gasoline |

diesel |

ethanol |

methanol |

electric vehicles |

By first shifting to methane a large step toward eliminating pollution will be realized. Thermodynamic Explanation HereUniversity of Minnesota, Physics; |

The heat generated is always adjusted to atmospheric pressure and 25 CThe (L) implies that the water is in the liquid form.^{o}. For methane one has

CH_{4 }+ 20_{2}= 2H_{2}0( L) + CO_{2}+ 212.0 kilocal/mole

When ethane is burned one hasFor the extra unit of CH2 the additional energy is (529-371) = 158 kilocal/mole.

CH_{2}H_{6}+ (7/2)0_{2}= 3H_{2}0( L) + 2 CO_{2 }+ 370.5 kilocal/mole

In this second case we have oxidized one extra unit of CH2 and received an extra amount of heat of

(370.5-212) kilocal/mole = + 158.0 kilocal/mole.

For propane

C_{3}H_{8}+ 5O_{2}= 4H_{2}0( L) + 3 CO_{2}+ 529.2 kilocal/mole

As one goes to longer and longer chain hydrocarbons, each additional unit of CH

In general one has the heat (^Q) for the burning of a hydrocarbon CnH2n+2 as

^Q = [212 + 158(n-1)] kilocal/mole.

For methyl alcohol one hasFor ethyl alcohol, one has an extra unit of CH

CH_{3}OH + (3/2)O_{2}= CO_{2}+ 2H_{2}0 + 183.0 kilocal/mole

In general for n carbons in the alcohol we have, the alcohol is described as CnHOne can see that the partially oxidized CH_{2}nHOH and the general equation for heat is

^Q = [183 + 158(n-1)] kilocal/mole

Amounts of fuel are most often reckoned in
terms pounds.

One pound is equal to 0.454 kilograms. The molecular weight of
CH_{4} is 0.016 kg/mole so that one pound contains 28.4 moles
which
can liberate 212 kilocal/mole. The molecular weight of CH_{2}HOH
is .032 kg/mole so that one pound contains 14.2 moles and each mole can
liberate only 183 kilocal/mole.

For long chain hydrocarbons and alcohols the molecular weight becomes almost the same so the number of moles per pound is about the same, and for each carbon in the chain the energy yield is about 158 kilocal/mole.

In the auto industry one is concerned about the volume required for a fuel tank.

Suppose that the tank has a volume V measured in cubic meters. Let me compare methanol, CH

The respective densities of the two substances are

Substance Density Mass No. of MolesThe heat of combustion of methanol is: = 183 kcal/mole.

Methanol 796 kg/m^{3}796V kg 24,875V moles

Octane 704 kg/m^{3}704V kg 6,175V moles

Ethanol 789 kg/m^{3}789V kg 17,152V moles

The heat of combustion of ethanol is [183 + (158)] = 341 kcal/mole.

The heat of combustion of octane is [212 + 7(158)] = 1318 kcal/mole.

For a volume V the energy then is

Substance EnergyFor a volume V there is more energy for octane than for methanol or ethanol.

Methanol 4.55 x 10^{6}V kilocal

Ethanol 5.8 x 10^{6}V kilocal

Octane 8.14 x 10^{6}V kilocal

When methanol burns, the equation isFor a volume ^V the number of moles of methanol is 24,875^V moles,

CH_{3}OH+(3/2)O_{2}= CO_{2}+2H_{2}O; ^H= - 183 kcal/mole

When ethanol burns the equation is

C_{2}H_{5}OH + 3O_{2}= 2CO_{2}+ 3H_{2}O ^H= - 341 kcal/mole

When octane burns the equation is

C_{8}H16 +(25/2)O_{2}= 8CO_{2}+9H_{2}O; ^H= - 1318 kcal/mole

for ethanol the number of moles is 17,152^V,

for octane the number of moles is 6175^V.

