The Carbon Lighthouse

The Carbon Lighthouse
The long answer involves a whole lot of fun with units, and a handful of different methods.
The solution requires that we understand how much carbon dioxide per mile a gasoline-powered car
emits compared with how much carbon dioxide per mile an electric car powered by coal releases.
To start, we calculate the carbon intensity per mile of a gas-powered car. Gasoline emits 19.4 pounds of
carbon dioxide per gallon. We will assume a car that gets 30 miles per gallon.
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19.4 pounds Carbon Dioxide per gallon / 30 miles per gallon = 0.65 pounds of CO2 per mile
Great, we found the answer for gasoline. We’re halfway there! … sort of.
The first pothole to consider is that electrically powered vehicles are much more efficient than internal
combustion engines (what gasoline powered cars use). The reason is pretty interesting: electric motors
drawing current are extremely efficient, around 90%, whereas internal combustion engines waste 75%
of their fuel as heat. So it’s important to find an effective miles per gallon equivalent for electric engines.
We start with the energy intensity of a gallon of gasoline. This is approximately 114 kBTU/gallon. We
convert this to the more friendly but also powerful-sounding MEGAjoules (MJ) to get 120 MJ of energy
per gallon. Assuming a gas-powered car gets 30 miles per gallon, then it requires:
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120 MJ per gallon / 30 miles per gallon = 4 MJ of energy to go 1 mile
An electric engine is much more efficient. Assuming the electric engine converts energy from its
batteries with 90% efficiency and that its batteries charge from the outlet with 90% efficiency, then an
electric car consumes energy with 0.9 * 0.9 = 81% efficiency. And therefore:
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81% electric efficiency / 25% gas efficiency = electric engines are 3.24 times more efficient with
their fuel than gas ones
4 MJ of energy for a gas car to go one mile / 3.24 = 1.24 MJ of energy to go 1 mile
So it requires about three times less energy for an electric car to go one mile than a gasoline car. Now
we need to convert units again. This time to convert from MJ per mile to kWh per mile. Kilowatt-hours
(kWh) is how we measure electricity drawn from the grid. There are 3.6 MJ per kWh so:
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1.24 MJ per mile / 3.6 MJ per kWh = 0.344 kWh per mile = 2.91 miles per kWh
In other words, for every kWh of electricity that the electric car draws from the energy grid, it can power
the car for 2.91 miles.
How does this compare with real-world data points?
Let’s look at the latest stars of the electric vehicle world, the Chevy Volt and Nissan Leaf.
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The EPA rates the Volt as being able to go 100 miles on 36 kWh of energy. That’s equal to 2.78
miles per kWh. And the EPA puts the Nissan Leaf in at 100 miles on 34 kWh of energy. That’s
equal to 2.94 miles per kWh. The real world average is thus 2.86 miles per kWh, and our 2.91
miles per kWh pure science number above is looking pretty good!
That smoothes out the first pothole. Now comes the hard part.
How much carbon does the coal power plant emit in order to get those kWh of electricity into your
home so you can charge your car?
The value we are after is pounds of carbon dioxide (CO2) per kWh of electricity produced.
A key point to understand is that the type of coal you are combusting, makes a big difference in
calculations. Different coal types – anthracite/bitumen vs. sub-bituminous/lignite – all have different
carbon contents per pound as well as varying amounts of energy released per pound. This tends to even
out, because at the end of the day combusting carbon is what makes energy. But it’s extremely
important to be consistent and to be careful with our assumptions. Sub-bituminous coal, for instance,
releases about half the energy per unit mass as bituminous coal, but it also has half the carbon content.
We thought of three ways to approach this problem. We tried them all.
Method 1: Use the energy power plants say is released when combusting a pound of coal
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According to information plucked from the Arizona Public Utility Commission via the
Department of Energy, a low end estimate for the energy released when a pound of coal is
combusted is 8,800 BTU.
