Deductions from the preceding Experiments.

1. That the quantities of water discharged in equal times by the same orifice, from the same head of water, are nearly as the areas of the orifices.

2. That the quantities of water discharged in equal times by the same orifices, under different heads, are nearly as the square roots of the corresponding heights of the water in the reservoir, above the surface of the orifices.

3. That the quantities of water discharged during the same time by different apertures, under different heights of water in the reservoir, are to one another in the compound ratio of the areas of the apertures, and of the square roots of the heights in the reservoir.

4. That, on account of the friction, small orifices discharge proportionally less fluid than those which are larger and of similar figure, under the same altitude of fluid in the reservoir.

5. That, in consequence of a slight augmentation which the contraction of the fluid vein undergoes, in proportion as the height of the fluid in the reservoir increases, the expenditure ought to be a little diminished.

6. That circular apertures are most advantageous, as they have less rubbing surface under the same area.

7. That the discharge of a fluid through a cylindrical horizontal tube, the diameter and length of which are equal to one another, is the same as through a simple orifice.

8. That if the cylindrical horizontal tube be of greater length than the extent of the diameter, the discharge of water is much increased, and may be increased with advantage to four times the diameter of the orifice.

TABLE

Of Comparison of the Theoretic with the Real Discharges per Minute through an Orifice One Inch in Diameter; also through a Tube of a Cylindrical Form, One Inch in Diameter and Two Inches long.

Constant altitude of the Theoretical discharges Real discharges in the Real discharges in the same water in the reservoir through a circular ori- same time, and through time by a cylindrical tube, above the center of the fice one inch in diam- the same orifice. one inch in diameter, and two inches long.

Cubic Inches.

orifice.

eter.

3,539

The line

TABLE

Of the Heights corresponding to different Velocities, in French Metres, per

Second.

NOTE. The metre equals 39-37023 inches, or 3.281 English feet.
Practical Rules, according to the Formula of Torricelli,
BY M. MORIN.

To obtain the velocity due to a given height of water, above the center or middle of an orifice, or the height due to a given velocity,

Rule.-Multiply the height of the water above the center of the

orifice by 19 62, and the square root of the product equal the velocity due to that height. Or: Divide the square of the velocity by 19.62, and the quotient equals the height, each being expressed in French

metres.

Practical Formulas, by M. PRONY, as deduced from the Experiments of
Bossut, on Simple Orifices and Additional Tubes.

Let A =the area of the orifice, in square feet.

Tthe time, in seconds.

H=the altitude above the middle of the orifice, in feet.
Q=the quantity of water discharged, in cubic feet.

Q

Q

Q

1. d=

49438 TH 2. T=

(4·9438 d2 T)2.

4-9438 TH 3. H=; d, being the diameter of the tube.

On the Discharge of Water by Horizontal Conduit, or Conducting Pipes. From the Experiments of M. Bossut.

1. The less the diameter of the pipe, the less, proportionally, is the discharge of fluid.

2. The greater the length of conduit pipe, the greater the diminution of discharge.

3. The discharges made in equal times, by horizontal pipes of different lengths, but of the same diameter, and under the same altitude of water, are to one another in the inverse ratio of the square roots of the lengths.

4. That, in order to have a perceptible and continuous discharge of fluid, the altitude of the water in the reservoir, above the axis of the conduit pipe, must not be less than 148 inches for every 180 feet of the length of the pipe.

5. That, in the construction of hydraulic machines, it is not enough that elbows and contractions be avoided, but also any intermediate enlargements, the bad effects of which are proportionate, as in the following

TABLE.

Practical Rules and Examples for ascertaining the Diameters of Pipes, and Quantity of Water discharged in any given time.

MR. TREDGOLD'S RULE, FROM EYTELWEIN'S EQUATION.

Rule. Multiply 2500 times the diameter of the pipe, in feet, by

the height in feet, and divide the product by the length in feet, added to 50 times the diameter, and the square root of the quotient will be the velocity of discharge, in feet, per second.

Example-Suppose the diameter of a pipe to be 375 feet; the height of the water in the reservoir, above the point of discharge, 51.5 feet, and the length of pipe 14637 feet. Required the velocity of the water, and the quantity discharged in cubic feet, per second? 2500 × 375×51.5 48281.25 14637+(50x375) 14655 75

velocity;

=

= √33=1816 feet per second,

And 3752x7854 × 1·816=20057 cubic feet per second.

Or, multiply 1542, 133 times the fifth power of the diameter of the pipe, in feet, by the height, in feet, and divide the product by the length, in feet, added to 50 times the diameter, in feet, and the square root of the quotient equals the discharge, in cubic feet, per second. These rules apply only to the case where the pipe is straight. If there be bends in it, the velocity (found by the rule as above,) must be diminished, by taking the product of its square, multiplied by the sum of the sines of the several angles of inflection, and by 0038, which will give the degree of pressure required to overcome the resistance occasioned by the bends, and deducting this height from the height corresponding to the velocity, the corrected velocity is obtained.

If power is to be obtained by the issuing stream of water through a pipe, the whole of the power due to the height which is necessary to send the water through the pipe, at the required velocity, is not lost, for the power due to this velocity is still in the water. Thus, suppose five tenths or half a foot of head of water be given to maintain a required velocity of water through a pipe, and this velocity be

found, as per rule, to be 2-988 feet per second: then (2·988)=-1395

or the height that would give that water a velocity of 2.988 per second. Hence, 5-1395=3605 of a foot, the positive height lost by the resistance in passing through the pipe.

A pressure of 75 of a foot in height, will give a velocity of discharge equal to 3.5355 feet per second, at the lower end of a straight pipe of 2 feet diameter and 200 feet in length; but if there be a bend, or number of bends, in the pipe, the velocity of the water will not be so great. Say that there are three bends, one of 90°, and the other two of 150 each, the sum of the sines of those angles equal 2. Hence, 3 53552 × 2 × 0038095, and 75-095655, that must be used or taken instead of 75 in the rule for obtaining the velocity of discharge at the lower end of the pipe.

As an approximate rule for determining the diameters of pipes capable of conducting any required quantity of water, in cubic feet, per minute:

Multiply the square of the quantity in cubic feet per minute by 96, and the product equals the diameter of the pipe in inches.

Example. Required the diameter of a pipe large enough to conduct 625 cubic feet of water per minute?

√625=25, and 25 × 96=24 inches diameter.

TABLE

Of the Diameters of Pipes, sufficient in Size to discharge a required
Quantity of Water per Minute.

Of the Discharge of Water by Rectangular Orifices in the Side of a Reservoir Extending to the Surface.

The velocity of the water varies nearly as the square root of the height, and the quantity discharged per second two thirds of the velocity due to the mean height, allowing for the contraction of the fluid, according to the form of the opening, which renders the coefficient, in this case, equal to 5.1. Hence:

Suppose, in the side of a lake or in the side of a reservoir, a rectangular opening is made, without any oblique lateral walls, 3 feet wide, and extending 2 feet below the surface of the water. Required the quantity that will be discharged, in cubic feet, per second?

The area of the opening=3×2=6 feet; and of √2×51×6= 28.85 cubic feet.

The same co-efficient is applicable to the finding of the dimensions of a weir in the side of a lake, or in the side of a still river. For example:

In the side of a lake is a wier of 3 feet in breadth, and the surface of the water stands 5 feet above it. It is required how much the weir must be widened, in order that the water may be a foot lower, and an equal quantity discharged?