the volume hA would be entirely discharged equal to

2A√h mS√28

Now, the time t, that in which the surface descends a height equal to a, is evidently the difference between the two expressions given above, that is—

90. Volume discharged in a given time.-The above expression for the time which the water requires to descend any given height, by a simple transformation, gives both the value of a, and also the volume of water discharged during the given time: thus we have from the equation (a)

Hence, substituting for √h its value given above, and

multiplying both sides by A, we have the discharge Q' for the given time

Q' = (H− h) A = tmS√2g√H

NH

tmS√28

44

28

91. Mean hydraulic Charge.-A prismatic vessel re

Fig. 31.

H

ceiving no supply, discharges through an orifice S, during T seconds, having at the commencement the head H, at the end h; required the mean hydraulic charge H', by which, cæteris paribus, the quantity is discharged: we have (§ 14)

same

Q' = mS√2gT√ H= (H-h) A; also § 89 (a)—

substituting this value of Tin (6), we have

Supply while discharging.-Let q be the volume received per second (less than that discharged), and x the space the water surface lowers in the time t: then dx will be its descent during the indefinitely small interval dt, and thus Adx will express the volume flowing out during dt, if no supply entered; but as it receives q in one sec., and, therefore, qdt in dt, the actual discharge will be Adx + qdt. From § 14 we thus have

putting H - x = h, and therefore dx = dh, we have

qdt - Adh = mSdt √2g √h, which gives

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―

2A

(mS√2g)2

(mS√2g√H-q+q hyp.log mS √2g✓H-q).

It is evident from this expression that when q = 0, 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 receives no supply, we shall have, from what has been laid down in § 46 and § 55

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

Adx=mSdt√2g (H− x),

from whence we have

dt=

Adx mS√2g (H-x)

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, the levels remaining constant, the quantity discharged is the same as if it had been into the 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,―—

2ndly. Let the level remain constant in the upper