site to that number in Col. A is the required velocity. Thus 62.5 × 1 for the case in (65) we have = 31.2 x (1760 × 36) ⚫0000316, the nearest number to which is 00003043 opposite ·425 foot By interpolation we may obtain a nearer approxi per second. 4252 × 0000316 ⚫00003043 •180625 × 316 6)$ or ( ⚫3043 = 4331 foot per = second, hence the discharge comes out ·4331 × 60 × 62·5 1624 cubic feet per minute, or practically the same as by the rule (65). TABLE 18.-For the DISCHARGE of CANALS, RIVERS, &c., by 2nd. To find the Fall.-Divide the given discharge by the given area, and by 60, which will give the mean velocity in feet per second; find the nearest number to that in Col. A, which, multiplied by the border in feet and by the length of the channel in inches, and divided by the area in square feet will give 3000 Thus, for the case in (66) we have the fall in inches. 48 = 62.5 48 feet per minute, or = 8 foot per second, the tabular num 60 ber for which is ·00009072; then •00009072 × 31.2 × (2000 × 36) – 3.26 inches fall, as before. 62.5 = 68. "Case of a Mill-stream."—As an example of the practical application of the rules, we will take a case in which it is desired to utilize a stream of water for driving a corn-mill. Say we have a stream 1500 yards long, with a total fall of 6 ft. 6 in. from the tail of the preceding mill. We have first to ascertain the quantity of water at disposal: selecting a spot where the current appears to be tolerably uniform for some 100 feet, and a season when the quantity is an average one according to local authorities, say we take it at a point 24 feet wide as in Fig. 41. We have then to obtain the area of the stream, and to do that, may divide the width into eight equal spaces of 3 feet each, as in the Figure, which may be done conveniently by stretching a tape across the stream: then we measure the depths midway between those divisions or at 1.5 foot, 4.5, 7.5 feet, &c., &c., using a measuring rod with a flat board about 7 or 8 inches square at the end of it, to prevent penetrating the soft bottom; and thus we obtain the series of measurements given in the figure, the mean of which we find to be 1.156 foot, the area is therefore 1.156 × 24 27.74 square feet. To find the velocity, two lines may be stretched across the stream near the surface, and say a "chain" or 66 feet apart, and a float being placed a few yards above the highest one, and in the centre of the width, or rather where the velocity is observed to be greatest, the exact time in passing from line to line is carefully noted. This float should be a small piece of thin wood, say only 1-inch thick, so = as to be almost wholly immersed, and thus expose little surface to the action of the wind. Say that the float travels the 66 feet in 20 seconds, in one minute therefore it would be 66 × 60 = 198 feet. This being the maximum velocity, the mean (50) over the whole area would be 198 × the discharge is 166 × 27.74 = 84 = 166 feet per minute, hence 4600 cubic feet per minute. (69.) The total fall is 6 feet 6 inches; allowing 6 inches for the fall of the stream itself, the net fall at the wheel will be 6 feet; a cubic foot of water weighing 62.3 lbs.; the horsepower being 33,000 foot-pounds; and allowing that a breastwheel yields 50 per cent., or 5 of the gross power of the water, 4600 × 62.3 × 6 × 5 we have =26 horse-power. A pair of 33000 4-feet stones, grinding 4 bushels of corn per hour, requires = 12.8, say 13 feet. With other kinds of water-wheel the duty would be different: a good overshot wheel would give from 70 to 80 per cent., a breast-wheel from 45 to 60, and an undershot, in which the water acts only by its impulse, from 27 to 30 per cent. (70.) The channel must now be altered, so as to deliver 4600 cubic feet per minute, with a fall of 6 inches in 1500 yards, or 1760 × 6 When altered to the form 1500 A, B, C, D, the area will be = 7 inches per mile. 24+ 12 x 3 = 2 54 square feet, the 4600 mean velocity to discharge 4600 cubic feet will be = 85.2 54 ! AROMA A. feet per minute; the border is 6.7+12+6·7 = 25·4 feet, and the hydraulic radius 54 25.4 = 2.126 feet. Then by Table 30 between 2 and 2.2 radii, the velocity 85·2 feet is found to be under the fall of 7 inches per mile, the fall we allowed. It should be observed that it is imperative that the slope shall be uniform from end to end, at least where the area of the channel is uniform. CHAPTER IV. ON WEIRS, OVERFLOW-PIPES, &c. (71.) " Weirs."—Fig. 36 shows a weir arranged for the purpose of gauging experimentally the quantity of water passing down the stream. A is a plate of thin iron with a notch cut out of it wide enough by estimation to carry off the water with a moderate depth of overfall; this is screwed to a thick plank B, to obtain the requisite stiffness for the plate, and the whole is fixed in the stream as shown. C is a stake with a flat and level top, which is driven into the bed of the stream to such a depth that its top is exactly level with the lip of the weir, and the depth of water flowing over is measured by a common rule held on its summit. The proper distance of the stake from the weir depends on the quantity of water to be dealt with; in small weirs it may be from 1 to 2 feet, in very large ones 20 to 25 feet; the object is to place it far enough away to avoid the curvature of surface which the water suffers as it approaches the weir, as shown by the Figure. There is some uncertainty in measuring by a rule in the manner indicated, arising from the capillary attraction causing the water to adhere to the rule and to rise above its true height. A more correct method is to use Francis's hook-gauge, a rough modification of which is shown by Fig. 36. The stake J is, in this case, driven to such a depth that its top is at a height convenient to the eye, say 30 inches above the level of the lip of the weir; then a rough hook-gauge D, formed of wood about 1 inch thick, is cut in the form shown, the end E being flat and level, and the length E F made exactly equal to GH or 30 inches. The hook-gauge is held against the stake, and carefully adjusted, by the hook at E being first immersed, and then raised until it just coincides with the surface of the water; the depth of overflow is then given by the distance from the top of the stake to the top of the gauge at F, measured by a rule, &c. (72.) With a thin plate, and depths thus measured from still water, we have the following rules: Thus, with 2 inches overflow, a weir 72 inches long discharges 2 × 1.4142 × 72 × 2.67 = 543.7 gallons per minute; again, to discharge 694 gallons per minute, with 3 inches overflow, we 694 should require a length of = 50 inches; and 3 x 1.732 x 2.67 again, to find the depth of overflow to carry 1282 gallons, with a length of 60 inches, we have 1282 = 8, then 3/8 = 2, and 22 = 4 inches, the depth required. Table 19 has been calculated by these rules, and its use may be illustrated by the examples just given; thus with 2 inches overflow the Table gives 7.552 gallons per inch, and a weir 72 inches wide will discharge 7.552 × 72 = 543.7 gallons; a weir with 3 inches overflow discharges 13.87 gallons per inch of width, and for 694 gallons 694 we require a length of 13.87 = 50 inches; a weir 60 inches |