You’ve probably seen the news – it been reported everywhere: the White House has installed 6.3 kilowatts of solar PV. How much energy will that system yield in practical terms?
The rated power of a PV system is the power that it will yield when the sun shines straight on to it with an irradiance of 1000 W/m2, and the temperature is 25 degrees C . If those conditions are met for one hour, then the white house PV system will yield 6.3 kilowatt-hours. If they are met for two hours, then it will yield 12.6 kilowatt-hours, etc.
But those conditions are rarely met exactly. In the United States those conditions will be approximately met around noon time on a cold sunny day if the panels are mounted at the proper angle. Your results will vary.
There is another way to approximate the average daily energy that solar panels will yield: multiply the rated value of the panels by the average daily insolation. The accuracy of this approach depends on the angle that the panels are mounted and several other variables, but this very simple approach will give an approximation good enough for our purposes
The average daily insolation in Washington DC is 4.23 kilowatt-hours/m2/day. So the White House PV system would yield about 27 kilowatt-hours per day.
How much energy is 27 kilowatt-hours in practical terms?
If you add up all he energy that is consumed in the United States each year and divide it by the population and 365 days, then you will get the average daily per capita energy consumption in the United States. Many people have a vague (and wrong) idea of what their energy consumption is, perhaps based on their monthly electricity bill. They are often shocked when they learn the truth.
Here are the numbers…
Total US yearly energy consumption: 97.4 Quads
This is equal to 28,560,000,000,000 kilowatt-hours (since one Quad is 293,000,000,000 kilowatt-hours).
This translates to a daily per capita consumption is 247 kilowatt-hours per day
Therefore, the solar PV on the roof of the White House provides about 10% of the 247 kilowatt-hours consumed by the average American each day!
In all fairness though, we have to understand that a large portion of that 247 kilowatt-hours per day per capita does not do any useful work. I hesitate to say that this large portion is “wasted.” Rather, it is lost mostly due to the inescapable consequences of thermodynamics. Look at the energy flow chart from Lawrence Livermore National Laboratory. On the right side you can see that out of the 97.4 Quads of energy used yearly in the United States, 38.4 Quads goes to “energy services” and 59.0 Quads are “rejected energy.” Effectively, 97.4 Quads of energy are consumed in order to yield 38.4 Quads of “energy services.” That is only about 40% efficient.
Almost all of the energy generated by the Solar PV system on the roof of the White House can be used for “energy services.” With this in mind we could fairly claim that the PV system on the roof of the White House provides about 30% of the energy needs (or “energy services”) of a single average American.
Some more perspective on 27 kilowatt-hours of energy
A gallon of gasoline has an energy content of about 32 kilowatt-hours. If you drive a truck that gets 15 miles per gallon, then you are consuming about 2.1 kilowatt hours per mile [ (32 kilowatt-hours / gallon) / (15 miles / gallon) ]. If you are driving down the highway at 60 miles per hour, then you will consume 27 kilowatt-hours in a mere 13 minutes!
If you are driving down the road at 60 mph in the President’s limo, which gets only 8 mpg, then you will burn up your 27 kilowatt hours in only 7 minutes. Of course, the President travels with an entourage of about 45 vehicles along with his limo. It is a pretty good bet that most of these vehicles are heavy-duty, low mileage vehicles. I think a good approximation would be an average of 20 miles per gallon. At 60 miles per hour this entourage would burn about 135 gallons an hour, or about 4320 kilo-watt hours per hour. At that rate they would burn the allotted 27 kilowatt-hours every 22.5 seconds
Consider the Boeing 747-200B (or its militarized version: the VC25, such as Air Force One). It has a range of 6,100 miles on 48,445 gallons of fuel and a typical cruising speed of 555 mph. That works out to about 8 gallons per mile (that is “gallons per mile” not “miles per gallon!”) and about 9 miles per minute, resulting in a fuel consumption rate of about 72 gallons per minute! The energy content of the jet fuel is about the same as gasoline, about 32 kilowatt-hours per gallon. So, in one minute of cruising the 747 consumes about 2300 kilowatt-hours of energy.
At that rate, the 747 will consume a days worth of the energy produced by the White House solar PV system in about 0.7 seconds (after traveling only about 500 feet), and a years worth of energy in about 4 minutes and 15 seconds. A round trip from, say, Washington DC to Hawaii and back is about 9540 miles. At 550 mile per hour that would be about 17.3 hours of flight. How long would the White House Solar PV array have to operate to produce enough energy for that round trip time? Answer: 243 years.
I have been working on solar PV research for about 17 years and believe any energy source is worthy of research. But religious devotion to one source or another does not advance the human race. Consequently, I am a supporter of both nuclear energy and solar energy (if and when it is affordable and competitive on its own merits – not massive subsidies).