When we represent the length of an airway by 1, the rubbing-surface by 8, the perimeter by o, we have THE LAWS AFFECTING AIR IN MINES. 22. THE laws affecting the circulation of air through mines or confined passages, such as gangways, etc., have been ascertained principally by such eminent men as Magnus, Regnault, Gay-Lussac, Daubisson, Peclet, and others, and are as follows: 1. The volume assumed by a given weight of air is inversely proportional to the pressure on each unit of surface under which it exists, so long as the temperature remains unaltered. Consequently, if we take a cubic foot of air under a pressure of five pounds, it will only be one-half a cubic foot under a pressure of ten 54 MINE VENTILATION. § 22. pounds, and one-third of a cubic foot under a pressure of fifteen pounds. 2. When the pressure is constant, the volume is uniformly increased in the ratio of part for each additional degree of heat, Fahrenheit's scale. 3. When air is discharged through orifices offering no sensible frictional resistance, the result is sixty-five per cent of the quantity due to the velocity multiplied by the area in case of a thin plate; ninety-three per. cent in case of a short cylindrical tube; and ninety-five per cent when the tube is conical, and the area taken from the small end. This contraction of the flowing air, which is similar to that which takes place when water is discharged through pipes under the same conditions, has the effect of reducing the quantity discharged in a given time below that which would be due to the velocity, if it existed, over an area equal to that of the orifice or tube. This contraction of the flowing air is termed the "vena contracta." 4. When air is impelled through a confined passage, the pressure or head of air-column required for its propulsion is proportional to the square of the velocity; so that to double this velocity there must be four times the head; to treble it, nine times the head; etc. (a) Ventilating pressure, p, or ("head of air," or "motive-column," reduced to pounds) total pressure, pa or ha, and ventilating power, P, are separate and distinct terms. Ventilating pressure, or simply pressure, is the force applied to each square foot of area of section to produce ventilation. That this pressure varies as the square of the velocity, as stated above, may be illustrated by the following: Problem. When a mine is passing 20,000 cubic feet of air per minute with a pressure of 2.6 pounds per square foot, as indicated by 0.5-inch water-gauge, what will be the pressure if the mine pass 40,000 cubic feet per minute? This principle may be resolved simply into finding a fourth proportional; thus, v2: (2V)2::p:? (20000): (40000)2 :: p:? (2)2: (4)2:: 2.6: 10.4 pounds. Ans. (b) The total pressure is the ventilating pressure, p, multiplied by the area of section a of the airway in square feet. Thus, if the sectional area of the above airway were 64 square feet, and the pressure 0.5-inch water-gauge, the total pressure would be pa = 64 × 2.6166.4 pounds. (c) Ventilating power is power used to obtain ventilating pressure. This power, P, varies as the cube of the velocity of the air-current. By this we mean, that, if we can circulate a quantity of air with 2 HP, we 56 MINE VENTILATION. § 22. must use 8 HP if we wish to circulate a double quantity. This, as can be readily seen, is a very important factor in fiery mines; as the engine working the fan or the furnace may be called upon at any time to do double and maybe threefold duty, in case blowers or barometrical differences allow an unusual amount of gas to be given off. Problem. Suppose we have 20,000 cubic feet of air passing with a water-gauge of 0.5 inches, equal to a pressure of 2.6 pounds per square foot, what will be the ventilating power, P, if we double the ventilation? Solution.-20,000 x 2.6 units of work = 52,000 × = foot-pounds. As there are 33,000 foot-pounds in 1-horse power, we have Again: from 22 (a) we find that to double the quantity we must employ four times the pressure: hence P 40000 × 2.6 × 4 = 416000 foot-pounds =2 or eight times the power in the first case. 5. In airways of the same sectional area, the pressure required to propel air is proportional to the length of the passage, or, in other words, there must be double pressure for double distance. Problem.- Suppose we have an airway 2,000 feet long, and another 4,000 feet long, of the same area; the pressure being 2.6 pounds per square foot in the shorter. What will be the pressure necessary to overcome the resistance in the longer airway? 2000 4000 :: 2.6: ? 1:22.65.2 pounds. Ans. 6. The pressure required to propel air through confined passages is proportional to the perimeter of the passages; the length and other data remaining constant. Thus, if we have an airway 4 feet square, and another 8 feet square, with a pressure of 2.6 pounds per square foot for the 4-foot airway, we shall have 5.2 pounds for the 8-foot airway, or 1:2:2.6: 5.2. Ans. 7. The pressure required on each unit of surface, square inch or square foot, to propel air through a confined passage, is inversely proportional to the sectional area of the passage, when all other things are equal; so that, the greater the area exposed to the pressure, the less is the amount of pressure required for each |