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.)
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