Scott Adams has a post up that’s really too dumb to go into. It does, however, raise an interesting question: What would a typical home’s energy usage look like, expressed in terms of pumping water? I think it’s a novel way to visualize the matter.
According to the DOE:
In 2008, the average annual electricity consumption for a U.S. residential utility customer was 11,040 kWh.
There are two problems with this statistic: It uses kWh, one of my least-favorite units, and it’s per year, which is bigger than I’m interested in. That’s easily fixed, though.
A kilowatt-hour is an energy rate of 1000 J/s delivered for 3600 s; that’s equal to 3.6 MJ. Therefore, if a typical home consumes 11,040 kWh over a year, it must consume about 109 MJ per day.
Near the earth’s surface, the potential energy of a mass can be estimated as 9.81*m*h J, where m is the mass in kilograms, and h is the altitude. Therefore, to raise 1 kg of water 1 meter requires about 9.81 J of energy.
A typical home’s daily energy usage is sufficient to raise 1.1 million kg of water by 1 meter. Okay, that’s not very helpful.
Let’s say a typical 1-story house has a 15-foot roofline. A typical home’s 109 MJ daily energy consumption could raise 2,430,252 kg of water from ground level to the roofline. (9.81*2,430,252*15*.3048 ~= 109 million). Water has a density of 1 kg per liter, and there are about 2.5 million liters in an olympic-sized swimming pool.
Therefore, a typical home’s daily energy consumption is roughly enough to lift all the water in one of the largest swimming pools made from ground level to roof level. Assuming perfect efficiency.