I know mAh tells how much milliamperes a battery can deliver in an hour. But does that also tell how many hours the battery would last? Sorry but I don't really get it. If we're talking about a water tank, to my impression, mAh is like how big the faucet is and not how much water there is in the tank. I'm really confused as to why we measure battery capacity in mAh if my understanding about it is correct.
Does mAh measure how long a battery would last
batteries
Related Solutions
There are vast resources on the web re lead acid batteries. Key parameterrs are provided below a loom around the internet and sorting the good references from the not so good would be a helpful part of your necessary education if you are going to do what you suggest.
Energy able to be stored in your water tank if it was converted to a large lead acid battery can be roughly determined from the Wh/l figure that I give further down.
- Energy density = 60 - 75 Wh/l
VANADIUM REDOX BATTERY - Energy stored in liquid !!!
- The main advantages of the vanadium redox battery are that it can offer almost unlimited capacity simply by using larger and larger storage tanks,
For a battery where the liquid IS the energy store and where adding more liquid adds more capacity see Vanadium Redox battery.
They note:
The vanadium redox (and redox flow) battery is a type of rechargeable flow battery that employs vanadium ions in different oxidation states to store chemical potential energy.
The present form (with sulfuric acid electrolytes) was patented by the University of New South Wales in Australia in 1986.
There are currently a number of suppliers and developers of these battery systems including Ashlawn Energy in the United States, Renewable Energy Dynamics (RED-T) in Ireland, Cellstrom GmbH in Austria, Cellennium in Thailand, and Prudent Energy in China.
The vanadium redox battery (VRB) is the product of over 25 years of research, development, testing and evaluation in Australia, Europe, North America and elsewhere.
The vanadium redox battery exploits the ability of vanadium to exist in solution in four different oxidation states, and uses this property to make a battery that has just one electroactive element instead of two. The main advantages of the vanadium redox battery are that it can offer almost unlimited capacity simply by using larger and larger storage tanks, it can be left completely discharged for long periods with no ill effects, it can be recharged simply by replacing the electrolyte if no power source is available to charge it, and if the electrolytes are accidentally mixed the battery suffers no permanent damage.
The main disadvantages with vanadium redox technology are a relatively poor energy-to-volume ratio, and the system complexity in comparison with standard storage batteries.
LEAD ACID:
Lead acid voltage per cell, as in any battery chemistry that you will probably encounter, is very largely a function of the battery chemistry, with other factors making a vey small difference to the cell voltage.
The example battery cited here on the Wikipedia Lead-Acid battery page gives values of key parameters which you would achieve if you implemented a competent design. For a battery of the size you suggest this would be at best impractical and liable to be near to impossible. So consider these as values you can aim at but will not achieve.
Note that a number of these values are somewhat dependant on sub technologies or mechanical construction methods.
Voltage per cell:
Open circuit fully charged 2.10 - 2.13 V / cell.
Open circuit, fully discharged 1.95 V - 2.0 V / cell
Loaded, fully discharged 1.75 V/cell Gassing threshold 2.35 V / cellSpecific energy 30-40 Watt.hour/kg ~= 0.10 - 0.15 MJ/kg
Energy density = 60 - 75 Wh/l
Specific Power = 180 W/kg
Charge efficiency 40% - 98% very much dependant on application circumstances.
Cycle life 100 - 1000+ cycles very much dependant on construction and usage patterns.
Useful:
- paper on 3.3 amps/100Ah battery capacity charge efficiency
Note: real world solar charge currents go up to 35 amps/100Ah. - http://en.wikipedia.org/wiki/Lead-acid_battery
- http://wattsupwiththat.files.wordpress.com/2011/11/ridley_rsa.pdf
The discharge rate dependes of several factors and will vary from manufacturer to manufacturer.
A good Li-Poly typically has 2-5% self-discharge rate per month. If it has a PCM (protection circuit module), that module will draw some current, expect less than 1uA for such a tiny cell. Parts of lesser quality may draw 30uA or more.
Self-discharge increases with temperature. I believe it also varies with cell construction (cylinder, prismatic etc.) Quality varies a lot from manufacturer to manufacturer so you have to ask for the exact specs.
UPDATE
I don't have much experience working with the current levels you mention. Wouldn't the ground current in the charger consume most of the 10-20uA? There is a lot of interest in energy harvesting and someone else can probably give a better answer. Slow-charging the battery is OK provided you cut off at 4.2V. My guess is that 10-20uA is not enough to offset self-discharge and other losses, but I'm frequently wrong.
Best Answer
mAh (or mA·h) is not how many milliamperes a battery can deliver in an hour. That would be mA/h. Current, measured in amperes, is already a rate of stuff. Specially, one ampere is one coulomb per second. So, if current is like speed, then mA/h is like acceleration, and mAh is like distance.
Rather, mAh it is a unit of charge. It is what you get when you multiply current by time. By multiplying by time, the "per time" part of the ampere is cancelled, and you get back to charge.
If an ampere is a coulomb per second, then:
$$ \require{cancel} 1~\mathrm{mAh} = 1\cdot10^{-3}~\mathrm{\frac{C}{s}h} $$
and by dimensional analysis:
$$ \require{cancel} \frac{1\cdot10^{-3}~\mathrm{C\cancel{h}}}{\cancel{\mathrm{s}}} \frac{60\cancel{\mathrm{s}}}{1\cancel{\mathrm{min}}} \frac{60\cancel{\mathrm{min}}}{1\cancel{\mathrm{h}}} = 3.6~\mathrm{C}$$
For example, if you draw 1 mA for 1 hour from a battery, you have used 1 mA · 1 h = 1 mAh of charge. If you draw 2 mA for 5 hours, you have used 2 mA · 5 h = 10 mAh.
You can approximate how long a battery will last by dividing its total charge (in mAh) by your nominal load current (in mA). Say you have a 1800 mAh battery, and you connect it to a 20 mA load:
$$ \require{cancel} \frac{1800~\mathrm{mA\cdot h}}{20~\mathrm{mA}} = \frac{1800\cancel{\mathrm{mA}}\cdot\mathrm{h}}{20\cancel{\mathrm{mA}}} = 90~\mathrm{h} $$
This is an approximation because:
The charge capacity (the number measured in mAh) is determined by measuring how much charge can be removed from the battery before voltage drops to some arbitrarily selected level where the battery is considered "discharged". This may or may not be the threshold at which your circuit no longer functions. Battery manufacturers, wanting to make their batteries seem as good as possible, typically select a very low threshold voltage.
Assuming you are considering charge available only down to some voltage threshold, the actual charge available from the battery depends on temperature, and the rate at which you discharge it. Lower temperatures slow the chemical reaction in the battery, making it harder to extract charge. Higher rates of discharge increase losses in the battery, decreasing the voltage, thus hitting the "discharged" voltage threshold limit sooner.
The electric potential difference provided by the chemicals in the battery is actually constant; what makes the voltage decrease is the depletion of the chemicals around the electrodes and degradation of the electrodes and electrolyte. This is why battery voltage can recover after a period without use. So, the point at which the threshold voltage is reached can actually be quite complex to determine.
If you can find a good datasheet for your battery, it may give some insight into the parameters under which these calculations were made.