1.Scope of Irrigation science: Extends from the watershed to the farm on the drainage channel. The watershed yielding irrigation water, the stream conveying the water, the arrangement and distribution of water and the drainage problems arising from irrigation practices are all concern of irrigationist. 2. Methods of recharging ground-water reservoir: a.) Systematic flooding of land surfaces overlaying or draining into underground reservoirs. b.) Increasing infiltration rate by using organic residues and grasses to condition ground surfaces to prolonged periods of submergence. c.) Use of shaft, pits, trenches or wells to convey water to the gravel and rock materials underground. 3. General ground water equation: AS = (Qi + Si + P) – (Qo + So + W + C) Where: AS = change in storage Qi = surface inflow Qo = surface outflow Si = Sub-surface inflow So = sub-surface outflow P = precipitation W = withdrawal C = consumptive use 4. Approaches for groundwater investigations: a.) Geological observations and investigations b.) Physical aspect of existing groundwater such as pressure, movements and temperature. c.) Geophysical method. d.) Logs of existing wells 5. Methods of drilling wells: a.) Cable tool method – cutting tools is suspended from a cable and vertical oscillation of the tool cuts and looser materials at the bottom of the holes. b.) Hydraulic rotary – a bit is rotated rapidly on the bottom of the string of drill pipe. A mud of clog and water, the consistency of which is varied with the drilling condition is forced through the drill pipe and carries the cutting around the outside of the pipe ti the ground surface. c.) Reverse rotary – when drilling mud is forced down the outside of the drill pipe and lower up outside the pipe carrying the drilling with it. 6. Requirement for slot opening of sewere for well: The slot opening should pass about 70% of the surrounding materials. The reduced velocity through the sewwere should be about 0.2 of a foot per second or lower if feasible. 7. Purpose of developing a well: 1. to increase the water yield. This is accomplished by increasing the permeability of the farmstead through which water moves toward the well. 2. to determine the water supply available and the needed characteristic of pump and power unit to be installed. 8. Methods for development of wells: a) pumping (3) Laminar flow (4) In the case of confined flow.7 IPP Israelson c) Assumption in the above two equation: (1) radial horizontal flow through uniform material perpendicular to vertical cylindrical surface. b) surging – the pluger is moved up and down opposite the perforated casing causing water to move alternately into and out of the well. c) Used of compressed air d) Chemical method 9. Whp = 69. in feet.3 log10 (re/rw) fig. the thickness of the acquifer is constant. 12.3 log10 (re/rw) fig. 3. Water yield of wells: a) Confined wells: Q = 2. Specified speed – an index use regarding the operating characteristic of pumps. Pump characteristic (Pump affinite laws) Q = N_ = D = W Q0 N0 D0 W 0 Where: Q = discharge N = speed D = width of impeller . hs = RPM x gpm H3/4 13. (2) Steady flow (no watershed with tree) is assumed.4 Qh = Qh 550 8. Water horsepower – the theoretical horsepower needed to lift a given quantity of water.11kt (he – hw) 2. 3.6 IPP Israelson b) Unconfined wells: Q = 11k (he2 – hw2) 2. It express the relationship between the speed in rpm. Pump characteristic: Characteristic performance curve – shows the characteristic of a pump under different operating condition.8 where: Q = discharge in cfs h = vertical lift in feet 11. discharge in gpm and head h. 10. Flow equation: a) continuity equation: Q = A1V1 = A2 V2 Where: A = cross sectional area V = velocity of flow b) Bernoulli’s equation: V12 + P1 + Y1 = V22 + P2 + Y2 + hL 2g W 1 2g W2 Where: P = pressure intensity at any point Y = elevation of the point above common point W = weight of unit volume of water h L = head loss. energy loss per unit weight of fluid between point 1 and 2. Earth canals: a) maximum velocity is 5 ft/sec b) best hydraulic cross section under favorable structural condition b = 2d tan Ө 2 . g = acceleration due to gravity c) Head loss equation h L = f LV2 d2g Where: f = coefficient of friction L = length between the two parts D = diameter of the conduit d) Relationship between head and pressure intensity h=P W Where: h = head P = pressure W = density e) Hydraulic radius R= A P Where: A = cross sectional area P = wetted perimeter of flow in contact with the channel 15. 14. H = head P = power Subscript zero refer to equal condition. collecting of sand and silt b) Weir Q = 3. Submerge flow – when the gage of the lower gage is greater than 70% of that of the upper gage. collecting of floating debris 2. 