We know it takes about 10,400 BTU of coal to make 1 kWh of electricity (thank you EIA)
So one pound of coal produces 8,800 BTU / 10,400 BTU = 0.846 kWh/lb of coal
But a pound of coal is not a pound of carbon dioxide. For starters only 78% of bituminous coal,
for example, is actually carbon. Additionally, carbon only has a weight of 12 grams per mole, but
carbon dioxide has a weight of 44 grams per mole. So one pound of coal, when burned
produces, 44 / 12 * 0.78 = 2.86 pounds of CO2 per pound of coal
And to calculate the CO2 per kWh produced, we do: 2.86 lbs CO2 per lb of coal / 0.846 kWh per
lb of coal = 3.38 pounds of CO2 per kWh produced.
So our final number for method 1 is: 3.38 pounds of CO2 per kWh
Method 2: Use the measured energy released from combusting a kilogram of coal
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There is a range of data points about the energy released from a kilogram of coal, but 26 MJ per
kg is a good average for bituminous coal. Bituminous coal is commonly used for electricity
production.
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Next, we convert this number to kWh. 26 MJ per kg = 7.2 kWh per kg.
We convert this to pounds. 7.2 kWh per kg = 3.3 kWh per pound
Unlike Method 1, where the efficiency of the power plant was accounted for by the fact that we
said 10,400 BTU generates 1 kWh of electricity, in this case we need to explicitly account for the
efficiency of the power plant. 3.3 kWh per pound is the energy released in the reaction, not the
useful energy harnessed. We will assume the power plant only harnesses 30% of the energy
released (note that plain old combustion, whether it’s a coal plant or gasoline car engine is
energy inefficient). So we multiply: 3.3 kWh released per pound of coal * 30% efficiency = 0.98
kWh harnessed per pound of coal
But recall a pound of coal is not a pound of carbon dioxide. So we use the number 2.86, which
we found above: 2.86 pounds of CO2 per pound of coal / 0.98 kWh per pound of coal = 2.91
pounds of CO2 per kWh
So our final number for method 2 is: 2.91 pounds of CO2 per kWh
Method 3: Use the standard chemical energy released from combusting a chunk of solid carbon
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Coal is not 100% carbon. Anthracite coal is about 90% carbon. Bituminous coal is 70-80%. Subbituminous is 40% carbon. And lignite is 30% carbon. But assuming we’re using bituminous coal
for electricity production, there are worse starting points than the chemical reaction:
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C(s) + O2(g) –> CO2(g) + energy
That “energy” released is 94 kcal/mol or 393 kJ/mol. See here and here.
There are 12 grams of carbon per mole (a mole is a number of carbon atoms, a gargantuan,
enormous number, specifically 1 mole of carbon = 6.02 x 10^23 carbon atoms). Since
combusting pure carbon releases 393 kJ per mole we divide by 12 grams per mole and obtain:
32.8 kJ of energy released per gram of carbon
Since there are 453 grams per pound, this number becomes 32.8 kJ per gram * 453 grams per
pound = 14,847 kJ per pound = 14.8 MJ per pound = 4.12 kWh per pound
Then we multiply by the coal plant’s combustion efficiency of 30% (newer plants do better than
30%; older plants are a little worse). 4.12 kWh released per pound of coal * 30% efficiency =
1.24 kWh harnessed per pound of coal.
But a pound of coal (or pure carbon in this case) is not a pound of carbon dioxide. We use our
friendly 2.86 number and get: 2.86 pounds of CO2 per pound of coal / 1.24 kWh harnessed per
pound of coal = 2.31 pounds of CO2 per kWh
So our final number for method 3 is: 2.31 pounds of CO2 per kWh
So we now have three different numbers for the carbon intensity of electricity coming from a coal
power plant: 3.38, 2.91, and 2.31 pounds of CO2 per kWh.
This range is to be expected: there are many varieties of coal, many varieties of power plants, and many
different metrics for assessing energy content and therefore carbon intensity. It is a very good sign of
course, that these numbers are all relatively close together. The real world gives us ranges and that is
exactly what we’ve found.
So what’s the answer:
Which produces more carbon dioxide: a gas-powered car or an electric vehicle charged from 100%
coal power?
Well, we’re not done. We need to convert the carbon intensive energy coming from the grid into a
carbon intensity per mile driven in an electric car. And we’ll do that in the next and final installment,
coming on July 22, 2014.