Green Stack: Hydroelectric Power

Because of the wide variety of climates within the United States.  Our group will be figuring out the hydroelectic potential of Iowa and applying that to the entire U.S.  Our reasoning is the higher than average rainfall in Iowa will be countered by it's relativity low elevation.

As shown by the map below (american-states.info) the average rainfall in Iowa is about 34 in./yr.



To obtain the kwh per day per person we use the following:

E = mgh, m = pV,

where E is the energy potential due to gravity.

We will assume that 20% of the rainfall gets collected and used to extract power, so the rainfall rate we will use is (34 * 0.2) = 6.8 in./yr

Land area in Iowa is 2.259e14 in.^2, so the volume of rainwater collected is:

6.8 in./yr * 2.259e14 = 1.53e15 in^3/yr or 2.5e10 m^3/yr

density of water is 1000 kg/m^3 so mass of total collected water is 2.5e13 kg.

Plugging into the Energy equation, using 335 m as the average elevation of Iowa:

E = (2.5e13)(9.81 m/s^2)(335 m) = 8.3e16 J

4.3e17 J / 86400 s per day = 9.6e11 W/day

(9.6e11 W/day)*(1/3,000,000 people in Iowa) = 320 kw/person/day

320*(1/24 hr) = 13.3 kwh/person/day