(z) ksv2 u = x Q = px Q. a With an additional airway of the same length, a and 8 will be doubled; and, as the velocity decreases as the cube root of the power, we have the formula to obtain the quantity of air in cubic feet circulating per minute. (c) When we have splits of different lengths, and wish to know the size of regulators so as to allow the same quantity of air to each division, we may find them when we know the size of the first regulator. Problem.-Suppose we have five different splits in a mine, 200, 300, 400, 700, and 800 yards long respectively, and the regulator placed at the entrance of the 200 yards' airway be 9 square feet. What will be the area of the other regulators so as to allow the same quantity of air to each split? This gives us the rate at which the friction increases when compared with the first regulator; and hence, if we multiply each of the above quotients by 9 respectively, it will give us the area of the required regulators, thus: (d) Again: if we wish to divide a given quantity of air so that it will be distributed, as it is needed, into. unequal quantities in different divisions of the mine, we may do so in the following simple manner. Problem. Suppose we have 50,000 cubic feet of air to be distributed in five divisions, each to obtain respectively 15,000, 12,000, 10,000, 7,000, and 6,000 cubic feet per minute, to travel at a velocity of 5 feet per second; which velocity is sufficient to render any discharge of fire-damp harmless, unless it happens to be a very exceptional case. Taking the first case as an example, 15,000 cubic feet to be distributed at a velocity of 5 feet per second, 5′ x 60 = 300' per minute, 15000 =50 feet area, and √50 7.07 feet, size of the regulator necessary for the first split. Proceeding in the same manner, the regulators to admit the above quantities of air may be found. 32. Airways are seldom of the same area in the same mine, but are subject to the same pressure in ventilation; and, as each airway takes up its part of the pressure to overcome the resistances which the air encounters while passing through it, we may find the pressure necessary to overcome the resistances of the whole mine by adding the several pressures, or the pressure for each airway, together. If a mine be ventilated by a number of airways of different areas and lengths, they must all be considered as subject to one common pressure, and the quantity of air passing in each may be found by the formula, k and p are common factors for each airway, and hence need not be considered. Problem. Suppose we have three airways, A, B, and C. A has an area of 30 square feet, and a rubbingsurface of 66,000 feet. B has an area of 36 square feet, 86 MINE VENTILATION. § 33. and a rubbing-surface of 96,000 feet. C'has an area of 25 square feet, and a rubbing-surface of 40,000 feet. What quantity of air will pass along each, if the total quantity passing be 50,000 cubic feet per minute? For the sake of brevity, the rubbing-surfaces may be reduced to the lowest whole numbers, and still remain in the same proportion to each other by dividing by 2,000; then The proportional part passing in each airway may now be found from the simple proportions: A = 87.728: 28.602 :: 50000 : 16301.5 cubic feet. B C 87.728: 27.950 :: 50000 : 15930.0 cubic feet. 33. Find the quantity of air which will pass through a mine of the dimensions as given in the following table, with a total pressure of six pounds per square foot. In this case, the upcast and downcast shafts are given. A, B, C, and D are splits, subject to the same pressure. We may assume any quantity of air we please — say, 50,000 cubic feet to pass through each division, and then find the actual quantity which will pass under six pounds' pressure. Before we can do this, however, we must calculate the pressure for the circulation of 50,000 cubic feet; then, by adding the several pressures together, place them in proportion, as shown in column VIII. of the table. obtain the actual quantities. Next, by the use of the formula q = Columns VI. and VII. are |