Chap. XVII.

THEORY OF TIIE MOTION OF FLUIDS.

1133

of the level of the water above the summit of the orifice; add these two products together;thirdly, multiply the greater velocity, All, by ten times the rectangle comprised by thewhole height, 11 A, of the water, and by the part, B E, which marks its level above the sum-mit of the orifice ; subtract this latter product from the sum of the two preceding, mul-tiply the difference by the base, C A, of the orifice, and divide this last product by five timesthe height, E A, of the orifice.

On the Measurement of Jf'ater flowing through vertical and circular Oriflces _The solid

under consideration must be regarded as formed by the sum of the products of the elementsof a semicircle, and by the corresponding ordinates of a parabola, because the diameter ofthe semicircle being vertical, this solid will be exactly half that which expresses the dis-charge of the entire circle: given a semicircle A E B, and a semi-parabola A F D, whoseaxis is the diameter A B ; re-quired the solid formed by thesum of all the planes comprisedbv the elements, I, P, of thesemicircle, considered as thebreadth of the planes of water,and by the corresponding or-dinates P XT, which express thevelocity of these planes duringa determinate time ; by calcu-lation we find that the solid is Fig. 1720.

{5 of the parallelepipedon com-prised by the square of the radius and the greater ordinate, or, which is the same thing,of the parallelepipedon comprised by the square of the diameter and the greater ordinate :therefore, to have the entire discharge per second through a circular orifice whose summitis on a level with the water, take T 8 3 of the product of the square of the diameter, multipliedby the velocity of which a body* is capable per second, and which it has acquired by a fallequal to the diameter of the orifice.

To have the discharge from an orifice in the form of a quarter circle, as A E C, multiplythe square of the diameter by four times the velocity, B D, answering to the extremity ofthe same diameter ; subtract from this product that of the square of the radius, multipliedby 14 times the velocity of the centre, and ^ part of the difference. The square B D beingtwice C F, B D will be to C F as the diagonal of a square to its side, or nearly as 7 to 5;wherefore the discharge of the upper quarter circle is | the volume of water, which has thesquare of the radius for its base, and the greatest velocity during the time of flowing for itsheight, and the discharge from the lower circle is two-thirds the same volume; thus thesetwo are as 5 to 3.

To find the discharge from a quarter circle whose diameter is at the level of the water,multiply the square of the radius by half the greatest velocity of the water during the timeof flowing, and by the entire velocity if that of a semicircle be required : in the first casethe fall is the radius, and in the second the diameter of a circle; thus the two velocities arcnearly as 7 to 5, or, by more accurate calculation, when two equal semicircles are disposed asFK II and PLM, the expenditure of the first is to that of the second in equal times, as25 to 28.

To find the discharge per second from a vertical and circular orifice above the level ofwater, multiply the least height, E A, of thewater by 29, and divide the product by 37 for afirst quotient; multiply the diameter of theorifice by 18, and divide the product by 85 fora second quotient: cube the diameter of theorifice, and divide the cube by the product of72, multiplied by the square of the least heightof the water for a third quotient; square thediameter of the orifice, and divide the productby 28 times the least height of the water for afourth quotient: subtract the fourth from thesum of the three other quotients; multiplytheir difference by the product of the squareof the diameter of the orifice multiplied by60; divide this product by the velocity ac-quired by a body in a fall equal to the least height of the water; the quotient will be theexpenditure required.

On Fluids flowing through Oriflces under a constant Pressure. — At the bottom of a vesselabed, is a horizontal orifice, gh, through which the liquid contained in the vase flows out:required the quantity that will flow out under a constant given pressure era. To resolvethis question, adapt the given vessel to a reservoir ABDC, whose height, A C, is less than

Fig.1721.