Ex. 6. A stay is fastened by a double eye to a T angle iron, the bolts are 1" diameter, the plates thick: what space must be left between the hole and the edge of the plate? The area of the hole should be equal to the area of section of plate between the hole and the edge. Here length X} = I XIX 7854, .. Length = I XIX '7854 ÷ 5 = 15708 inches. Answer. Ex. 7. If 8000 lbs. is the strain allowed on iron plates, what would be the pressure on a boiler 12 feet in diameter, double rivetted, " rivets, and 24" pitch, thickness of plate ", crossings not to be taken into account? Ex. 8. The stays of a boiler are 12′′ apart and one of them breaks, thus throwing more strain on the four stays that surround the broken one; they are 18" diameter, and the pressure is 70 lbs. per square inch: what extra strain per square inch will be on the stays if we allow more than each had at first? Ex. 9. 1 = 1.625 12 X 12 X 70 + $ =6480.5 lbs., strain per square inch. How far apart should the stays be pitched when the pressure is 48 lbs. per square inch and the plates Formula. 8= thick ? (T+1 S surface supported by I stay. T= thickness of plate in 16ths. Caution.-Remember that T must be sixteenths, then if the thickness of plate were, then Ex. 11. .'. /9869°273 inches, the pitch. Answer. What is the least diameter of one of these stays? Total pressure Area of stay The stays of a boiler are to be fixed by collars, the depth of the collars being 3 times the thickness, the diameter of stays 13": what size must the end be swelled out to to have uniform strength of bar? 08 Rule. D=1+(+) d. Where n = 4, d = 1}". D=1+(+4) 1·875 = 1 + ('03 + 2 =) 0′22 = 1′22. Then 122 X 1·875=2°28750. Answer. 4)*08 02 ..√42 i8='2, and 1.875 1.22 3750 22500 2*28750 quired to know ix following ex N. pressure. 24 " 25 39 39 99 en the diam ow a free esc when close ised to the nfinite num The same hei ylinder is ec of a cylinder e XCI, page But since .. hei of valve rcumference ..heig the area of d divide the divide the di alve must be what height Height= -We cancel below: also 6.5 divided by ft. 3 in. high: square yard? inches. and this must lap ad weighs 8 lbs. ? at is, 14' 6" + 1' 3" <41441*5625 with lead, and this lead square foot of lead weighs broader and longer than the 12 ft. 8 in. + 1 ft. 5 in. = SX 85 X 4144 = 2'0069 covered with lead, which has to be will be the cost of the lead? Length in.; weight of load per sq. ft. 8 lbs.; According to the Board of Trade instructions, circular furnaces with the longitudina joint welded or made with a butt strap, 90000 x the square of the thickness of plates in inches the working pressure per sq. in. Without the Board's special approval of the plates the pressure in no case to exceed 8000 x thickness in inches diameter in inches Ex. 12. Find the working pressure when the diameter is 37", length 6′ 9′′, and thickness of plates ". Ex. 13. Upon every square foot of fire grate is burnt 11 lbs. of coal per hour, each lb. of coal evaporates 71 lbs. of water, the superheater has 31 cubic feet of capacity for every square foot of fire grate, and the volume is 512 times that of water: what is the average time each portion of steam remains in the superheater? Evaporation, in cubic feet of steam = 1'2496 × 512 = 639'7952 volume, .*.639795: 3 :: 3600" : 17'44 seconds. Answer. Ex. 14. If 1 lb. of coal evaporates 75 lbs. of water when the gauge is at 55 lbs.: how many cubic feet of steam will this be for 1 lb. of coal ? 410 +P÷4 (where P= gross pressure) may be taken as what 1 lb. of water will expand to. P+1 Here 410+70 ÷ 4 410 X 17'5 71 6021 cubic feet of steam for 1 lb. of coal. .. 6.021 X 7'545175 cubic feet of steam. Answer. Ex. 15. The boiler contains 60 tons of fresh water on starting on a voyage. The water in the hot-well, through some leak, is 1 of the saltness of the sea: how many tons of water must have been evaporated during the voyage, the water in the boiler at the end of the voyage being 2-2 times the saltness of the sea ? = 22 times as much salt as at starting, then 22 X 601320 tons evaporated. Answer. Ex. 16. The horse-power of an engine is 140, the boiler 15 feet long, diameter 16 feet, the water shows in the glass 7 inches. The engine uses 20 lbs. of steam per H.P. per hour: how long will the water remain in the glass after the feed is shut off? 7 inches of 1 foot. Time water remains in boiler = 15 X 16 X X 64 =3m 1998. Ex. 17. A boiler with 140 cubic feet at water level and 60 lbs. pressure; a f' rivet is out of the boiler bottom: how long will it take to blow 8" down? Rule. 2 do VP = cubic feet blown out per minute. Cubic feet blown out per minute = }2 × 2} √/60 = 875 X 875 X 2'5 X 7'746 =14826328125, .. Time to blow out contents of boiler = 93°333, &c., ÷ 14°826328125 = 6m 18" nearly. Ex. 18. A rectangular boiler is 16 ft. 2 in. long, II ft. I in. wide, and 9 ft. 3 in. high: what would be the cost of cementing the two sides and the top at 88. 4d. per square yard ? Area of top 16′ 2′′ × 11′ 1′′ = 194′′ X 133′′ = 25802 sq. inches. 2 sides 2 (16′ 2′′ × 9′ 3′) = 2 × 194 X 111 = 43068 Ex. 19. A boiler 14 ft. 6 in. by 13 ft. 9 in. is to be covered with lead, and this must lap over 7 inches: what will be the cost at 5d. per lb. if I sq. foot of this lead weighs 8 lbs. ? The sheet of lead must be twice 7" broader and longer than boiler, that is, 14′ 6′′ + 1′ 3′′ = 15′ 9′′ long, and 13′ 9′′ + 1′ 3′′ = 15′ broad. .*. 15'75 X 15 = 236°25 square feet. But from this we have to deduct 4 corners 7" square square feet. 75 X 75 × 4 ÷ 144 = 1°5625 Ex. 20. A boiler 13 ft. 9 in. by 12 ft. 8 in. has to be covered with lead, and this lead must overlap 8 inches: what will be the cost at 34d. per lb. if I square foot of lead weighs 8 lbs. ? It is evident that the sheet of lead must be twice 8 inches broader and longer than the boiler, that is, 13 ft. 9 in. + 1 ft. 5 in. 15 ft. 2 in. long; and 12 ft. 8 in. + 1 ft. 5 in. = 14 ft. 1 in. broad=213'5972. But from this deduct 4 corners 8 inches square, that is, 8.5 X 8.5 X 4144 = 2*0069 square feet. Ex. 21. The top of a rectangular boiler is to be covered with lead, which has to be turned down over the sides a certain distance: what will be the cost of the lead? Length of lead 13 ft. 9 in.; breadth 12 ft. 8 in.; over-lap 8 in.; weight of lead per sq. ft. 8 lbs.; cost 34d. per lb. |