5. durability Disadvantage: 1. Where: b = bottom width d = depth Ө = angle between the side slope and horizontal 16. crest should be sharp.5 ft/sec c) Parshall flume 1.49 H5/2 – triangular Advantages: 1. sand and silt above the weir Rule for setting and operating weir: 1. C = 0. durability Disadvantage: 1.8 Advantages: 1. Throat and flow is level. fare of weir should be vertical 2. depth of water over crest should not be less than 2 inches 7. Devices for measuring flow: a) Orifice Q = CA 2gh . velocity of approach of not over 0.33 LH3/2 – for rectangular without contraction Q = 3.5 to 0. e) Current meter . Accuracy 2. depth over crest should be not more than one-third the length of the crest 6. 2. requirements of considerable fall 2. crest should be level 3. simplicity and ease of contraction\ 3. Free flow – lower gage is less than 60% of that of the upper gage. non-obstruction floating materials 4.33 (L – 0. 4.37 H3/2 – trapezoidal Q = 2.2) H3/4 – for rectangular with contraction Q = 3. d) Cut throat flume – similar in operation to that of parshall flume. distance of crest above the bottom of face should be about three times the depth of water flowing over crest. small loss of heads or difference in elevation of water surface and downstream 2. simplicity 4. accuracy 3. deposition of gravel. it upstream edge. ) real specific gravity d. high moisture 3.8 of the depth f) Float method Vs = S t v = Vsf where f ranges from 0.) soil structure c.8 to 0. Classes of soil water a) gravitational b) Capillary c) Hygroscopic water 19. Q = av Velocity is measured 1.) Porosity f. Consumptive use of water (evapotranspiration) A) Direct measurement of consumptive use a) tank and lysimeter b) field experimental plot . average of the represented by the average of the velocity at 0. at about 0. Soil Physical propee affecting storage and movement of water a. Soil moisture content a) saturation point – tention = 0 b) Field capacity 1/10 – 1/3 atmosphere tention c) Permanent wilting point – 15 atmosphere 20. 22. Readily available water moisture (RAM) 75% of the available range is readily available 21. fiber glass – low tention. 23.5 feet. if over 1.) soil texture b.) Permeability 18. nylon cloth 2. Leaching requirements – leaching requirements is defined as the fraction of the irrigation water that must be lacked through the root zone to central solivity at any specific level.6 of the depth from the surface when average depth is not over 1.8) d) Use of thermal properties e) Neutron method – fast neutron are entitled from a source into the surrounding soil and the number of showed down neutron is registered in the counter. Methods of soil moisture measurement: a) gravimetric b) use of electrical properties such as 1. 2.2 and 0.95 17. gypsum blocks – 1 to 15 atmosphere c) Tensiometer – less than 1 atmosphere (0.5 ft.) apparent specific gravity e.) Infiltration or intake g. and hours. solar radiation and humidity b. c) Soil moisture studied d) Integration method e) Inflow-outflow method B) Indirect method 1) Use of climatic factors a. 26. day time. Blanney-Criddle – temperature. piche 24. Three major consideration that influence the time of irrigation and how much to apply: a) water needs of the crops b) availability of water with which to irrigate. Irrigation efficiencies a) water conveyance = Wf x 100 Wr b) Application efficiencies = Wf stored x 100 Wr c) Water-use efficiency = Wu x 100 Wd d) Storage efficiency = water stored at one root zone Water needed at the root zone e) distribution efficiency = ( 1 – y/a) 100 f) Consumptive use efficiency = Wu x 100 Wd . color and moisture content b) Climatic factors – partially all c) Plant factor – foliage characteristic and root development d) Management practices 25. temperature. Factors affecting consumptive use a) soil factor – texture. c. Thornwaite method – temperature 2) Evaporation a. evaporimeter b. Root Extraction Pattern % of depth % of water extraction 0-25 40 25-50 30 50-75 20 75-100 10 27. Penman method – wind speed. c) Capacity of the root zone soil to absorb water. gpm s = slope of land in percent L= 1000____ . where Q = maximum size of stream. ft. 5. 8. Draw-down – is the difference in elevation between the ground water table and the water surface at the well when pumping. Seepage face – the difference in level of water inside the well and the water table immediately outside the well. Pumping plant efficiency – rate of power output to power input. cm (I – A) WS I = rainfall intensity A = absorption or infiltration rate of soil. Confined well – when the flow is restricted by impervious layer above the flow. 7. where: L = length of run. percent 30. 