Saturday, April 08, 2006

 

day six

The water in Lake Logan Martin leaves Pell City and passes through the power generating dam at the end of the lake.

So how much energy does a drop of water release when it drops through Logan Martin Dam? To calculate that number, we'll need a few approximations.
First we note that the water level in Lake Logan Martin is about 64 feet higher than the level in Lay Lake, but we will convert that distance to meters since in the US we use the International System of units (Watts = Joules per Second) when we talk about electrical power.

64 feet = 19.5072 meters

Next we will say there are 20 drops of water in a cc (cubic centimeter) and a cc of water has a mass of one gram. Converting that to kilograms gives: 1 drop of water = 0.00005 Kg mass.

So, the energy (in Joules) of a drop of water which has the potential to fall 19.5 meters is given by the formula:
PE = mhg
Potential Energy = mass x height x (the acceleration due to the Earth's gravity (9.8 meters/sec/sec)).

For our drop of water, the energy release is

0.00005 x 19.5 x 9.8 = 0.009555 J

If Alabama Power were 100% efficient getting that drop's energy from the dam to your 100 watt lightbulb, that bulb would burn brightly for 0.009555 / 100 = 0.00009555 seconds.

Wow! Thanks little drop of water, and thanks Alabama Power.

Comments:
HUH???
no, no, that's actually very interesting :)
 
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