| abstract
| - I. Air. Any burning or volatization of fossil fuels contaminates the air with emissions that would not naturally be there, and some uses create a worst problem than others. The change that would have the largest positive impact is the one most obvious, and per the peak oil commentators coming whether we want it or not, which is to cease use of fossil fuels. Avg Air Human Exhale 21.0% 16.3% CO2 .03% 4.0% N 78.0% 79.7% Inert/Other 1.0% NH3 5-25g/m3 4-9g/m3 Overt Pollution Example: 2 Stroke Engines - Do you own or operate any two-stroke gasoline engine? The issue of noise and pollution produced by a typical two-stroke gasoline engine (i.e. lawnmower) vs four-stroke (i.e. car engine) is potentially significant. The two-stroke gasoline engine generally puts out 10 times (more of some) as many pollutants per amount of fuel burned. The operation of these engines, in general, initiates and forcefully imposes upon others the operator's fouled air and excess noise. The two-stoke system is used because it provides the lightest fuel burning engine for the power produced, but paradoxically the two-stroke is significantly LESS fuel efficient than a four-stroke engine. The fuel in-efficiency of these engines leads to the pollution problem. But the pollution from these engines is not limited to transportation. These engines are extensively used on lawnmowers, weed whackers, portable blowers, etc. The California Air Resources Board has calculated that 2% of the smog generated by all engines originates from lawn mowers. I don't have a gas mower, so I'm guessing, say a mower runs an hour per gallon of gas? (Corrections anyone?) This would mean that in one hour of mowing you pollute at least equal to burning 10 gallons of gasoline in your car. A further discussion of fuel and engine types is in the transportation presentation. What is there that uses a 2 cycle engine that cannot be done with a less polluting and more efficient engine, done manually, or is so essential that the pollution is justified? Embedded Pollution Example: Food - In peak oil discussions, it is frequently presented that food production in the industrial world consumes 10 calories of oil for every calorie of food produced. (Transportation or cooking of the food NOT included in this estimate.) In general, a human needs 2000 calories of energy per day. Although they are normally spelled the same, a food calorie is in fact 1,000 "heat" calories. A gallon of gasoline contains energy equal to around 36,000 food calories. If a person needs 2,000 calories per day, then to produce those 2,000 calories of food 20,000 calories of oil were used. (55% of a gallon) If you eat commercially produced food, your daily meals require the consumption of fuel and production of pollution equal to a 30 mpg vehicle driving 16 miles. Food Item Calories times ten divided by 36,000 equals the fuel consumption embedded in producing the food. (Processing & shipping fuel not included) For a city with a population of a million, producing food represents the external daily use of 550,000 gallons of fuel. The brewing, canning, shipping, etc. all consumed additional fuel. As a local driving estimate for a modest city, the Pima Association of Governments estimates that 23,000,000 miles are driven every day in Tucson. At an average of 30 mpg for vehicles that would be over 760,000 gallons of gasoline per day. Given the above estimates of food production and local transportation for an example city of a million, the life-support infrastructure under current fossil fueled design requires 1.3 gallons of fuel per day. (1.3 gallons per person.) One gallon of gas weighs about 6.25 pounds. When burned the hydrocarbons combine with oxygen from the air. The result per gallon is exhaust with a CO2 aspect of 19.3 pounds and around 8 pounds (1 gallon in liquid form) of water vapor. (Both greenhouse gases, that would not naturally have