28. 2. Requirement for application of irrigation water for surface irrigation: a) length of run must be based on the slope and soil-intake rate b) The allowable stream size is run down the row and the reduced to the soil-intake rate after the water reaches the end of furrow. Owning water – water is considered as other natural reservoir which may be expanded in the interest of current development. 31. water harvesting – collection of rainfall by means of serve of ponds. Condition favoring sprinkler irrigation: a) soil too porous for good distribution by surface methods b) shallow soils the topography of which prevents the proper leveling for surface irrigation purposes. For surface irrigation (empirical equations) Q = 10/s . 4. S = slope of furrow.Definition of terms: 1. Criteria to be satisfied in water application: a) uniform distribution b) maximum erosion in damage to the land c) maximum efficiency in the use of water d) practical and economic performance from the standpoint of crop. iph W = furrow spacing. 3. c) Land having steep slopes and easily erodable soils d) Irrigation stream too small to distribute water efficiently by surface irrigation e) Undulating land too costly to level sufficiently for good surface irrigation . 29. c) Water should be applied only as necessary to fill the soil to the depth of root zone. 6. water hammer – the phenomenon that happens due to sudden closing of gate valve of a close conduit. watershed – area of land contributing to the flow of stream or the area where the water impounded by the reservoir is coming from. Sprinkler system can be designed and installed quickly. the pressure needed for sprinkling can be obtained with a minimum of additional investment. 33. h) Frequent and small applications of water can be applied readily by sprinkler irrigation.) Fertilizer application 34. e) When domestic and irrigation water comes from the same source. 32. Other points to consider when comparing sprinkler and surface method of irrigation: a) Water measurement is easier with sprinkler than with surface method. a common distribution line can frequently be used. d) Uniformity of application – not lower than 80% e) Water losses – not to exceed 10 – 15% f) Economical pipe size g) Crop damage – minimal 37. f) For areas requiring infrequent irrigation. b) Sprinkler systems can be designed so that less interference with cultivation and other farming operations occurs less land is taken out of production than with surface method. during the season b) portable set – put in place only during irrigation period. Parts of sprinkler system: a) sprinkler head or nozzle b) riser c) lateral line d) main or submain lines e) pumping plant 36. Designing the sprinkler layout: a) system must have capacity to supply enough laterals b) lateral length should be short enough so that practical size will result to pressure variation not to exceed 20% c) laterals should be placed across the slope. sprinkler irrigation is particularly attractive. sprinkler irrigation can be provided at a lower capital investment per acre of land.water should not be applied at a rate faster than the soil will absorb b) depth of application – not to exceed field capacity c) system capacity – must have sufficient capacity to apply the needed water. .) Frost protection b. c) Higher water-application efficiency can be attained by sprinkler irrigation. Factors to be checked to determine the adequacy of design and operation of sprinkler system: a) application rate. Other uses of sprinkler system: a. d) When water is already being pumped to the point of use. f) Land needs to be brought into top production quickly. g) Whenever water can be delivered to field under gravity pressure. 35. Types of sprinkler systems: a) solid set – main and lateral lines remain fixed. 8 inch per foot Clay – 2. Arrangement of Main and Laterals: The arrangements selected should allow a minimal investment in irrigation pipe. Irrigation must be started soon enough to enable the field to be covered before plant in the last portion to be irrigated suffer moisture deficiency.4 Depth of water to be pumped at 70% efficiency = 2.5”/hr Crop: Corn Rooting depth = 3 ft Peak use rate = 0.0 inch per foot Example: Areas – 40 Acres Soil type: silt loam Water holding capacity = 1. x 3 ft.42 gpm/acre 8 x 10 39. = 3. e) Lateral pipe should be only one size or two sizes at the most. = 5.30 = 8 days Area to be irrigated per day = 40/8 = 5 acres/day Application rate assuming 10 hours operation per day.4/0. d) Changes in pipe size necessary for pressure control should be made on main line wherever possible.30 iph Application efficiency = 70% Solution: Total water holding capacity of the soil = 1.5 inch per foot Silt loam – 1. and an additional 5 psi for each added 1/32” of nozzle size. but also the design should be adapted to the availability of labor for moving the sprinklers and the pipe.34” Irrigation period (internal) = 2. 