Hence, multiplying both sides by A, we have the discharge Q' for the given time— tmS √29 44 91. Mean hydraulic charge.-A prismatic vessel (Fig. 27), receiving no supply, discharges water through an orifice, S, during the time T seconds, having at the commencement the head H, and at the end that represented by h; required the mean hydraulic charge H', by which, ceteris paribus, the same quantity of water would have been discharged: we have ($14) substituting this value of Tin (b), we have— This proposition gives a method of determining by experiment the coefficient of contraction at an orifice, and has been used by Dr. Young for that purpose (Transactions, Royal Irish Academy, vol. ii., p. 81, &c.). Vide EXAMPLES to Chap. II. 92. Case in which the basin receives a constant stream during the time it is discharging.-Let us suppose that a prismatic basin receives a constant supply (less than that discharged), and that we have to determine, in these circumstances, the time that the surface will require to fall a given height. Continuing the same letters in the same significations, let us, in addition, call the volume of water arriving at the basin in one second q, and let ≈ be the height which the water descends in the given time t, and dr the height which the surface descends in the indefinitely small time dt, Adx will then express the volume of water which flows out during dt, if the basin received no supply; but as it does receive q in one second, and consequently qdt in dt, the volume of water actually discharged will be Ada+qdt. This same volume, from the formula for the discharge of water through orifices (§ 14), is expressed by m Sdt 2g (H-x); we shall there fore have and if we make H- x = h, and therefore dx = dh, we have In order to integrate this equation, we may put— Determining the value of C for the commencement of the motion, when to and x = 0, and H also being equal to h, we have, substituting for y its value above,— C equal to 2 A (mS√ 29)2 (mS√ 2 g√ H-q+qhyp.log.mS√2g√ H−q). Hence t is equal to— 2 A (mS√29):{mS√2g(VH-√/h)+ghyp.log. mS√2gH-q mS√ 2gh-q It is evident from this expression that when go, that is, when no supply is flowing in, it becomes identical with that in $ 89. If we had to determine the height which the level of the water would descend in a given time, the question would be reduced to this other--namely, to find the charge h at the end of this time, and subtract it from H, the head, at the commencement of the discharge. To obtain h we must substitute successive values of it; i. e. of (H- x), in the equation given above, and thus tentatively determine that which satisfies the equation. 93. Case when the water is discharged over a weir.-In the case when the water issues from the basin by an overfall, supposing that it receive no supply, we shall have, from what has been laid down in § 46 and § 55, 2 3 2g whence, by a method analogous to that which has been used above, we have 94. Reservoirs not being prismatic. We have hitherto considered only the particular case of prismatic basins or reservoirs: the determination of the time of discharge for any other form is much more complicated, and is even impossible in most cases which present themselves. The fundamental equation is, however, always But here A is variable, and we must, in order to integrate, express A in terms of x, which can only be effected when we know the law by which A decreases, and in the cases where the basin itself is a solid of revolution, whose generatrix is known. In every other case it will be necessary to proceed by approximations and by parts. To this end, we must divide. the basin or reservoir into horizontal sections of small depth. Each of these may be taken as prismatic, and we can determine the time it takes to be discharged by the aid of the formula given above. The sum of these partial times will give the time that the surface of the water takes to descend a height equal to the sum of the heights of the prisms. 95. Flow of water when it is discharged from one reservoir into another. Ist. In the case when the orifice is covered with water on both faces or sides, the levels remaining constant, the quantity discharged is then the same as if it had taken place into the open air under a charge H-h, equal to the difference of the charges upon each face; thus we have, representing by Q the discharge per second, Q = mS √ 2g (H – h). 2ndly. Let us suppose that the level remains constant in the upper basin, and that the lower of a given area receives it without any loss, required the time which must elapse before it has reached the level of the upper basin or a given height. This problem is the exact inverse of that in § 89, in which the discharge took place into the air, and the surface of the water above the orifice descended with a motion uniformly retarded. In the present case, the surface of the basin, urged from below upwards with a force measured by the difference of the level between the two basins,-which decreases in the same proportion as the charge decreased in the former case,-rises with a motion uniformly retarded, and it will require the same time to traverse the same space under these similar pressures. Let H represent the pressure or charge AC (Fig. 28) at the commencement, and h the charge AD at the end of the time t, A the horizontal section of the vessel being filled, and S and m as before,—we shall have, for filling up to DE, and for that of filling up to AF we shall have These latter formulæ are of some importance: they serve to determine the time in which canal locks, &c., may be filled, and to assign the area of sluice-way required to fill in a given time. 96. 3rdly. The level of the water being variable in each vessel -We come now to the third case that can arise, namely, when two reservoirs of different level communicate with each other, each being limited in area and receiving no supply, and thus one surface descends as the other rises. Such is the case of the two basins K and L (Fig. 29), communicating by a wide pipe EF, provided with a sluice-door or cock at G. Before the opening of this sluice-door the level of the water is at AB in the first reservoir, and CO in the second. At the end of a certain time after the opening of the communication it has descended to MN in the first, and has ascended to PQ in the second. It is required to determine the relation between these two heights at a given time, or vice versa, from the given difference in the respective heights, to determine the time corresponding to a given discharge. = = = = = Let t equal this time, BE H, CF h, NE = x, PF = y, A horizontal section of the first vessel, and B that of the second, section of the pipe of communication: in the coefficient m we must include the resistance of the water passing through this pipe. Whilst the water has risen in the second basin by the quantity dy, during the instant dt, it will have lowered in the other by da; and remembering that a diminishes while y and t increase, we have Adx = - Bdy and (§ 14) (a) ms√2g (x − y). dt, from whence Adx == The first equation being integrated, remembering that when x = H, y = h, becomes (c) dt = 2g (x − y) solving for y we have and substituting this value of y in the preceding equation (b), integrating and observing that a = H when to, we have t = 24 VB − − – AH – Bh mS √ 29 (A + B) {√B (H − h) − √ (A + B) x − ̧ 2g |