### Would Champ Car’s Switch to E85 Be Green?

Before we can begin to answer that we have to define green, a term we hear a lot about in reference to preservation of the planet. It is sometimes used in reference the amount of man made CO2, which is often blamed for the current climate change with respect to global warming. There’s no doubt that CO2 is a green house gas and that it can cause the earth to warm. But how much is the subject of much debate depending on which side you take. Frankly I think the IPCC estimates are way off but that is another discussion. I’m for anything that will reduce the amount of pollutants we humans generate.

So for this discussion let’s determine the amount of CO2 produced by the fuels currently used and those that are proposed to be used by Champ Car.

If you don’t care to see the math or the method skip to the bottom for the results, you may be surprised!

Methanol – currently used by Champ Car.

E85 – 85% Ethanol 15% Octane (gasoline) proposed to be used.

When these fuels are burned the by product is carbon dioxide and water. For this discussion we will consider only a stoichiometric combustion.

First let’s determine the weight of a gallon of each of the fuels. I’ll use gallons and pounds here for the metrically challenged. If you search the internet for this information you will find many different answers so in order to be as accurate as possible I will use the information given in the handbook of Chemistry and Physics from CRC press.

Conversion factors

3.785 l/gallon

453.59 g/lb

Octane C8H18 density at 25degrees C = .6986 g/ml

Ethanol C2H6O density at 25C = .7873 g/ml

Methanol CH4O density at 25C = .7872 g/ml

When converted to grams per gallon:

Octane) = 2644.201 g/gal

Ethanol) = 2979.9305 g/gal

Methanol) = 2979.5520 g /gal

Divide grams per gallon by 453.59 to get pounds per gallon

Octane) 2644.201/453.59 = 5.829 lbs/gal

Ethanol) 2979.9305/453.59 = 6.5696 lbs/gal

Methanol) 2979.5520/453.59 = 6.5688 lbs/gal

Next find ratio of CO2 molecule mass to fuel molecule mass.

Hydrogen = 1

Carbon = 12

Oxygen = 16

Carbon Dioxide [C = 12 and 2O’s = 32] 12 + 32 = 44

Octane C8H18 [C8 (8X12) = 96 [H18 (18X1) = 18] 96 +18 = 114

Using the same process for Ethanol and Methanol gives:

Ethanol = 46

Methanol = 32

To find out how many CO2 molecules are produced from each fuel simply count the carbon atoms in one molecule of each fuel.

Octane = 8 for a total mass of 8 X 44 = 352

Ethanol = 2 for a total mass of 2 X 44 = 88

Methanol = 1 for a total mass of 1 X 44 = 44

So the ratio of CO2 molecule mass to fuel molecule mass is:

Octane = 352/114 = 3.0877

Ethanol = 88/46 = 1.913

Methanol = 44/32 = 1.375

We're almost there, multiply these ratios by the weight per gallon of each fuel to see how many pounds of CO2 each one produces when combusted.

Octane - 5.829 X 3.0877 = 17.998 lbs CO2/gal

Ethanol - 6.5696 X 1.913 = 12.567 lbs CO2/gal

Methanol - 6.5688 X 1.375 = 9.058 lbs CO2/gal

To find CO2 in lbs/gal for E85 find the sum of .85 X 12.567 and .15 X 17.998

E85 produces 13.380 lbs of CO2 for each gallon combusted, Over 4 lbs more than Methanol.

There you have it, by our standard E85 is not green………….but if CCWS decides to use it a big oil company might pony up a different kind of green as a series sponsor! Like I said first you have to define green.

So for this discussion let’s determine the amount of CO2 produced by the fuels currently used and those that are proposed to be used by Champ Car.

If you don’t care to see the math or the method skip to the bottom for the results, you may be surprised!

Methanol – currently used by Champ Car.

E85 – 85% Ethanol 15% Octane (gasoline) proposed to be used.

When these fuels are burned the by product is carbon dioxide and water. For this discussion we will consider only a stoichiometric combustion.

First let’s determine the weight of a gallon of each of the fuels. I’ll use gallons and pounds here for the metrically challenged. If you search the internet for this information you will find many different answers so in order to be as accurate as possible I will use the information given in the handbook of Chemistry and Physics from CRC press.

Conversion factors

3.785 l/gallon

453.59 g/lb

Octane C8H18 density at 25degrees C = .6986 g/ml

Ethanol C2H6O density at 25C = .7873 g/ml

Methanol CH4O density at 25C = .7872 g/ml

When converted to grams per gallon:

Octane) = 2644.201 g/gal

Ethanol) = 2979.9305 g/gal

Methanol) = 2979.5520 g /gal

Divide grams per gallon by 453.59 to get pounds per gallon

Octane) 2644.201/453.59 = 5.829 lbs/gal

Ethanol) 2979.9305/453.59 = 6.5696 lbs/gal

Methanol) 2979.5520/453.59 = 6.5688 lbs/gal

Next find ratio of CO2 molecule mass to fuel molecule mass.

Hydrogen = 1

Carbon = 12

Oxygen = 16

Carbon Dioxide [C = 12 and 2O’s = 32] 12 + 32 = 44

Octane C8H18 [C8 (8X12) = 96 [H18 (18X1) = 18] 96 +18 = 114

Using the same process for Ethanol and Methanol gives:

Ethanol = 46

Methanol = 32

To find out how many CO2 molecules are produced from each fuel simply count the carbon atoms in one molecule of each fuel.

Octane = 8 for a total mass of 8 X 44 = 352

Ethanol = 2 for a total mass of 2 X 44 = 88

Methanol = 1 for a total mass of 1 X 44 = 44

So the ratio of CO2 molecule mass to fuel molecule mass is:

Octane = 352/114 = 3.0877

Ethanol = 88/46 = 1.913

Methanol = 44/32 = 1.375

We're almost there, multiply these ratios by the weight per gallon of each fuel to see how many pounds of CO2 each one produces when combusted.

Octane - 5.829 X 3.0877 = 17.998 lbs CO2/gal

Ethanol - 6.5696 X 1.913 = 12.567 lbs CO2/gal

Methanol - 6.5688 X 1.375 = 9.058 lbs CO2/gal

To find CO2 in lbs/gal for E85 find the sum of .85 X 12.567 and .15 X 17.998

E85 produces 13.380 lbs of CO2 for each gallon combusted, Over 4 lbs more than Methanol.

There you have it, by our standard E85 is not green………….but if CCWS decides to use it a big oil company might pony up a different kind of green as a series sponsor! Like I said first you have to define green.