I think using solar energy to heat water is one of the most elegant uses of energy possible. Rather than generating heat in a central power plant and immediately throwing away 60% to run the steam turbines(in the electric hot water heater case), light comes from the sun directly to its destination(untaxed!!) and is immediately harnessable for its intended purpose. There are no big conversion efficiency penalties and it can be quite potent even at only 700 Watts/meter^2.
I would like to share with you an example that was taught to me by Professor John Wilson of the University of New Hampshire. Let’s image we take the same amount of energy required to heat a forty gallon tank of water from ground water temperature to a typical tank temperature of 140F and apply it to lifting an automobile. How high could we lift the car? A foot? Two meters?
First let’s calculate the energy in the tank:
Energy = mass * specific heat of water * rise in temperature
Energy = volume * density * specific heat of water * rise in temperature
Energy = (40gallons)*(8lbs/gallon)*(1 Btu/(lbm*deg F))*(140F-50F)
Energy = 28,800 Btu = 30385609.92 Joules
Now let’s take a vehicle that most people around the world will recognize, the 2005 Holden Monaro or Pontiac GTO or Vauxhall Monaro VXR or Chevy Lumina Coupe. Yes, this poor car has an identity crisis. It also weighs in at a burly 4491 lbs or 2005 kg. The energy required to lift an object is:
Energy = mass * gravitational constant * height raised
So reshuffling with a little algebra:
Height = Energy / (mass * gravitational constant)
Height = 30385609.92 Joules / (2005 kg * 9.81 meters/sec^2)
Height = 1545 meters = 5,068.9 feet
To perform this experiment you would need permission from Air Traffic Control since general aviation’s pattern altitude is only 1500 feet above the ground! It does make you think twice about taking long showers!! So how big would a solar collector have to be to collect this amount of energy in four hours?
Energy = Power*Time
Energy = Insolation*Area*Time
Area = Energy/(Insolation*Time)
Area = 30385609.92 Joules /( 700 W/m^2 * 14400 seconds)
Area = 3.01 meter^2 = 32.4 ft^2
This is about the area of a typical sheet of plywood in the US. Of course I have left out seasonal variations in incidence angle, correction for cloud cover, correction for the sun’s movement, heat losses from the solar collector, reflectance from the collector…..but you get the basic idea. Now think about the energy hitting an entire roof! I will let your minds wander…..