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If hy is the depth in feet of the top of a weir below the surface of the water, and h is the depth in feet of the bottom of the weir below the surface of the water, the actual discharge Qa, in cubic feet per second, is
·.615 × } b √2 g (√ h3 — V√ h3).
FLOW OF WATER IN PIPES.
= mean velocity of discharge in feet per second; total head in feet = vertical distance between the level of water in reservoir and the point of discharge;
length of pipe in feet;
= diameter of pipe in inches;
coefficient of friction.
Then, for straight cylindrical pipes of uniform diameter, the mean velocity of efflux may be calculated by the formula, hd
Vm =2.315 Vƒl + 125 ď
NOTE. The head is always taker as the vertical distance between the point of discharge and the level of the water at the source, or point from which it is tak and is always measured in feet. It matters not how long the pipe iswhether vertical or inclined, whether straight or curved, nor whether any part of the pipe goes below the level of the point of discharge or not-the head is always measured as stated above.
EXAMPLE.-What is the mean velocity of efflux from a 6" pipe, 5,780 ft. long, if the head is 170 ft.? Takeƒ= .021. SOLUTION.
Vfl + .125 d
170 X 6 021 5,780 + (.125 × 6)
-6.69 ft. per sec.
When the pipe is very long compared with the diameter, as in the above example, the following formula may be used:
√ ƒ l'
in which the letters have the same meaning as in the prece ding formula. This formula may be used when the length of the pipe exceeds 10,000 times its diameter.
The actual head necessary to produce a certain velocity Vm may be calculated by the formula
If the head, the length of the pipe, and the diameter of the pipe are given, to find the discharge, use the formula
fl+ .125 d'
that is, the scharge in gallon per second equals .09445 times the square of the diameter of the pipe in inches, multiplied by the square root of the head in feet, multiplied by the diameter of the pipe in inches, divided by the coefficient of friction times the length of the pipe in feet, plus .125 times the diameter of the pipe in inches.
To find the value of f, calculate Vm by formula (b) assu ming that ƒ = .025, and get the final value of from the following table:
+ .0233 V.
215 X 8
EXAMPLE.-The length of a pipe is 6,270 ft., its diameter is 8 in. and the total head at the point of discharge is 215 ft. How many gallons are discharged per minute?
7.67 ft. per sec., nearly.
8 (see table), Q =
22.03 gal. per sec. =
215 X 8
.09455 X 82.
.0205 X 6,270+ (.125 X 8)
If it is desired to find the head necessary to give a discharge of a certain number of gallons per second through a pipe
whose length and diameter are known, calculate the mean velocity of efflux by using the formula
find the value of ƒ from the table, corresponding to this value of Vm, and substitute these values of ƒ and V in the formula for the head.
EXAMPLE.-A 4" pipe, 2,000 ft. long, is to discharge 24,000 gal. of water per hr.; what head is necessary? 6 gal. per sec. Vm
60 X 60 10.2 ft. per sec. From the table, f .0205 for Vm = 12; assume that f .02 for Vm 10.2. .02 X 2,000 X 10.22 Then, h 5.36 X 4
+.0233 X 10.22 = 196.53 ft.
To find the diameter of a pipe that will give any required discharge in gallons per second, the total length of the pipe and the head being known, find the value of d by formula (ƒ); substitute this value in formula (e), and find the value of Vm. Then find from the table the value of ƒ corresponding to this value of Vm Substitute the values of d and f just found in the righthard member of formula (g) and solve for d; the result will be the diameter of the pipe, accurate enough for all practical purposes. 51 Q2 d=1.229 5 (fl+fd) Q2 d = 2.57 S (9) h EXAMPLE.-A pipe 2,000 ft. long is required to discharge 24,000 gal. of water per hr. The head being 195 ft., what should be the diameter of the pipe?
60 X 60
- 63 gal. per sec. 5/2,000 X (63)2 195 Substituting this value in formula (e), Vm 9.352 ft. per. sec. From the table, the value off for Vm = 9.352 is .0201. Substituting this value of ƒ and the value of d, found above, in formula (g),
4.18+ in. 24.51 X 6 4.182
5(.0201 X 2,000 + X 4.18) X (63)2
ting in formula (ƒ), d
8, and .0193 for V
24.51 X 6 42
APPROXIMATE WEIGHT OF FIREPROOF FLOORS. (Exclusive of Partitions.)
Weight. Lb. per Sq. Ft. Including Beams.
Dense-tile flat arch.
WEIGHT OF FIREPROOFING MATERIALS.
Name of Wind.
Thickness. Weight. Lb.
per Sq. Ft.
Pressure. Lb. per Sq. Ft.
The wind pressure on a cylindrical surface is usually taken as one-half that on a flat surface of the same height and width.