Why Liquefied Methane is a better vehicle fuel than:
electric vehicles
The production of sufficient hydrogen in a non-polluting way to run nations cars must be considered. 
By first shifting to methane a large step toward eliminating pollution will be realized. 
Thermodynamic Explanation Here
George D. Freier 
University of Minnesota, Physics;



 Energy of Combustion

 The Oxidation of Hydrocarbons*.

Example HR

The following note is to compare some common hydrocarbon fuels.

The heat generated is always adjusted to atmospheric pressure and 25 Co.  For methane one has
               CH4 + 202 = 2H20( L) + CO2                   + 212.0 kilocal/mole
The (L) implies that the water is in the liquid form.
When ethane is burned one has
               CH2H6 + (7/2)02 = 3H20( L) + 2 CO      + 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
              C3H8 + 5O2  =  4H20( L) + 3 CO2              + 529.2 kilocal/mole
For the extra unit of CH2 the additional energy is (529-371) = 158 kilocal/mole.
As one goes to longer and longer chain hydrocarbons, each additional unit of CH2   = 158 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.
Example HR


     One can similarly proceed with the burning of alcohol, which is a partially oxidized hydrocarbon.
For methyl alcohol one has
               CH3OH + (3/2)O2 = CO2 + 2H20               + 183.0 kilocal/mole
For 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.
In general for n carbons in the alcohol we have, the alcohol is described as CnH2nHOH and the general equation for heat is
                            ^Q = [183 + 158(n-1)] kilocal/mole
One can see that the partially oxidized CH2HOH with 183 kilocal/mole is not as good as CH4 with 212 kilocal/mole.

    Number of Moles.

     The CH2 groups added to either a hydrocarbon or an alcohol contribute 158 kilocal/mole.  As either chain gets longer and longer the heat available from either kind of fuel becomes the same.  For short chains the difference becomes greater.

     Amounts of fuel are most often reckoned in terms pounds.
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.

All of these numbers show the advantages of methane

Volume Considerations

     The above considerations are for mass.
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, CH2OH, and octane C8H18.
The respective densities of the two substances are
      Substance  Density        Mass       No. of Moles
      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
The heat of combustion of methanol is:                       =   183 kcal/mole.
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  Energy
      Methanol    4.55 x 106V kilocal
      Ethanol        5.8  x 106V kilocal
     Octane        8.14 x 106V kilocal
For a volume V there is more energy for octane than for methanol or ethanol.


     When a car piston gets fuel by an injector, the same volume ^V is injected on each stroke.
          When methanol burns, the equation is
               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
For a volume ^V the number of moles of methanol is   24,875^V moles,
                            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     Heat
     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
All reactions produce CO2 and H2O.
   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 then
     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
If both the CO2 and H2O have the same specific heat, Cp, then the temperature rise of the final gasses should be
^T = [Heat added]/[ No. of moles ]Cp
This 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.

Example HR


In the above cases I have been comparing liquid fuels.  Methane burns as a gas with the reaction equation:
CH4 + 2O = CO + 2H2 O        ^H = -212 kcal/mole
In 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/Cp
This is better than methanol or ethanol but not so good as octane.
Example HR

Car Performance


With known fuel values one can consider the performance of a typical car with various kinds of fuels.
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 10-3mg3
      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

Amounts of Fuel in the 20-Gallon Tank

I will compare all fuels with octane, C8H10.  This has a density  p = 704 kg/m3, and a molecular weight,  U = 0.114kg/mole.
In a 20 gallon tank one then has:
  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
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          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
With octane at $1.00/gal the competitive cost of the other fuels would be
      Fuel       Price $/gallon
      Octane           1.00
      Methanol        0.56
      Ethanol           0.72
      Methane         0.67
The most obvious result is the decrease in CO2 emissions if one switches to methane.

Electric Cars versus a 20-gallon tank of gas

One can compare all of this with using an electric car.
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 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][12x10 J/batt] = 215 batteries
This many batteries would give the performance of a car with a 20 gallon gas tank full of gas.
Even if these batteries were as light as 10 lb/bat the battery weight would be over one ton.
If the batteries are lead batteries, the reaction during discharge is
          Pb + H2SO4 = PbSO4 + 2H + 2e-
On recharging the battery the reaction is
          PbSO4 + 2H20 = PbO2 + 2H + 2e-
In 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.
     Each battery provides
          (300 coul/sec)(3600sec) = 106 coul.
One notes from the reaction equations that there is an atom of hydrogen for each electronic charge, one on charging and one on discharging.
On charging one will then produce
          10 x 106  coul/1.6 x 10-19 coul/atom = 6.25 x 1024 atoms
On one charging cycle there will be twice this number or 12.5 x 1024 atoms or 6.25 x 1024 molecules of hydrogen.
Since one mole contains 6.02 x 1023 molecules, the cycle will produce 10.4 moles of molecular hydrogen.
The 215 batteries will produce
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 is to be compared with the 617 x 103 kcal of energy used to drive the car 600 miles.
This must come from a power plant while charging the batteries at about 25% efficiency.
This means the generation of
617 x 103  kcal/0.25 = 2,468 x 103  kcal.
These numbers indicate that the use of electric cars is more or less a ridiculous concept.

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.

Example HR


The fuels which have been considered are all hydrocarbons which contain hydrogen and carbon.

The reaction for burning hydrogen was

H2(g) + (1/2)O2  =  H2O -57.6 kcal/mole

with a combustion product that produces no pollution.

The reaction for burning carbon is

    C(s) + O2(g) =  CO2(g) -94.1 kcal/mole

with a combustion product that is all pollution.
The burning of carbon produces more energy per mole of carbon so that carbon rich fuels produce more energy.
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.
    The molecular weight of hydrogen is u = .002 kg/mole
while the molecular weight of carbon is u = 0.012 kg/mole.
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
Obviously, the energy per unit mass is much greater for hydrogen than for carbon.

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.
*Dr. George Freier
University of Minnesota
Department of Physics
May  2005