3. What is the absolute (total, ambient) pressure in bars at 17 meters of fresh water? Answer: Pa = (17 mfw x 1 atm + 1 atm )x 1.013 bars = 2.68 bars 10.33 m 1 atm U.S./Imperial 1. What is the water pressure in pounds per square inch at 112 feet of seawater? Answer: Pw = (112 fsw x 1 atm ) x 14.7 psi = 49.8 psi 33 fsw 1 atm Or, alternatively: Pw = 112 fsw x 0.445 psi/fsw = 49.8 psi 2. What is the absolute pressure in pounds per square inch at 112 feet of fresh water? Answer: Pa = (112 ffw x 0.432 psi/ffw) + 14.7 psi = 63.1 psi When finding absolute pressure, remember that the barometric pressure is one atmosphere (101.3 kPa/14.7 psi) only at sea level. As altitude increases, barometric pressure decreases. For instance, at an altitude of 1,000 meters/3,300 feet the pressure is only about 90 kPa/13 psi. A table of barometric pressures at altitude can be found in the Decompression chapter. Sample Problems: SI/metric: What is the absolute pressure at a depth of 14 meters in a high altitude lake if the air pressure is 950 millibars? Answer: Pa = (14 mfw x 1.013 bars/10.33 mfw) + 0.95 bar = 2.32 bars The barometric pressure is converted from millibars to bars by moving the decimal point three places to the left. U.S./Imperial What is the absolute pressure at a depth of 45 feet in a high altitude lake if the barometer reads 27.3 inches? Answer: Pa = (45 ffw x 0.432 psi/ffw) + (27.3 in x 14.7 psi/29.92 inches) = 32.8 psi The barometric pressure is converted from inches of mercury to psi by using the equivalency: 14.7 psi equals 29.92 inches of Hg. Important: At altitude, analog depth gauges are inaccurate. They read shallower than actual depth, and corrections must be applied to obtain an accurate depth. A correction table for gauges is also in the Decompression chapter. Most dive computers will adjust for altitude. Divers who regularly dive at altitude, especially using analog gauges, would do well to purchase or prepare a slate with a table of observed depths, true depths, and sea-level equivalent depths. Partial Pressure In any mixture of gases, such as air, the total pressure of the mixture is equal to the sum of the individual pressures exerted by each component gas. This was first observed in the early 19th century by the English chemist John Dalton. Dalton’s law states: “The total pressure exerted by a mixture of gases is equal to the sum of the pressures that would be exerted by each of the gases if it alone were present and occupied the volume.” In other words, the whole is equal to the sum of the parts. The pressure exerted by each component gas is termed the partial pressure of that gas. Expressed mathematically, Ptotal = P1 + P2 + P3 +...Pn where “n” is the total number of component gases. The air we breathe is approximately 78% nitrogen and 21% oxygen. Another way to express this is to say that the fraction of nitrogen in air is 0.78 (FN2 = 0.78), and the fraction of oxygen is 0.21 (FO2 = 0.21). The partial pressure of any component gas in a mixture can be found by multiplying the fraction for that gas by the total pressure of the gas mixture. At sea-level, the total air pressure is 1 ata, so the partial pressure of the nitrogen in the air is 0.78 ata, and the partial pressure of oxygen is 0.21 ata. Partial pressure is usually abbreviated “P” or “Pp.” So for atmospheric oxygen at sea level, PO2 = 0.21 x 1 ata = 0.21 ata. In practice, nitrox and technical divers will usually use FN2 and PN2 to mean entire inert gas component of air (including argon, which is almost 1% of the air by volume), in which case FN2 = 0.79 for air. NAUI Master Scuba Diver 84 Diving Physics
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