Posts Tagged ‘photovoltaics’


How much photovoltaics to provide 100 kilowatt hours per person per day?

November 8, 2015

Suppose you wanted to power the world at the level that each human being can enjoy the same level of energy abundance as the average American. And suppose we wanted to do it all with photovoltaic solar energy. What would it take?

There are an average of 250 kilowatt hours consumed per person per day in the United states. Maybe that seems like a lot to you because you occasionally look at your home electric bill and see less than 1000 kilowatt hours used in a entire month for a home that houses four people. That 1000 kilowatt hours for four people in a month works out to only about eight kilowatt hours per person per day. But that electric bill is a very poor indicator of how much energy is actually expended for your benefit. That is why claims that some energy source will power X number of homes is incredibly misleading.

Here is the reality.  According to Lawrence Livermore National Laboratory the United States consumes 98.3 quads of energy every year.


That works out to about 250 kilowatt hours per person per day

(98.3 quads/year) x (2.933 x 1011 kw-hr/quad) / (year / 365 days) / (3.2 x 106 people) = 247 kilowatt hours/person/day

Fortunately, this daunting amount of energy is also somewhat misleading.  Look at the right side of the graph from Lawrence Livermore.  Notice that the two final energy outputs on the right side of the graph are “Energy Services” and “Rejected Energy.”  “Energy Services” is energy that actually does some useful work.  “Rejected Energy” is energy that is lost, mostly in form of waste heat.  For example, if you burn a lump of coal in a steam generator and get a kilowatt of energy out in the form of electricity, but lose two kilowatts in the form of heat to the atmosphere, then you got one kilowatt hour of Energy Service but two kilowatt hours of Rejected Energy.  As you can see from the graph, only 40% of the energy that is input comes out the the system as Energy Services (38.9 quads / 98.3 quads).

One of the big advantages of solar photovoltaics is that you don’t lose 60% of your energy to heat.  Electric cars put far more of their stored electric energy into useful work (Energy Services) and far less into “Rejected Energy” than do blazing hot internal combustion engines.

Let’s make the assumption for now that every possible efficiency is applied, so that we only need to produce 40% of the 250 kilowatt hours per day per person, or 100 kilowatt hours per day per person.  Still a lot of energy, but more manageable than 250 kilowatt hours.

So, for 7 billion people we need 700 billion kilowatt hours per day (100 kilowatt hours per person x 7 billion people).  If we got all that energy from solar photovoltiacs, how much land would be required for solar arrays, how much would it cost?

Topaz Solar Farm

To get estimates of these values, we can look at some of the world’s biggest solar arrays.  Consider the Topaz Solar Farm in California.  It is one of the biggest and one of the newest in the world and in an area of very high solar insolation.  It is expected to generate 1,100 GWh of energy per year while occupying 25 km2 with a cost of $2.5 billion.  Therefore it would generate the energy consumed by about 30,000 people at 100 kWh per person per day.

(1100 GWh/year)x(1×106 kWh/GWh)x(year/365 days)/(100 kWh/person/day) = 30,136 people

From this it is clear that it would take about 6 million km2 of solar photovoltaics of the Topaz Solar Farm density to generate all the energy consumed by 7 billion adequately powered people.

(7×109 people) / (30,136 people/25 km2) = 5.8×106 km2

Keeping in mind that the Topaz Solar Farm cost $2.5 billion and yields enough energy for 30,136 people, then the cost for 7 billion people would be about $580 trillion.

(7×109 people) / (30,136 people/$2.5×109) = $5.8×1014 .

For the sake of comparison the, the gross domestic product of the United States is about $17 trillion, or less that 3% of that $580 trillion.  The gross product of the entire world  is about $78 trillion, or about 13% of that $580 trillion.  So, if every penny or mark or yen, etc. of world product for about 7.5 years were dedicated to this project, it could be accomplished.

Some points to consider

What would be the consequences of covering 6 million square kilometers of land with PV?  This would be like completely covering an area the combined size of Arizona, Nevada, Colorado, Wyoming, Oregon, Idaho, Utah, Kansas, Minnesota, Nebraska, South Dakota, North Dakota, Missouri, Oklahoma, Washington, Georgia, Michigan, Iowa, Illinois, Wisconsin, Florida, Arkansas, Alabama, North Carolina, New York, Mississippi, Pennsylvania, Louisiana, Tennessee, Ohio, Virginia, Kentucky, Indiana, Maine, South Carolina, West Virginia, Maryland, Vermont, New Hampshire, Massachusetts, New Jersey, Hawaii, Connecticut, Puerto Rico, Delaware, Rhode Island with solar panels.  Of course, this would be spread out over the about 100 million square kilometers of land at latitudes lower than about 50 degrees.

