In these formulas v is in each case equal to 8.02/h. The head on a vertical orifice is the average head. In the case of a circle this would be the center.

The average head upon a vertical orifice is 16 feet; what is the theoretical velocity of water flowing from that orifice? In the formula v = 8.02√h we may substitute for h its value, then, v = 8.02 × √16

= 32.08

the required velocity in feet per second. Suppose in the above problem that the area of the orifice is 3 square inches and the time of flow 10 seconds, then the theoretical quantity of flow will be: 3 X 32.08 X 10 X12 11548.8 cubic inches.

Had the orifice been a circle of 2 inches diameter the theoretical discharge would have been ;

3.1416 X 32.08 x 10 x 12

= 12093.9 cubic inches.

NOTE. 12 occurs as a factor to reduce the velocity from feet per second to inches per second.

EXAMPLES FOR PRACTICE.

1. Find the theoretical velocity of water flowing from an orifice under an average head of 81 feet. Ans. 72.18 feet per second. 2. What is the theoretical discharge from a circular orifice of 4 inches diameter, under a head of 36 feet, flowing for 22 seconds? Ans. 159639.52 + cubic inches.

NOTE. Reduce head to inches.

3. What is the theoretical discharge from a square orifice two inches on a side, if the velocity is 40 feet per second and the time of flow is 30 seconds. Ans. 57,600 cubic inches. standard orifice is used.

In the measurement of water the This term signifies that the opening is so arranged that the water in flowing from it touches only a line. To secure this the inner edge of the opening is a square corner which alone is touched by the water. In precise experiments the orifice. may be in a metal plate whose thickness is really small, as may be seen in Fig. 19

J

FIG. 19.

at A. More commonly it is cut in a board or plank, care being

taken that the inner edge is a definite corner.

It is usual to bevel the outer edges of the orifice as at J, so that the escaping jet shall not touch except at the inner edge. We observe by the figures a certain contraction of the jet, and owing to this contraction the discharge is always less than the theoretic discharge. In order to obtain the actual discharge we multiply the theoretic discharge by a coefficient c, i. e., in the formula Q = avt, we place another factor called the coefficient of discharge c. Thus our formula becomes: Q =cart,

where c is the coefficient of discharge, a the area of the orifice, v the velocity (theoretical) and t the time. Our units would ordinarily be square feet, feet per second, and seconds. The result would be a total discharge in cubic feet. The following tables give the coefficients for circular, square, and rectangular (one foot wide) vertical orifices.

Table III. Coefficients for Circular Vertical Orifices.

This table shows that the coefficient of discharge decreases as the size of the orifice increases, and that for diameters less than 0.2 feet it decreases as the head increases. The coefficients above the horizontal lines in the last three columns may not be used in actual problems where the actual discharge is to be computed. The error in these coefficients depends upon the ratio of the diameter of the orifice to the head.

Table V.

Head H in Feet.

This table shows similar variations to Table IV. If the head is less than 4 times the side of the orifice an error is committed in finding the coefficient by the same method as in finding it for a greater head. The coefficients above the horizontal lines are not to be used in actual problems.

From this table we see that the coefficient of discharge for a rectangle of constant breadth increases as its depth decreases. In selecting a coefficient for use with an orifice whose size falls without the limits of the table, it should be borne in mind that large orifices have a smaller value for the coefficient than small orifices.

One way of measuring is by the use of the miner's inch which may be roughly defined as the quantity of water which will flow from a vertical standard orifice 1 inch square if the head on the center of the orifice is 6 inches. From the fourth table the coefficient of discharge is seen to be about .623. discharge in cubic feet per second would be

Hence the actual

Q=

.623 X 8.02
144

6.5
12

= .0255 cubic feet.

The discharge in 1 minute would be 60 x .0255 = 1.53 cubic feet. The mean value of the miner's inch is therefore roughly about 1.5 cubic feet per minute. It differs considerably, however. When water is bought for mining or irrigating purposes, a much larger quantity than the miner's inch is required and various methods of measurement are used.

to be,

FIG. 20.

A weir is a notch in the top of the vertical side of a reservoir through which water flows. The notch is usually rectangular, and is shown in the following illustration, Fig. 20. In measurements of water flowing over weirs the same precaution should be taken as for orifices, namely, that the inner edge shall have a definite angular corner. The theoretic

discharge of a rectangular weir is found √2gb H√Í.

In this formula b is the breadth of the notch, commonly called the length of the weir, g the acceleration of gravity, and H the depth of water on the lower edge. This depth must not be measured in the plane of the weir, but the height must be taken a considerable distance back from the weir on account of the decrease in head where the water flows over; H is ordinarily determined by the head a certain distance back from the weir, and determining

an extra head to be added to this by ascertaining the velocity with which the water approaches the weir. In order to get the actual discharge, a coefficient of discharge must be used, and its computation is rather beyond the province of this work.

If water flows through a pipe the diameter of which varies, the theoretical velocity varies inversely as the square of the diameter of the pipe, because the same quantity passes through each section of the pipe in the same time, but in some places the area is greater, therefore the velocity need not be so great. Friction causes a loss of effective head as we go along the pipe. The longer the pipe, as we know, the more slowly water flows. If small tubes were inserted along the pipe, the water would rise in them, but as we approached the end of the pipe from which the water flows the heights of the water in the tubes would decrease. The head acting to cause flow throughout the remainder of the pipe decreases as we go towards the end from which the water flows. Let us find the probable discharge in 10 seconds from a vertical orifice, .2 feet in diameter under a head of 10 feet. The velocity is 25.4 feet per second from the formula v = 8.02 √h. The time t, is ten seconds, the area a of the orifice is .031416 square feet, and the coefficient of discharge e from Table III. is .597. If we substitute these values in the formula Qca v t, our equation vt, becomes,

Q = .597 × .031416 × 25.4 × 10

4.76 cubic feet.

EXAMPLES FOR PRACTICE.

1. Find the probable discharge in 5 seconds from a circular vertical orifice .1 foot in diameter if the head is 20 feet.

Ans. .839 cubic feet.

2. The theoretical discharge from a square, vertical orifice is 4.96 cubic feet per minute. If the coefficient of discharge is .602 what will be the probable amount of flow in 15 seconds? Ans. .746 cubic feet.

3. What is the probable discharge from a rectangular orifice one foot wide, and .75 feet deep if the theoretical flow for 10 seconds is 12.03 cubic feet and the head is 4 feet?

Ans. 7,3 cubic feet.