been in the atmosphere.) You also get carbon monoxide and other nasty stuff. Now, let's see, if we burn at a minimum 1.3 million gallons each day in each city of a million..... Carbon monoxide (CO): Replaces oxygen in the red blood cells thus reducing the amount of oxygen that can reach the brain, heart and other tissues. CO can cause dizziness, slowed reaction times, headaches, an increased risk of heart disease and may promote the development of arteriosclerosis. Carbon monoxide (CO) is a colorless, odorless gas produced by the incomplete combustion of fuels. The major source of CO in our community is motor vehicles, which release over 85 percent of the CO emissions in Pima County. Stagnant weather conditions coupled with reduced engine efficiency associated with cold temperatures cause increased levels of CO in the winter months Hydrocarbons (also known as volatile organic compounds (VOC)): These are compounds made of hydrogen and carbon. They are released from gasoline engines and the evaporation of paint and solvents and are also produced naturally from the decomposition of organic matter and by certain types of plants. Ozone (O3): This pollutant can impair lung function and irritate the mucous membranes in the nose and throat causing coughing and choking. It aggravates chronic respiratory diseases like asthma and bronchitis, and can irritate the eyes, reduce lung capacity over time and increase sensitivity to allergens. Ozone is a highly reactive form of oxygen. At normal concentrations it is colorless and odorless. At high concentrations (often associated with thunderstorms or arching electric motors) it is an unstable bluish gas with a pungent odor. Ground level ozone in high concentrations is considered an air pollutant, while stratospheric ozone in the upper atmosphere (12 - 30 miles above the ground) is critical for absorbing cancer-causing ultraviolet radiation. Ozone is a secondary pollutant formed when nitrogen oxides and volatile organic compounds (VOC) react in the presence of sunlight. Volatile organic compounds come from automobile exhaust, gasoline vapors, and chemical solvents (and also some vegetation). Nitrogen oxides come from burning fuel. The reactivity of ozone causes health problems because it damages lung tissue, reduces lung function, and increases the sensitivity of the lungs to other irritants. Symptoms of decreased lung function include chest pain, coughing, sneezing and pulmonary congestion. Ozone can reduce immune system capacity. In high concentrations, ozone causes damage to plants and deteriorates materials such as rubber and nylon. Particulate matter (PM10 and PM2.5): May cause breathing difficulties and respiratory pain, irritations to the nose, throat and ear canal which are often mistaken for allergic reactions. PM can also weaken the immune system, diminish lung function and increase the incidence and severity of acute bronchitis, pneumonia, asthma and emphysema. Particulate matter (PM10 and PM2.5) is comprised of solid particles or liquid droplets tiny enough to remain suspended or floating in the air for up to weeks at a time. Of greatest concern to the public health are the particles small enough to be inhaled into the deepest parts of the lung. These particles are less than 10 microns in diameter--about 1/7th the thickness of a human hair--and are known as PM10. This includes fine particulate matter known as PM2.5. PM2.5 has a specific range of particles 2.5 micrometers or less. PM10 is a major component of air pollution that threatens both our health and our environment. General PM composition can include everything from fine dust to carbon (soot), and can be microscopic or visible to the naked eye. Particulate matter is generated from a variety of sources including traffic on paved and unpaved roads, combustion, and earth-moving activity such as mining, farming and construction. Fine particles present in the