For the smaller airway, s = 20000, and v2 = = 0.0808082 0.26881 × 20000 5913.82 .. v=V0.0808082 x 1000 = 284.27 feet per minute; and, as the quantity passing is found by (9), § 21, we have 284.27' 30 8528 cubic feet per minute. = Problem (b).-Suppose, now, we take two airways whose lengths are each 35 units, and which have the same area; the quantity of air passing through each being 9,000 cubic feet. What will each circulate of the total amount, if their lengths be in the ratio of 3 to 4? Solution. From (1) and (4), § 23, we deduce two proportions, one for the longer, and one for the shorter airway. Now, if these airways were in the ratio of 3 to 4 in length, then one would be 30 units and the other 40 units in length, and the quantities of air which would flow through them under the same conditions may be computed thus: 64 MINE VENTILATION. For the shorter airway √30:√35:: 9000: 2, or √6: √7:: 9000: x and hence 6:√42:: 9000: 9721 cubic feet. Ans. Total, 8418 +9721 = 18139 cubic feet. Proof, √3: √4:: 8418: 9721 $ 24. √4:√3 :: 9721 : 8418. (c) Suppose we have two airways of the same sectional areas and lengths, each passing 9,000 cubic feet as before. Suppose each to have 4 units of lengths. If one of these airways be shortened to unity, it will have but the rubbing-surface of its former length; and the volume of air will be, according to (1), found thus:— VI: √4:: 9000: 18000. Again: let the length of one of the airways be increased fourfold, then the volume of air will have four times the rubbing-surface, and by (4) we have √16: √4:: 9000: 4500. (The above illustrates in a striking manner the effect that rubbing-surface has of diminishing the flow of air through a gallery.) The total volume will be 22,500 cubic feet; and, notwithstanding the rubbing-surface is more than doubled, the volume is only increased by twenty-five per cent. The pressure required to circulate the air under each set of conditions is precisely the same; for the smaller rubbing-surface multiplied by the square of the highest velocity is equal to the greater rubbing-surface multiplied by the square of the lower velocity; thus X 4202500, and X 16 202500, and (9400) 3 4500\2 = (18000) × 1 = 202500, if we assume the areas of the 40 above to be 40 square feet. That the pressure remains the same may be shown by Atkinson's method: Let the above unit of length be assumed as 100 feet, then equal airways are 400 feet in length; and, if we assume the area and perimeter to be respectively 25 and 20 feet, we may find the pressure, thus:— 0.26881 x 100 x 20 x 0.5184 0.26881 x 1600 × 20 × 0.0324 = 11.148 ft. head. 25 11.148 ft. head. 66 MINE VENTILATION. $24. (d) If, through two airways 6 feet square, 9,000 cubic feet of air flow per minute, and one of them be altered in section to a circle (its area being unaltered), then the quantity of air circulating may be computed, thus: No21.27:24 :: 9000: 9560. Ans. The volume flowing through both airways would now be 18,560 cubic feet; but, in the event of this quantity being reduced to 18,000 cubic feet, the circular airway would have more flowing through it than the other. (e) What volume of air would flow through an airway 5 feet square, if 6,000 cubic feet flow through an airway 10 feet square, the pressure and length being the same? Solution. The volume varies as the square root of the rubbing-surface (§ 23, 4), and directly as the area ($23, 2): hence we have a compound proportion, It may be reasoned thus: as the area was reduced onefourth, the resulting volume would be one-fourth also, a to namely, 1,500 cubic feet; but, instead of the perimeter being reduced to one-fourth, that is, 10 feet, it is only reduced to 20 feet. The volume may now be found, thus: or √20:10:: 1500: x √2: VI:: 1500: 1061 cubic feet. Ans. (h) Suppose we have a pressure equivalent to 40 HP, giving a circulation of 120,000 cubic feet per minute: what quantity will a pressure equivalent to 32 HP give? From $23, 6, we have V40: V32: 120000: 111398 cu. ft. per minute. Ans. Proof. (111398)3: (120000)3:: 32:40. Ans. (i) Suppose we have a pressure equivalent to 32 pounds per square foot, circulating 107,350 cubic feet per minute, what pressure will circulate 120,000 cubic feet? Solution. (107350)2: (120000)2:: 32:40. Ans. 25. From the proof of the last problem we see that air may be measured by the pressure, or, what amounts to the same, the water-gauge, and we can say the quan |