45 psi for 9/64 in. Recommended pressures are 40 psi for nozzle sizes up to 1/8”.43 x 453 = x 1 = 19. have a low water requirement and provide for an application of water over the .4” Net depth of application = 45% of 5. Design Procedure: Not only should the sprinkler system to be properly designed hydraulically and economically in cost.43/10 = .8”/ft. The three basic facts to be established before the design of sprinkler irrigation system is initiated are the following: a) rate of application b) irrigation period c) depth of application Rates of application of one half the infiltration rate of the soil combined with nozzle pressures which provide a fine spray have resulted in improved maintenance of soil structure and minimization of soil compaction. 38. f) Whenever possible the source of water should be located in the field center for the most economical size of pipe. g) Split operation should be used whenever possible. Irrigation period is the time required to cover an area with one application.34”/hr System capacity = 3. It is suggested that irrigation should be done when the moisture level of the field reaches 55% of the available moisture capacity of the soil typical moisture holding capacities are as follows: Sandy loam – 1. through 3/16.4/peak use rate = 2.4” = 2. nozzles.70 = 3.4”/ . in.8”/ft Limiting application rate = 0. . the laterals are moved 60 feet at each setting and the sprinkler are spaced every 40 feet of each lateral. Maximum Spacing for Low and Medium Pressure Sprinklers Lateral spacing in % of the diameter of coverage in ft. pressure available and capacity of the sprinkler.98.85 Cdn2p1/2 where: Sl = sprinkler spacing along the lateral.95 to 0. 40. the total discharge is the combined capacity of both. inches P = pressure at the nozzle. iph q = nozzle discharge. The most suitable layout can be determined only after a careful study of the condition to be encountered. along the lateral Sm. Distribution pattern of sprinkler depends on the following: a) nozzle pressure b) wind velocity c) speed rotation 43. the lower is the coefficient. wph SI. 42. Sprinkler selection and spacing: The actual selection of sprinkler is based largely upon design information furnish by the manufacturers of the equipment. It depends primarily upon the diameter of coverage. Size of laterals and mains: ASAE recommends that the total pressure variation in the lateral. along the main 0 50 65 4 or less 45 60 4-8 40 50 8 or more 30 30 * for laterals normal to wind direction 44. Where the sprinkler has two nozzles. small nozzles varies from about 0. ft r = rate of application. when practical should not be more than 20 percent of the higher pressure. total area in the required period of time. feet Sm = sprinkler spacing between lines or along the main. Sprinkler capacity: q = (Sl) (Sm) (r) /96. In many systems.3 q = discharge of each sprinkler q = 29. Normally the larger the nozzle. Wind Velocity. Capacity of the sprinkler system: Q = Da where: t Q = dfs d = inches a = acres t = hours (total) 41. psi The coefficient of discharge for well-designed. rpm c = coefficient of discharge dn = diameter of the nozzle on fire. 338 40 .335 Q of lateral = q x no.500 4 .9 x (1.365 .000 2 .¾ He + Hr Where He = Maximum difference in elevation between the 1st and last sprinkler.350 .338 50 .376 . .370 . of sprinkler per lateral The position of the lateral that gives the highest friction loss in the main should be used for design purposes.338 .000 1. Hn = Ho + Hf where Hn = Head at the main Hn = Ha + ¾ Hf +. The average operating in pressure is used in design.345 . Scobey’s equation : Hf = KsLQ1. Hm = maximum friction loss in the main and in the suction line.337 100 .349 14 .343 . Pump and Power unit Ht = Hn + Hj + Hs + Hm Where Ht = total design head against which the pump is working Hn = maximum head required at the main to operate the sprinkler on the lateral at the required average pressure.342 25 .359 .361 .369 8 .340 30 . CORRECTION FACTOR FOR FRICTION LOSSES IN ALUMINUM PIPES WITH MULTIPLE OUTLETS Correction Factor F st 1 sprinkler one. Hj = elevation difference between the pump and the function of the lateral and the main.353 12 .354 .469 . one 1st sprinkler sprinkler Interval from Main one-half sprinkler No.625 . Average pressure (Ha): If Hf is with in 20% of average pressure Ha = Ho + ¼ Hf where Ho is pressure at the farthest end if lateral is nearly level land.347 .398 .339 35 .385 .393 6 . of Sprinkler Interval from Main 1 1.45 x 10-8) For sprinkler lateral the friction loss obtained from the above equation should be multiplied by a factor F as adjustment for conduit with multiple outlet as shown in the following table.358 10 .345 18 .421 .347 16 . Hr = the riser height 45.343 20 . 