air even though it might seem invisible. Their size alone makes them a danger, as they easily reach deep into our lungs. But what they are made of can make the situation worse. Moving air - The maximum theoretical power that can be tapped from a moving mass of air is 57% of the energy in any given mass passing thru a given area. In general, a windmill should be located 30 feet above the ground, and 10 feet higher than any other object, to obtain a clear air flow. As you consider the needs of maintenance, perhaps you do not want to climb and work on a generator on a high fragile tower. Although gearing can "waste" 15% of your power, envision the generator on the ground, spun by gear and shaft from on high. The “rule of thumb” formula for power from a typical windmill is V = the cube root of (P/.02). That is a given velocity of wind in miles per hour cubed, then multiples by .02 should calculate out to watts of electricity potential. II. Water. One inch of rain per square foot is around ½ gallon of H2O . Vacuum's Affect on Water Vacuum PSIA Microns Water Boil Point 0 14.696 760,000 212 °F 10.24"Hg 9.629 500,000 192 °F 22.05"Hg 3.865 200,000 151 °F 25.98"Hg 1.935 100,000 124 °F 27.95"Hg .968 50,000 101 °F 28.94"Hg .481 25,000 78 °F 29.53"Hg .192 10,000 52 °F 29.72"Hg .099 5,000 35 °F 29.84"Hg .039 2,000 15 °F 29.82"Hg .019 1,000 +1 °F 29.901"Hg .010 500 -11 °F 29.917"Hg .002 100 -38 °F 29.919"Hg .001 50 -50 °F Vacuum = Inches Mercury (Hg) PSIA = lbs. per sq. in. Absolute Pressure Microns = A Special Unit of Vacuum Water Boil Pt. = Temperature That Water Boils at. Frozen water - We have (2005) around 6 million cubic miles of ice located on 10% of the Earth's land mass. 86% is in Antarctica, 10% in Greenland, 4% "other". Readily circulated claims are that if the bulk of this ice melted, water volume would raise the sea level around 200 foot. Further warming of sea water would result in expansion due to expanding water molecules. A 1920 Serbian physicist indicated the ice cover seems to follow a 40,000 year cycle, within which he put us at early to mid "summer" of the cycle. As part of the theory, it seems that open water in the Arctic is to be a signal of the start of cooling, not further melting. Global annual evaporation. Ocean – 74,000 cubic miles. Land – 18,000 cubic miles. Total 92,000 cubic miles. The averaged global rainfall is 28”, with overall around 25,000 cubic miles of rain falling on land. III. Food: Basil metabolic rate - An estimate of the daily number of calories to keep a sedatory person of a given weight alive without a loss in weight. Calories = 70 x (kg) 3/4 That is, take the persons weight in kilograms to the 3rd power (weight x weight x weight) then find the 4th root of that number. Take this 4th root times 70. An example: A person who weighs 60 kg (around 120 lbs). 60 to the 3rd power is 216,000. The 4th root of this is around 21.55. 70 times 21.55 is 1508.5, so this person needs around 1509 calories per day. Planning for a storage program requires knowing the properties of foods the family likes, or at least will eat. Below is information on a variety of grains, nuts, fruits, canned foods, etc., for use in calculating a food storage program with sufficient calories. In the storage program, I do not address vitamin content directly, trusting that for storage purposes a multivitamin, or better, sprouted or some minimum garden area can address the minimum vitamin needs. Grains - Misc. Item Cal.Lb. Pkg. Servings Carb. Protein Calorie Barley, pearl 1632 30 oz. 17 39 5 180 Millet, Whole 1285 28 oz. 15 33 4 150 Flax, seed 2380 16 oz. 17 9 6 140 Sesame, seed 1866 12 oz. 10 8 6 140 Rice, Jasmine 1955 20 lb. 230 39 3 230 Rice, black sweet 1600 16 oz. 8 46 4 200 Rice, sweet white 1530 16 oz 9 39 3 170 Oats, processed 1714 42 oz. 30 27 5 150 Beans Item Cal.Lb. Pkg. Servings Carb. Protein Calorie Pinto 750 64oz. 50 22 7 60 Mung 1554 14 oz. 4 58 23 