The heat from each stroke will be

Substance No. of moles HeatAll reactions produce CO

Methanol 24,875^V (24,875^V)(183) = 4.55 x 10^{6}^V kcal

Ethanol 17,152^V (17,152^V)(341) = 5.85 x 10^{6}^V kcal

Octane 6,175^V (6,175^V)(1318) = 8.14 x 10^{6}^V kcal

Each mole of methanol produces 3 moles of reaction gasses,

each mole of octane produces 17 moles of reaction gasses,

and each mole of ethanol produces 5 moles of reaction gasses.

The number of moles of reaction gases is thenIf both the CO

Substance Moles of reaction gas Temperature change

Methanol 3(24,875^V) = 7.5 x 10^{4}^V moles 6.1 x 10^{4}/Cp

Ethanol 5(17,152^V) = 8.6 x 10^{4}^V moles 6.8 x 104Cp

Octane 17( 6.175^V) = 10.5 x 10^{4}^V moles 7.7 x 10^{4}/Cp

^T = [Heat added]/[ No. of moles ]CpThis is then shown in the above table.

Since the octane yields a higher temperature it should then be a more efficient engine fuel than methanol or ethanol and ethanol should be better than methanol.

CHIn this case, there are 3 moles of exhaust gases so that the temperature rise should be_{4}+ 2O_{2 }= CO_{2 }+ 2H_{2 }O ^H = -212 kcal/mole

^T = 212 x 10This is better than methanol or ethanol but not so good as octane.^{3}/3Cp = 7.1 x 10^{4}/Cp

For a typical car, I will use a four cylinder car going at 60 mi/hr while the engine speed is 2000 revolutions per minute getting 30 miles per gallon and having a 20 gallon gas tank.

I will assume that all cars perform in the same way, which means that lower grade fuels will go a shorter distance on a tank of gas. In other words, use fuel faster to get a certain level of performance.

I will work things out in the metric system and then convert to English units at the end of the calculation. In the metric system

60 mi/hr = 26.8 m/sec

2000 rpm = 33 rev/sec

20 gal. = 75.8 x 10^{-3}mg3

where mg3 of gasoline.

In the metric system one then has the relation between engine speed, r, and car speed, v, as

v = [(26.7m/sec)/33.3rev/sec)]R = (0.802m/rev)RIf the 4 cycle engine has N cylinders injecting a volume ^V of fuel every other revolution the rate of burning fuel, dF/dt, is

dF/dt = NR^V/2 = [N^V/2][1.25rev/m]v

If D is the total distance traveled at velocity, v, and V is the volume of the fuel tank, then

D/v = V/(dF/dt) or dF/dt = Vv/D

Equating the expressions for dF/dt gives

V/D = [N^V/2][1.25rev/m] or D/V = [1.60m/rev][1/N^V]

Since D/V = 30 mi/gal = 12.74 x 10^{6}mg3 one has

N^V = [1.60m/rev][1/12.74 x 10^{6}/mg3] = 0.125 x 10^{-6}mg3/rev

In a 20 gallon tank one then has:For the same performance as octane which takes one 600 miles one can simply take 600 miles times the ratios of the corresponding energies.

Fuel p M U n

Octane 704kg/m^{3}53.4kg 0.114kg/mole 468moles

Where M is the mass of the fuel in the tank and N is the number of moles in the tank.For the other fuels under consideration one has

Fuel p M U n

Methanol 796 kg/m^{3}60.3 kg 0.032 kg/mole 1884 moles

Ethanol 789 kg/m^{3}59.8 kg 0.046 kg/mole 1300 moles

Methane 416 kg/m^{3}31.5 kg 0.016kg/mole 1969 molesIn this table, I assume that methane gas is refrigerated to become a liquid. One can now calculate the energy, W, from a 20-gallon tank for each of the fuels

Fuel n ^H W nCO2

Octane 468 moles -1318 kcal/mole 617 x 10^{3}kcal 3744 moles

Methanol 1884 moles -183 kcal/mole 345 x 10^{3}kcal 1884 moles

Ethanol 1300 moles -341 kcal/mole 443 x 10^{3 }kcal 2600 moles

Methane 1969 moles -212 kcal/mole 417 x 10^{3}kcal 1969 moles

Fuel miles miles/gallon CO_{2}/mile

Octane 600 30 6.24 moles/mile

Methanol 335 16.8 5.62 moles/mile

Ethanol 431 21.5 6.03 moles/mile

Methane 406 20.2 4.84 moles/mile

The most obvious result is the decrease in COFuel Price $/gallon

Octane 1.00

Methanol 0.56

Ethanol 0.72

Methane 0.67

Suppose that one has a 300 ampere hour battery at 12 volts.

The energy from this battery is

W = (300coul/sec)(3600sec)(12volts) = 12.00 x 10^{6 }Joules.

The 617 x 10^{3}kcal expressed in Joules is

W = (617x10^{3}kcal)(4.18x10^{3}J/kcal) = 2.58 x 10^{9}Joules

It would require

[2.58x10^{9}J][12x10^{6 }J/batt] = 215 batteries

If the batteries are lead batteries, the reaction during discharge isIn cycling the battery one produces 4 atoms of hydrogen which is the same amount of hydrogen as found in a methane atom, but in battery use, this hydrogen is lost.

Pb + H_{2}SO_{4}= PbSO_{4}+ 2H + 2e-

On recharging the battery the reaction is

PbSO_{4}+_{2}H_{2}0 = PbO_{2}+ 2H + 2e-

Each battery providesOne notes from the reaction equations that there is an atom of hydrogen for each electronic charge, one on charging and one on discharging.

(300 coul/sec)(3600sec) = 10^{6}coul.

On charging one will then produceOn one charging cycle there will be twice this number or 12.5 x 10

10 x 10^{6}coul/1.6 x 10^{-19}coul/atom = 6.25 x 10^{24}atoms

Since one mole contains 6.02 x 10

The 215 batteries will produceThis is to be compared with the 617 x 10

215(10.4moles) = 2.236 x 10^{3}moles of hydrogen

The reaction equation for burning hydrogen is

H_{2}(g) + (1/2)O_{2}= H_{2}O(g) -57.6 kcal/mole.

The wasted hydrogen could produce

(57.6 kcal/mole)(2.236 x 10^{3}moles) = 128 x 10^{3}kcal

This must come from a power plant while charging the batteries at about 25% efficiency.

This means the generation ofThese numbers indicate that the use of electric cars is more or less a ridiculous concept.

617 x 10^{3}kcal/0.25 = 2,468 x 10^{3}kcal.

One can argue that the performance of an electric car need not be so
great as a gasoline powered car.

However, if one lowered the performance of a gasoline powered car by
the same amount, then the gasoline car would still produce less overall
pollution than the electric car.

with a combustion product that produces no pollution.## The reaction for burning hydrogen was

## H

_{2}(g) + (1/2)O_{2}= H_{2}O -57.6 kcal/mole

with a combustion product that is all pollution.## The reaction for burning carbon is

## C(s) + O

_{2}(g) = CO_{2}(g) -94.1 kcal/mole

The burning of carbon produces more energy per mole of carbon so that carbon rich fuels produce more energy.The molecular weight of hydrogen is u = .002 kg/mole

However, if we are concerned with a mass M, of any fuel, there will be more energy per unit mass from hydrogen than from carbon.

while the molecular weight of carbon is u = 0.012 kg/mole.

Obviously, the energy per unit mass is much greater for hydrogen than for carbon.The energy from a mass M of the two kinds of fuel is

Hydrogen [M/(.002)] x (57.6 kcal/mole) = 28,800M kcal

Carbon [M/(.012)] x (94.1 kcal/mole) = 7,842M kcal

Of the fuels considered, namely, octane methanol ethanol and
methane,
one finds that methane is the most hydrogen rich and will produce the
least
pollution.

Methane and hydrogen are excellent fuels but both must be refrigerated in order to carry them as liquids in a car.*

The refrigeration problem is much greater for hydrogen than for methane.

It seems that the way to proceed is to first convert cars, trucks, and railroad engines to burn methane.

As the technology for refrigeration develops in the future, we should be able to shift to hydrogen and reduce car pollution to zero.

May 2005