Glow fuel mixing

                                           

   Recent nitromethane problems have resulted in limited availability of specific mixtures. You may want to do some fuel
 mixing or changing of existing mixes. Also, you may want to get some idea of what these fuels do in an engine. The following
   may help. All mixes are described on a percentage by volume basis which seems to be standard in the glow engine fuel
 business. Standard American volume equivalents are: 1 U S gallon= 4 quarts= 128 fluid ounces= 231 cubic inches. A typical
 mix, called 10% fuel, is 10% nitromethane, with 25% castor oil and 65% methanol- all by percentage volume so that a gallon
     of 10% looks like the tabulation below. 12.8 oz. nitromethane 10% 32.0 oz. castor oil 25% 83.2 oz. methanol 65%
  --------------------- ----- 128.0 oz. total (gallon) 100% The calculation for oil, as an example, is 25% of a gallon
 which equals 0.25 x 128 oz. or 32 oz., and the calculation for nitromethane at 10% is 0.10 x 128 oz. or 12.8 oz. The nitro and
  oil are carefully mixed with about half gallon of methanol and then the gallon is completed with methanol- thus avoiding
 directly measuring the methanol. Stir some more and cap the mixed fuel. Mixed fuel, well capped, will last for years with no
    change in quality. I personally do two and a half gallons at a time. The materials discussed in this paper are methanol,
  nitromethane and castor oil. Alternates or options, such as nitroethane and synthetic oils, will be discussed briefly. More
 exotic and dangerous materials should not be used. Methanol (wood alcohol, methyl alcohol, and meths in the UK) is the first
  oxidation product of methane, CH4, and has the chemical name CH3OH. Further oxidation produces formaldehyde, then
  formic acid, and finally carbon dioxide and water (CO2 & H2O), which is ultimate--you cannot burn water! All of these
 godies occur in your engine and produce heat which does the work. Methanol is a colorless and almost oderless liquid which is
 a violent poison and is highly flameable. It burns with a pale blue flame which is not visible in daylight. You could be on fire
 with methanol and nobody could see the flames. Methanol produces 55,550 net BTU per gallon, a 3450 OF flame temperature
  in air, has a near 43 OF flash point (does not evaporate below 43 OF), and a 725 OF ignition temperature. The addition of
   castor oil and nitromethane will slightly lower the ignition temperature of the mix. 1300 OF to 1500 OF temperature is
 indicated by the "color" of a hot glow plug coil. The flammable limits of methanol (% vapor in air by volume) are 6.7% lower
   and 36.0% upper. So if lean or rich on a cold day (43 0F flash point), your engine will be a real bear to start. A gallon of
 methanol requires about 560 standard cubic feet of air for combustion, resulting in exhaust gasses composed of 78.4 standard
 cubic feet of carbon dioxide (CO2), 156.9 cu ft of water (H2O) at 212 OF (vapor), and 445.5 standard cubic feet of nitrogen
 (N2). The nitrogen comes from air, 21% oxygen (O2) and 78% nitrogen (N2), which must be pushed through the engine to
 provide oxygen for combustion of the fuel. Good methanol is both anhydrous (free of water), and hydroscopic (absorbs water
 from air and is soluable in water). It should never be exposed to humid air (most air is humid) for more than a few minutes.
  Dry methanol is a good fuel, while wet methanol is a dog- will still burn, but not well. Good dry methanol has a gravity of
  0.7914 and anything higher shows water. Pure water has a gravity of 1.0000. Castor oil is the "classsic" oil for glow fuel; a
 vegitable oil from the beans of a plant called RICINUS COMMUNIS. Chemically it is C11H10O10, viscous, clear to yellow,
  and about the same weight as water. Medically it was an old fashioned laxitive and if 75 year old memory serves was pretty
 potent! Castor oil as a lubricant is exceptional because of heat resistance, cling, and lubricity. The cooeficient of frictiom for
 steel om steel, using castor oil as lubricant is 0.095, which is better than all other oils listed in the handbooks. Compare to wet
  ice on wet ice with a cooefficient of friction of 0.05 to 0.15, where the lower number is the most slippery. Castor oil at 10%
  would lubricate an engine, but it is raised to 25% so that the oil can also act as a coolant. It does not enter into combustion,
   just pumps through the engine the same as nitrogen. The castor "smoke" you see in the air is mostly fine droplets of oil.
 Traces of true smoke may be produced in a very hot engine. Castor oil smoke has a pleasant odor, while most of the synthetic
 oils have a perfume-like odor associated with their smoke. The two together produce a definite and unique flying-field odor.
 The main difficulty of the synthetic oils is that there are so many of them, each manufactured for a specific purpose, but not
 for glow engines. I tried one by NATIONAL CARBON, which was manufactured as a substitute for castor oil, and used as a
   lubricant for rubber conveyor belts. Using 1/3 synthetic and 2/3 castor as a 25% fuel component, I have experienced no
  problems and will continue as long as the keg of synthetic oil lasts. The big negative with castor is that it polymerizes to a
  tacky varnish when left on surfaces and exposed to air- makes messy airplanes. Synthetic oils seem to reduce or soften this
   polymerization when mixed with castor oil. Nitromethane (CH3NO2) is a colorless liquid, sweet smelling, soluable in
  methanol, and normally stable- but with the chemical potential to detonate. When mixed in methanol it gives the mixture
 the "alcohol" smell and provides some oxygen for combustion beyond the induced air. The OSHA exposure limit is 100 ppm
  over an eight hour period, so don't just sit there and sniff the stuff. At my purchasing level nitro is about $30.00 per gallon.
 The "old hands" say that the nitromethane "activates" the methanol, but as a nasty old combustion engineer, I do not know