340 Great Northern 1170 32 oz. 26 22 8 90 Kidney, red 840 16 oz. 12 22 9 70 Black 948 16 oz 12 23 9 79 Peas, blackeye 1080 16 oz. 12 23 9 90 Lentils 1040 16 oz. 13 20 10 80 Tian Jin Red 1554 14 oz. 4 63 21 340 Soybean 2240 16 oz. 16 10 12 140 Fruit, dried Item Cal.Lb. Pkg. Servings Carb. Protein Calorie Apricot 1142 7 oz. 5 24 1 100 Pineapple 1493 6 oz. 4 34 0 140 Mango 2080 4 oz. 3 32 0 130 Nectarine 1173 6 oz. 4 25 2 110 Peach 1066 6 oz. 4 25 2 100 Plum 1320 12 oz. 9 25 1 110 Dates 1440 8 oz. 6 30 1 120 Figs 1280 9 oz. 6 28 1 120 Cranberry 1466 6 oz. 5.5 25 0 100 Cherry 1280 6 oz. 4 32 0 120 Raisin 1456 15 oz. 10.5 31 1 130 Processed Pasta Item Cal.Lb. Pkg. Servings Carb. Protein Calorie Kanton-wheat noodle 4160 16 oz. 16 31 9 260 Bean thread-special 1371 10.5 oz. 5 4.7 0 180 Nuts Item Cal.Lb. Pkg. Servings Carb. Protein Calorie Peanut-roasted 2720 16 oz. 16 5 8 170 Peanut-raw Cashew-roasted 2940 16 oz. 14 10 6 210 Almond 2560 14 oz. 14 5 7 160 Pistachio 1520 16 oz. 8 9 8 190 Sunflower-roasted 2720 14 oz. 14 7 6 170 Peanut-spanish 2800 12 oz. 10 4 5 210 Walnut 3024 10 oz. 9 3 5 210 Misc. Canned Food Item Cal.Lb. Pkg. Servings Carb. Protein Calorie Hormel chili no bean 448 15 oz. 2 17 16 210 Peanut butter 7448 16 oz. 28 (tbsp) 84 112 266 Tuna 384 6 oz. 2.5 0 13 60 Spam 480 12 oz. 6 1 7 180 Root Calorie Crops Item Cal.Lb. Pkg. Servings Carb. Protein Calorie Carrot 239 16 oz. 3 oz. 9 1 45 Onion, yellow 184 Ea. (148 gram) 14 2 60 Potato, russet 337 5 lb. (ea) 23 3 110 Potato, red 337 5 lb. (ea) 23 3 110 Potato, gold 337 5 lb. (ea 48 gram) 23 3 110 If you stored one pound of each of the above items, the calories would add up to 742343. If you therefore planned on eating from your storage program an equal weight of each of the above food items, your would need roughly 10 pounds of each item. The author has available a spreadsheet that includes the above items that can be used to estimate the calorie value of an input storage selection. Greens Item Cal.Lb. Pkg. Servings Carb. Protein Calorie Spinach 106 10 oz. 3.3 3 2 20 Lettuce 80 10 oz. 3.3 3 1 15 Special Concentrated Items Item Cal.Lb. Pkg. Servings Carb. Protein Calorie Soy Protein 3027 22.2 oz. 35 17 14 120 Whey Protein 1813 12 oz. 17 1 16 80 Vegetable Protein 1760 15 oz. 15 0 24 110 IV. Shelter. The material and design of your clothing, vehicle, home, etc., can, merely thru the natural characteristics of the materials and their orientation have a significant effect on your comfort. Weight and Thermal Conductivity of Sample Materials. Conduction is heat transfer by agitation of the molecules in a material without any observed motion of the material. If one side of metal or concrete surface is at a higher temperature, energy will be transferred thru the material toward the cooler side. The formual to use is: Q = kA(T hot minus Tcold) t d Q = heat transferred in time = t k = thermal conductivity of barrier A = Area T = Temperature d = Thickness of barrier Density Conductivity Specific Heat at 68 F BTU in/hr ft2 F BTU/lb. Degree F Lb/ft3 Air, Still - .0169-.215 Aluminum 168.0 1404-1439 Asbestos board w/cement 123 1.7 Asbestos, wool 25.0 .62 Brass, red 536 715.0 Brick Common 112.0 5.0 .2 Face 125.0 9.2 .2 Fire 115.0 6.96 .2 Bronze 509 522 Cabots 3.4 .25 Cellulose, dry 94 1.66 Celotex (sugar cane fiber) 13 .34 Charcoal Coarse 13.2 .36 6 mesh 15.2 .37 20 mesh 19.2 .39 Cinders 40 1.1 Clay Dry 63 3.5-4.0 Wet 110 4.5-9.5 Concrete Cinder 97 4.9 .2 Stone 140 12.0 .2 Corkboard 8.3 .28 Cornstack insul board 15 .24-.33 Cotton 5.06 .39 Foamglas 10.5 .40 Glass wool 1.5 .27 Glass Common thermometer 164 5.5 Flint 247 5.1 Pyrex 140 7.56 Gold 1205 2028 Granite 159 15.4 Gypsum, solid 78 3.0 Hair felt 13.0 .26 Ice 57.5 15.6 Iron, cast 442.0 326 Kapok 1.0 .24 Lead 710 240 Leather, sole 54 1.1 Lime Mortar 106 2.42 Slaked 81 - Density Conductivity at 68 F BTU in/hr ft2 F Lb/ft3 Limestone 132 10.8 Marble 162 20.6 Mineral wool Board 