43. Those which he made use of carried a mouthpiece, ABCD, not unlike the form of the contracted vein, AB being equal to 0.1332 feet, and CD equal to 0.1109 feet. The body of the adjutage varied in length and in its divergence: this last was measured by the angle contained between the sides EC and FD, supposed pro

longed until they meet. These adjutages were attached to a reservoir maintained at a uniform level; the flow took place under a constant charge of 2.89 feet; and the time required to fill a vessel of 4.838 cubic feet was observed.

The following Table gives the result of his principal observations, premising that the time corresponding to (unity as a coefficient, that is to) the theoretic velocity, was 25.49 seconds :

TABLE showing the Variation of the "Coefficient" of the Discharge with Conical Diverging Adjutages at different Angles and Lengths.

Venturi has drawn the conclusion that the adjutages of maximum discharge should have a length of nine times the diameter of the smaller base, and an angle of divergence equal to 5° 6' : it is represented in the woodcut, Fig. 16. This, he adds, would give a discharge 2.4 times greater than the orifice in a thin plate, and 1.46 times greater than the theoretic discharge.

44. Flow of Water under very small Charges.-When the charge over the centre of the orifice is very small compared with the vertical depth of the orifice itself, the mean velocity of the different threads of the fluid vein—that is to say, the velocity which, being multiplied by the area of the orifice, gives the actual discharge-is no longer that of the central thread. It differs from it in proportion as the charge is less: its true value will be about the hundredth part less when the charge is equal to the depth of the orifice, and by about the thousandth part when equal to three times that depth. Let us examine what theory teaches on this point; and first, of the law which it assigns for the velocity of the fluid threads in proportion as their depth below the surface of the water increases.

The italic capital H is used for the depth from the surface to the sill or bottom line; the italic h for the depth to the top of any rectangular orifice; and the capital H for the mean depth, or H+h

45. Velocity of any Fluid Thread whatever.-Let AA, Fig. 18, be the level of the surface of water in a vessel, and upon the face AB-which, for greater simplicity, we suppose vertical-let us imagine a series of very small orifices placed one below the other, and of which that at B is

the lowest, and putting H for the height AB, the velocity of the jet issuing from B will be expressed by √2gH; making BC equal to this quantity, it will represent this velocity; for any other point P, taken at a depth equal to AP or x, the velocity of issue will be represented by the line PM = √2gx, and calling this y, we shall have y = √2gx, or y2 = 2gx.

Through all the points M so found drawing a curve line, it will, from the above equation, be a parabola having 2g or 64.4 feet for its parameter; and thus we have this proposition:-The velocity of a fluid thread issuing at any depth is equal to the ordinate of a parabola whose parameter is equal to 2g, and the depth the abscissa.

46. Let us next suppose that this series of orifices over each other was continuous, forming a rectangular slit, whose width was /, and seek now the Discharge in

this case, omitting at present the "contraction." Suppose this opening divided into elementary rectangles by horizontal lines, the volume of water which will issue from each of them in one second will be equal to the

E

volume of a prism whose base is the elementary rectangle, and height the velocity, or ordinate, corresponding. The sum of all the prisms will also be equal to a single prism whose base is the parabolic segment ABCMA, and height 7, the width of the opening.

From a property of the parabola,this segment is 3rds of the rectangle ABCK, whose area AB × BC, as shown above, is equal to H√2gH; thus the discharge for the rectangular opening, whose height is H and width 7, is

47. Let us now seek to determine the discharge through a rectangular orifice opened in the same face, but only from B to D, and with the same width 7. Let h = AD, then the discharge of the opening from A to D will be

Now the discharge from the rectangular orifice, BD × 1, will be the difference of those from the openings AB and AD, each into /, and therefore

That which has been established in § 13-namely, Q = S√2gH, is, substituting for S its value, S=1(H − h),

Q=1√2g. ✓H. (H-h),

on the supposition, very nearly correct, that the velocity at the mean depth was the mean velocity.

48. Mean Velocity.-Let us now determine the mean

velocity (§ 44), and first, that of the rectangular opening up to the surface. Let G be the point from which issues the thread with this mean velocity. If we make AG = 2, it will be expressed by√2g, and this being multiplied into the area of the opening, l × H, must give the total discharge, which we have already found equal

dividing each side by the common factors / × H× √2g, we have, squaring both sides,

4
2= H;
9

and, therefore, V =

√ 28 21 H =

✓ 2gH.

3

Thus it appears that the mean velocity is equal to twothirds of the velocity of the lowermost fillet; and so GH, which represents the former, is 3rds of BC, which, in like manner, represents the latter.

For the case of the rectangular orifice, whose depth is BD or H – h, we shall, in like manner, have the area expressed by (H − h) × l, and the discharge, making z' the height due to the mean velocity, by—

(H − h) × 1 × √/2g7' = 2 1× √ 28 (H √ H - h√/h).

l

3

Dividing both sides by x 2g, we have

(H − h) × √ 7 = 2 (H√ H-h√h),

3