If a person when breathing were perfectly efficient at extracting oxygen from the air, how much air would supply the person's daily requirement of oxygen?
From our One-page Handbook (January 1983) I take for the heat of combustion, food, or fuel, 10 kcal/g. (Remember that a "food calorie" is 1 kcal.) Then to maintain a 3000-kcal diet the daily requirement is 300 grams of fuel. If that were pure carbon, burning it to CO2 would require 25 moles of О2, or approximately 3000 liters of air. That says we need the oxygen in about one liter of air for every kcal of energy obtained by oxidizing food. But of course we don't eat plain carbon. How does it work out for some reasonable assortment of fats, carbohydrates, and protein? If you look up the enthalpy of combustion to CO2 and H2O of these materials you will find that the yield in energy is remarkably uniform if it is expressed, not in kcal per gram of fuel, but in kcal per mole of oxygen used. I can remember it most easily as 5 eV per molecule of oxygen. This is correct within 25 percent for the complete combustion of carbon, hydrocarbons, fats, proteins, and carbohydrates. It is a useful "universal" figure and as such ought to replace the entry for heat of combustion of fuel in our One-page Handbook. I am indebted to Ralph Baierlein, who suggested the problem, for pointing this out. We conclude that the minimum requirement for almost any diet is one cubic meter of air for every thousand food calories.
«Мы считаем, что минимальные требования для практически любой диеты – один кубический метр воздуха на каждую тысячу калорий пищи.»
Removing a 70-kg body from Earth costs at least 70 Rg, or 4 × 109 J. It is now orbiting the Sun at a speed of 30 km/s, with kinetic energy (70/2) (3×104)2, or 3.1×1010 J. This equals the minimum energy required to remove the body from the Solar System. The total investment required, starting with the body on Earth, is 3.5 × 1010 J. That is almost enough to provide a person with 2500 food calories (kilocalories) per day for 10 years.
«Суммарно необходимо 3,5 × 1010 Дж. Это почти достаточно, чтобы обеспечить человека пищей (из расчета 2500 калорий (ккал) в день) в течение 10 лет.»
Many readers will have heard this problem before in one version or another, sometimes with one of the breaths that of Julius Caesar. There are 800 g of nitrogen atoms over each cm2 of the Earth's surface, or 4× 1021 g in the entire atmosphere, with negligible amounts anywhere else. Let n0 be the number of nitrogen atoms in the atmosphere; it is 1.7 × 1044. Let n1 and n2 be the number of atoms in the two breaths. The probability that a certain atom in breath 1 is not in breath 2 is 1 — n2/n0. The probability that no atom in breath 1 is in breath 2 is then (1 — n2/n0)n1 ~e— n1n2/n0, which has the value 0.5 if nln2/n0 = 0.69. One liter of air contains 5×1022 nitrogen atoms. If v1 and v2 are the two breath volumes in liters, "likely" is the correct answer if the product v1v2>(1.7×l044×0.69)/(5×l022)2 = 0.05 liter2. Make your own estimate of the breath volumes. Molecule cannot be substituted for atom in this problem without examination of atmospheric chemical processes which could result in atoms changing partners.
How long would it take to transmit over a video channel the information contained in the human genome, approximately 1 meter of DNA?
Each base pair along the double helix counts as 2 bits of information, equivalent to one letter from a four-letter alphabet, and there is a base pair every 3.4 A. Thus 1 m of DNA contains about 3 × 109 base pairs, or 6 ×109 bits (not quite a gigabyte!) A video channel, nominally 4 MHz wide, could transmit that much information in about a half hour.
Given that the heart of a resting person pumps about 75 gallons of blood per hour at a systolic pressure of 120 Torr, estimate the minimum pump power required, in watts.
The pressure 120 Torr is approximately 1.6 ×105 dyne/cm2 and a flow rate of 75 gallons per hour is equivalent to 80 cmVs. The product of pressure and flow rate is then 1.3 watts.
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