15.0 .33 Fill type 9.4 .27 Nickel 537.0 40 Paper 58 .9 Parafin 55.6 1.68 Plaster Cement 73.8 8.0 Gypsum 46.2 3.3 Redwood bark 5.0 .26 Rock wool 10.0 .27 Rubber, hard 74.3 11.0 Sand, dry 94.6 2.23 Sandstone 143.0 12.6 Sawdust 8 .41 Sil-O-Cel (power diatomaceous) 10.6 .31 Silver 656 2905 Soil Crushed quartz (4% H2O ) 100 11.5 Dakota sandy loam (4% H2O ) 110 6.5 (10% H2O ) 110 13.0 Fairbanks sand (4% H2O ) 100 8.5 (10% H2O ) 100 15.0 Healy clay (10% H2O ) 90 5.5 (20% H2O ) 100 10.0 Steel 1% C 487 310.0 Stainless 515 200 Tar, bituminous 75 - H2O , fresh 62.4 4.1 1.0 Wood Balsa 7.3 .33 Fir 34.0 .8 Maple 44 1.2 Red Oak 48 1.1 White Pine 32 .78 Wood fiberboard 16.9 .31 Wool 4.99 .264 Solar Absorption and Re-radiance Rates of Selected Materials Metal Absorb Emits Al , pure .1 .1 Al , anodized .12 .65 Cr .4 .2 Cu , polished .15 .03 Au .2 .025 Fe .44 .07 Metal Absorb Emits .36 .1 Ag , polished .035 .02 Zn .5 .05 The emissivities of many materials change with wavelength of the radiation being emitted. For example, silicon is an excellent emitter of visible light, but is essentially transparent to infrared radiation. We find below that good emitters are also good absorbers. Good absorbers are good emitters: An ideal absorber is often called a black body. It absorbs all the radiation that hits it. The absorptivity () is the complement of reflectivity (r = 1 - A good reflector is a poor absorber. Shiny aluminum is a such a good reflector. It is easy to see that an ideal absorber of a particular wavelength of radiation is also the best possible emitter at that wavelength. That is, no object at temperature T can emit more radiation than a black body. Proof: Start, for example, with two objects A and B that are close to each other and at the same temperature. Suppose that A is an ideal black body and B is not. While A absorbs all radiation that hits it, B does not. It reflects some. The question is whether B can emit more radiation than A. Since A absorbs all of the radiation emitted by B it would get hotter than B if B really could emit more radiation than A. But, this situation is a direct violation of the Second Law of Thermodynamics. We are not allowed to start with two bodies at the same temperature and find that one heats while the other cools! This means that a black body, the perfect absorber is not only the best absorber but also the best emitter. In summary, excellent absorbers are also excellent emitters. Example 1: Radiators. In the old days homes often used water or steam radiators to heat rooms. From a practical point of view, you really wouldn't want to make your radiator out of shiny aluminum. The low emissivity of aluminum means that it is both a poor emitter and a poor radiator and would give out less heat than one made of cast iron, for example. Example 2: Home Insulation. Aluminum foil on insulating panels has very small emissivity, It is therefore a very poor emitter of infrared radiation, a desirable feature. Example 2: Hot black roads. On a clear summer day a black asphalt road in the sun gets hot as it absorbs radiation from the sun. Most of this radiation has short wavelength as it comes from the sun's surface with a temperature of some 5800 K. To the extent that the hot black road is a "black body", it absorbs all the sun's incident radiation. It's emissivity is nearly 1 for this incident short wavelength light. The warm road also emits infrared radiation and continues to heat up until the power emitted, Pout = AT4 , balances the power absorbed from the sun, Pin = I0A. Here, I0 is the sun's intensity at the hot black road, typically 1000 W/m2. The hot black road's emissivity, , is also nearly 1 for these longer infrared wavelengths. With Pin = Pout we solve for T and find that a hot black road has a temperature of 364 K (91 0C), hot enough to fry an egg, but not hot enough to boil water! Example 3: You standing in a bathroom. With no clothes, taking