33

C—3a

one having sides 8-093 ft. ? Now, the area of an airway 8-093 square feet is 65-496, and this divided by 7, the height of the proposed airway, gives 9-3566, which is only the approximate width of the proposed airway, as, although the areas are equal, the mean depths are not equal, and they would not have the same carrying-capacity. This, however, can be corrected by the rule (Formula No. 5) — the carrying-capacity of airways is proportional to the areas of the respective airivays multiplied by the square roots of their respective mean depths (length and pressure being the same). Thus — Proposed Rectangular Airway. Airway 8-093 ft. square. Area, 65-496 Log. 1-8162148 Area, 65-496 Log. 1-8162148 Perimeter, 32-713 „ 1-5147204 Perimeter, 32-372 „ 1-5101695 Mean depth „ 0-3014944 Mean depth „ 0-3060453 yMean depth „ 0-1507472 VMean depth „ 0-1530226 Area, 65-496 „ 1-8162148 Area, 65-496 „ 1-8162148 92-675 = 1-9669620 93-162 = 1-9692374 Now, as 92-675 : 93-162 :: 65-496 : the required area— 93-162 Log. 1-9692374 I Bequired area, 65-840 65-496 „ 1-8162148 7 x 9-3566 = 65-496 3-7854522 Second area too small by 00-344 92-675 „ 19669620 = 0-0491 extra width. „„ „,.. , 010 , nm • j 9-3566 approximate width. 65-840 = 1-8184902 required area. 0 . 0491 width 9-4057 required width. 7 65-8399 required area. The working-out of this question in the "Miners' Guide" is correct, but the formula is too complicated. The formula here given is more simple, less liable to error, and the result shows it to be practically correct; it also shows the use of the mean depth, and the rules for determining the relative air-carrying capacity of various forms of airways. Notes on Mine-ventilation (comparing Water Formula with Air Formula). Water Formula. ■\/RiriO,s6o = v. theoretical or 102-76 VRS = v. or 4 ,\/2gRS =v. practical. Air Formula No. 1. 113 = v. practical. What would be the velocity of water in a waterway 1,000 ft. long, 4 ft. square, with a head of 100 ft. of water ? Area, 4 ft. x 4 ft. = 16 ft. = A. Formula y/BS 10,560 = theo. vel. A 16 ■ B = =5 Ift. = E. = 102-76 yRS. s = ro = °' 1 =s - = 102 7 6 yFI Coefficient = 032 =C. = 10276 x 0-316 = 32-47216 = theo. vel. 32-47216 x 0-32 = 10-3910912 = vel. Air-pressure, 1-5524 in. on water-gauge = 8-07248 lb. per square foot. Formula 113 \/BS = v. Area, 4 ft. x 4 ft. = 16 ft. =A. RxS =Ixo-008072 = 0008072 R = =*| = 1 ft. =B. = v/0 T 0()8072 = 009 S = = 0 0 08072 =s- °' 09 x 113 = 10 1 7 = vel Water Formula. yBS 10,560 = v. theoretical or 102-76 VBS = v. or 4 =v. practical.

5—C. 3a.