# Lead Acid Battery Discharge Info



## ohio (Jul 25, 2007)

say what ?


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## rbgrn (Jul 24, 2007)

I think ohio could use a little terminology here. xrotaryguy - care to explain DOD and AH?


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## xrotaryguy (Jul 26, 2007)

Sorry, I was just basically copying and pasting. You know, it takes real insight to regurgitate what someone else said. 

Ok, DOD means "depth of discharge". As I understand it, depth of discharge basically refers to how much of the battery's energy has been used. If a battery has sustained a 50% DOD, then half of the battery's energy is gone and 50% still remains for use. 

However, according to the source I was plagiarizing, if a lead acid battery pack is going to be discharged beyond 50% DOD, then a battery management system is required. I'm guessing this means that the individual batteries within the pack will begin to discharge unevenly which could damage individual cells and ultimately cause the entire pack to go bad.

 Ah means "amp hours". An "amp hour" is amps multiplied by hours. If a battery pack is discharged at a rate of 20 amps, and that rate of discharge is sustained for 2 hours, then the number of amp hours used is 40Ah. 

I believe that amps can be thought of as current.


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## rbgrn (Jul 24, 2007)

To get technical, if you think of electricity as a river of moving electrons, then voltage is the speed of the water and amperage is the width/depth of the water. Multiply the two together and you have wattage, which is the total amount of water moving at a certain rate

Add in a time factor and you have watt-hours, which represents how much total electricity was moved.


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## Mr. Sharkey (Jul 26, 2007)

> voltage is the speed of the water


I know what you are trying to say, but a much better analogy would be that voltage is the _pressure_ of the water, and current is the flow. Making reference to speed when comparing the two is likely to confuse anyone who doesn't recognize that electrons flow at a fixed speed no matter the voltage or current.


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## KiwiEV (Jul 26, 2007)

That's interesting! I didn't know electrons flowed at the same speed, no matter what. It's been a good day for learning. 

While on the topic of volts and amps equalling wattage, does anyone have the exact formula for calculating wattage?
When complete, my cheap and gentle charging system of 12 seperate small chargers will use 42 amps at 12 volts (basically 3.5A x 12V x 12).
What amperage at 240 volts (New Zealand's domestic voltage) would that equal?
Also, for the reeeeally technically-minded, what wattage would that equal?

Go on. Someone have a stab.


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## Mr. Sharkey (Jul 26, 2007)

Disregarding the efficiency of the chargers:

42 amps x 12 volts = 504 Watts

504 watts / 240 volts = 2.1 amps

-=BUT=-

You aren't going to be charging at 12 volts, you are going to have to take your 12 volt batteries up to something more like 14.4 volts to completely replenish their charge, so:

42 amps x 14 volts = 588 watts

588 watts / 240 volts = 2.45 amps

Of course, the more completely charged the batteries get, the lower the charging current will be, so using 588 watts would be considered "worst case" power draw from the AC mains.

Now, if you knew the efficiency of the chargers, you could multiply the wattage by that factor, and come up with a real-world value for your AC supply.

As for the speed of electricicty:

A lot of things can affect the speed of electricity, the type of conductor, the temperature, the frequency (if AC current), etc. but the commonly accepted "rule of thumb" seems to be about 2/3 the speed of light. Since the actual delivery of power is accomplished by the electrons knocking around inside the conductors, the real-world speed of the electrons moving through the wire is something really slow like .1mm/second.

One of the example I read was: "If you had a piece of pipe filled with tennis balls and you stuffed one too many into one end, how fast would one be pushed out the other end? How far did the ball you put in the tube move? If the tube was 100 miles long (ignoring friction), how fast were the balls moving to deliver one to the opposite end?"

Electric power delivery is similar, and especially so in AC systems. The electrons that the power company is disturbing at their end will never actually arrive at your mains receptacle, they will only cause "pressure" that moves the electrons in the wires at your end.

Confusing enough?


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## KiwiEV (Jul 26, 2007)

Fantastic!
Thank you Mr Sharkey. That approximation (while disregarding charger efficiency) was just was I was after. 

A big southern hemisphere thumbs up to you mate.

Cheers!


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