valve rod being in position and connected to the rocker arm by means of the short link, the nut or nurs securing the valve to the rod should be so adjusted as to be equidistant from the lugs on the valve, say or

of an inch according to the amount of lost motion. desired, which latter factor governs the length of stroke in some makes of duplex pumps, while in others it is controlled by tappets on the valve rod outside of the valve chest. Care should be taken while making these adjustments that the valve be retained exactly in its central position.

Having set the valves correctly, move one of the pistons far enough from mid-stroke to get a small opening of the steam port on the opposite side, then replace the valve chest covers and the pump will be ready to run. As these valves are generally made. without any outside lap, a slight movement of one of the pistons in either direction from its central position will suffice to uncover one of the ports on the other cylinder sufficiently to start the pump.

Sometimes duplex pumps "work lame," that is, one piston will make a quick full stroke while the other piston will move very slowly and just far enough to work the steam valve of the opposite side. In the majority of cases this irregular action is due to unequal friction in the packing of the rods, or the packing rings on one of the pistons may be worn out.

If one side of a duplex pump becomes disabled from any cause, as breaking of piston rod in the water cylinder, for instance, which is liable to happen, the pump may still be operated in the following manner until duplicate parts to replace the broken ones have been secured. Loosen the nuts or tappets on the valve stem of the broken side and place them far

enough apart so that the steam valve will be moved through only a small portion of its stroke, thereby admitting only steam enough to move the empty steam piston and rod, and thus work the steam valve. of the remaining side. The packing on the broken rod should be screwed up tight, so as to create as much friction as possible; there being no resistance in the water end. In this way the pump may be operated for several days or weeks and thus prevent a shut down.

Hydraulics for Engineers. Among the many difficult problems that are continually coming up for engineers to solve, there is none more perplexing than the correct calculation of the quantity of water which will be discharged in a given time from pipes of various sizes and under the many different heads or pressures. Problems in hydraulics, as given by the majority of writers on engineering, are usually in elaborate algebraical equations, which, to the ordinary working engineer, are very perplexing, at least the author has found them to be so in his experience. Therefore with a view of assisting his brother engineers in the solution of problems along this line which they may be called upon to solve, the author has spent considerable time and labor in searching for and compiling a few rules and examples for hydraulic calculations in plain arithmetic which he hopes may be of benefit.

First, to find velocity of flow in the pump, or in other words, piston speed.

Rule. Multiply number of strokes per minute by length of stroke in feet, or fractions thereof.

Second, the velocity of flow in the discharge pipe is in inverse ratio to the squares of the diameters of the pipe and the water cylinder of pump.

Thus, a pump cylinder is 6 in. in diameter, and the piston speed is 100 ft. per minute; the discharge pipe being 3 in. in diameter. What is the velocity of flow in the pipe? Example. 3x3 4. In this case the velocity in the pipe is four times that in the pump, and 100 × 4 = 400 ft. per minute, velocity for water in the discharge pipe.

6×6

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Third, to find velocity in feet per minute necessary to discharge a given quantity of water in a given time.

Rule. Multiply the number of cubic feet to be discharged by 144 and divide by area of pipe in inches. Fourth, to find area of pipe when the volume and velocity of water to be discharged are known.

Rule. Multiply volume in cubic feet by 144 and divide by the velocity in feet per minute.

Fifth, one of the first requisites in making correct calculations of the quantity of water discharged from any sized pipe is to obtain the velocity of flow per second. There are several rules for doing this, among which the following appear to be the plainest and most simple:

Rule 1 Multiply the square root of the head in inches by the constant 27.8. For instance, assume the head to be 100 ft. 1200 in. The square root of 1200 is 35 nearly, then 35 × 27.8 = 973 in. = 81 ft. per second velocity.

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Rule 2. Multiply the square root of the head in feet by the constant 8, as follows: The square of 100 = 10 and 10 x 8 80 ft. velocity per second.

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Rule 3. Multiply twice the acceleration of gravity by the head in feet and extract the square root of product. The acceleration of gravity may be considered the constant number 32, neglecting decimals.

32 × 2 × 100 = 6400. Square root of 6400 = 80 ft. per second.

In many instances it is more convenient to use the pressure in pounds per square inch as shown by gauge instead of the height or head, and we can then apply Rule 4.

Rule 4. Multiply the square root of the pressure. in pounds per square inch by the constant number 12.16 as follows: Pressure due to 100 ft. head 44 lbs., nearly. Square root of 44 = 6.6, which multiplied by 12.16 80.2 ft. velocity per second.

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Having ascertained the velocity of flow, we may now proceed to calculate the weight of water in pounds per second discharged from any size of pipe, neglecting for the time being the loss in pressure caused by friction from elbows and bends in the pipe and also the peculiar shape assumed by a stream of water flowing through pipes or conduits when there is no resistance except the pressure of the atmosphere and friction caused by long distance transmission.

We will take for our calculation a four-inch pipe from which the water has a free flow under a head of 100 ft., which gives a velocity of 80 ft. per second.

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Rule 5. Divide the velocity in feet per second by the constant 2.3, and multiply the quotient by the area of discharge pipe in square inches. 802.3 34.7. Now the area of a four-inch pipe is 12.57 sq. in., and 34.7 x 12.57 436 lbs. discharged per second.

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In order to get the matter clearly before us, let us assume that we have a section of four-inch pipe just 80 ft. in length and that it lies in a horizontal position and is filled solidly full of water. It will contain area, 12.57 sq. in. x length, 960 in. 12,067.2 cu. in. of water, and as one pound of water occupies a space

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of

27.7 cu. in., we therefore have 1,2067.2 + 27.7 = 436 lbs. of water, and at a velocity of 80 ft. per second our pipe will be emptied and refilled continuously each. second. We have also Rule 6 to find the number of cubic feet discharged per minute when the velocity per minute is known.

Rule 6. Multiply the area of pipe in square inches by the velocity in feet per minute and divide by the constant 144.

Example. Area of 4 in. pipe = 12.57 sq. in. Velocity of flow = 80 ft. per second = 4,800 ft. per minute. Then 12.57×4,800 419 cu. ft. per minute = 6.99 cu. ft. per second, which multiplied by 62.3 lbs. (weight of 1 cu. ft.) 435.4 lbs. per second.

144

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As stated before, no allowance is made by the above rules for friction or other retarding influences, but for ordinary purposes in connection with a steam plant a deduction of 25 per cent. is probably sufficient. Of course if the water is being discharged into an elevated tank or against direct pressure of any kind, the resistance in pounds per square inch or the height in feet must be deducted from the impelling pressure or head. Let us assume, for instance, that our 4 in. pipe is discharging water into a tank at an elevation of 75 ft. above the level of the pump, and that to reach the tank requires 100 ft. of pipe with two 90° ells and one straight-way valve. We wish to discharge 500 gal. per minute into the tank and will therefore require a velocity of about 13 ft. per second, which will necessitate a pressure of a little more than 1 lb. per square inch to be maintained at the pump over and above all resistance. Now the resistance to be overcome in this case will be: