By first shifting to methane a large step toward eliminating pollution will be realized.
Thermodynamic Explanation Here
University of Minnesota, Physics;
The heat generated is always adjusted to atmospheric pressure and 25 Co. For methane one hasThe (L) implies that the water is in the liquid form.
CH4 + 202 = 2H20( L) + CO2 + 212.0 kilocal/mole
When ethane is burned one hasFor the extra unit of CH2 the additional energy is (529-371) = 158 kilocal/mole.
CH2H6 + (7/2)02 = 3H20( L) + 2 CO2 + 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.
C3H8 + 5O2 = 4H20( L) + 3 CO2 + 529.2 kilocal/mole
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 CH2, which adds 158 kilocal/mole so that for ethyl alcohol the heat of combustion is 341 kilocal/mole.
CH3OH + (3/2)O2 = CO2 + 2H20 + 183.0 kilocal/mole
In general for n carbons in the alcohol we have, the alcohol is described as CnH2nHOH and the general equation for heat isOne can see that the partially oxidized CH2HOH with 183 kilocal/mole is not as good as CH4 with 212 kilocal/mole.
^Q = [183 + 158(n-1)] kilocal/mole
Amounts of fuel are most often reckoned in
One pound is equal to 0.454 kilograms. The molecular weight of CH4 is 0.016 kg/mole so that one pound contains 28.4 moles which can liberate 212 kilocal/mole. The molecular weight of CH2HOH 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.
Substance Density Mass No. of MolesThe heat of combustion of methanol is: = 183 kcal/mole.
Methanol 796 kg/m3 796V kg 24,875V moles
Octane 704 kg/m3 704V kg 6,175V moles
Ethanol 789 kg/m3 789V kg 17,152V moles
Substance EnergyFor a volume V there is more energy for octane than for methanol or ethanol.
Methanol 4.55 x 106V kilocal
Ethanol 5.8 x 106V kilocal
Octane 8.14 x 106V kilocal
When methanol burns, the equation isFor a volume ^V the number of moles of methanol is 24,875^V moles,
CH3OH+(3/2)O2 = CO2+2H2O; ^H= - 183 kcal/mole
When ethanol burns the equation is
C2H5OH + 3O2 = 2CO2 + 3H2O ^H= - 341 kcal/mole
When octane burns the equation is
C8H16 +(25/2)O2 = 8CO2+9H2O; ^H= - 1318 kcal/mole
Substance No. of moles HeatAll reactions produce CO2 and H2O.
Methanol 24,875^V (24,875^V)(183) = 4.55 x 106^V kcal
Ethanol 17,152^V (17,152^V)(341) = 5.85 x 106^V kcal
Octane 6,175^V (6,175^V)(1318) = 8.14 x 106^V kcal
The number of moles of reaction gases is thenIf both the CO2 and H2O have the same specific heat, Cp, then the temperature rise of the final gasses should be
Substance Moles of reaction gas Temperature change
Methanol 3(24,875^V) = 7.5 x 104^V moles 6.1 x 104/Cp
Ethanol 5(17,152^V) = 8.6 x 104^V moles 6.8 x 104Cp
Octane 17( 6.175^V) = 10.5 x 104^V moles 7.7 x 104/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.
CH4 + 2O2 = CO2 + 2H2 O ^H = -212 kcal/moleIn this case, there are 3 moles of exhaust gases so that the temperature rise should be
^T = 212 x 103/3Cp = 7.1 x 104/CpThis is better than methanol or ethanol but not so good as octane.
60 mi/hr = 26.8 m/sec
2000 rpm = 33 rev/sec
20 gal. = 75.8 x 10-3mg3
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)R
If 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 106mg3 one has
N^V = [1.60m/rev][1/12.74 x 106/mg3] = 0.125 x 10-6mg3/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/m3 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/m3 60.3 kg 0.032 kg/mole 1884 moles
Ethanol 789 kg/m3 59.8 kg 0.046 kg/mole 1300 moles
Methane 416 kg/m3 31.5 kg 0.016kg/mole 1969 moles
In 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 103 kcal 3744 moles
Methanol 1884 moles -183 kcal/mole 345 x 103 kcal 1884 moles
Ethanol 1300 moles -341 kcal/mole 443 x 103 kcal 2600 moles
Methane 1969 moles -212 kcal/mole 417 x 103 kcal 1969 moles
Fuel miles miles/gallon CO2/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 CO2 emissions if one switches to methane.
Fuel Price $/gallon
The energy from this battery isThis many batteries would give the performance of a car with a 20 gallon gas tank full of gas.
W = (300coul/sec)(3600sec)(12volts) = 12.00 x 106 Joules.
The 617 x 103 kcal expressed in Joules is
W = (617x103 kcal)(4.18x103 J/kcal) = 2.58 x 109 Joules
It would require
[2.58x109 J][12x106 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 + H2SO4 = PbSO4 + 2H + 2e-
On recharging the battery the reaction is
PbSO4 + 2H20 = PbO2 + 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) = 106 coul.
On charging one will then produceOn one charging cycle there will be twice this number or 12.5 x 1024 atoms or 6.25 x 1024 molecules of hydrogen.
10 x 106 coul/1.6 x 10-19 coul/atom = 6.25 x 1024 atoms
The 215 batteries will produceThis is to be compared with the 617 x 103 kcal of energy used to drive the car 600 miles.
215(10.4moles) = 2.236 x 103 moles of hydrogen
The reaction equation for burning hydrogen is
H2(g) + (1/2)O2 = H2O(g) -57.6 kcal/mole.
The wasted hydrogen could produce
(57.6 kcal/mole)(2.236 x 103 moles) = 128 x 103 kcal
This means the generation ofThese numbers indicate that the use of electric cars is more or less a ridiculous concept.
617 x 103 kcal/0.25 = 2,468 x 103 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
H2(g) + (1/2)O2 = H2O -57.6 kcal/mole
with a combustion product that is all pollution.
The reaction for burning carbon is
C(s) + O2(g) = CO2(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.
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
one finds that methane is the most hydrogen rich and will produce the
Methane and hydrogen are excellent fuels but both must be refrigerated in order to carry them as liquids in a car.*Dr. George Freier
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.