your area to be ~2 m2, and your skin temperature to be ~300 K, with an emissivity of 1.0, you would radiate a power of Pout = AT4 = 919 watts, clearly an unsustainable value. In empty space you would indeed radiate heat away at this value. However, suppose you are in a bathroom with walls at 20 0C (293 K), ones with their own emissivity of 1. Then, the net heat radiated by you is given by Pout = A(Tyou4 - Twall4) = 83 watts, a much more reasonable number. The general expression for power exchanged between two parallel surfaces with emissivities and temperatures {1, T1} and {2, T2} is P = A(T14 - T24)/ [1/1 + 1/2 - 1]. Ref: Kraushaar & Ristinen, Energy and Problems of a Technological Society, p 156. With this equation, you can calculate the power exchange between two surfaces with different emissivities and temperatures. Building materials with aluminum foil come to mind. Selective Surfaces: It would be great fun to find a way to create a surface that could get hotter than a hot black road in the sun, maybe even one that could boil water. To do this we need either to absorb more of the sun's radiation coming in or emit less. We assumed that our hot black road was a perfect absorber of short wavelength light (short wavelength = 1) and a perfect emitter of infrared (long wavelength = 1). If we make the emissivities less, then we reduce both the absorbed radiation from the sun and the radiated radiation from the road. Are we stuck? The trick lies in creating a surface that has short wavelength > long wavelength . This does not violate the second law. To create our selective surface, we start with a layer of stainless steel and add a thin layer of gold and, on top of that, a thin layer of silicon. The silicon layer looks black to visible light and has short wavelength ~ 1. Since silicon is essentially transparent to infrared light, our selective surface behaves as a gold surface for infrared. Gold has an emissivity of only 0.10 for infrared wavelengths. This combination, then, is an excellent absorber of short wavelength light from the sun and a poor emitter of infrared light. Repeating our calculation, we find that this selective surface can rise to a temperature of 648 K (375 0C)! silicon gold stainless steel Thermal Storage Capabilities of Selected Materials. Officially 1 BTU is the amount of heat energy needed to raise the temperature of one pound of water one degree F. Media Melts Latent Heat Specific Heat Density BTU/lb C BTU/lb-F lt/ft3 ICE 32 144 .49 58 Water - - 1.0 62 Steel (scrap iron) - - .12 489 Basalt (lava rock) - - .2 184 Limestone - - .22 156 Paraffin wax 100 65 .7 55 Salt Hydrates NaSO4-10H2O 90 108 .4 90 NA2S2O3-5H2O 120 90 .4 104 NA2HPO4-12H2O 97 120 .4 94 Fire Brick - - .22 198 Ceramic oxides - - .35 224 Fused salts - - .38 140 C - - .2 140 Transmission Percentage of Light Thru Glass at Selected Angles to Solar Intercept Incident Solar Angle Intercept Percent 0 100 5 99.5 10 98.5 15 96.5 20 94.0 25 90.6 30 86.6 35 81.9 40 76.6 45 70.7 50 64.3 55 57.4 60 50.0 65 42.3 70 34.2 75 25.8 80 17.4 85 8.7 90 0.0 The Human Factor Postulate an average human of around 150 pounds, who needs 2000 food calories per day. A food calorie is 1,000 "heat" calories, so this person operates on 2,000,000 calories of heat, or in other energy terms. Energy and Our Bodies Around 8,000 BTU (1 BTU = 251.995761 heat calories) Around 2.4 kilowatt hours (1 watt = 859.8452279 heat calorie) One BTU is the amount of heat required to raise the temperature of one pound (1 pint) of water 1 degree F (144 BTU to melt 1 lb or pint of ice, 970 BTU to evaporate a pint of water). One calorie is the amount of heat required to raise the temperature of one gram of H2O 1 degree C. (80 calories per gram to melt a gram of ice, 540 to evaporate a gram of H2O ). In theory the daily heat of a person could melt around 56 pounds of ice (7 frozen gallons - 25 liters) or cause the