  "what the watermelon hell", "activates" means. Let's look at the combustion to see if we can figure it out. The molecular
 balance for the combustion of methanol (CH3OH) is: 2(CH3OH) + 3O2 --> 2CO2 + 4H2O or two mols of methanol require
 3 mols of molecular oxygen (two volumes of molecular methanol vapor require 3 volumes of molecular oxygen for complete
 combustion), but oxygen comes in a package called air which is 21% molecular oxygen and 78% molecular nitrogen. There is
 3.76 times as much N2 as O2 in air. 3O2 + (3.76 x 3N2) or 14.26 "mols air" are required to burn 2 mols of methanol. Fuel/air
 ratio for methanol is 1/7.14. The molecular balance for the combustion of nitromethane, (CH3NO2) is: 4(CH3NO2) + 3O2
     --> 4CO2 + 6H2O + 2N2. 4 volumes of nitromethane vapor require 3 volumes of molecular oxygen for complete
  combustion, which is 14.28 volumes of air as above. The fuel/air ratio for nitromethane is 1/3.57. Nitroethane (C2H3NO2)
 will provide available oxygen the same as nitromethane, but will not provide as much extra oxygen per unit volume. Compare
 methanol fuel/air ratio with nitromethane fuel/air ratio, 1/7.14 for methanol and 1/3.57 for nitromethane. The nitromethane
  is providing some oxygen for combustion and needs less induced air to burn; in fact, twice as much output is available from
  pure nitromethane as from pure methanol for the same amount of air. So the nitro is a chemical form of super-charging.
 With an increase in hp there is a corresponding increase in rpm, with an additional increase in induced (carbuerator) air, with
a compound increase in output hp. I am working on an algorithum to relate % nitro to % increase in engine output-- realizing
 that no two of these little engines are the same or will experience the same operating conditions. How much output increase is
 very difficult to calculate, but we know with increase in % nitro we gain in output, engine operating temperature, and the cost
  of the fuel. The conclusion is that nitromethane provides extra oxygen above that provided by combustion air and therefor
 improves hp output. The practical rule is to provide no more nitromethane than needed for your plane, your engine, and your
 flying wants. Most of us can get along on 5% or 10% nitromethane fuel. 10% nitro is actually 13.3% nitro if you look only at
 combustibles- the oils do not enter into combustion so can be thrown out of the calculation for effective percent of nitro. A
 twenty year old ringed SUPER TIGRE "60" size engine running at 11500 rpm pumps 6900 cu in of air per minute or 3.99 cu ft
of air per minute- on a good day! Round this out to 3,5 cu ft of air per minute because of nominally poor air, porting losses and
  blow-by. Apply the old combustion rule that a cu ft of air can burn a 100 btu, and the old engine burns about 350 btu per
 minutr. The engine may "suck up" a lot more fuel than this, but wastes it; these engines are rich even when "leaned out". 1 1/8
 oz of 0-25-75 fuel runs about one minute in the "60", but this is only 0.844 oz combustibles since the oil is not to be burned
 or about 366 btu of methanol per minute. Note that there is more btu of fuel than btu of combustion air (366 to 359), so the
 air is limiting the input of the engine to 350 btu per minute. 350 btu per minute is roughly equivament to several hp, but a lot
  of energy is lost to exhaust temperature and very low efficiency in these small engines. The result is that engine output is
 estimated to be less than 3/4 hp- would hardly pull a sick river-boat gambler out of church! Your engine output may be more
  or less when running on 0% nitro. Back to mixing fuels. If we have a gallon of 10% and wish to chaange it to 5%, we first
 realize that the 10% has 12.8 oz nitro per gallon and we need only 6.4 oz per gall for 5%. There is no way to take nitro "out" of
  the 110% gallon, so the 10% gallon must be diluted to reduce the nitro to 5%. The extra nitro can build another gallon of
   mixture at 5%. The second gallon requires 32 oz (25%) of oil. Using a two gallon container, dump in the gallon of 10%
 mixture, add the 32 oz of oil, and enough methanol to complete the 2 gallons. Stir well and tightly cap the new 5% fuel. The
new 2 gallons of 5% mixture is: 12.8 oz nitromethane 5% of 256 oz (2 gal) 64.0 oz oil 25% of 256 oz 179.2 oz methanol 70% of
  256 oz --------------------- ---------------------- 256.0 oz total 100% of 256 oz (2 gal) When the added
  methanol completed the two gallons, 6.4 oz of the added methanol filled in for the 6.4 oz nitro that was "taken out" of the
 10% gallon to start the new 5% gallon. To move from 5% fuel to 10% fuel, start with the 5% gallon which holds 6.4 oz nitro.
 Now add enough nitro to make a 10% mixtue (larger container now needed). The new mixture will have a volume of 128 oz
 plus whatever nitro is required to make a 10% mix. This sets up a small algerbra problem-- algerbra at the RC field-- what is
  this world comming to? If "x" is the added nitro, then (128 oz + "x" oz)(0.10) = "x" oz + 8.4 oz. This equation translates to
   10% of the new volume is equal old nitro plus the added nitro. Now you hot shot math purists realize that adding nitro
requires that we add a little more oil and a little more methanol. We are approaching a true 10% mix by simultaneously adding
 oil and methanol. The above algebraic approxamation puts us within 2% of being "right on", so do not worry about it unless
 you enjoy simultaneous equations. Presume a 5% mixture and a 10% mixture; the challange being to produce a 10% mixture. I
 would not waste my time thinking about this one- the practical approach is to correct the 5% to 10% and the 20% to 10% and
 throw the two into a container marked 10%. References are the North American Mfg Co combustion engineering book and
                        Chemical Rubber Co chemistry and physics handbook. 

 
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