"These formulas comprise the pressure due to resistance, and not that necessary for final velocity: they are, therefore, more correct for long than for short airways. The pressure required for final velocity becomes a smaller fraction of the whole drag as the airways extend. If it be required to take into account the pressure to create velocity, 124 MINE VENTILATION. $45. APPENDIX B. PROBLEMS. 45. 1. What is the area of an airway 6 feet by 5 feet? What is its perimeter? a = 30; o = 22. Ans. 2. What is the area of a shaft 14 feet diameter? 153.98feet. Ans. 3. What is the perimeter of a shaft 16.5 feet diameter? Rule. Diameter X 3.1416. 58.83 feet. Ans. 4. The long axis of an elliptical shaft is 14 feet, its short axis 6 feet: what is its area? Rule. AX a X 0.7854. 65.9736 feet. Ans. 5. Find the perimeter of an elliptical shaft whose axis A is 16 feet, and its axis a 8 feet. 6. An air-course is 500 yards long, 6 feet high, and 7 feet wide: what is its area, perimeter, and rubbingsurface? a 42; 26; 8 = 39000 square feet. Ans. 7. What is the rubbing-surface of a shaft 15 feet diameter, 1,200 feet deep? 56548.8 feet. Ans. 8. In an airway 8 feet by 9 feet, when the current has a velocity of 15 feet per second, what quantity of air is passing per minute? Rule. ax v × 60". 64800 cubic feet. Ans. 9. When the water-gauge is 1.85 of an inch, what pressure per square foot does it indicate? Rule.wx 5.2. 9.62 pounds. Ans. 10. When the quantity of air passing is 60,000 cubic feet, with a water-gauge of 1.5 inches, what are the units of work producing ventilation? 11. What horse-power is there in 468,000 units of work? 12. The pressure producing ventilation is 7.8 pounds: what is the water-gauge? 13. There are 50,000 cubic feet of air passing, having a rubbing-surface 24,000 feet, and an area of 20 square feet what is the water-gauge? ksv2 Rule. W = α 3.12 inches water-gauge. Ans. ་ 126 MINE VENTILATION. $45. 14. The rubbing-surface of an airway is 25,000 feet, its area 25 square feet: what is its length? 1250 feet. Ans. 15. What units of work are necessary to overcome friction of an airway 6 feet by 6 feet, 1,000 feet long, when the quantity passing is 7,200 feet per minute? ksv2 u = x q. a 4166 units. Ans. 16. Let a 36; s= 24,000, to find the value of 17. With 0.9 of an inch water-gauge, 16,000 cubic feet of air are passing: what quantity will pass when there is a water-gauge of 1.6 inches? V0.9: V1.6::16000: 21333+. Ans. 18. With a fan and furnace combined, 46,706 cubic feet are produced; the furnace produces alone 42,670 cubic feet: what will the fan produce by itself? √467062 — 426702 18993. Ans. 19. If, with a water-gauge of 0.65 of an inch, 20,000 cubic feet of air are obtained, what height will the water-gauge be when there is a quantity of 75,000 cubic feet of air passing? 20. How much must we increase the pressure to double the quantity of air? 4 times. Ans. 21. How much has the ventilating power to be increased to treble the quantity of air? 33 27 times. Ans. 22. If we obtain 25,000 cubic feet of air by 5-horse power, what horse-power will be required to circulate 60,000 cubic feet in the same mine? (9′)3: (q)3::5HP : Ans. or :q:: 5: Ans. 23. There are two air-courses through which a total quantity of 100,000 cubic feet of air is passing; the resistances of the airways are in the proportion of 4 to 1: what quantity will pass along each? √1 x 100000 √4 + √1 √4 × 100000 √t + VI = = 100000 = 3 333331 cubic feet for airway having greater resistance. 666663 cubic feet for airway having only one-fourth the resistance. 24. Find the motive-column where the upcast and downcast shafts are 540 feet deep, the temperature of the upcast being 129°, and that of the downcast 43°. 129-43 Rule.-MDX 92.51. Ans. |