On Saturday 10 May 2008 00:45, LPS wrote: > Hi there. I have some questions about properly stating capacity of a > battery. I like running numbers as you will see below. :-) > > I have a battery that is hooked up to a bicycle to assist in getting me > up hills and such. It is represented as a 42 volt 20 ampere hour battery. > > I figure that a·h is really not using SI properly. The hour is not the > unit of time -- it is the second that would be correct. > > So to properly indicate the capacity of the battery should I be using > Coulombs?
True, but the symbol for ampere is A, not a. The hour isn't metric, but it is allowed to be used with SI. > Today I finished a ride of 54 km and my battery analizer showed that I > consumed 21 a·h of power since my last recharge. My battery is rated at > 20 a·h, however it has a 2 a·h reserve. > > I assume that 21 a·h is 75600 a·s or 75600 C (Coulomb). So my current > use between charges is 75.6 kC? To me a Coulomb has always been an > enormous number of electrons. > > I guess the battery should be rated at 72 kC (20 a·h = 20 x 3600 = 72000 > C). Is this right? My reserve power is 7.2 kC. Right? Yes. > I thought it would be interesting to represent the battery in Joules, > but that gets tricky. In order to represent it, I need a voltage > potential. In the case of my battery, it is 42 volts. The problem is > that the voltage drops as the battery is used. The battery drops to > about 36 - 37 volts at the lowest end of its capacity before the battery > management unit shuts off the current (this is a lithium polymer > battery). My calculation is not an easy one from my perspective. I do > have a Watt meter attached to the charger, so I could show how much > power was required to charge the battery, and run the fan on the > charger. The unit shows 132 watts while it is charging, and I guess the > battery will charge for 8 hours. 132 w for 8 hours = 132 x 8 x 3600 = > 38016000 J or 3.8016 MJ. I will know more in the morning because the > watt meter will indicate how many kW·h were consumed to charge the > battery. 1 kW·h would be 3.6MJ. Right? Yes. Again "W" is always capitalized when it refers to watts. To figure out how much energy you can get from the battery, you would need to integrate the power versus time. So you'd need to measure the voltage and the current as it's discharging. The energy you can get out of it is less than the energy you put into it. > A nice little side note... Let's assume that I used 3.6 MJ recharge my > battery after I travel 54 km. I guess I am in good shape in terms of > energy use? For a car with a gasoline engine, each liter of fuel > represents 34.8 MJ. My car has a 7 liter / 100 km fuel economy rating. 7 > x 0.54 (54 is 0.54 of 100) = 3.78 liters. 3.78 x 34.8 = 131.544 MJ! =-O > My Bike with a battery gets me 54 km with only 3.6 MJ > My car using 3.78 liters gets be 54 km with 131.544 MJ. I looked at the last sentence and at first I thought "That's awfully small for a car's gas tank. Sounds more like a moped." ;) > Using these figures, one could say that the bike gets 172 ml per 100km > if you wanted to represent it in gasoline terms. There is also the fact > that my bike is a hybrid. I can power it with the batteyr, or pedal, or > do both. So I can complicate this with my food intake required to keep > going. That would be an interesting study. How much sunlight energy > would go into growing vegetation to feed me and the occasional animals > that I eat so that I can travel on my bike? Let's not get into what it > took to manufacture the bike and the battery or the car! It doesn't make sense to express the bike's consumption in milliliters, since it's not consuming a fluid. It does make sense to express both consumptions in megajoules. Pierre
