896 



THE IRRIGATION AGE. 



This land should be worth at least 100 dollars per acre when 

 ready for cultivation, and provide homes and means for a 

 livelihood for 100,000 families. 



The experience of the State of Missouri in reclaiming a 

 large area of swampland, which formerly sold at from 1 to 

 5 dollars, and which now readily brings from $50 to $100 per 

 acre after drainage, should impel all States having such land 

 to take steps toward reclamation. 



Many an irrigation plant has been spoiled 

 Important and is operated at a loss because the plan- 



Points for ning of the works was done in a haphaz- 



Pumping zard manner and without considering im- 



Requirements. portant points which affect the economy of 



the work. The first consideration is the 

 amount of water actually needed for the land to be provided 

 for. When this quantity is settled, we will say in a certain 

 number of acre feet per annum, then divide this amount by 

 the number of days the pumping plant will be in operation 

 during the year, which will give the requirement per day. 

 As such a plant rarely runs day and night a certain number 

 of hours per day should be assumed, say 10 hours; hence 

 divide the amount required per day by 10x60x60 or 

 36,000 will give the flow in cubic feet per second. The next 

 important point is the height of the lift; this height is taken 

 from the surface of the water, at the suction end to the 

 center of the stream at the discharge end plus the friction 

 in the pipes and pump measured in feet. For instance, one 

 pump may pump through a height of 40 feet with 50 feet of 

 piping while another pumps the same amount of water the 

 same height through a length of piping 500 feet long; then 

 it stands to reason that the latter case requires more power 

 for its pumping than the former on account of the frictional 

 resistance of the long pipe; this resistance increases with 

 the length and decreases as the diameter increases, so that 

 it is good economy when water has to be pumped consider- 

 able distances to increase the size of the discharge pipe. The 

 horse power of the engine required to do the pumping can 

 be found approximately by taking the weight of the water 

 pumped per minute, multiply this by the total head in feet, 

 including resistance, divide by 33,000, which gives the actual 

 water horse power, then double this and the result will be 

 pretty close the requirement of the case. 



To show in an assumed problem the application of the 

 above principle let it be required to calculate the size of 

 centrifugal pump, piping and horse power of engine required 

 for the following problem : The number of acres to be irri- 

 gated is 20 ; water requirements 24 inches per annum, which 

 is equal to 2 times 20 times 43,560, or 1,742,400 cubic feet per 

 annum. Divide by 120 as the number of days the plant is 

 operated gives 14,520 cubic feet per day ; divide this by 36,- 

 000 which is the number of seconds in 10 hours, makes .403 

 cubic feet per second. Let the speed of the water in the 

 centrifugal pump be 8 feet per second. Divide .403 by 8, 

 gives .0504 square feet, as the cross sectional area of the 

 pump; consult a table of areas of circle or compute the di- 

 ameter of the required circle by first dividing area by 3.14, 

 extracting square root and multiplying by 2 the diameter of 

 pump is found to be .252 feet or slightly more than 3 inches, 

 say y/ 2 inches. The discharge pipe should be of a much 

 larger diameter in order to reduce the velocity which reduces 

 the resistance to flow. Assume the length of discharge pipe 

 200 feet, 6 inches in diameter. The area of a 6-inch pipe is 

 .196 square feet; divide this into .403 cubic feet gives a ve- 

 locity of 2.05 feet per second in the discharge pipe. A table 

 of resistancies for a six-inch cast iron pipe with a velocity 



of 2.0."> feet per second gives a loss of only 2.67 feet; let the 

 suction lift be IS feet and the discharge head 32 feet, then 

 the total lift plus the resistance is IS + .12 + 2.67 ft. = 52.67 

 ft. To find the horse power multiply the weight of the 

 water pumped per minute with 52.67 and divide by 33,000 

 times 2. .403 c. f. s. multiplied by 60 gives 24.18 cubic feet 

 of water per minute ; multiply this by 62.5 gives the pounds 

 of water pumped, 1511.25 per minute; multiply by the lift 

 plus resistance 52.67 gives 79, 598 foot pounds of work. 

 Divide by 33,000 gives 2.412 net water horse power; 

 multiply this by 2 gives 4.824 gross horse power or say, it 

 requires a 5 horse power engine to run this pump. 



Suppose now some one in solving the above problem in- 

 stead of using a 6 inch pipe wanted to save money by using 

 a 4 inch pipe. In this case the area of a 4 inch pipe is .087 

 square feet which makes the velocity in the pipe 4.63 feet per 

 second, and gives a resistance equal to 17.5 feet ; add this 

 to the 50 foot head, makes 67.5 in all, multiply this with 

 1511.25 pounds of water pumped per minute gives 102,010 

 feet pounds, which divided by 33,000 gives 3.09 net water 

 horse power; double this gives 6.18 horse power, which is 

 considerable more than the 5 horse power engine required 

 with the 6 inch pipe. As in this case a 6 ! /2 horse power en- 

 gine would be required, the extra cost of the engine would 

 probably balance the greater cost of the 6 inch pipe not to 

 talk about the greater cost of operating the larger engine. 



The foregoing considerations show the necessity of figur- 

 ing on each irrigation problem the various elements entering 

 into a concise solution of such problem. The "Primer of 

 Hydraulics," which is now in process of evolution and which 

 will be ready for the market by the end of 1911, will just 

 qualify any irrigationist to make his own computations and 

 get satisfactory results. 



CONSERVING MOISTURE. 



C. W. GRANDEY, MONTANA. 



Dry farming, as it is practiced in this section of eastern 

 Montana means no more nor less than following those meth- 

 ods in tillage that will conserve all the moisture for the grow- 

 ing crop. It requires the same amount of water to produce 

 a bushel of corn in Montana that it does elsewhere. Our 

 rainfall is fourteen to eighteen inches. We cannot waste 

 the moisture the Illinois or Iowa farmers do. On the bench 

 lands, where the soil is best adapted to conservation of mois- 

 ture, the plow is followed immediately by a roller or sub- 

 soil packer, then disked and harrowed, after seeding with 

 a disk drill. While the crop is growing, the harrow is used 

 as often as a crust is formed by rains until the grain is 

 eight to ten inches high. 



As June is usually a month of abundant rainfall, much 

 breaking is done during this period for winter wheat. Where 

 the steam and gas tractors are used, much summer fallow- 

 ing is resorted to, raising two crops in three years. That 

 the bench lands of Montana will yield abundant harvests 

 of wheat, flax, oats, barley and rye, and even corn, has 

 been demonstrated in the last three years by the homesteader 

 and for many years by the state experiment station. What 

 failures we have here are due to loose farm methods, neglect 

 to harrow and poor seed. 



Some results obtained in 1910 were exceedingly good. 

 A neighbor started his steam plow in April on the raw prairie, 

 breaking, subsoil packing, working the ground, seeding with 

 a double disk drill and following with three sets of harrows. 

 all in one operation. Loss of moisture was very slight. In 

 this manner some 800 acres were put in on the bench land. 

 His harvest yielded him better than $10,000. Another neigh- 

 bor went through the same operation, but at different times, 

 and had almost a total failure. 



One summer, fallow oats yielded forty bushels, and in 

 the same section this variety of oats on spring breaking, 

 with a few days between disking and harrowing, made five 

 bushels. I certainly have faith in the bench land farming 

 of Montana. 



