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.
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?
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.
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!