17 ft at 80°F 1. Suction lift Determination Refer to figure 1 54. 46. 50. Factors to consider: 1. 52.5 x 10-8) D4.5 ft at 60°F.31 = head (ht) 51. Static head (Ht) – it is the vertical distance that the pump must raise water. a) brake horsepower required b) Initial cost c) Availability and cost of energy d) Depreciation e) Dependability of unit f) Portability of unit g) Maintenance and convenience h) Labor availability and quality 47. Ks = Scobey’s coef.9 x (1. safety factor Friction loss: (see table 1) Scobey’s Hf = KsLQ1. 2. Hs = elevation difference between the pump and the water supply after draw down. TDH = Ht + Hf + Hv 49. Of retardation . Pressure head (Hp) – the pressure that water is to be discharge expressed in terms of height of water. Friction loss in suction pipe 2. Theoretical suction lift is 34.62 ft at 90°F. vapor pressure of liquid at pumping temperature 90. Net positive suction head (NPSH) 4. 0. Power requirement Whp = Q (gpm) x TDH 3960 Where: whp = water horsepower Q = discharge in gallons per minute TDH = Total dynamic head in feet 48. Features to be considered in the selection of pumps. Velocity head (Hv): Hv = V2 2g 53.21 ft at 100°F) 3. friction head (Hf) and velocity head (Hv). 1. Total Dynamic Head (TDH) is the sum total of the static head (Ht) pressure head (Hp).84 ft at 70°F.9 Where: Hf = total friction loss on line in ft. It is the total of the static suction head (Hs) plus static discharge (Hd). Friction head (Hf) – the amount of friction also expressed in terms of head of water between the pipe and the fluid. Pressure (psi) x 2. Example: The NPSH at 310 gpm at 1750 RPM is 3. Capacity much higher than capacity of peak efficiency of pump.32 for new transite pipe 0.6 = 495 gpm NPSH – (1. This head causes flow from either the pressure of the atmosphere or from static head plus atmosphere. Q = total discharge in gpm D = inside diameter of pipe in ft. This must be supplies by the manufacturer. Required NPSH – is a function of pump design. Q = 310 x 1.6 The capacity will increase directly as this ratio therefore. 4. This increase in velocity is of course accompanied by a reduction in pressure. NPSH is usually assumed to bay as the square of the speed ratio. In centrifugal pump this maybe explained as follows: when a liquid flow through the suction line and enters the eye of the pump impeller an increase in velocity takes place.2 Net positive Suction Head (NPSH) – is defined as the head that causes liquid to flow through the suction piping and finally enter the eye of the impeller. 54. 54.3 Suction Lift NPSH A suction characteristics of irrigation pump are very important. the liquid will vaporize and the flowing stream will consist of liquid plus pockets of vapor. For centrifugal pumps avoid as much as possible the following conditions: 1.4 ft. Ks value: 0. L = length of pipe in ft.42 for portable galvanized steel pipe and couplers. It can br calculated for every installation. Safety Factor: Usually an allowance of 2 ft is used as a safety factor.6) (1. Cavitation – a term used to describe a rather complex phenomenon that may exist in a pumping installation. What will be the gpm if the RPM is increased to 2800 RPM? What is the corresponding NPSH? Speed ratio = 2800/1750 = 1.40 for steel pipe and portable aluminum 0. 3. the NPSH at increased engine speeds must be checked to be certain pump will operate at the capacity required without cavitations. NPSH – is a function of a system in which the pump operates. If the pressure falls below the vapor pressure corresponding to the temperature of the liquid.4 = 8. 2.6) x 3. Liquid temperature higher than that for which was originally designed. Head much lower than head at peak efficiency of pump. 5. When the . Speed higher than manufacture’s recommendation. Parallel and Series Operation When the pumping requirements are available.7 ft 55. it may be more desirable to install several pumps in parallel rather than use a single one. Any installation. Friction lift higher or positive head lower than recommended by manufacturer. to operate successfully must have an available NPSH equal to or greater than the required NPSH at one desired pump conditions. It varies between different make of pumps as well as pumps of the same make and varies with the capacity and speed of any one pump. otherwise cavitation will result. Check with gauge. 