and. by similar ▲, we get { = = = = α 2 : ༤ ༩ or = the defective side of the rhombus; wherein as the descending surface is > or < than Hence this descending surface is a2√6 X = + 2 = a2√6 × x = 2a26 a a 2 Let xa a according as or is taken, x = a to x = T= (a = x)} :: ( a 2 a2√6 2 4.x a x x2 to x = 0. Now supposing the section a√6 2 2 a a: √ a 2.x2 a2 where the descending sur face first changes its form from the segment of a rhombus to an isosceles A to be an orifice, the time of the descent of the fluid to that section is (696) 1 = √0 x (+√xdz - 2rlds t= 2.x dx Ag a2 × {1 − (1 − F and the latter from 4 1) 5 is used according half the rhombus. ....... 2x-a ; then 0 and C = × (13 — 2√2). a (a) (b) .. when a a, the time of the descent from xato x = 13 15/2 ....... (c) Again, the time of the fluid's descent in the pyramidal portion of the vessel is (696 and (b)) = 2a√6 T = Sx dx 4a√6 5a Ag and down the whole of this part of the vessel, or for x = a2 √3× √α 4a6 295a A√9 Hence the whole time required is T+T= a2 √ a 5A√9 202 √ a 5A 3g a 2√3 x (8 - √2). , it is × 13−2√2 + √3} 716. In the first place let us seek the locus of the intersections of the parabolas described by the fluid issuing through the holes in any given vertical line of the surface. Let y be the variable ordinate of any of these parabolas, x the corresponding ordinate measured from the top of the vessel, and not from the vertex or hole, and p the variable parameter of all the parabolas; then their general equation is (Vince, Hyd. prop. XXXIX.) y2 = p × (x - 1 ) Now differentiating this equation on the supposition that p alone is variable, and putting the result = 0, and thence deducing p in terms of x, substitute for p in the equation, and the transformed equation will belong to the required locus. See Appendix to Vol. II. of Simpson's Fluxions, pp. 339, &c., or pp. 47, 4S, 67, 68, of Vol. II. of this work. We thus get for the locus the equation which belongs to a straight line inclined to the horizon at an of 45°, and meeting the top of the cylinder. Hence it follows, that the boundary of the water issuing from the cylinder is the surface of the frustum of a cone, the radii of the upper and lower ends being HYDRODYNAMICS. r and r+a where is the radius of the base of the cylinder, and a its altitude. 717. Let A be the area of any section parallel to the horizon, a that of the orifice, and h the altitude of the vessel; then the time of emptying is (Vince's Flux. p. 268,) 2A t= √ 1. a 718. The area of any section parallel to the horizon is жу2 = ржx p being the parameter of the generating parabola, and x the abscissa measured from the vertex. Also if A denote the area of the orifice, and h the whole height of the vessel; then the velocity at the orifice is 2gh when the vessel is quite full; and the quantity discharged in each second is (by the question) A√2gh = nQ Q being the given supply. Hence 14). n2 and since the water descends with the velocity Q = A √(2g X x = h no Let t0, when x h; then 98 n√x h n2 + 2h n√x - √hr 23 (n-1)√hS + 2h x + √ + 21(n−1 √ h). n3 x + (2) Now when the descending surface becomes stationary, equation (1) gives us and the time of arriving at this position, equation (2) informs us is infinite, unless when n = 1, and then this time is zero. Hence we conclude that unless the influx is equal to the efflux at the very beginning of the discharge, the surface of the fluid will never be quiescent. Although this conclusion is deduced after the methods expounded in all the Elementary Treatises (see Prony's Architecture Hydraulique, Tom. I. p. 342, and Bossut, Vol. I.; also Leybourne's Mathematical Repository,) yet it must be confessed it is far. from satisfactory. If the velocity of the descending surface be thus rightly estimated, it may easily be shewn that the expression for the time of reaching the state of quiescence will always involve log. (0), or be oo, whatever may be the form of the vessel; in other words, no such state of quiescence is possible. This seeming contrary to experience, we intend prosecuting the subject somewhat farther at our leisure, and shall probably insert the results in some Scientific Journal. 719. Let r be the radius of the sphere, x the altitude of the fluid at any epoch, and A the area of the orifice; then the velocity at the orifice being A√29 3 412 T A 29 - x x = maximum, whence the position of the surface when its velocity is a minimum is obtained, and this minimum velocity is 2A √39 Although this velocity is a minimum, yet it is not the least velocity of all, that at the orifice being 0. The reader must observe, that at the orifice the descending surface is orifice; for otherwise it might appear from equation (1) that this is oo instead of 0. |