(XXXVII.) Again, suppose the descent of the surface to be 5 ft., calculate the time, h = 6-5 = 1, and VH-√h-1.449, so that = (XXXVIII.) §91. Mean hydraulic charge. Let us suppose in any prismatic vessel receiving no supply, that the head, at the instant of opening the orifice of discharge, was 6 ft. = H, and at closing it had decreased to 5 ft. = h, calculate the mean constant charge at which, in the same time, the orifice would discharge the same volume of water; the vessel being now, necessarily, supposed to receive that same constant quantity which it discharges with a uniform velocity. = If h be taken equal to 4, then H' = 4.96; if equal to H': 3, 4.376; if h = 2, then H' 3.732; and when ho, we have H' = 1.5. If in 10" we observe the surface to fall 2 ft., determine the coefficient of discharge. If A6 ft., S.01, and T= 10", then H being = 6, and h = 4, we have = 12 cb. ft., and Q = 1.2, and H' = 2.227. Hence (XXXIX.) § 92, p. 67. A reservoir, half an acre in area, with sides nearly vertical, so that it may be considered prismatic, receiving a stream which yields 9 cb. ft. per second, discharges through a sluice 4 ft. wide, which is raised 2 ft.; calculate the time required to lower the surface 5 ft., the charge upon the centre of the sluice, when opened, being 10 ft. From the formula given at the end of $92, we have, substituting the numerical values, A = 21780 sq. ft. the acre, being 43560 sq. S8 sq. ft, m being found 0.70, and h = 10-5 = 5, also q=9 cb. ft. per second. 2 × 21780 (.7 × 8 × 8.024)= { ·7 × 8 × 8.024 (√10 - √5) + 2.303 × 9 × log ft.; .7 × 8 × 8.024 √10 In this we have .7 × 8 × 8.024 = 44.9, and √TO - √5 = 3.1622.236.926. Hence t = 43560 { 44.9 × .926 + 20.7 log 1.455} 2016 = 21.607{41.6 + 3·37} = 972′′ = 16′, 12′′. If q, the constant supply received by the reservoir, had been 20 cb. ft. per second, then— the log of which is 0.1809856 (in the former case subtracting 9 we had 132.97 = 1.455, the log being 0.1628630), and the value of t is now 21.607 (41.6+ 2.303 x 20 x .181) 1759 29′ 19′′ to lower the surface 5 ft. 90.4 (XL.) Referring to the latter part of § 92, in order to determine the depth which the surface would descend in a given interval of time, the formula must be arranged so as to separate the factors of VH from Vh, then transposing, so as to make the left-hand side = O, we have t + 2A (mS√2g)2 {mS√2g√H+2.303×q×log(mS√zg√H−q)} 2A (mS√2g)={mS√2g√h+2.303 × q× log. (mS√/2g√h−q} =0. Let us suppose all the letters to have their former values, t being taken at 20 minutes, calculate the value of h— -43560 (t =) 1200′′ – 2016 {44.9 × 3.162 + 20.73 × log 133} and thus we have = 1200 - 4020=-2820, 21.61 × {44.9 √ĥ + 20.73 × log (44.9 √h−9)} − 2820 = 0, when the true value of h is substituted. To further prepare this last expression for the tentative determination of h, we must multiply out by 21.61, hence 970.3 √h+448 log (44.9 √h − 9) – 2820 = 0. If we take at first The surface, therefore, descends 5.9 feet in 20'. (XLI.) § 93. A pond, whose area is 12000 square feet, has an overfall outlet 3 feet wide, and at the commencement of the discharge has a head of 2.8 feet, calculate the length of time required for the surface to descend 1 foot, it being supposed that no supply is received. We have then H = 2.8, and h of m being taken at 0.61. The formula = 2.8 1 = 1.8, the value being put into numbers for this question, we have— Calculate the time in which the surface descends 0.5 feet. I I In this case h = 2.8 − .5 = 2.3, and√1.516. Hence— Again, if we suppose the depth descended to be 1.5, and all the other quantities remain the same, we shall thus have the depths then being 0.5, 1, 1.5 feet, the corresponding intervals are 2′ 31′′, 6′ 4′′, 11′ 5′′. If h = o, it is evident that t becomes infinite, as 2452 = infinity, and so also of any finite number in the numerator, arising from any other data. If the depth sunk had been nearly equal to the whole charge at the commencement, as, suppose, 2.4, so that h = 2.8 - 2.4 = 0.4, then (XLII.) § 95. In question XIII., page 77, Fig. 31, taken from D'Aubuisson, the time of filling the lower part of a canal lock, on the Canal du Midi, is calculated, i. e. up to the level of the centre of the sluices, placed in the upper pair of gates; we can now, by the 2nd case of § 95, calculate the time of filling up to the level of the upper reach, from the centre of the sluice doors, which, added to the 25" as determined in XIII., will give the total time. Substituting in the formula2A T= mS √ 2g the several numerical values given at page 77, we shall have2 × 3503.6 × √6.3945, T= that is .548 × 13.532 × 8.024 to which adding 25", we have 5′ 23′′ as the total time of filling a lock of such dimensions. (XLIII.) The locks on the Montgomeryshire canal have a length of 81 and width of 7.75 feet; and at one, named the Upper Belun Lock, the lift or rise was 7 feet. A pipe leads the water from the upper level, and discharges below the surface of the lower level in the lock chamber, the diameter of which is 2 feet. As the mouth of this pipe is a square, 2 feet in the side, gradually altered into a circular pipe, 2 feet in diameter, we may take m=1, a result which is justified by comparing the observed time of filling this lock with that calculated by the formula when m is put equal to unity, for 2 x 81 x 7.75 × 2.645 = 132′′ = 2′ 12′′, 1 × 22 × 7854 × 8.024 the observed time being 2' 10". CHAPTER III. FLOW OF WATER THROUGH PIPES, ARTIFICIAL CHANNELS, AND RIVERS. 110. Gravity is the sole force that acts upon a mass of water left to itself in a bed of any form; it produces all the motion which takes place, the inclination of the surface of the water in the channel is the immediate cause of motion, being that which enables gravity to act: and thus the measure of this force is in ft. per second, gx sin i. If, then, water flowing in a channel or pipe, and subject to this constant accelerating force, meet with no resistance, it will descend with an increasing velocity which would never be found uniform. But observation and experience show that in open channels and pipes, even those of very great slope, the flow very soon becomes uniform. Bossut made the following experiment to prove this truth directly: Having constructed a canal in wood, 650 ft. long, with a slope of 1 in 10, and marked off equal spaces of 108 ft. each, it was found that the water traversed each space, except the first, in equal times. There must then exist a retarding force, which destroys at each instant the effect of the accelerating force, and which is necessarily equal to it. But in pipes, channels, &c., there can be no retarding force but that which arises from the resistance of the sides or bed: and of its existence we cannot doubt, for the simple experiment of observing the discharge through a tube in a certain time, and again when the tube has been lengthened-all else remaining the same proves that the time required to give a certain volume of water has been increased also; and this can only arise from the fact that the tube, or other channel, by reason of its increased length, offered a greater resistance to the velocity. The surface thus opposed motion. To these retarding forces the name of Friction has been applied though, from the difference between the laws of friction of water flowing over its resisting bed, and the friction of solid bodies sliding upon each other, we must look upon it as the application of an old word in a new sense, in preference to adding a new term to express this peculiar resistance. It may be useful to state here briefly the laws of friction of solid bodies, with the view of showing this contrariety. P |