4. pump too much water c) mechanical defects 1. Moreover. Pump works for a while and then quits a) Looking suction line b) Stuffing box packing worm or water seal plugged – allowing leakage of air into pump casing. h) Stuffing box packing worm or water seal plugged – allowing leakage of air into pump casing. 2. thus allowing the remainder to operate at or near peak efficiency. causing excessive friction loss). i) Air leak in line. shaft bent 2. c) Air pocket in suction line. pump and driving unit misaligned 4. e) Suction lift too high (suction pipe may be too small or long. e) Wrong direction of rotation f) Be sure pressure gauge is in correct place not on top of pump case. d) Not enough suction head e) Air or gasses on liquid f) Suction lift too high 5. rotating element bends 3. Not enough pressure a) Speed too low b) Air in water c) Wrong impeller (diameter) d) Mechanical defects as mentioned in No. f) Wrong direction of rotation g) Air pocket in suction line. a) packing worm or not properly lubricated b) packing incorrectly inserted or not properly run in . Similarly. Pump leaks excessively at stuffing box. demand drops. one or more smaller pumps maybe shut down. multiple pumps in series maybe used when liquid must be delivered at high heads. 2. No water delivered a) Pressing – casing and suction pipe not completely filled with water. Pump takes too much power a) speed too high b) head lower than racing. Trouble Check List for Centrifugal Pumps 1. 56. wrong direction of rotation 6. Not enough water delivered (same as above plus the following): a) Foot value too small b) Mechanical defects: Wearing ringworms Impeller damaged Shaft packing defective 3. when smaller units are used opportunity is provided during slack demand periods for repairing and maintaining each pump in turn thus avoiding plant shut-downs which would be necessary with single units. b) Speed too low c) Discharge head too high – check vertical head (particularly friction loss) d) Impeller suction pipe or opening completely plugged up. b) Mechanical defects: 1. rotating parts bind. bearing worn out 4. check with gauge. loose or broken 3. c) packing not right kind of liquid handled. shaft bent 2. Pump is noisy a) hydraulic noise – cavitations. suction lift too high. . d) Shaft scored 7. pump and driving unit misaligned . Problems: 1. 200 gpm – 40 ft.3 Seepage .02) d) TDH assuming that the point of discharge is on the level of the ground.00 3 + 00 3. design for an average pressure of 40 psi at the sprinkler nozzle.2 cm/day db – 1. Design a sprinkler irrigation system for a square 40-acre field to irrigate the entire field within a 12-day period.50 99.02. Two 620-ft sprinkler lines are required. with hard pan Crop to be raised – lowland rice Source of water – river with unlimited capacity Allowance fro drawdown – 2 ft.00 5 + 00 4. A 75 ft well located in the center of the field will provide the following discharge-drawdown relationship..00 104.76 Stations are measured in meters and rod readings in feet.00 99. 2. (f = 0. Remarks 0 + 00 4. *assumed Find: a) capacity of pump needed in gpm b) velocity head if pump diameter is 10 inches c) friction head if total length of pipe used including strainer is 60 ft. BS HI FS Elev. 100.00 6.7 cm/day 3.00 4 + 00 4. Use the following data (per ha.00 101. Duty of water – 60 inches at 100 pumping days Slope of centerline of the probable diversion ditch was recorded as flows: Sta. Calculate the water requirement of lowland rise with full irrigation.1 cm/day Pfc – 40% Evapotranspiration – 0. Basis) Evaporation – 0. Not more than 16 hours per day are available for moving pipe and sprinkling. 4. .28 96.00 . Highest point in the field is 4 feet above the well site and 3 feet risers are needed for the sprinkler.0.00 98.00 97. e) Bhp of the power unit at 60% efficiency f) Design of a rectangular weir needed g) Slope of center line canal h) Cross sectional dimension of the canal if n = 0.1 cm/day ds – 1 ft Percolation – 0. A 100 hectare farm in Central Luzon was surveyed and the following data were gathered Soil – clay loam. 250 gpm – 50 ft.. Determine the required capacity of a sprinkler system to apply water at a rate of 0. determine the rated output for an internal combustion engine. and side slope = 1:1.00* River 1 + 00 4. 300 gpm – 65 ft. Sixteen sprinklers are spaced at 40 ft interval on each line and the spacing between lines is 60 feet. Two inches of water are required at each application to be applied at a rate not to exceed 0.50 2 + 00 5. Assuming a pump efficiency of 60% ans assuming the engine will furnish 70% of its rated output for continuous operation.5 iph.35 iph.