Jesse, In reality, the discharge of *any* electro-chemical battery into a constant load is accurately represented only by a non-linear function of current (or terminal potential) versus time.
These non-linear functions degrade with age, with the number of charge and discharge cycles, and depend on temperature. They are empirical (experimental), not easily calculated. Efforts to represent available energy *accurately* by just a few numbers are hopelessly futile. A family of non-linear curves for each particular battery type in question and for the history of each particular specimen is much more informative. Your essay below states some of these facts, but fails to mention the unavoidable non-linearity of every cycle of battery's discharge. Gene. ---- Original message ---- >Date: Sat, 10 May 2008 12:16:00 -0700 (PDT) >From: "Ziser, Jesse" <[EMAIL PROTECTED]> >Subject: [USMA:40884] Re: SI power questions and a lot of random thoughts :) >To: "U.S. Metric Association" <[email protected]> > >OK, this has now been analyzed to death (and quite correctly) by the other >replies, but I'll add a >few things not said so far: > >--- LPS <[EMAIL PROTECTED]> 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? >> >> 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. > >Please understand that battery ratings can be very approximate. As you and >others have pointed >out, you have to do some serious math to find out the details. The total >amount of energy OR >charge that the battery can push through your circuit will probably vary a lot >depending on the >load and on various environmental factors. > >> 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. > >To me as well, but you are perfectly correct. Bear in mind that there are an >enormous number of >electrons in a small segment of uncharged metal wire. The wire is >electrically neutral because it >also contains an enormous number of protons. If you were to get rid of the >protons, leaving that >much negative charge, the amount of potential energy represented by all those >electrons repelling >one another would be insanely vast, as you imagine. But fact is that the >protons are there, so >there's nothing strange or unstable about that much charge being present. > >Since you like to run numbers, I'll run you some more. Imagine that your >wires are copper and >have a cross-sectional diameter of 1 mm. Then their cross-sectional area is >about 3.14 mm^2, or >3.14 x 10^-6 m^2. Copper has a density of 8960 kg/m^3 and an atomic weight of >0.0635 kg/mol, >giving an "atom density" (is there a better name for this?) of 141 000 >mol/m^3. Since a copper >atom has just one conduction electron, this means there are 141 000 mol/m^3 of >conduction >electrons, which is 8.49 x 10^28 electrons per cubic meter. For the wire in >this example, that >gives 2.67 x 10^23 conduction electrons per meter of wire. Since an electron >has a charge of >-1.60 x 10^-19 C, we have -42 700 C/m, or -42.7 kC/m of free charge in the >wire. For a current of >1 A, which is 1 C/s, the average drift velocity of the free charges in the >wire must then be 2.34 >x 10^-5 m/s, or about 0.2 mm/s. Although this sounds very slow, I think it is >actually about >right. Individually, the electrons are moving much more quickly than this of >course, but mostly >in random directions. > >If the devil had visited upon us a set of "Imperial units for electricity", I >would imagine that >the above calculations would be difficult enough to drive an engineer to >drinking. Perhaps we >could invent such a set of units as a rhetorical device? I shall begin with >the "youch"... > >> 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? > >Careful with your terminology. I'd say "reserve charge pumping capacity" or >something. Power, >strictly speaking, is a different quantity. > >> 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? >> >> 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. > >Right, but also remember there are a lot of inefficiencies in any system. The >electricity you use >to charge your battery may come to you through many kilometers of power lines, >losing some power >all along the way, and it was probably generated using fossil fuels to start >with, and I can just >about guarantee that the power plant has far less than 50% efficiency itself. >And let's not even >get into the energy costs of operating and maintaining (a fraction of) a power >plant versus a car! > >> 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! > >One fact that may interest you in this regard: the recommended average human >power intake is about >100 watts. You can discover that just that by converting the familiar but >ugly "2000 kilocalories >per day" recommendation into "joules per day" and then finally into joules per >second, which are >equivalent to watts. > >Also, you may or may not be aware that the percentage of your diet that >consists of unlucky >animals is very relevant, since eating critters is a lot less energy efficient >than eating plants. > (No flames necessary. I'm not trying to promote vegetarianism or anything > here. I'm rather fond >of meat myself, actually.) > > > > > ____________________________________________________________________________________ >Be a better friend, newshound, and >know-it-all with Yahoo! Mobile. Try it now. >http://mobile.yahoo.com/;_ylt=Ahu06i62sR8HDtDypao8Wcj9tAcJ >
