18 MINE VENTILATION. $7. Let the lines below represent a gallery from which some of the top has fallen. Fis the floor, RR the roof, and H the hole caused by the fall. If a lamp be raised about halfway between R and F, an explosive mixture may be encountered; but, if the lamp is held near the cracks in RR, not a trace of gas will be indicated by it, provided there is sufficient ventilation. Gradually stop the ventilating current, so that the air scarcely travels, and then apply the lamp, and the fire-damp issuing from the cracks may be detected. Active ventilation sweeps away all the gas which escapes from cracks in the line of the top rock or roof; but as the current can find no outlet through the hole at H, and encounters a pressure equal or superior to that along the roof, it travels onward, heedless of any gas situated out of the line RR. When fire-damp accumulates near the roof or top line of a gallery or heading, it indicates that the gas is given off in greater quantity than can be carried off by the ventilation; and, where any such accumulation takes place, there must be either a deficiency of ventilation, or an unusual inpour of gas. EXPANSION OF GASES. 8. One of the chief characteristics of any gas is its expansive property. To calculate the expansion of any volume of air, the starting-point must always be taken at 0° on Fahrenheit's scale; for air at that temperature will expand of its volume for every degree of heat added. Therefore 459 cubic feet of air at 0° will become 460 cubic feet at 1° F. Careful experiments show that 459 cubic feet of air at 0° F. weigh 39.76 pounds, when the pressure is 30 inches of mercury of the density due to 32°, a pressure equal to 14.7 pounds per square inch; but, when the pressure is one inch, it weighs only part of this, or 1.3253 pounds. To find the weight of a cubic foot of air at any temperature or height of the barometer, let then B = (1) height of the barometer in inches, t temperature by Fahrenheit's thermometer; = Problem. What is the weight of a cubic foot of air, the temperature of which is 96°, under a barometric pressure of 29.5 inches of mercury? By substituting 29.5 in formula 1 for B, and 96° for t, and then performing the operations indicated, we have W = 1.3253 x 29.5 = 0.07044 pound as the weight of a cubic foot of air under the above. conditions. The following table has been made out to facilitate calculations. It gives the weight of 100 cubic feet of air in pounds at different barometrical pressures. 22 MINE VENTILATION. $9. 9. As we have seen, air has weight, and therefore becomes subject to the "physical laws" that govern liquids and falling bodies; i.e., air is acted on by gravity in the same manner as a solid. Leth = the distance fallen through in feet. 9 the velocity acquired at the end of the fall, in feet per second. = the distance in feet which an unresisted gravitating body falls in the first second of time; which distance has been found by experiment to be 16 feet near the earth's surface. Since a body falls 16.08′ in one second, it gains a velocity of 32.16' at the end of the first second: hence we have √4gh = 8.0208 √π. In this equation h represents the necessary height, in feet, of a vertical air-column which will produce by its weight a velocity equal to v. If this velocity be represented in feet per minute, we shall have |