THE IEEIGATION AGE. 



907 



CORRESPONDENCE 



adjacent to the flume GE, which would facilitate the irriga- 

 tion work. 



The problem under consideration shows how necessary it 

 is, however, to study each individual problem in irrigation 

 in order to determine the best and most economical layout. 



AN IRRIGATION PROBLEM. 



EDITOR IRRIGATION AGE : 



I want you to please assist me if you will in regard to 

 a plan we wish to carry out in regard to irrigation, see ac- 

 companying drawing. We have ordered a six horsepower 

 engine and a centrifugal pump with a capacity of 185 gallons 

 per minute. Tell us how to apply it to field. We want to 

 use as little pipe as possible so we will ditch A, C and D, if 

 you think best. If we ditch those lines how far apart should 

 they be placed. If you can think of a better way please tell us. 



As to the soil. It is 8 to 12 in. deep, under which is 

 a tough clay that holds water like a jug. 



We wish to raise potatoes, cabbage, beans, etc. How 

 many acres can we irrigate per day with the outfit above 

 named? We were expecting 2 to 3 acres per day. 



Yours truly, 



J. F. D. 



P. S. Herewith is a complete drawing of a 45-acre 

 field. We want to irrigate this season between line D and 

 creek, containing from 12 to 15 acres. The soil is dark and 

 reaches 10 to 15 inches down ; clay or light soil is 10 to 15 

 inches under ground. 



The problem is outlined in the accompanying diagram 

 showing the land surrounded by water on 2 sides. The con- 

 tour line A is 40 ft. above the surface of the water, line C 

 is 30 ft., line D is 20 ft. above the water level. At point E 

 it is proposed to install a centrifugal pump raising 185 gals, 

 of water per min. ; the lines A, C and D divide the land in 

 about 3 equal parts of 12 to 15 acres each ; a pipe line along 

 the ridge B would be about 900 ft. long, and if the pump at E 

 was to supply water for the level at A it would have to over- 

 come not only the 40 ft. lift but also the resistance of 900 

 lin. ft. of pipe; this friction in the case under consideration 

 amounts to about 3.85 Ibs. pressure per 100 ft. or 34.65 Ibs., 

 this is equal to an additional head of 80 ft., making the whole 

 work of the pump 120 X 1541 Ibs. = 5.61 net water horse 

 power, but as only about 60 per cent of the engine horse 

 power is converted into useful work, 9.35 indicated horse 

 power would be required. A far better proposition would be 

 to install the plant on the side of the bluff B opposite A at 

 an elevation of about 20 feet above the water level indicated 

 by point F. An intake ditch should be dug from the pond 

 to the pump suction pipe along the low flat at the base of the 

 bluff. From this point the length of the y/2 inch discharge 

 pipe to lift the water to the 40 foot level will be quite short 

 and the heavy loss in the proposed case will be avoided. As- 

 suming that the length of the 3V 2 inch diameter pipe from 

 pump F to point G at the 40 foot level is 100 feet then the 

 loss due to this pipe line will be only 1/9 of 80 feet, say 9 

 feet, which added to the head of 40 feet makes 49 feet of 

 total head required. There is also a considerable saving in 

 the pipe purchased as the water from the point G can be % 

 taken care of by ditches or a flume. For good work a 

 wooden flume should be built from G to E with outlets for 

 the lateral ditches indicated by the dotted lines GG, HH, II. 

 JJ, etc. These ditches should not be made too deep and 

 should be built with a slight grade so as to carry the water 

 near the farther side of the field. The number of these 

 lateral ditches depends of course a great deal on local con- 

 ditions. 



The flow of 185 gallons per minute is about .4 cubic foot 

 per second, exactly .412 c. f. s. ; if it is kept up for 24 hours 

 it will mean a pumpage of 266,400 gallons or 35,615 cubic feet : 

 this would be enough to cover one acre 10 inches deep, or 

 which is the same as to cover 10 acres 1 inch deep, and as 

 the land is located in Kentucky, where there is a great deal 

 of rain, the writer cannot see any reason why this plant 

 should not be able to furnish all the irrigation needed for 

 these 45 acres by alternately turning on the water for the 

 upper 15 acres, say for 24 hours, then turn the water on the 

 middle 15 acres and then on to the lower 15 acres as required. 

 In this way different fields could be laid out, arranging it so 

 that the crops requiring the most water be planted in fields 



WANTS A DEFINITION FOR HYDRAULIC 

 RADIUS. 



EDITOR IRRIGATION AGE : 



Permit me to offer you my compliments on the splendid 

 paper which the IRRIGATION AGE is getting to be, and I would 

 not do without it if it cost five times as much. Your latest 

 improvement, the addition of a Correspondence Department, 

 in which readers may ask and answer questions, is a long step 

 in the right direction and will make your paper more popular 

 than ever. 



There is a question that I would like to be enlightened 

 on, it is used in hydraulic works, but I never heard a clear 

 definition for the expression "hydraulic radius." I would 

 appreciate it very much if you would explain this in your 

 paper, as there are undoubtedly others who will be benefited 

 thereby. 



Thanking you in advance for a reply in the columns of 

 your paper, I remain 



A FAITHFUL READER. 



The inquiry in the above letter relates to one of the 

 principal elements in hydraulics and will receive complete 

 attention in the "Primer of Hydraulics" in the course of a few 

 months. However, there is no reason why the subject should 

 not be editorially explained at this time. 



The hydraulic radius, also termed hydraulic depth of a 

 water channel, is a fraction obtained by dividing the flow 

 area of a channel by the length of the wetted perimeter. To 

 illustrate, let in the accompanying diagram, A B C D, be 



a cross section of a 



A m M-Z' rectangular flume, 



the flow line reach- 

 ~KT -gr // ing the line E F; 



*& let the width, B C, 



for instance, be 

 equal to 3 ft, and 

 // W, the depth, E B, 

 equal to 2 ft. ; then 

 the flow area will 

 be 3 X 2 = 6 square 

 ft. ; the wetter peri- 

 meter will be EB -f- 

 BC + CF or 2 + 

 3 + a = 7 ft; 



rr. /iOe, hence the hydraulic 



** radius will be found 



by dividing 7 into 6, which is 6/7, or changing to a 

 decimal fraction the hydraulic radius eqnals .857 feet. As 

 the depth of the flow decreases the hydraulic radius decreases 

 also, for instance : let BG equal 1.5 ft., then the flow area 

 will be 1.5 X 3 = 4.5 square ft. ; the wetter perimeter will be : 

 1.5 + 3 + I- 5 = 6 ft ; hence the hydraulic radius equals 

 4.5 -f- 6 = .75 ft. 



The hydraulic radius is usually given the symbol r and 

 is an important factor in determining the sizes, grades, etc., of 

 irrigation channels. EDITOR. 



