedances, blah, blah, blah.

It will run WAY past the stated 55Ah capacity if you're only running a .5W LED or something with a small current drain.

You might want to look more towards a deep cycle marine battery that's designed to run a trolling motor. It's a better match for your application.

I vaguely remember a web site from a power experimentor. This guy was picking up "dead" golf cart batteries for next to nothing, putting them through a de-sulphurizing charge cycle to revive them some, and using them in all sorts of home grown applications.
Seemed like there could be lots of opportunity there, but I have too much going on to screw with that right now.

Whatever you put together, you owe us pics and a report now.
Link Posted: 7/26/2007 8:11:38 PM EDT
[#16]
 We need the capacity of the battery, usually measured in Ampere Hours (AH),
the battery voltage, and the light source current draw (the light source voltage
requirement should be about the same as the battery output.  I would highly
recommend a DEEP CYCLE battery.  How long is the wire run from the
battery to the light source?  Is the wire stranded or solid?  What gauge is the
wire (AWG)?  What temperature range will the battery be utilized in?  
Do you plan to constantly recharge the battery when it is used up?
Batteries lose most of their capacity in very cold temperatures.
If you live in a northern state, you should have it in an area that is
not below freezing, if possible.

ETA- I bought a cheap incandescent 12vdc inspection lamp,
and replaced the incandescent bulb with an LED accessory light bulb.
Bright, efficient, and uses VERY LITTLE POWER.  And if SHTF,
you can scrounge a replacement bulb if need be from vehicles
that are out of fuel.  LEDs are a SF members friend if you need
to conserve battery power.  I still prefer incandescent or halogen
light for close work though.
Link Posted: 7/26/2007 8:19:47 PM EDT
[#17]
Link Posted: 7/27/2007 3:21:51 AM EDT
[#18]


Quoted:
OK, the math is simple! But you need to know the Amp Hour rating of the battery, not the cold cranking amps.


the AH rating of the battery is specified in the original post.  no one is using CCA for calcs.


Quoted:
We need the capacity of the battery, usually measured in Ampere Hours (AH),
the battery voltage, and the light source current draw


all the data you are asking about is specified in the original post, see the top of this thread.

ar-jedi

Link Posted: 7/27/2007 5:11:59 AM EDT
[#19]
There are many variables, and a car battery is a poor choice for electricity storage (they are meant to run a HIGH CURRENT load for a short period of time).  A golf cart battery (Trojan T-105) is generally estimated as 2 kWH of reserve electric power.  In other words, it will run 2000 watts for one hour, 200 watts for 10 hours, and 20 watts for 100 hours, 2 watts for 1000 hours.

HTH
Link Posted: 7/27/2007 6:58:44 AM EDT
[#20]
A car battery is not a deep cycle battery -- As others have indicated.

'Car batteries' are designed to be kept fully charged, and respond to high current demands, such as starter motors.  Frequent discharging will kill them.
Link Posted: 7/27/2007 12:09:07 PM EDT
[#21]

Quoted:
A golf cart battery (Trojan T-105) is generally estimated as 2 kWH of reserve electric power.  In other words, it will run 2000 watts for one hour, 200 watts for 10 hours, and 20 watts for 100 hours, 2 watts for 1000 hours.


no.

as noted in my post above, the discharge curve is NOT linear.  doubling the load does not half the time the battery supports the load -- the battery actually supports the doubled load less than half the time.

to wit, from
www.trojanbattery.com/Products/ProductSpec.aspx?Name=T-105

we see that the Trojan T-105 has the following characteristics:

Capacity Minutes, 447 @ 25A
Capacity Minutes, 115 @ 75A

so the "apparent capacity" with a 25A load is (447/60) x 25A = 186 AH
and the "apparent capacity" with a 75A load is (115/60) x 75A = 143 AH

continuing on, we see that the manufacturer also states:

5 Hr Rate, AH = 185
20 Hr Rate, AH = 225.

this tells us that the battery will deliver,

185AH/5h = 37A for 5h
225AH/20h = 11.25A for 20h

IF the discharge curve were linear, it would show that the battery could supply 4 times the current for 1/4th the time.  but, 11.25Ax4=45A and 20h/4=5h.   so ideally the battery would put out 45A for 5h.  this is not the case; from the manufacturer's data (linked above), the battery can only supply 37A for 5h.

putting all of the above into a table, we have a total of 4 datapoints:

11A for 20h  (~220AH)
25A for 7.5h  (~187AH)
37A for 5h  (~185AH)
75A for 1.9h   (~142AH)

as you can see, the bigger the load, the lower the capacity of the battery.  this is just physics, and every lead-acid (wet or gelled electrolyte) battery has a similar characteristic.

ps:
keep in mind that the T-105 is a 6V battery; you need two connected in series to drive a 12V inverter or to power 12V loads.  using two in series does not double the AH capacity, it simply doubles the output voltage.

ar-jedi


Link Posted: 7/27/2007 7:34:14 PM EDT
[#22]
Wow!

Thanks for all the input guys.

What batteries would you recommend for a home battery bank?
Link Posted: 7/27/2007 8:33:25 PM EDT
[#23]

Quoted:
Wow!

Thanks for all the input guys.

What batteries would you recommend for a home battery bank?


Well, if money is no object, a rack of Surrett cells is top drawer.

I run 16 Trojan T-105's with an Outback 2kw charger/intverter (24 volt input).  The trojans stike a real nice price performance balance.

By adding a small 2-3kw genset (I use a Yammi 2400 efi unit), you'll have a dandy set-up to get you trough times of no power.  Run off the batteries, and fire up the genny when the power reserve drops.  Then shut the genny off and go back to bateries.

Frankly, I couldn't be happier.
Link Posted: 7/27/2007 8:51:18 PM EDT
[#24]
tagged to review arjedi's calcs till i fully understand them.
Link Posted: 7/28/2007 6:27:26 AM EDT
[#25]

Quoted:
tagged to review arjedi's calcs till i fully understand them.


What arjedi is talking about is the peukert effect with batteries en.wikipedia.org/wiki/Peukert's_law.
Basically just because you can bench press 135 lbs 10 times doesn't mean you can bench press 1350 lbs once.
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