evaporation of just over a gallon of water (3.7 liters). Therefore if your body was in a microenvironment isolated from thermal energy exchange with the surrounding environment by a perfectly insulating suit, hourly inside your suit your would need to melt ice of just over a pint / just under a liter, or evaporate (and vent) 1/3 of a pint or about 150 ml. In a cold climate, you would just need some ventilation. In a warm climate, something more might be appropriate. Air to Sustain Life Atmospheric CO2 levels today average 383 parts per million (PPM). Human exhaled breath is around 378 PPM. As a human breathes, starting from less than 1% in "fresh" air, the upper "safe" CO2 level is around 3%. When the concentration exceeds 3%, even though there is still oxygen in the air, humans are adversely affected. An average person produces around .67 cubic ft. (5 gallon volume) per hour of CO2 , so the 3% limit represents a starting volume of 22.5 cubic feet of air (about 1 cubic yard, around 168 gallon). If you for example needed to be sealed in for a year, you need to start with 197,100 cubic feet, or a cube 58 feet on a side. In say a 10 foot ceiling commercial building, it's an area 140 feet on a side. Water absorbs it's own volume of CO2 , so for every (.67 cubic ft. or 5 gallon) of water that your air is filtered thru, you gain an hour on the CO2 limit. You are of course still using up the oxygen. Symptoms Starting Volume for 1 Hour Duration Concentration With CO2 Absorb Cubic Ft. / Gallon 21% None - normal O2 air level N/A 15% No immediate effects 11.6 - 86.8 14% Fatigue, impaired judgment 9.6 - 71.8 10% Dizziness, shortness of breath, 6.1 - 45.6 deeper and more rapid breathing 7% Stupor sets in 4.8 - 35.9 5% Minimum amount to support life 4.2 - 31.4 2%-3% Death within 1 minute 3.7 - 27.7 Experiments show that approximately 8 gallons of well aerated algae in sunlight balances the breathing of a typical human. (Remember, you need enough "extra" air volume to carry you past periods of dark/dim light.) If you're not bubbling the air thru the algae, set up a "surface area" of water for the 8 gallons at about 8 meters square. (A square about 9 feet on a side) Since the water alone weighs 64 lbs., if this is to be a portable unit, you'll want some type of cart. Water for Our Bodies In addition to the earlier discussed minimum gallon/day water need to provide for evaporative cooling, a human needs water for other metabolic processes. Water is lost from the body mainly via the lungs, skin, intestine, and kidneys. The Pacific Institute for Studies in Development, Environment, and Security puts the minimum daily intake at 3 liters. They recommend 20 liters for hygiene, 15 per bathing, 10 for food preparation, or an overall average of 50 liters. (Around 13.195 gallon) If you had to store it all for a year, it's 4,800 gallons, 644 cubic feet, or a tank 8.635 feet on a side. Water purification. 1/3 cup chlorine bleach per 1,000 gallons. OR 5 drops of tincture of iodine per quart. The more “cloudy” the water, the more disinfectant, as the chemical may bind to the surface of “dirt” particles, or the germs may be hiding inside the dirt. (Filter first.) If there is no water, do not eat. It takes “excess” water to metabolize food. Remember though, between water and food is the need to maintain the electrolyte balance in our bodies. If you just drink water you may experience growing symptoms of chemical imbalance. Short of food, an approach to reintroduce electrolytes is the following home-brew of things such as Pedialyte. 1 - Liter/quart of H2O ½ - Teaspoon sale ½ - Teaspoon baking soda 3 – Tablespoon sugar Resources - Portal - - Images - Village cinema - Random facts - - Department of FUN! - Image:Img13713.jpg Village pump SCA Wiki - Places, projects & networks - Ideas Bank - News - Diary - Resources - Community / Avoid adverts
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