This plan would also require a distribution system that could move energy from daytime areas to nighttime areas, or at least a few days of storage for every person on the planet.  Such a distribution system is not feasible at this time, and the massive amount of storage is prohibitively expensive.

Two days of storage would be 200 kilowatt hours of stored energy per person.  Probably the best mass storage option today (2015) is with Tesla’s Powerwall, which stores 7 kilowatt hours, costs $3,000, and weights 220 pounds.  So we would need about $90,000 and about 6,600 pounds of storage for each of the 7 billion people.  That adds another $630 trillion to the cost.

These calculations serve simply to give a feel for what could be done with solar photovoltaics and what the limitations might be.  I am not suggesting that the world should be powered solely with PV.  With other energy sources in the mix less money and land would need to be devoted to PV (but more to those other sources).  For example, if you did the same calculations for wind, then you would find that about twice as much area  (about 12 million square kilometers) would have to be covered by wind farms to get the same amount of energy.  But at least you can grow corn are graze cattle below the turbines in a wind farm.

I have led you to water.  It is up to you to drink up your own conclusions about the viability of using solar energy to bring the world up to a reasonable level of energy consumption.


News: White house installs 6.3 kW solar PV system

May 28, 2014

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.

243 years

My position

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).


Bjorn Lomborg on German solar subsidies

February 18, 2012

Bjorn Lomborg (The Skeptical Environmentalist and Cool It) has an interesting article (Goodnight Sunshine) describing the late realization in Germany that their massive investment in solar energy is “a massive money pit” and “a threat to the economy.”   They have subsidized 1.1 million solar photovoltaic installations to the tune of $130 billion, which provide a whopping 0.3% of the nation’s energy.

If you are worried about anthropogenic CO2 induced global warming (a.k.a. climate change, a.k.a. climate disruption), he points out

“This sizeable investment does remarkably little to counter global warming. Even with unrealistically generous assumptions, the unimpressive net effect is that solar power reduces Germany’s CO2 emissions by roughly 8 million metric tons—or about 1 percent – for the next 20 years. To put it another way: By the end of the century, Germany’s $130 billion solar panel subsidies will have postponed temperature increases by 23 hours.”

And as to “green jobs” boosting the economy of Germany…

“[E]ach job created by green-energy policies costs an average of $175,000… And many ‘green jobs’ are being exported to China, meaning that Europeans subsidize Chinese jobs, with no CO2 reductions.”

I have one large quibble with his Lomborg’s numbers.  He says

“Even with the inefficiency of current PV technology, we could meet the entire globe’s energy demand with solar panels by covering 250,000 square kilometers.”

It would really take closer to a million square kilometers.

Why a million square kilometers?

Because even good solar panels (say, rated for 150 Watts/m2) will yield much less than their name plate power for a variety of reasons.

1. A one square meter solar panel in a huge array actually requires a land area of at least 1/cos(latitude) meters.  This is the land area shaded by the one meter panel by the sun at noon.  It is larger at other times of day.

2. Extra area is required for infrastructure, such as service roads

3. The capacity factor for solar energy is about 20%.  This means that if a solar panel’s name plate says that it is 150 watts,  then on the average it will yield about 30 Watts because sometimes it is night, sometimes it is cloudy, and even during a sunny day the incident angle is optimal for only several hours before and after noon.

In Berlin, for example, where the latitude is 52 degrees, a one square meter panel in a huge array will require closer to 1.7 square meters of land.  A typical good quality one square meter panel with a name plate wattage of  150 Watts at peak averages out to about 30 watts over the course of a year because of the 20% capacity factor.  This translates to an average of about 18 Watts per square meter (30 Watts divided by 1.7 meters).

World energy consumption is about  5×1017  BTUs per year (about 1.5×1017 watt-hours/year).  That is an average power consumption of  about 1.7×1013 watts (1.5×1017 watt-hours/year divided by 8760 hours/year)

Therefore, it would take about a million square kilometers of solar arrays (1.7×1013 watts divided by 18 Watts per square meter).

This all assumes that the energy impoverished of the world are happy to stay that way.  Oh well, we all have our crosses to bear.

You can see Lomborg’s complete article here.