N. K. Ludlow, Mobile, Ala., writes:

"Will you kindly answer this question and give me the rule to work out the same, to wit: A 1-inch iron pipe is attached to an 8-inch cast-iron water main in the street and is run in this shape a distance as is shown on the inclosed slip. [Sketch described in answer - Ed.] Now/, how much water would this pipe discharge per hour with a pressure of 75 to 80 pounds per square inch on the main in street? Please answer and send rule to work it out, and oblige."

[If an elastic ball is thrown against a hard substance we should expect it to rebound with the same force or velocity with which it struck, and when we find that a ball of glass or ivory dropped on a smooth flagstone rebounds nearly to the height from which it fell we are prepared to believe that, making allowance for the resistance of the air, a ball or other body thrown upward with a given velocity will rise to the height from which it would have had to fall to acquire that velocity. When, therefore, we further observe that a vertical jet of water will under favorable conditions rise nearly as high as the surface of the reservoir from which it is supplied we see that if our former supposition is correct it must issue with a velocity as great as it would have acquired in falling freely through a distance equal to the amount that the surface of the reservoir is higher than the orifice of the jet. This is in theory exactly true and would be equally true in practice but for the effect of the friction of the water through the pipe and nozzle before escaping.

Let us then first see how much the theoretical discharge of our pipe would be and then how much the friction is likely to reduce it.

To find the theoretical velocity of the flow we must know what head of water will give a pressure of 75 or 80 pounds. A cubic foot of water weighs about 62½' pounds, consequently a column of water 1 foot square and 2.3 feet high will weigh 144 pounds and give a pressure of one pound per square inch on the bottom; therefore, to get a pressure of, say 76 pounds per square inch, requires a head of 76 X 2.3= 175 feet. A body in falling acquires a velocity per second equal to about eight times the square root of the distance fallen. The square root of 175 is I3¼ which, multiplied by 8, is 106 feet per second.

A "1-inch" pipe is a little more than 1 inch in diameter, but rust and roughness make it unsafe to count on more, and therefore as the area of a 1-inch circle is.7854 of a square inch, that amount multiplied by 1,272, the number of inches in 106 feet, and divided by 231, the number of cubic inches in a gallon, equals 4.32 gallons per second, or 15,600 gallons per hour, the theoretical discharge through a 1-inch pipe with 76 pounds pressure per square inch. How much the friction of the pipe will reduce this discharge is uncertain. The problem of the flow of water in pipes is most difficult and complicated, and though the ablest hydraulic engineers have long endeavered to devise a formula for it at once simple, accurate, and generally applicable, their efforts have so far been only partially successful. Your sketch shows your pipe to be 132 feet long with two elbows and a stopcock in it. The frictional resistance of an ordinary screwed elbow is estimated to be equal to that of a length of pipe equal to 100 diameters, which in this case would be about 8 feet. A stop-cock, with its reduced opening, offers perhaps twice as much resistance as an elbow, or altogether the resistance of stop cock and elbows may be assumed equal to that of 32 feet of pipe, so your question practically becomes what will be the discharge per hour through 165 feet of 1-inch pipe with a pressure of 76 pounds per square inch. To this, some standard formulas give answers varying from 1,747 to 2,605 gallons, a difference of 50 per cent., which well illustrates the uncertainty of our present knowledge of the subject. On August 11,1888, we published some "Notes on Simple Methods of Calculating the Flow of Water Through Pipes," by Edward Murphy, in which occurs the formula \/ (425 X d X p) ÷ I = V; or, in other words, multiply the diameter in inches by the pressure in pounds per square inch and multiply the pro duct by 425, divide this new product by the lengthan feet and extract the square root of the quotient and the answer will be the velocity in feet per second, from which the discharge can be found as before. Applying the rule to this case 1 X 76 X 425 = 32,300, which divided by 165 = 195.76, of which the square root is 14. This velocity in feet per second multiplied by .7854, the area of the pipe, and by 3,600, the number of seconds in an hour, and then by 12 to reduce it to inches and divided by 231, the number of cubic inches in a gallon, equals 2,056, the discharge in gallons per hour. As this is reasonably between the extremes mentioned before and agrees very closely with the discharge found by the rule in Box's hydraulics, it is probably as near the truth as we can expect to get by any process of calculation. From this it appears that the friction will in this case reduce the discharge to about one eighth of the theoretical amount.

We have taken more than usual space to answer what seems a very simple question, but it is one of very general interest, and the answer here given may serve for many similar questions.]